Electromagnetics for Engineering Students 9781681085043, 9781681085050

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Electromagnetics for Engineering Students Authored by Sameir M. Ali Hamed Nile Valley University Atbara, Sudan


Electromagnetics for Engineering Students Author: Sameir M. Ali Hamed eISBN (Online): 978-1-68108-504-3 ISBN (Print): 978-1-68108-505-0 ©2017, Bentham eBooks imprint. Published by Bentham Science Publishers – Sharjah, UAE. All Rights Reserved.

DEDICATION  

To my father, mother, brothers, and sisters To my wife To Mohannad, Lama, and Mohammad

i

FOREWORD The subjects of electromagnetics and electromagnetic fields theory constitute a fundamental base in the study of electrical and electronic engineering and systems. It has been noticed that the concepts pertinent to these subjects are difficult to grasp by the average university student. Because of this difficulty these subjects need to be exposed in a way that is conducive to their understanding by these students. With this in mind the author presented the student with an elucidated revision of vector theory, one of the mathematical bases of the electromagnetic fields theory and developed the subsequent chapters in an easyto-follow style. The intricacy of the mathematics normally used in the exposition of these concepts and the analyses necessary to present them in a logically graspable manner call for many solved examples and unsolved exercises for student training and familiarization. The author, being an experienced university professor, managed to cater for both aspects, and has presented the concepts in an easy-tofollow, streamlined style supported by many diverse solved examples besides a large collection of problems to be tried by the student. Besides the excellent style and the many and diverse solved examples and large collection of exercise problems, the author illustrated the concepts with well prepared and nicely and clearly presented figures to facilitate understanding of the underlying mathematics and comprehension of the subject concepts. Although the book is targeting mainly undergraduate university students the rich manner in which it is prepared makes it extremely useful and helpful to graduate students, researchers and university instructors who shoulder the responsibility of teaching these subjects Mohammed Ali H. Abbas, PhD Associate Professor Ph.D. Department of Electrical and Electronics Engineering University of Khartoum Sudan

ii

PREFACE This textbook bridges the gap between the advanced and the introductory books in engineering electromagnetics. Although many excellent texts cover in details the introductory topics suitable for junior undergraduates, some topics, which may be important to higher-level undergraduates and beginning graduates, are ignored or not given the same attention. These topics include the application of the multipole expansion in electromagnetic fields problems, the solution of electrostatic boundary value problems in cylindrical and spherical coordinate systems, the detailed analysis of waveguides including circular waveguides, the topic of metamaterials, the refraction through metamaterials and the concept of negative refractive index. The book exposes to these topics in sufficient details while retaining the usual introductory topics in electromagnetics, which makes it useful textbook for both undergraduate and introductory graduate courses in electromagnetics. The book presumes that the student is acquainted with the general knowledge in differential and integral calculus, differential equations, vector analysis, and complex numbers. Chapter 1 has been devoted to review vectors analysis, since it is an essential topic to understand the subject of electromagnetics. The electrostatic field and related concepts and laws are covered in Chapter 2. Chapter 3 discusses the electric properties of the materials. In Chapter 4, more advanced techniques are introduced to solve the boundary-value problems in electrostatics. The magnetostatic field and related concepts are covered in Chapter 5. Chapter 6 discusses magnetic materials and introduces metamaterials and their properties in an appropriate depth for the level of undergraduate students. In Chapter 7, Maxwell’s equations are presented in integral and point form for time-varying fields, in addition to the concept of vector potential, boundary conditions for timevarying fields and Poynting vector. The wave propagation and associated parameters in different media are discussed in Chapter 8. The reflection and transmission of plane waves at the interfaces between two different media in addition to the concept of negative refractive index are considered in Chapter 9. The metallic waveguides and resonant cavities of rectangular and circular cross sections are analyzed in Chapters 10 to 12. The behavior of transmission lines in frequency domain is discussed in Chapter 13. Basic concepts of antennas and radiation from some metallic structures are presented in Chapter 14. The intended audience of the book are the undergraduate students in the field of electrical, electronics, telecommunications, and computer engineering and related fields. The topics of the book are presented in sufficient details, such that the

iii

students can follow the analytical development easily. Each chapter is supported by numerous illustrative examples, solved problems, and end of chapter problems to explain the principles of the topics and to enhance the knowledge of the student. Sameir M. Ali Hamed, PhD Associate Professor, Nile Valley University Atbara, Sudan

CONFLICT OF INTEREST The editor declares no conflict of interest, financial or otherwise. ACKNOWLEDGEMENTS I would like to express my profound gratitude to Associate Professor Dr. Mohammed Ali H. Abbas form the University of Khartoum who read the draft of the book and gave useful and constructive comments that improved greatly the contents of the book. 


CONTENTS

FOREWORD PREFACE CONFLICT OF INTEREST ACKNOWLEDGEMENTS

i ii iii iii

CHAPTER 1 VECTOR ANALYSIS 1.1. 1.2. 1.3. 1.4.

1.5.

1.6. 1.7.

1.8.

1.9. 1.10. 1.11. 1.12. 1.13. 1.14. 1.15. 1.16.

INTRODUCTION NOTATION OF SYMBOLS AND UNITS SCALAR AND VECTOR THE UNIT VECTOR THE ORTHOGNOLAL COORDINATE SYSTEMS 1.4.1. The Rectangular (Cartesian) Coordinate System 1.4.2. The Cylindrical (Circular) Coordinate System 1.4.3. The Spherical Coordinate System TRANSFORMATION BETWEEN COORDINATE SYSTEMS 1.5.1. Cylindrical-to-Rectangular 1.5.2. Rectangular-to-Cylindrical 1.5.3. Spherical-to-Rectangular 1.5.4. Rectangular-to-Spherical 1.5.5. Spherical-to-Cylindrical 1.5.6. Cylindrical-to-Spherical THE GENERALIZED ORTHOGONAL COORDINATE SYSTEM VECTOR ALGEBRA 1.7.1. Vectors Addition and Subtraction 1.7.2. Vectors Multiplication 1.7.3. Triple Product 1.7.4. Vectors Properties THE POSITION VECTOR, THE LINE, AND THE PLANE 1.8.1. The Position Vector 1.8.2. The Straight Line 1.8.3. The Plane SCALAR AND VECTOR FIELDS THE LINE INTEGRAL THE GRADIENT OF A SCALAR FIELD SURFACE INTEGRAL AND THE FLUX OF A VECTOR THE DIVERGENCE OF A VECTOR THE CURL OF A VECTOR FIELD THE DIVERGENCE THOEREM STOKES' THEOREM SOLVED PROBLEMS PROBLEMS

1 2 2 4 4 4 8 11 16 16 18 19 21 22 24 27 30 30 30 33 34 37 37 39 40 41 41 43 45 47 50 53 55 59 70

CHAPTER 2 ELECTROSTATIC FIELDS INTRODUCTION SOURCES OF ELECTROSTATIC FIELDS COULOMB'S LAW FORCE ON A CHARGE DUE TO SYSTEM OF POINT CHARGES ELECTRIC FIELD INTENSITY 2.4.1. Electric Field due to a Point Charge 2.4.2. Electric Field due to a Discrete System of Point Charges 2.4.3. Electric Field due to a Continuous Distribution of Charge 2.5. ELECTRIC FLUX AND ELECTRIC FLUX DENSITY 2.6. GAUSS'S LAW FOR ELECTRIC FLUX 2.7. APPLICATIONS OF GAUSS'S LAW 2.7.1. Electric Field due to a Point Charge 2.7.2. Electric Field due to an Infinite Line of Charge 2.7.3. Electric Field due to a Uniform Distribution of Surface Charge 2.7.4. Electric Field due to a Uniform Distribution of Volume Charge 2.8. ELECTROSTATIC POTENTIAL 2.8.1. Electrostatic Potential due to a Point Charge 2.8.2. Electrostatic Potential due to System of Charges 2.9. THE ELECTRIC FIELD AS A GRADIENT OF THE POTENTIAL 2.10. ELECTROSTATIC ENERGY 2.10.1. Electrostatic Energy due to a Continuous Distribution of Charge 2.10.2. Energy Stored in an Electrostatic Field 2.11. THE ELECTRIC DIPOLE SOLVED PROBLEMS PROBLEMS 2.1. 2.2. 2.3. 2.4.

76 77 79 80 83 83 85 89 98 99 101 102 103 104 106 107 111 113 122 125 126 127 130 133 147

CHAPTER 3 CONDUCTING AND DIELECTRIC MATERIALS INTRODUCTION CONDUCTION CURRENT AND CONDUCTIVITY PROPERTIES OF PERFECT CONDUCTORS RESISTANCE DIELECTRIC MATERIALS 3.4.1. Polarization 3.4.2. Relation between the Polarization Vector and the Potential 3.4.3. Permittivity and Relative Permittivity 3.4.4. Linear, Isotropic, and Homogenous media 3.5. CONSERVATION OF CHARGE AND RELAXATION TIME 3.6. BOUNDARY CONDITIONS FOR ELECTROSTATIC FIELDS 3.6.1. Interface between Two Dielectrics 3.6.2. Interface between Two Dielectrics of Finite Conductivities 3.7. CAPACITANCE 3.7.1. Parallel-Plates Capacitor 3.7.2. Spherical Capacitor 3.7.3. Cylindrical Capacitor 3.7.4. The Energy Stored in the Capacitor 3.1. 3.2. 3.3. 3.4.

155 156 162 164 167 167 171 175 178 180 182 182 186 189 189 193 194 195


3.8. DIELECTRIC STRENGTH 62/9('352%/(06 352%/(06

197 198 214

CHAPTER 4 ELECTROSTATIC BOUNDARY-VALUE PROBLEMS INTRODUCTION 4.1. THE METHOD OF IMAGES 4.1.1. A Point Charge above a Conducting Ground Plane 4.1.2. A Point Charge in Front of a Conducting Grounded Sphere 4.1.3. A Point Charge in Front of a Conducting Isolated Sphere 4.1.4. A Conducting Sphere in a Uniform Electric Field 4.1.5. A Point Charge in Front of a Dielectric Planar Interface 4.2. MULTI-POLE EXPANSION 4.3. POISSON'S AND LAPLACE'S EQUATIONS 4.4. UNIQUENESS OF THE SOLUTION 4.5. SOLUTION OF LAPLACE'S EQUATION IN ONE DIMENSION 4.5.1. Rectangular Coordinate System 4.5.2. Cylindrical Coordinate System 4.5.3. Spherical Coordinate System 4.6. SOLUTION OF LAPLACE'S EQUATION IN TWO DIMENSIONS 4.6.1. Rectangular Coordinate System 4.6.2. Cylindrical Coordinate System 4.6.3. Spherical Coordinate System 4.7. NUMERICAL METHODS 4.7.1. The Method of Moments 4.7.2. The Finite Difference Method 62/9('352%/(06 352%/(06

221 222 222 224 226 227 229 233 237 239 241 241 241 243 248 248 257 265 272 272 284 292 298

CHAPTER 5 MAGNETOSTATIC FIELDS INTRODUCTION 5.1. SOURCES OF THE MAGNETOSTATIC FIELDS 5.2. AMPERE'S CIRCUITAL LAW 5.2.1. The Magnetic Field due to an Infinite Filamentary Wire Current 5.2.2. The Magnetic Field due to an Infinite Current Sheet 5.2.3. The Magnetic Field due to a Long Current-Carrying Conductor 5.2.4. The Magnetic Field inside a Long Solenoid 5.2.5. The Magnetic Field inside a Toroid 5.2.6. Ampere's Law in Point Form 5.3. MAGNETIC FLUX AND MAGNETIC FLUX DENSITY 5.4. MAGNETOSTATIC POTENTIALS 5.4.1. Magnetic Scalar Potential 5.4.2. Magnetic Vector Potential 5.5. BIOT-SAVART'S LAW 5.6. MAGNETIC FORCE 5.6.1. Lorentz Force Equation 5.6.2. The Force on a Current Element in a Magnetic Field

304 305 307 310 311 312 314 315 319 320 323 323 324 331 338 338 341

5.6.3. The Force between Two Current Elements 5.7. MAGNETOSTATIC ENERGY 5.8. MULTI-POLE EXPANSION FOR THE VECTOR POTENTIAL 5.8.1. Thin-Wire Current-Carrying Conductor 5.8.2. Thin-Wire Circular Loop of a Uniform Current 5.9. THE MAGNETIC DIPOLE 5.10. MAGNETIC TORQUE SOLVED PROBLEMS PROBLEMS

345 351 353 355 358 361 365 368 387

CHAPTER 6 MAGNETIC MATERIALS AND METAMATERIALS INTRODUCTION 6.1. MAGNETIZATION 6.1.1. Relation between the Magnetization and Vector Potential 6.1.2. Permeability and Relative Permeability 6.2. MAGNETIC MATERIALS 6.2.1. Materials with a Small Susceptibility 6.2.2. Materials with a Large Susceptibility 6.3. MAGNETIC HYSTERESIS 6.4. BOUNDARY CONDITIONS FOR MAGNETIC FIELDS 6.5. INDUCTANCE 6.5.1. Self Inductance 6.5.2. Internal Inductance of a Long Conductor 6.5.3. Mutual Inductance 6.5.4. Energy Stored in the Inductance 6.6. MAGNETIC CIRCUIT 6.7. FREQUENCY-DEPENDENT PERMITTIVITY AND PERMEABILITY 6.7.1. Dispersion 6.7.2. Models of Materials 6.8. METAMATERIALS AND ARTIFICIAL MAGNETIC CONDUCTORS 6.8.1. Material Classifications 6.8.2. Artificial Magnetic Conductors 62/9('352%/(06 352%/(06

395 396 400 404 408 408 409 410 412 417 417 419 422 429 433 439 439 442 443 444 447 449 458

CHAPTER 7 TIME-VARYING ELECTROMAGNETIC FIELDS 7.1. 7.2. 7.3. 7.4. 7.5.

7.6.

INTRODUCTION FARADAY'S LAW OF INDUCTION DISPLACEMENT CURRENT MAXWELL'S EQUATION IN INTEGRAL FORM CONSTITUTIVE RELATIONS AND PARAMETERS MAXWELL'S EQUATION IN POINT FORM 7.5.1. General Time-Varying Fields 7.5.2. Time-Harmonic Fields 7.5.3. Static Fields TIME-VARYING VECTOR AND SCALAR POTENTIALS .6.1. General Time-Varying Fields

465 466 471 472 475 476 476 477 479 482 482


.6.2. Time-Harmonic Fields THE ELECTRIC AND MAGNETIC HERTZ VECTORS HELMHOLTZ EQUATION AND ITS SOLUTION MULTI-POLE EXPANSION FOR THE TIME-HARMONIC VECTOR POTENTIAL 7..1. Thin-Wire Conductor with a Uniform Time-Harmonic Current 7..2. Infinitesimal Electric Dipole with a Time-Harmonic Current 7..3. Thin-Wire Circular Loop with a Uniform Time-Harmonic Current 7..4. Magnetic Dipole with a Time-Harmonic Current 7.10. BOUNDARY CONDITIONS 7.11. ELECTROMAGNETIC POWER 7.11.1. Poynting Theorem and Poynting Vector 7.11.2. Average Power for Time-Harmonic Fields SOLVED PROBLEMS PROBLEMS 7.7. 7.8. 7.9.

483 486 488 495 497 498 499 501 502 505 505 507 509 515

CHAPTER 8 PROPAGATION OF UNIFORM PLANE WAVES 8.1. 8.2. 8.3.

8.4.

8.5.

8.6.

8.7.

8.8.

INTRODUCTION THE WAVE EQUATION THE SOLUTION OF THE WAVE EQUATION IN UNBOUNDED MEDIA CHARACTERISTICS OF HOMOGENOUS MEDIA 8.3.1. Loss Tangent 8.3.2. Propagation, Phase, and Attenuation Constants 8.3.3. Intrinsic Impedance 8.3.4. The Phase Velocity and Wavelength 8.3.5. Skin Depth UNIFORM PLANE WAVES IN UNBOUNDED MEDIA 8.4.1. Lossless Media 8.4.2. Lossy Media 8.4.3. Good Dielectrics 8.4.4. Good Conductors THE DIFFUSION EQUATION 8.5.1. Current Flow in a Semi-Infinite Conducting Slab 8.5.2. Surface and A.C. Resistances of a Conductor POWER FLOW 8.6.1. Lossless Media 8.6.2. Lossy Media 8.6.3. Good Dielectrics WAVE POLARIZATION 8.7.1. Linear Polarization 8.7.2. Circular Polarization 8.7.3. Elliptical Polarization 8.7.4. Axial Ratio GROUP VELOCITY SOLVED PROBLEMS PROBLEMS

521 522 523 527 527 528 529 530 531 533 533 538 541 545 548 549 550 553 555 555 555 560 561 561 562 563 566 567 576

CHAPTER 9 REFLECTION, TRANSMISSION, AND NEGATIVE REFRACTION 9.1. 9.2. 9.3.

9.4.

9.5.

9.6.

9.7.

INTRODUCTION OBLIQUE PLANE WAVE INCIDENT, REFLECTED, AND TRANSMITTED WAVES OBLIQUE INCIDENCE 9.3.1. Interface between Arbitrary Media 9.3.2. Dielectric – Dielectric Interface, σ1 = σ2 = 0

583 584 586 591 591

9.3.3. Dielectric – Conductor Interface, σ1 = 0 and σ2 ≈ ∞ SNELL'S LAW AND TOTAL INTERNAL REFLECTION 9.4.1. Snell's Law 9.4.2. Total Internal Reflection TOTAL TRANSMISSION AND BREWSTER ANGLE 9.5.1. Perpendicular Polarization Case 9.5.2. Parallel Polarization Case NORMAL INCIDENCE 9.6.1. Reflection and Transmission Coefficients 9.6.2. Standing Wave Ratio NEGATIVE REFRACTIVE INDEX SOLVED PROBLEMS PROBLEMS

605 608 608 610 614 614 615 617 617 620 629 631 640

598

CHAPTER 10 RECTANGULAR WAVEGUIDES INTRODUCTION 10.1. PROPAGATION MODES 10.1.1. Transverse Electric and Magnetic Fields (TEM) 10.1.2. Transverse Electric Field (TE) 10.1.3. Transverse Magnetic Field (TM) 10.1.4. Hybrid Mode 10.2. GENERAL ELECTROMAGNETIC FIELDS EQUATIONS 10.2.1. TEM Mode Fields 10.2.2. TE Mode Fields 10.2.3. TM Mode Fields 10.3. E L E C T R O M A G N E T I C F I E L D S I N T H E R E C T A N G U L A R W A V E G U I D E 10.3.1. TE Mode Fields in the Rectangular Waveguide 10.3.2. Characteristics of the TE Mode Fields in the Waveguide 10.3.3. TM Mode Fields in the Rectangular Waveguide 10.3.4. Characteristics of the TM Mode Fields in the Waveguide 10.4. DOMINANT MODE 10.5. POWER LOSS AND ATTENUATION IN WAVEGUIDES 10.5.1. Attenuation Constant αd 10.5.2. Attenuation Constant αc 10.6. POWER LOSS IN THE RECTANGULAR WAVEGUIDES 10.6.1. Power Loss in the Waveguides Supporting TE Mode Fields 10.6.2. Power Loss in the Waveguides Supporting TM Mode Fields SOLVED PROBLEMS PROBLEMS

645 647 648 649 649 650 651 655 664 666 667 668 673 682 685 690 695 695 696 700 701 705 705 710


CHAPTER 11 CIRCULAR WAVEGUIDES INTRODUCTION 11.1. ELECTROMAGNETIC FIELDS 11.1.1. TE Mode Fields in the Circular Waveguide 11.1.2. Characteristics of the TE Mode Fields in the Waveguide 11.1.3. TM Mode Fields in the Circular Waveguide 11.1.4. Characteristics of the TM Mode Fields in the Waveguide 11.2. DOMINANT MODE IN THE CIRCULAR WAVEGUIDE 11.3. POWER LOSS AND ATTENUATION 11.3.1. Power Loss in the Circular Waveguide 11.3.2. Waveguides Supporting TE Mode Fields 11.3.3. Waveguides Supporting TM Mode Fields SOLVED PROBLEMS PROBLEMS

715 715 716 721 729 732 736 740 740 740 741 743 747

CHAPTER 12 RESONANT CAVITIES INTRODUCTION 12.1. ELECTROMAGNETIC FIELDS 12.1.1. TE Mode Fields in a Resonant Cavity 12.1.2. TM Mode Fields in a Resonant Cavity 12.2. QUALITY FACTOR 12.3. RECTANGULAR CROSS-SECTION CAVITIES 12.3.1. TE Mode Fields in the Rectangular Cavities 12.3.2. TM Mode Fields in the Rectangular Cavities 12.4. CYLINDRICAL CAVITIES 12.4.1. TE Mode Fields in the Cylindrical Cavities 12.4.2. TM Mode Fields in the Cylindrical Cavities SOLVED PROBLEMS PROBLEMS

751 752 753 754 755 755 755 760 767 767 770 773 780

CHAPTER 13 TRANSMISSION LINES INTRODUCTION 13.1. PROPERTIES OF TEM MODE TRANSMISSION LINES 13.2. TRANSMISSION LINE PARAMETERS 13.2.1. Parallel-Plates Lines 13.2.2. Coaxial Cables 13.2.3. Two-Wires Lines 13.2.4. Microstrip Lines 13.3. VOLTAGE AND CURRENT EQUATIONS 13.4. REFLECTION AND TRANSMISSION COEFFICIENTS 13.5. STANDING WAVE RATIO 13.6. THE INPUT IMPEDANCE 13.7. THE LOSSLESS TRANSMISSION LINE 13.7.1. Voltage, Current, and Input Impedance 13.7.2. Voltage and Current Maxima and Minima 13.7.3. Power Flow in Lossless Lines

785 786 791 792 795 799 810 814 818 821 822 824 826 827 833

13.8. MATCHING TECHNIQUES 13.8.1. Stub Matching 13.8.2. Quarter-Wave Transformer 13.9. SMITH CHART FOR TRANSMISSION LINES CALCULATIONS 13.9.1. The Smith Chart 13.9.2. Applications in the Usage of the Smith Chart SOLVED PROBLEMS PROBLEMS

837 837 840 843 843 847 855 866

CHAPTER 14 PRINCIPLES OF ANTENNAS INTRODUCTION 14.1. ANTENNA TYPES 14.1.1. Wire Antennas 14.1.2. Aperture Antennas 14.1.3. Slot and Microstrip Antennas 14.1.4. Array Antennas 14.1.5. Reflector Antennas 14.1.6. Lens Antennas 14.2. BASIC ANTENNA PARAMETERS 14.2.1. Radiated Power and Radiated Power Intensity 14.2.2. The Input Impedance of the Antenna 14.2.3. The Radiation Efficiency 14.2.4. The Directivity 14.2.5. The Gain 14.2.6. The Antenna Effective Aperture 14.2.7. The Relation between the Effective Aperture and the Directivity 14.3. FAR-FIELD APPROXIMATIONS 14.4. BASIC PARAMETERS OF THIN-WIRE LINEAR ANTENNAS 14.4.1. The Infinitesimal Dipole 14.4.2. The Thin-Wire Dipole Antenna of a Finite Length 14.4.3. The Half-Wave Dipole Antenna 14.4.4. The Quarter-Wave Monopole Antenna 14.4.5. Long Wire Antennas 14.5. LOOP ANTENNAS 14.5.1. The Magnetic Dipole 14.5.2. Large Circular Loop Antennas 14.5.3. The Uniform Current Circular Loop 14.6. ARRAY ANTENNAS 14.6.1. The Array Factor 14.6.2. The Uniform Linear Array 14.6.3. Maxima, Nulls, and Half-Power Beam Points 14.6.4. The Visible Region 14.6.5. Examples of Arrays 14.6.6. The Directivity of an Array of Isotropic Elements 14.7. RECIPROCITY THEOREM FOR ANTENNAS 14.8. FRIIS TRANSMISSION FORMULA

874 875 875 876 878 879 880 881 881 882 891 893 894 902 904 905 907 911 911 916 918 921 923 928 928 931 933 936 936 938 939 943 943 944 947 950


SOLVED PROBLEMS PROBLEMS

953 960

APPENDIX A: BESSEL FUNCTIONS APPENDIX B: ASSOCIATED LEGENDRE FUNCTIONS APPENDIX C: USEFUL INTEGRALS AND MATHEMATICAL FORMULAS APPENDIX D: FREQUENCY BANDS APPENDIX E: PHYSICAL CONSTANTS

966 970 973 977 979

REFERENCES

980

SUBJECT INDEX

986

Electromagnetics for Engineering Students, 2017, 1-75

1

CHAPTER 1

Vector Analysis Abstract: The electromagnetic field theory is a fundamental subject for the electrical engineering students. The student should be acquainted with a quite knowledge in differential and integral calculus, differential equations, vector analysis, matrix algebra, and complex numbers in order to understand the subject of electromagnetics and be able to analyze the behavior of electromagnetic fields in various media. The knowledge of the vector analysis is essential in the analysis of electromagnetic problems. This first chapter of the book is focusing on the important concepts and techniques in the vector analysis that are required to understand the topics of the electromagnetic fields in the later chapters of the book. The chapter covers the topics and concepts in the vector analysis including vector algebra, orthogonal coordinate systems, vector differentiation operators, and vector integration. The chapter also introduces the scalar and vector field concepts, in addition to the divergence and Stokes’ theorems. The chapter includes numerous illustrative examples covering each topic in the chapter, in addition to miscellaneous solved problems supported by illustrative figures to enhance the student problems solving skills. The chapter is ended by numerous homework problems covering the topics of the chapter.

Keywords: Curl, curvilinear orthogonal coordinate, divergence, divergence theorem, flux, gradient, Laplacian, position vector, right-hand rule, scalar, scalar field, scalar product, Stokes’ Theorem, vector, vector field, unit vector, vector product. INTRODUCTION The knowledge of vector analysis is essential to understand the subject of electromagnetic fields theory. This why we start the book by a comprehensive review of the vector analysis. This first chapter of the book covers the topics and concepts in the vector analysis including vector algebra, orthogonal coordinate systems, vector differentiation operators, and vector integration. In addition to the divergence and Stokes’ theorems, which are discussed at the end of the chapter. The book presumes that the student is acquainted with the engineering applied mathematics. In particular the topics of differential and integral calculus, differential equations, vector analysis, matrix algebra, and complex numbers that are required to understand the topics of the electromagnetic fields. The student may refer to many excellent books covering these topics, e.g. [1-9]. Special functions, such as Bessel’ functions, associated Legendre polynomials, and other mathematical formulas that are necessary for understanding some topics of the book are available in [10-15] and some of them are included as appendices. Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

Electromagnetics for Engineering Students

2

Sameir M. Ali Hamed


1.1. NOTATION OF SYMBOLS AND UNITS The symbols and notations of basic electromagnetic quantities used in this book are the same as those used in similar textbooks of electromagnetics [5-21]. Table 1.1 lists examples of notations and symbols used in the book. Table 1.1. Notation of the symbols in the book. Quantity

Description

Scalars

Indicated in ordinary typeface

Vectors

Indicated in bold typeface

Time-Varying scalars Time-Varying vectors

Indicated in ordinary typeface monotype corsiva font Indicated in bold typeface monotype corsiva font

Rectangular coordinate variables Cylindrical coordinate variables

Indicated in italic typeface

Spherical coordinate variables Unit vectors

Represented by the letter a in bold typeface with a subscript

Examples

l , Φ , Ex A , r , Ta V ,
, Ex A, E,Ta

( x, y , z ) ( ρ ,φ , z ) ( r ,θ ,φ ) a x, a n, aA

The SI system of units is used throughout the book. 1.2.

SCALAR AND VECTOR

Physical quantities such as electric potential, temperature, length, mass, volume, speed, pressure, and resistance can be specified fully by their magnitudes only. Any quantity that is fully described by its magnitude only such as aforementioned quantities is called a scalar quantity. On the other hand, the physical quantity that is fully described only when both its magnitude and direction are known such as force, velocity, acceleration, and momentum, is called a vector quantity. A vector quantity is represented by a straight line segment with arrowhead as shown in Fig. (1.1) for vectors A and B . The arrowhead end is called the tip of the vector and the other end is the base of it. The length of the line is proportional to the magnitude of the vector and the arrow denotes the direction of the vector. This straight-line representation is called a directed straight-line segment. Shifting the directed straight-line segment of a vector parallel to itself such that its length and direction remain the same is known as a translation of the vector. The translation of a vector does not change the vector as illustrated in Fig. (1.1b).

Vector Analysis


Electromagnetics for Engineering Students

B

3

B

A Vector B

Translation of vector B

(a)

(b)

Fig. (1.1). Translation of a vector.

To add vectors A and B shown in Fig. (1.2a), we translate B as shown in Fig. (1.2b) and the sum is the vector C connecting the base of A and the tip of B as shown in Fig. (1.2b). A similar result can be obtained if we translate A as illustrated in Fig. (1.2c). Thus, the vectors addition is commutative, or C = A + B = B + A . The above concept can be used to add any number of vectors. Parallelogram rule can also be used to add vectors A and B as shown in Fig. (1.3b). A B

C

B C

B A

A (a)

(b)

(c)

Fig. (1.2). Vector addition. (a) A and B . (b) C = A + B . (c) C = B + A .

B

B

C A (b)

A (a) Fig. (1.3). Vector addition using parallelogram rule.

Any vector B can be scaled by a scalar quantity α to obtain the vector α B . The vector D = A − B can be obtained as D = A + α B by letting α = −1 , then translating − B as illustrated in Fig. (1.4).

B

B

D B A

Fig. (1.4). Vector subtraction D = A − B .

A

4

1.3.

Electromagnetics for Engineering Students

Sameir M. Ali Hamed


THE UNIT VECTOR

As explained in the previous section the vector quantity possesses both magnitude and direction. For a given vector A , the magnitude of A is its quantitative value which is a scalar quantity. The direction of A is described by a vector known as a unit vector of magnitude one with its direction along A . The magnitude of the vector A is denoted by A = A and the unit vector a A of the vector A is

aA =

A A = A A

(1.1)

Based on (1.1), the vector A can be written in terms of its magnitude and unit vector a A as

A =aAA 1.4.

(1.2)

THE ORTHOGONAL COORDINATE SYSTEM

For vector quantities to be manipulated algebraically, they must be defined analytically with respect to a specific coordinate system. The most commonly used coordinate systems are orthogonal rectangular (Cartesian), cylindrical (Circular), and spherical coordinate systems. The rectangular coordinate system is the simplest coordinate system that can be used as a reference for the vector representation. 1.4.1. The Rectangular (Cartesian) Coordinate System The rectangular coordinate system is represented by three mutually orthogonal axes x , y , and z as reference axes, and the positive sense along these axes is taken along the direction of increasing x , y , and z . 1.4.1.1. The Right-Handed Orthogonal Coordinate System The system of orthogonal axes is said to be a right-handed system when the orientation of the x-axis, y-axis, and z-axis is such that the positive direction along the z-axis is the one in which a right-handed screw aligned along the z-axis will advance when rotated from positive x-axis to the positive y-axis. Fig. (1.5) shows a right-handed orthogonal coordinate system.

Vector Analysis


Electromagnetics for Engineering Students

5

z

90° 90°

Direction of rotation

y 90°

x Fig. 1.5. The right-handed orthogonal coordinate system.

A point P ( x 0 , y 0 , z 0 ) is located in the rectangular coordinate system at the intersection point of the surfaces x = x 0 , y = y 0 and z = z 0 , as illustrated in Fig. (1.6). The distances x 0 , y 0 , and z 0 are in meter and their ranges are

− ∞ ≤ x 0 ≤ +∞ ,

z

y

− ∞ ≤ y 0 ≤ +∞ ,

− ∞ ≤ z 0 ≤ +∞

Surface y = y0

y0

Plane x =z x=0z0 Surface

P (x0, y0, z0)

z0

x

x0 Surface x = x0 Fig. (1.6). Locating a point P ( x 0 , y 0 , z 0

) in the rectangular coordinate system.

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1.4.1.2. Components of a Vector Let A x , A y , and A z be orthogonal vectors along the x , y and z axes as shown in Fig. (1.7), with magnitudes Ax , A y , and Az , respectively. The unit vectors in the direction of the positive x , y and z axes are denoted by a x , a y and a z respectively. Using (1.2), the vectors A x , A y , and A z can be written in terms of their magnitudes and unit vectors as

A x = a x Ax ,

A y = a y Ay ,

A z = a z Az

(1.3)

Now if A is the resultant of the vectors A x , A y , and A z , A can be obtained by translating A y

to obtain A x + A y , then translating A z to obtain

A = A x + A y + A z as shown in Fig. (1.7). Substituting A x , A y , and A z from (1.3) into this last expression of A , yields A = a x Ax + a y A y + a z Az

z

(1.4)

Az

A az Ax

Ay

ay

ax

Ax A y

y

x Fig. (1.7). The component of a vector A in the rectangular coordinate system.

The respective magnitudes A x , A y and Az of vectors A x , A y , and A z , are the components of the resultant vector A in the direction of x , y and z respectively. Equation (1.4) can be written also in matrix form as

A = A TR U R

(1.5a)

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where

⎡a x U R = ⎢a y ⎢ ⎢⎣ a z

⎤ ⎥, ⎥ ⎥⎦

⎡ Ax A R = ⎢ Ay ⎢ ⎢⎣ A z

⎤ ⎥ ⎥ ⎥⎦

(1.5b)

T stand for matrix transpose. Referring to Fig. (1.7), the magnitude of A is

A =

Ax2 + A y2 + Az2

(1.6)

and its unit vector in the direction of A is obtained using (1.1) as aA =

A a x Ax + a y A y + a z Az = A A2 + A2 + A2 x

y

(1.7)

z

1.4.1.3. Differential Length, Surface Area, and Volume It is important in the analysis of electromagnetic field problems to know the expressions for the differential length dl along a path, the differential surface area dS , and the differential volume dv , in different coordinate systems. In the rectangular coordinate system, the differential length dl along the path l in Fig. (1.8), can be expressed as

dl = a x dx + a y dy + a z dz

(1.8a)

az dx 

z

ax dl

dz

l

dy 

x Fig. (1.8). A differential length in the rectangular coordinate system.

y

a y

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and its length can be obtained as

dl = dl = dx 2 + dy 2 + dz 2

(1.8b)

The differential surface area dS is a vector quantity with magnitude dS and directed along the normal to the differential area. dS is positive when the normal points outwards of the differential surface and negative when it points inwards. The differential surfaces for the three faces of the differential cube in Fig. (1.9) are

dS xy = a z dxdy

dS xz = a y dxdz

dS yz = a x dydz

(1.9)

The differential volume d
is a scalar quantity. It can be shown from Fig. (1.8)

d
= dxdydz

(1.10)

dS xy = a z dxdy 

z

dx  dz

dy

dS xz = a y dxdz

d


dS yz = a x dydz

y x Fig. (1.9). Differential surface area and volume in the rectangular coordinate system.

1.4.2. The Cylindrical (Circular) Coordinate System A point in the cylindrical coordinate system is represented by ρ , φ , and z , where ρ is the radial distance from the z-axis and parallel to the xy plane, φ is the angle measured from the xz plane in the direction of y-axis, and z is the distance along the z-axis. The point P ( ρ 0 ,φ 0 , z 0 ) in the cylindrical coordinate system is located at the intersection point of the cylinder of the radius ρ 0 with the

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z-axis as its axis, the surface φ = φ 0 , and the surface z = z 0 , as illustrated in Fig. (1.10). The radial distance ρ 0 and z 0 are in meter and φ 0 is in radian, such that

0 ≤ ρ 0 ≤ ∞,

0 ≤ φ 0 ≤ 2π , z

− ∞ ≤ z 0 ≤ +∞

P (ρ0 , φ0 , z0 ) ρ0

Surface z Cylinder

y

φ0

Surface φ = φ0

x Fig. (1.10). The cylindrical coordinates system.

1.4.2.1. Components of a Vector In the cylindrical coordinates system, the vector A is written as

A = a ρ Aρ + a φ Aφ + a z Az

(1.11)

where a ρ , a φ and a z are the unit vectors in the direction of ρ , φ and z respectively as shown in Fig. (1.11). Aρ , Aφ and Az are the components of the vector A in the direction of ρ , φ and z respectively. Equation (1.11) can be written in matrix form as

A = A TC U C

(1.12a)

where

⎡a ρ U C = ⎢ aφ ⎢ ⎣⎢ a z T for matrix transpose.

⎤ ⎥, ⎥ ⎦⎥

⎡ Aρ A C = ⎢ Aφ ⎢ ⎣⎢ A z

⎤ ⎥ ⎥ ⎦⎥

(1.12b)

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z Az 

A

az 

aφ aρ

Aφ  A ρ + Aφ 

Aρ 

y x Fig. (1.11). The component of a vector A in the cylindrical coordinate system.

1.4.2.2. Differential Length, Surface Area, and Volume The differential length dl along the path l in Fig. (1.12) in the cylindrical coordinate system can be expressed as

dl = a ρ dρ + a y ρ dφ + a z dz

(1.13a)

z az l dz

aφ

aρ

dl

dρ  dφ 

y x

φ

ρ

ρ dφ 

Fig. (1.12). A differential length in the cylindrical coordinate system.

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11

dS ρφ = a z ρ dρ dφ  dS ρz = a φ dρ dz 

dz

dS φz = a ρ ρ dφ dz 

dρ dφ 

y x

φ

ρ

ρ dφ 

Fig. (1.13). A differential surface area and volume in the cylindrical coordinate system.

The magnitude of the differential length in (1.13a) is obtained as

dl = dl = dρ 2 + ρ 2 dφ 2 + dz 2

(1.13b)

The differential surfaces for the faces of the element shown in Fig. (1.13) are

dS φz = a ρ ρ dφ dz

dS ρz = a φ dρ dz

dS φz = a z ρ dρ dφ

(1.14)

The differential volume in Fig. (1.13) is a scalar quantity and can be written as

d
= ρ dρ dφ dz

(1.15)

1.4.3. The Spherical Coordinate System In the spherical coordinate system, any point in space is represented by r , θ , and φ , where r is a radial distance from the origin, θ is the angle that the radial distance makes with the z-axis, and φ is the angle measured from the xz plane in

the direction of the y-axis. The point P ( r0 ,θ 0 ,φ 0 ) in the spherical coordinate

system is located at the intersection point of a sphere of radius r0 centered at the origin, a right cone of apex angle 2θ 0 with its apex at the origin, and its axis as the z-axis, and the surface φ = φ 0 , as illustrated in Fig. (1.14).

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θ0

z Cone

r0 Sphere

P ( r0 ,θ 0 ,φ 0 )  Surface φ = φ0

y φ0 

x Fig. (1.14). The spherical coordinate system.

The radial distance r0 is in meter and the angles θ 0 and φ 0 are in radian. The quantities r0 , θ 0 and φ 0 are defined in the ranges

0 ≤ r0 ≤ ∞ ,

0 ≤θ0 ≤ π ,

0 ≤ φ 0 ≤ 2π

1.4.3.1. Components of a Vector A vector A in the spherical coordinate system is written as

A = a r Ar + a θ Aθ + a φ Aφ

(1.16)

where a r , a θ and a φ are the orthogonal unit vectors in the directions of r , θ and φ respectively as shown in Fig. (1.15). Ar , Aθ and Aφ are the r , θ and φ components of vector A respectively. In matrix form (1.16) can be expressed as

A = A ST U S

(1.17a)

where

⎡ar ⎢ U S = ⎢aθ ⎢⎣ a φ T for matrix transpose.

⎤ ⎥ ⎥ ⎥⎦

⎡ Ar ⎢ A S = ⎢ Aθ ⎢⎣ Aφ

⎤ ⎥ ⎥ ⎥⎦

(1.17b)

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z

Ar

A

ar

A

a

a

13

A

A A

y Fig. (1.15). The component of a vector A in the spherical coordinate system.

1.4.3.2. Differential Length, Surface Area, and Volume With reference to Fig. (1.16) the differential length dl and its magnitude dl along the path l in the spherical coordinate system can be obtained as

dl = a r dr + a θ r dθ + a φ r sin θ dφ

(1.18a)

dl = dl = dr 2 + r 2 dθ 2 + r 2 sin 2 θ dφ 2

(1.18b)

z r dθ


r θ


l

ar  dl 

dθ

aφ 

aθ  dr

r sin θ dφ 

y

r sin θ

x

dφ 

φ

Fig. (1.16). A differential length in the spherical coordinate system.

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The differential surface areas for the faces in the directions r , θ and φ of the differential element shown in Fig. (1.17) can be obtained as

dS θφ = a r r 2 sin θ dθ dφ , dS rφ = a θ r sin θ dr dφ , dS rθ = a φ rdr dθ (1.19) Similarly, the differential volume d
of the differential element shown in Fig. (1.17) is

d
= r 2 sin θ dr dθ dφ

z

(1.20)

dS θφ = a r r 2 sin θ dθ dφ  r dθ


θ


r

dθ 

dS rθ = a φ r dr dθ 

dr dS rφ = a θ r sin θ dr dφ 

r sin θ dφ 

y

r sin θ

x

φ

dφ 

Fig. (1.17). A differential surface area and volume in the spherical coordinate system.

Example 1.1 Find the length of each of the following curves using an appropriate coordinate system: (a) x 2 + y 2 = 8 z , x = y , 0 ≤ x ≤ 2 . (b) ρ = 1, 0 ≤ φ ≤ π 2 , z = 1 . (c) r = 3 , π 3 ≤ θ ≤ π 2, φ = 0 .

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Solution (a) Given that x 2 + y 2 = 4 z , x = y , 0 ≤ x ≤ 2 . Hence, on the curve we have

dy dy dz =1 =4 , dx dx dx dz 1 ⇒ = (x + y) = x dx 2

2x + 2 y

In the rectangular coordinate system, the differential length is obtained as

dl = dx 2 + dy 2 + dz 2 = 1 + ( dy dx ) + ( dz dx ) dx = 2 + x 2 dx 2

2

Thus, the length of the curve l in the range 0 ≤ x ≤ 2 is x=2

2

l

= ∫ dl = ∫ l

0

= 6 + ln

x 2 + x dx = ⎡⎢ 2 + x 2 + ln 2 2 + x 2 + 2 x ⎤⎥ ⎣2 ⎦ x =0 2

(

3+ 2

)

(b) Given ρ = 1 , 0 ≤ φ ≤ π 2 , z = 1 , ⇒ dρ = dz = 0 , then the differential length in the cylindrical coordinate systems is

dl = dρ 2 + ρ 2 dφ 2 + dz 2 = ρ dφ = dφ ⇒ l = ∫ dl = l

π 2

π

∫ dφ = 2 0

(c) Given r = 3 , π 3 ≤ θ ≤ π 2 , φ = 0 , ⇒ dr = dφ = 0 , then the differential length in the spherical coordinate system is

dl = dr 2 + r 2 dθ 2 + r 2 sin 2 θ dφ 2 = r dθ = 3 dθ Thus, the length of the curve is π 2

π π π l = ∫ dl = 3 ∫ dθ = 3⎛⎜ − ⎞⎟ = ⎝2 3⎠ 2 π 3 l

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1.5.

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TRANSFORMATION BETWEEN COORDINATE SYSTEMS

1.5.1. Cylindrical-to-Rectangular 1.5.1.1. Transformation of the Coordinates of a Point Given a point P ( ρ ,φ , z ) in the cylindrical coordinate system as shown in Fig. (1.18a), the corresponding point P ( x , y , z ) in the rectangular coordinate system can be obtained as

x = ρ cosφ y = ρ sin φ z=z

(1.21a) (1.21b) (1.21c)

Note that z is the same for both systems.

P ( ρ ,φ , z )

z

ρ az z

aφ

z

aφ

aρ

y

x x

ρ

φ

φ

φ

ax y

y

ay

φ  aρ

x (b)

(a)

Fig. (1.18). Transformation between rectangular and cylindrical coordinate systems.

1.5.1.2. Transformation of Unit Vectors The unit vectors a x , a y , and a z in the direction of x , y , and z axes in terms of the cylindrical coordinate system unit vectors a ρ , a φ , and a z can be obtained easily from Fig. (1.18b) as

a x = a ρ cosφ − a φ sin φ

(1.22a)

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a y = a ρ sin φ + a φ cosφ

(1.22b)

az = az

(1.22c)

In matrix form, the set of equations (1.22a) – (1.22c) becomes

⎡a x ⎢a ⎢ y ⎢⎣ a z

⎤ ⎡ cosφ ⎥ = ⎢ sin φ ⎥ ⎢ ⎥⎦ ⎢⎣ 0

− sin φ cosφ 0

0 ⎤ ⎡a ρ 0⎥ ⎢ aφ ⎥⎢ 1 ⎥⎦ ⎢⎣ a z

⎤ ⎥ ⎥ ⎥⎦

(1.23)

Using (1.5b) and (1.12b), (1.23) can be written in matrix notation as

U R = TCR U C

(1.24)

where TCR is transformation matrix for cylindrical-to-rectangular components and given by

TCR

⎡ cosφ = ⎢ sin φ ⎢ ⎢⎣ 0

− sin φ 0 ⎤ cosφ 0 ⎥ ⎥ 0 1 ⎥⎦

(1.25)

1.5.1.3. Transformation of Vector Components Let vector A given in the cylindrical coordinate as in (1.11), then its cylindrical coordinate components matrix A C in (1.12b) can be transformed to the rectangular coordinate components matrix

A R using the cylindrical-to-

rectangular transformation matrix TCR as

A R = TCR A C

(1.26)

Substituting A R , A C and TCR from (1.5b), (1.12b), and (1.25) respectively, the components of A in the x , y , and z directions in terms of the cylindrical coordinate components are obtained as

Ax = Aρ cosφ − Aφ sin φ

(1.27a)

A y = Aρ sin φ + Aφ cosφ

(1.27b)

Az = Az

(1.27c)

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1.5.2. Rectangular-to-Cylindrical 1.5.2.1. Transformation of the Coordinates of a Point Consider a point P ( x , y , z ) in the rectangular coordinate system, then using (1.21a) – (1.21c) the coordinates of the corresponding point P ( ρ ,φ , z ) in the cylindrical coordinate system can be obtained in terms of x , y , and z as

ρ = x2 + y2 φ = tan −1 ( y x ) z=z

(1.28a) (1.28b) (1.28c)

1.5.2.2. Transformation of Unit Vectors Multiplying both sides of (1.24) by the inverse of cylindrical-to-rectangular −1 transformation matrix TCR , yields −1 U C = TCR U R = TRC U R

(1.29)

where TRC is the rectangular-to-cylindrical transformation matrix. TRC is the inverse matrix of TCR ; hence from (1.25) we have

−1 TRC = TCR

⎡ cosφ = ⎢ sin φ ⎢ ⎣⎢ 0

− sin φ 0 ⎤ cosφ 0 ⎥ ⎥ 0 1 ⎦⎥

−1

⎡ cosφ sin φ 0 ⎤ = ⎢ − sin φ cosφ 0 ⎥ ⎢ ⎥ 0 1 ⎦⎥ ⎣⎢ 0

(1.30)

It follows from (1.5b), (1.12b), and (1.29) that the cylindrical coordinate unit vectors a ρ , a φ , and a z in terms of the rectangular coordinate unit vectors are

a ρ = a x cosφ + a y sin φ

(1.31a)

a φ = − a x sin φ + a y cosφ

(1.31b)

az = az

(1.31c)

1.5.2.3. Transformation of Vector Components The rectangular coordinate components matrix A R in (1.5b) of vector A can be transformed to cylindrical components matrix A C using TRC as

A C = TRC A R

(1.32)

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Substituting A R , A C and TRC from (1.5b), (1.12b) and (1.30) respectively, the cylindrical components of A in terms of the rectangular coordinate components are expressed as

Aρ = Ax cosφ + A y sin φ

(1.33a)

Aφ = − A x sin φ + A y cosφ

(1.33b)

Az = Az

(1.33c)

1.5.3. Spherical-to-Rectangular 1.5.3.1. Transformation of the Coordinates of a Point A point P ( r ,θ ,φ ) in the spherical coordinate system can be transformed to a corresponding point P ( x , y , z ) in the rectangular coordinate using Fig. (1.19a) as

x = r sin θ cosφ y = r sin θ sin φ z = r cosθ

(1.34a) (1.34b) (1.34c)

z r sin θ r cosθ 

θ

ar 

aφ 

a z = a r cosθ − a θ sin θ

z

r

aφ

y

ax x

φ x

(a)

ay

φ  a r sin θ + a θ cosθ

r sin θ sin φ 

r sin θ cosφ 

φ

φ

aθ

y

(b)

Fig. (1.19). Transformation between rectangular and spherical coordinate systems.

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1.5.3.2. Transformation of Unit Vectors The unit vectors a x , a y , and a z of the rectangular coordinate system in the direction of x , y , and z axes in terms of the spherical coordinate system unit vectors at a point P ( r ,θ ,φ ) can be obtained using Fig. (1.19b) as

a x = a r sin θ cosφ + a θ cosθ cosφ − a φ sin φ

(1.35a)

a y = a r sin θ sin φ + a θ cosθ sin φ + a φ cosφ

(1.35b)

a z = a r cosθ − a θ sin θ

(1.35c)

In matrix, the last set of equations becomes

⎡ a x ⎤ ⎡ sin θ cosφ cosθ cosφ − sin φ ⎤ ⎡ a r ⎢ a ⎥ = ⎢ sin θ sin φ cosθ sin φ cosφ ⎥ ⎢ a ⎥⎢ θ ⎢ y⎥ ⎢ 0 ⎦⎥ ⎢⎣ a φ − sin θ ⎣⎢ a z ⎦⎥ ⎣⎢ cosθ

⎤ ⎥ ⎥ ⎥⎦

(1.36)

Using (1.5b) and (1.17b), (1.36) can be written in a reduced from

U R = TSR U S

(1.37)

TSR is the spherical-to-rectangular transformation matrix, which is given by TSR

⎡ sin θ cosφ cosθ cosφ = ⎢ sin θ sin φ cosθ sin φ ⎢ − sin θ ⎢⎣ cosθ

− sin φ ⎤ cosφ ⎥ ⎥ 0 ⎥⎦

(1.38)

1.5.3.3. Transformation of Vector Components The spherical coordinate components matrix A S in (1.17b) of vector A can be transformed to the rectangular coordinate components matrix A R YLD the spherical-to-rectangular transformation matrix TSR as

A R = TSR A S

(1.39)

Substituting A R , A S and TSR from (1.5b), (1.17b) and (1.38) into (1.39), the components of vector A in the rectangular coordinate system Ax  A y , and Az in

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terms of the spherical coordinate system components Ar  Aθ , and Aφ  can be obtained as

Ax = Ar sin θ cosφ + Aθ cosθ cosφ − Aφ sin φ

(1.40a)

A y = Ar sin θ sin φ + Aθ cosθ sin φ + A z cosφ

(1.40b)

Az = Ar cosθ − Aθ sin θ

(1.40c)

1.5.4. Rectangular-to-Spherical 1.5.4.1. Transformation of the Coordinates of a Point Using (1.34a) – (1.34c), a point P ( x , y , z ) in the rectangular coordinate system can be transformed to a corresponding point P ( r ,θ ,φ ) in the spherical coordinate system, where

r=

θ = sin −1

(

x2 + y2 + z2

(1.41a)

)

x 2 + y 2 r = cos −1 ( z r )

φ = tan −1 ( y x )

(1.41b) (1.41c)

1.5.4.2. Transformation of Unit Vectors From (1.37) it can be shown that −1 U S = TSR U R = TR S U R

(1.42)

where T RS is the rectangular-to-spherical transformation matrix. T RS can be obtained from (1.38) as −1 SR

TRS = T

⎡ sin θ cosφ sin θ sin φ = ⎢ cosθ cosφ cosθ sin φ ⎢ cosφ ⎣⎢ − sin φ

cosθ ⎤ − sin θ ⎥ ⎥ 0 ⎦⎥

(1.43)

Substituting T RS from (1.43) into (1.42), yields

a r = a x sin θ cosφ + a y sin θ sin φ + a y cosθ

(1.44a)

a θ = a x cosθ cosφ + a y cosθ sin φ − a z sin θ

(1.44b)

a φ = − a x sin φ + a y cosφ

(1.44c)

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1.5.4.3. Transformation of Vector Components When A is given in the rectangular coordinate, its components matrix A R in (1.5b) can be transformed to the spherical components matrix A S using T RS as

A S = TRS A R

(1.45)

Substituting A R , A S and T RS from (1.5b), (1.17b), and (1.43) into (1.45), yields

Ar = Ax sin θ cosφ + A y sin θ sin φ + Az cosθ

(1.46a)

Aθ = A x cosθ cosφ + A y cosθ sin φ − A z sin θ

(1.46b)

Aφ = − A x sin φ + A y cosφ

(1.46c)

1.5.5. Spherical-to-Cylindrical 1.5.5.1. Transformation of the Coordinates of a Point Comparing the set of equations (1.21a) – (1.21c) with the set (1.34a) – (1.34c), the relation between cylindrical coordinates ( ρ , φ , z ) and spherical coordinates ( r ,θ ,φ ) can be obtained by eliminating the rectangular coordinate system

variables x , y , and z . Hence, the point P ( r ,θ ,φ ) in the spherical coordinate can be transformed to the corresponding point P ( ρ ,φ , z ) in cylindrical coordinate as

ρ = r sin θ z = r cosθ φ =φ

(1.47a) (1.47b) (1.47c)

1.5.5.2. Transformation of Unit Vectors Using (1.26) and (1.39), we can write the relationship between cylindrical and spherical components for the vector A as

TCR A C = TSR A S

(1.48)

Multiplying both sides of (1.48) by inverse of TCR , the cylindrical components matrix is obtained as −1 A C = TCR TSR A S = TSC A S

(1.49)

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where TSC is the spherical-to-cylindrical transformation matrix and can be expressed from (1.49) as −1 TSC = TCR TSR = TRC TSR

(1.50)

Substituting TRC and TSR from (1.30) and (1.38) into (1.50) and simplifying, gives the spherical-to-cylindrical transformation matrix TSC as

TSC

⎡ sin θ =⎢ 0 ⎢ ⎣⎢ cosθ

cosθ 0 − sin θ

0⎤ 1⎥ ⎥ 0 ⎦⎥

(1.51)

The unit vectors of the spherical coordinate system U S can be transformed to that of cylindrical coordinate system U C YLD the matrix TSC as

U C = TS C U S

(1.52)

Substituting U C , U S , and TSC from (1.12b), (1.17b), and (1.50) respectively into (1.52), after multiplication, the cylindrical coordinate system unit vectors a ρ , a φ , and a z can be expressed in terms of the spherical coordinate unit vectors as

a ρ = a r sin θ + a θ cosθ

(1.53a)

aφ = aφ

(1.53b)

a z = a r cosθ − a θ sin θ

(1.53c)

1.5.5.3. Transformation of Vector Components The cylindrical coordinate components of A in terms of the spherical coordinate components are obtained by substituting A C , A S and TSC from (1.12b), (1.17b), and (1.51) into (1.49), and after multiplying the matrices, the cylindrical components of the vector are obtained in terms of the spherical ones as

Aρ = Ar sin θ + Aθ cosθ

(1.54a)

Aφ = Aφ

(1.54b)

Az = Ar cosθ − Aθ sin θ

(1.54c)

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1.5.6. Cylindrical-to- Spherical 1.5.6.1. Transformation of the Coordinates of a Point Using the relations (1.47a) – (1.47c), the point P ( ρ ,φ , z ) in the cylindrical coordinate can be transformed to the corresponding point P ( r ,θ ,φ ) in the spherical coordinate, as

r = ρ2 + z2 θ = tan −1 ( z ρ ) φ =φ

(1.55a) (1.55b) (1.55c)

1.5.6.2. Transformation of Unit Vectors The spherical-to-cylindrical transformation matrix TCS can be obtained by inversion of TSC in (1.51). Thus,

−1 SC

TCS = T

⎡ sin θ =⎢ 0 ⎢ ⎣⎢ cosθ

cosθ 0 − sin θ

0⎤ 1⎥ ⎥ 0 ⎦⎥

−1

⎡ sin θ = ⎢ cosθ ⎢ ⎣⎢ 0

cosθ ⎤ 0 − sin θ ⎥ ⎥ 1 0 ⎦⎥ 0

(1.56)

The unit vectors of the cylindrical coordinate system U C can be transformed to that of the spherical coordinate system U S YLD TCS as

U S = TCS U C

(1.57)

Substituting U C , U S , and TCS from (1.12b), (1.17b), and (1.56) into (1.57), after multiplication, the spherical coordinate system unit vectors a r , aθ , and aφ can be expressed in terms of the spherical coordinate system unit vectors a ρ , aφ , and

a z as yields

a r = a ρ sin θ + a z cosθ

(1.58a)

a θ = a ρ cosθ − a z sin θ

(1.58b)

aφ = aφ

(1.58c)

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1.5.6.3. Transformation of Vector Components The cylindrical coordinate components matrix A C in (1.12b) can be transformed to the spherical coordinate components matrix A S using TCS as

A S = TCS A C

(1.59)

Substituting A C , A S and TCS from (1.12b), (1.17b), and (1.56) into (1.59), the components of A in the r , θ , and φ directions in terms of the cylindrical coordinate components are obtained as

Ar = Aρ sin θ + Az cosθ

(1.60a)

Aθ = Aρ cosθ − A z sin θ

(1.60b)

Aφ = Aφ

(1.60c)

Example 1.2 Convert the point coordinate systems.

(2 ,

5,4

)

from rectangular to cylindrical and spherical

Solution Given that x = 2 , y = 5 , z = 4 . Using (1.28) the coordinates of the corresponding point ( ρ ,φ , z ) in the cylindrical coordinate system is obtained as

ρ = x2 + y2 = 4 + 5 = 3 φ = tan −1 ( y x ) = tan −1 ( 5 2 ) = 48.17  z=4 Similarly, the corresponding point ( r ,θ ,φ ) in the spherical coordinate system is obtained using (1.41) as

r = x 2 + y 2 + z 2 = 4 + 5 + 16 = 5 θ = cos −1 ( z r ) = cos −1 ( 4 5 ) = 36.86 

φ = tan −1 ( y x ) = tan −1 ( 5 2 ) = 48.17 

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Example 1.3 Convert the vector A = ( y 2 x )a x + x 2 + y 2 a y + a z from rectangular to cylindrical and spherical coordinate systems. Solution Given the components of A are A x = ( y 2 x ), A y =

x 2 + y 2 , and Az = 1.

Using the rectangular-to-cylindrical transformation matrix TRC given by (1.30), the components of A in the cylindrical coordinate system are obtained as

⎡ Aρ ⎢A ⎢ φ ⎢⎣ A z

⎤ ⎡ cosφ ⎥ = ⎢ − sin φ ⎥ ⎢ ⎥⎦ ⎢⎣ 0

sin φ

0 ⎤ ⎡ Ax cosφ 0 ⎥ ⎢ A y ⎥⎢ 0 1 ⎥⎦ ⎢⎣ A z

⎤ ⎡ cosφ ⎥ = ⎢ − sin φ ⎥ ⎢ ⎥⎦ ⎢⎣ 0

sin φ cosφ 0 0

2 0⎤ ⎡ y x ⎢ 0⎥ ⎢ x 2 + y 2 ⎥ 1 ⎥⎦ ⎢⎣ 1

⎤ ⎥ ⎥ ⎥ ⎦

Thus,

Aρ = ( y 2 x )cosφ + x 2 + y 2 sin φ Aφ = − ( y 2 x )sin φ + x 2 + y 2 cosφ

Az = 1 Same results can be obtained by using (1.33). Substituting x = ρ cosφ and y = ρ sin φ , then A in the cylindrical coordinate system becomes

A = ρ sin φ ( sin φ + 1)a ρ + ρ (cosφ − tan φ sin 2 φ )a φ + za z Similarly, using the rectangular-to-spherical transformation matrix TR S given by (1.43), the components of A in the spherical coordinate system are

⎡ Ar ⎢A ⎢ θ ⎢⎣ Aφ

⎤ ⎡ sin θ cosφ ⎥ = ⎢ cosθ cosφ ⎥ ⎢ ⎥⎦ ⎣⎢ − sin φ

sin θ sin φ cosθ sin φ cosφ

2 cosθ ⎤ ⎡ y x ⎢ − sin θ ⎥ ⎢ x 2 + y 2 ⎥ z 0 ⎦⎥ ⎢⎣

⎤ ⎥ ⎥ ⎥ ⎦

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Thus,

Ar = ( y 2 x )sin θ cosφ + x 2 + y 2 sin θ sin φ + z cosθ

Aθ = ( y 2 x )cosθ cosφ + x 2 + y 2 cosθ sin φ − z sin θ

Aφ = − ( y 2 x )sin φ + x 2 + y 2 cosφ

Same components can be obtained by using (1.46). Substituting x = r sin θ cosφ , y = r sin θ sin φ , and z = r cosθ , then A in the spherical coordinate system becomes

A

= r (sin 2 θ sin 2 φ + sin 2 θ sin φ + cos2 θ ) a r

+ r sin θ cosθ (sin 2 φ + sin φ − 1) aθ + r sin θ (cosφ − tan φ sin 2 φ ) aφ

1.6.

THE GENERALIZED ORTHOGONAL COORDINATE SYSTEM

The coordinate systems that are discussed in the previous sections may be derived from a general orthogonal curvilinear coordinate system ( u , v , w ) as special cases. The general orthogonal curvilinear coordinate system ( u , v , w ) is represented by arbitrary curvilinear mutually orthogonal axes u , v , and w as shown in Fig. (1.20a).

z

u

ut

u

wt

w

au dSvw

Δu

awdSuv

(u, v, w) v

vt

(a)

Δw  av dSuw

y x

w

Δv

v (b)

Fig. (1.20). The general mutually orthogonal curvilinear coordinate system ( u , v , w ) . (a) Axes and tangent vectors. (b) Differential volume element.

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Let u t , v t , and w t be the tangent vectors to the axes u , v , and w respectively as shown in Fig. (1.20a). These tangent vectors can be expressed in terms of rectangular coordinate system variables x , y , and z , and unit vectors a x , a y , and a z as

dx +ay du dx vt = ax +ay dv dx wt = ax +ay dw ut = a x

dy dz +az du du dy dz +az dv dv dy dz +az dw dw

(1.61a) (1.61b) (1.61c)

The unit vectors in the direction of the tangent vectors are

a u = u t ut ,

a v = v t vt ,

aw = uw uw

(1.62)

where u t , v t , and w t are the scaling factors. These factors are related to the

rectangular coordinate variables ( x , y , z ) by

u t = ( dx du ) + ( dy du ) + ( dz du ) 2

2

v t = ( dx dv ) + ( dy dv ) + ( dz dv ) 2

2

2

(1.63a)

2

(1.63b)

wt = ( dx dw ) + ( dy dw ) + ( dz dw ) 2

2

2

(1.63c)

The differential length dl in the generalized coordinate system can be written as

dl = a u u t du + a v v t dv + a w wt dw

(1.64a)

and its magnitude is

dl = ( u t du ) + ( v t dv ) + ( wt dw ) 2

2

2

(1.64b)

Referring to Fig. (1.20b), the differential surface areas in the generalized coordinate system can be written in the planes vw , uw , and uv respectively as dSvw = au vt wt dvdw

(1.65a)

Vector Analysis


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dSuw = avut wt dudw dSuv = a wut vt dudv

29

(1.65b) (1.65c)

The differential volume d
in the generalized coordinate system is

d
== ut vt wt dudvdw

(1.66)

The rectangular, cylindrical, and spherical coordinates system quantities can be obtained from the generalized coordinate system formulas (1.61) – (1.66), by substituting u , v , w , u t , v t , w t , a u , a v , and a w by their corresponding values of the respective coordinate system listed in Table 1.2. Table 1.2. Coordinate systems variables, scaling factors, and unit vectors. Coordinate System

Coordinate Variables

Scaling Factors

Unit Vectors

u

v

w

ut

vt

wt

au

av

aw

z

1

1

1

ax

ay

az

Cylindrical

x ρ

y φ

z

1

x

1

aρ

aφ

az

Spherical

r

θ

φ

1

r

r sin θ

ar

aθ

aφ

Generalized Rectangular

Example 1.4 Derive the differential volume in rectangular, cylindrical, and spherical coordinate systems from the differential volume in the generalized coordinate system. Solution In the generalized coordinate system ( u , v , w ) d
is given by (1.66) as

d
== u t v t wt dudvdw In the rectangular coordinate system, from Table 1.2, u t = v t = wt = 1, x = u , y = v , and z = w . Hence,

d
== u t v t wt dudvdw = dxdydz In the cylindrical coordinate system, from Table 1.2, u t = 1 v t = ρ  wt = 1 u = ρ , v = φ , and w = z . Hence,

d
== u t v t wt dudvdw = ρ dρ dφ dz

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In the spherical coordinate system, we have  u t = 1 v t = r  wt = r sin θ  u = r ,

v = θ , and w = φ from Table 1.2 . Hence,

d
== u t v t wt dudvdw = r 2 sin θ dr dθ dφ 1.7.

VECTOR ALGEBRA

1.7.1. Vectors Addition and Subtraction Addition or subtraction of two vectors A and B is carried out component by component provided that both vectors are given in the same coordinate system. If A and B are given in rectangular coordinate system as

A = a x Ax + a y A y + a z Az

(1.67a)

B = a x Bx + a y B y + a z Bz

(1.67b)

The new vectors C = A + B and D = A − B are

C = A + B = a x ( Ax + Bx ) + a y (Ay + By ) + a z ( Az + Bz ) D = A − B = a x ( Ax − Bx ) + a y (Ay − By ) + a z ( Az − Bz )

(1.68) (1.69)

1.7.2. Vectors Multiplication There are two types of vector multiplication, the dot product, and the cross product. The dot product results in a scalar quantity, while the cross product results in a vector quantity. For this reason, the dot product is known as a scalar multiplication and the cross product as a vector multiplication. 1.7.2.1. The Dot Product For the vectors A and B given by (1.67a) and (1.67b), and shown in Fig. (1.21), the dot product is defined by

C = A ⋅ B = A B cosθ

(1.70)

where the result C is a scalar and θ is the smaller angle between A and B shown in Fig. (1.21).

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B A⋅B = |A||B| cosθ

θ A Fig. (1.21). The dot product of two vectors.

Since the magnitude of the unit vector is unity, and the unit vectors of the different coordinate system are orthogonal, then using (1.70) it can be shown that

a x ⋅a x = a y ⋅a y = a z ⋅a z =1

a x ⋅ a y = a x ⋅ a z = a y ⋅ a z = 0 (1.71a)

a ρ ⋅a ρ = aφ ⋅aφ = a z ⋅a z =1

a ρ ⋅ a φ = a ρ ⋅ a z = a φ ⋅ a z = 0 (1.71b)

a r ⋅a r = aθ ⋅aθ = aφ ⋅aφ =1

a r ⋅ a θ = a r ⋅ a φ = a θ ⋅ a φ = 0 (1.71c)

Referring to Fig. (1. 22) the projection of vector B along vector A is obtained as

B cosθ =

A ⋅B = a A ⋅B A

(1.72)

where a A is the unit vector in the direction of A . Similarly, the projection of vector A along B is obtained as

A cosθ =

B ⋅A =aB ⋅A B

(1.73)

where a B is the unit vector in the direction of B . Using (1.72) or (1.73) with (1.2) it can be shown that

cosθ = a A ⋅ a B B

 |B| cos Fig. (1.22). The projection of vector B along vector A .

A

(1.74)

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Substituting A and B from (1.67) into (1.70) and multiplying component by component, the dot product of the vectors A and B is obtained as

C = A ⋅ B = Ax B x + A y B y + Az B z

(1.75)

From (1.70) and (1.75) it can be shown that

cos θ =

A ⋅ B Ax B x + A y B y + Az B z = A B A B

(1.76)

1.7.2.2. The Cross Product For any two vectors A and B , the cross product is defined by

C = A × B = a n A B sin θ

(1.77)

where a n is a unit vector perpendicular to both A and B . The angle θ is the smaller angle between A and B as shown in Fig. (1.23). The result C is a vector perpendicular to both A and B , and oriented in the direction of advance of a right-handed screw aligned along C when rotated from the tip of A to that of B as illustrated in Fig. (1.23). From (1.77) the unit vector a n can be expressed as

an =

A×B A B sin θ

(1.78)

C = A×B =an |A||B| sinθ

B

θ C an A Fig. (1.23). The cross product of two vectors.

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Since the magnitude of the unit vector is unity, and the unit vectors of the different coordinate system are orthogonal, then using (1.77) it can be shown that

a x ×a x = a y ×a y = a z ×a z = 0

(1.79a)

a ρ ×a ρ = aφ ×aφ = a z ×a z = 0

(1.79b)

a r ×a r = aθ ×aθ = aφ ×aφ = 0

(1.79c)

a x × a y = a z  a y × a z = a x  a z × a x = a y

(1.80a)

a ρ × a φ = a z  a φ × a z = a ρ  a z × a ρ = a φ

(1.80b)

a r × a θ = a φ  a θ × a φ = a r  a φ × a r = a θ

(1.80c)

If A and B are given in rectangular coordinate system as in (1.67), then the cross product of the vectors A and B can be obtained by substituting them into (1.77) and multiplying component by component as

C = A×B (1.81) = a x ( A y B z − Az B y ) − a y ( Ax B z − Az B x ) + a z ( Ax B y − A y B x ) The right-hand side of (1.81) is the determinant of a 3-by-3 matrix with rows (a x , a y , a z ) ( Ax , A y , Az ), and ( B x , B y , B z ). Thus, (1.81) can be written as

ax

ay

az

C = A × B = Ax

Ay

Az

Bx

By

Bz

(1.82)

1.7.3. Triple Product For arbitrary vectors A , B , and C the scalar triple product is defined by

A ⋅ ( B × C ) = B ⋅ (C × A ) = C ⋅ ( A × B )

(1.83)

Assuming that A , B , and C are given in rectangular coordinate system, (1.83) may be written in the determinant form as

Ax

Ay

Az

D = A ⋅ B × C = Bx

By

Bz

Cx

Cy

Cz

(1.84)

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The vector triple product of A , B , and C is defined by

A × ( B × C ) = B ( A ⋅ C ) − C( A ⋅ B ) = ( A ⋅ C )B − ( A ⋅ B )C

(1.85)

1.7.4. Vectors Properties Let A , B , and C be any vectors and let α and β be arbitrary scalars. The properties of vectors algebra on vectors A , B , and C are: 1. Commutative law

A+B =B+A A⋅B = B⋅A

(1.86) (1.87)

A × B = −B × A

(1.88)

( A + B ) + C = A + (B + C ) ( αβ )A = α (β A ) = β ( α )A

(1.89) (1.90)

α(A + B ) = α B + α A A ⋅ (B + C) = B ⋅ A + A ⋅ C

(1.91) (1.92)

A × (B + C ) = A × B + C× A (α + β )A = α A + β A

(1.93) (1.94)

2. Anti-commutative

3. Associative Law

4. Distributive law

5. Linearly dependent vectors: the vectors A , B , and C are said to be linearly dependent if

( A × B )⋅ C = 0

(1.95)

( A × B )⋅ C ≠ 0

(1.96)

and linearly independent if

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6. Other properties

αA = α A

(1.97)

A⋅A = A A ⋅ (B × A ) = 0

(1.98)

2

A×A = 0

A × ( B × C ) ≠ ( A × B )× C

(1.99) (1.100) (1.101)

Example 1.5 Given that A = a x + 1.5 a y − 5.5 a z and B = 2 a x − 5 a y + a z , find (a) (b) (c) (d)

C = 2A + B . The smallest angle between B and C . The unit vector along B . The vector projection of C along B .

Solution (a) Given A = a x + 1.5 a y − 5.5 a z and B = 2 a x − 5 a y + a z , then

C = 2 A + B = 2 (a x + 1.5 a y − 5.5 a z ) + 2 a x − 5 a y + a z = 4 a x − 2 a x − 10 a z (b) Using (1.76), yields

cosθ =

B ⋅ C (2 a x + 5 a y − a z ) ⋅ ( 4 a x − 2 a x − 10 a z ) 8 − 10 + 10 2 = = = B C 60 15 4 + 25 + 1 16 + 4 + 100

⇒

θ = cos −1 ( 2 15 ) = 82.3 

(c) The unit vector along B is

aB =

B 2a x + 5a y − a z = = 0.37 a x + 0.91a y − 0.18 a z B 4 + 25 + 1

(d) Using (1.72), the scalar projection of C along B can be obtained as

C cosθ = 16 + 4 + 100 × 2 15 = 30 × 4 15

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Using the result of (c), the vector projection of C along B is

a B C cosθ

= 30 × =

4 4 ⎛ 2 a x + 5a y − a z a B = 30 × ⎜⎜ 15 15 ⎝ 30

4 (2 a x + 5 a y − a z 15

⎞ ⎟⎟ ⎠

)

Example 1.6 Given that A = a x − 4 a y − a z , B = 2 a x − 6 a y + a z and C = a x + 2 a z , find (a) A × B (b) ( A × B ) ⋅ C Solution (a) Given that A = a x − 4 a y − a z , B = 2 a x − 6 a y + a z . Using (1.82), then

ax ay az ax A × B = Ax A y Az = 1 Bx B y Bz 2 = −10 a x − 3a y + 2 a z

ay

az

− 4 −1 −6

1

(b) Similarly, substituting the components of A , B and C into (1.84) yields

Cx Cy Cz 1 0 2 ( A × B ) ⋅ C = C ⋅ ( A × B ) = Ax A y Az = 1 − 4 − 1 Bx B y Bz 2 −6 1 = −10 + 4 = −6 Alternatively, the result of A × B obtained in (b) can be used to find C ⋅ ( A × B ) as C ⋅ ( A × B ) = ( a x + 2 a z ) ⋅ (− 10 a x − 3 a y + 2 a z = −6

)

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THE POSITION VECTOR, THE LINE, THE PLANE

1.8.1. The Position Vector A vector that defining a point in space with respect to the origin of the coordinate system is called a position vector. The position vector of an arbitrary point in the rectangular coordinate system P ( x 0 , y 0 , z 0 ) shown in Fig. (1.24), is the vector r given by

r = a x x0 + a y y0 + a z z 0

z

(1.102)

P ( x 0 , y 0 , z 0 ) r z0 

az

ay

ax

x0 

x

y

y0 

Fig. (1.24). The position vector.

Let the position vectors of points A and B be ra and rb respectively, then the vector directed from point A to B with its tip at B as shown in Fig. (1.25a), is

rba = rb − r a

(1.103)

Similarly, the vector directed from point B to A with its tip at A , as shown in Fig. (1.25b), is

rab = ra − rb rba and rab are known as the distance vectors between points A and B .

(1.104)

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A

A

rba

rab

ra

ra B

Origin

B Origin

rb

rb

(a)

(a)

Fig. (1.25). The distance vectors.

Example 1.7 Let the position vectors of point A and B be ra and rb respectively. Find the position vector rc of the mid-point C of the line from A to B . Solution Referring to Fig. (1.26), we have rca = rbc since C is the mid-point of the line AB . We can write

rca = rc − ra ,

rbc = rb − rc

Since rca = rbc , then the position vector rc of point C is obtained by equating the right-hand sides of the above equations and solving for rc as

rc − ra = rb − rc

⇒ rc =

r a + rb 2

A

rca C

ra

rbc rc B

Origin Fig. (1.26). Geometry of the problem of Example 1.7.

rb

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1.8.2. The Straight Line Consider a straight line passing through a point A defined by the position vector ra as shown in Fig. (1. 27) and P ( x , y , z ) is an arbitrary point on the line with position vector r . If the line is parallel to a vector b , then the locus of point P ( x , y , z ) on the straight line can be written as

r = ra + mb

(1.105)

where m is a scalar quantity in the interval − ∞ ≤ m ≤ +∞ . Let ra = x a a x + y a a y + z a a z and b = b x a x + b y a y + b z a z , then the vector form of the straight line in (1.105) can be written as

x − xa bx

=

y − ya by

=

z − za bz

=m

(1.106)

If the line passing through the points (xa , ya , za ) and (xb , yb , zb ), the vector b will be b = ( x b − x a )a x + ( y b − y a )a y + ( z b − z a )a z and the corresponding equation of the line becomes

x − xa y − ya z − za = = =m xb − x a y b − y a z b − z a z Line

(1.107)

P

mb A

r ra

x Fig. (1.27). The equation of a straight line.

y

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1.8.3. The Plane A plane is fully defined when a fixed point on it and a vector normal to it are given. Consider the fixed point A defined by a position vector r a on the plane shown in Fig. (1.28), a normal vector to the plane N , and an arbitrary point P ( x , y , z ) on the plane. It is clear that from the Figure the vector ra − r is perpendicular to the normal N . Thus,

( ra − r ) ⋅ N = 0

(1.108)

This is valid for any point P ( x , y , z ) on the plane. Consequently, (1.108) is the locus of the point P ( x , y , z ) and can be written as

ra ⋅ N = r ⋅ N

(1.109)

N

z

P A

r

Plane

ra  y

x Fig. (1.28). The equation of a plane.

Example 1.8 Find the equation of a line in rectangular coordinate system passing through the points (1, − 1, 2 ) and ( 4 , 0 ,1) . Solution Let (1, − 1, 2 ) be point A and ( 4 , 0 ,1) point B and using (1.107), the equation of the line is

x −1 y − 0 z −1 = = 4 −1 0 +1 1− 2

⇒

1 ( x − 1) = y = 1 − z 3

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Example 1.9 Find the equation of a plane in a rectangular coordinate system containing the point (1, 0 ,1) with normal N = 2 a x − 3 a y + a z . Solution Since the plane contains the point (1, 0 ,1) , then ra = a x + a z . We have also r = a x x + a y y + a z z , and then the equation of the plane is

ra ⋅ N = r ⋅ N 1.9.

⇒

3= 2x −3y + z

⇒ 2x −3y + z −3= 0

SCALAR AND VECTOR FIELDS

A field is a function that defines a physical quantity everywhere and at any time in a certain region. Fields can be expressed as a physical entity that capable of carrying energy and momentum, and interact with the surrounding objects exactly like particles of matter [12]. This field concept is an essential to describe and understand the electromagnetic phenomenon. The field is said to be a scalar field when the physical quantity defined by the field is a scalar which possesses magnitude only. An example of a scalar field is the electric potential distribution in a certain region. The electric potential at any point ( x , y , z ) in a certain region at an instant of time is denoted by a function V ( x , y , z ) as a scalar field. The field is said to be a vector field when the physical quantity defined by the field is a vector which possesses both magnitude and direction such as the electric and magnetic fields. The electric field at any point ( x , y , z ) in a certain region at an instant of time is denoted by a function E ( x , y , z ) as vector fields. Similarly, the magnetic field H at a given region is denoted by the vector field H ( x , y , z ). 1.10. THE LINE INTEGRAL Consider a force F acting on a particle as shown in Fig. (1.29). It is known that from mechanics the work done by F to move the particle a distance Δl along path l is given by ΔW = F Δl cosθ = F ⋅ Δl . When Δl → 0 , then the work is dW = F ⋅ dl

(1.110)

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A l

dl

θ

Fdl cosθ F B

Origin



Fig. (1.29). The work done by a force F to move a particle from point A to point B .

Hence, the total work done by F in moving the particle along l from point A to B is obtained by adding all differential work elements dW at the points along the path l from point A to B . This can be accomplished by integrating both sides of (1.110) and the work is the line integral of F along the path l from point A to B , or B

W = ∫ F ⋅ dl

(1.111)

A

Generally, the line integral of a vector field A along a path l is given by

∫ A ⋅ dl

(1.112)

l

If the path l is a closed loop, the integral in (1.112) is called the circulation of A around l and is written as (1.113) ∫ A ⋅ dl l

The vector field A is said to be a conservative field if its circulation is zero. Example 1.10 Find the work done by a force F = ( x 2 − y )a x + ( y 2 − z )a y + ( z 2 − x )a z in moving its point of application from ( 0 , 0 , 0 ), (1,1,1) along (a) The straight line between the two points. (b) The curve x = t , y = t 2 , z = t 3 , 0 ≤ t ≤ 1 .

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Solution (a) The equation of the straight line between ( 0 , 0 , 0 ) , (1,1,1) is x = y , y = z , z = x . Hence, the force is F = ( x 2 − x )a x + ( y 2 − y )a y + ( z 2 − z )a z . Substituting F and dl = a x dx + a y dy + a z dz in (1.111), yields B

x =1

1 ⎡ x3 x2 ⎤ =− W = ∫ F ⋅ dl = ∫ ( x − x )dx + ∫ ( y − y )dy + ∫ ( z − z )dz = 3 ⎢ − ⎥ 2 ⎦ x =0 2 ⎣ 3 A 0 0 0 1

1

2

1

2

2

(b) Along the curve x = t , y = t 2 , z = t 3 , we have dx = dt , dy = 2 tdt , dz = 3t 2 dt and F = (t 2 − t )a x + (t 4 − t 2 )a y + (t 6 − t 3 )a z . Substituting F and

dl = a x dt + 2 a y tdt + 3a z t 2 dt in (1.111), yields B

1

1

1

W = ∫ F ⋅ dl = ∫ (t 2 − t )dt + 2 ∫ (t 4 − t 2 )t dy + 3 ∫ t 2 (t 6 − t 3 )dz A

0

0

t =1

0

t =1

t =1



t ⎤ t ⎤ t ⎤ 1 ⎡t ⎡t ⎡t ⇒ W = ⎢ − ⎥ + 2 ⎢ − ⎥ + 3⎢ − ⎥ = − 2 ⎣ 3 2 ⎦ t =0 ⎣ 6 4 ⎦ t =0 ⎣ 9 6 ⎦ t =0 3

2

6

4

9

6

Results along the two paths in (a) and (b) are the same, which verifies that the work done by the given force is path independent. 1.11. THE GRADIENT OF A SCALAR FIELD Consider a scalar field
in a generalized orthogonal coordinate system. If the field at any point P1 (u , v , w ) is
( u , v , w ), at a point P2 ( u + Δu , v , w ) very close to P1 in the direction of u will be
(u + Δu , v , w ) . Let the gradient from

P1 to P2 along the path dl u and in the direction of u be Fu , then

Fu = lim


( u + Δu , v , w ) −
( u , v , w ) Δl u

Δu →0

=

∂
∂l u

(1.114)

From (1.64a) we can write ∂l u = u t ∂u . Thus, (1.114) becomes

Fu =

1 ∂
u t ∂u

(1.115a)

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Similarly, if we moved small distances in the directions of v and w , the gradients in the directions of v and w respectively become

Fv =

1 ∂
, v t ∂v

Fw =

1 ∂
wt ∂w

(1.115b)

Note that since Fu , Fv , Fw depend on the direction, they represent vectors in the directions of u , v , and w respectively. Hence, in terms of generalized orthogonal coordinates unit vectors a u , a v , and a w , the resultant gradient of the scalar field
at the point P1 ( u , v , w ) can be written as

1 ∂ 1 ∂ 1 ∂ ⎞ ⎛ + aw + av ∇
= a u Fu + a v Fv + a w Fw = ⎜ a u ⎟
(1.116) v t ∂v wt ∂w ⎠ ⎝ u t ∂u It follows that the del ( ∇ ) operator in the generalized orthogonal coordinate system is

∇ = au

1 ∂ 1 ∂ 1 ∂ + av + aw u t ∂u v t ∂v wt ∂w

(1.117)

Equation (1.116) gives the gradient of the scalar field
in terms of the generalized orthogonal coordinate system. In the rectangular coordinate system, we have u t = v t = wt = 1, x = u , y = v , z = w , a x = a u , a y = a v , and a z = a w . Accordingly

∇
= a x

∂
∂
∂
+ay +az ∂x ∂y ∂z

(1.118)

In the cylindrical coordinate system  u t = 1 v t = ρ  wt = 1 u = ρ , v = φ ,

w = z , a u = a ρ , a v = a φ , and a w = a z . Hence, the gradient of
in cylindrical coordinate system becomes

∇
= a ρ

∂
∂
1 ∂
+ aφ +az ∂ρ ρ ∂φ ∂z

(1.119)

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In spherical coordinate system we substitute  u t = 1 v t = r  wt = r sin θ  u = r , v = θ , w = φ , a u = a r , a v = a θ , and a w = a φ to get

∇
= a r

∂
1 ∂
1 ∂
+ aθ + aφ ∂r r ∂θ r sin θ ∂φ

(1.120)

For a vector in space with a unit vector a the rate of change of
in the direction of a is known as the directional derivative of
along a and given by ∇f ⋅ a  Example 1.11 If
= − xy 2 − 4 yz 2 , find the gradient of
and the directional derivative of it along the unit vector a = ( 2 a x + a y − 2 a z ) 3 at the point ( 2 ,1, 0 ) . Solution Since
= − xy 2 − 4 yz 2 , then the gradient is

∇
= a x

∂
∂
∂
+ay +az = − y 2 a x − 2 ( xy + 2 z 2 )a y − 8 yza z ∂x ∂y ∂z

At the point ( 2 ,1, 0 ) the gradient of
is ∇
= − a x − 4 a y . Hence, the directional derivative of
along the unit vector a is obtained as

∇
⋅ a = − (a x + 4 a y ) ⋅ (2 a x + a y − 2 a z

)

3 = − ( 2 + 4 ) 3 = −2

1.12. SURFACE INTEGRAL AND THE FLUX OF A VECTOR Suppose we have a surface S and let a unit vector a n acting normal to S at any point P on S . If the point P changes position over S without crossing its border and returns back to its original position for all movements of P , then S is said to be an oriented surface. Note that all closed surfaces are oriented. If the equation describing the surface S is f , the normal unit vector a n is

an = ±

∇f ∇f

(1.121)

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The positive sign is taken when a n is directed outwards of the surface and the negative sign for inward direction. Let A be a vector field on an oriented surface S , then the integral of A ⋅ a n over S is called the flux of the field A across S in the positive direction, or

ψ = ∫ A ⋅ a n dS = ∫ A ⋅ dS S

(1.122a)

S

For closed surfaces, (1.122a) is written as

ψ = ∫ A ⋅ dS = ∫ A ⋅ a n dS S

(1.122b)

S

For any plane defined by f ( x , y , z ) the differential surface area is determined in the xy , yz , or zx planes as the projections of differential surface area on these planes, or

⎧ ∇f dxdy , xy plane ⎪ dS xy = ∇f ⋅ a z ⎪ ⎪ ∇f dS ′ = ⎨ dS yz = dydz , yz plane ∇f ⋅ a x ⎪ ⎪ ∇f dzdx , zx plane ⎪ dS zx = ∇ f ⋅ a y ⎩

(1.123)

Example 1.12 Find the flux of the vector field A = xza x + yza y outwards across the surface S cut from the cylinder x 2 + y 2 = a 2 , y ≥ 0 by z = 0 , z = a . Solution The geometry of the surface is shown in Fig. (1.30). The equation of the surface is f = x 2 + y 2 − a 2 , then

an = ±

2 xa x + 2 ya y xa x + ya y ∇f = =± a ∇f 2 x2 + y2 

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Since the flux is outwards, then a n is positive and the flux is

ψ

= ∫ A ⋅ a n dS S

⎛ xa x + ya y = ∫ ( xza x + yza y ) ⋅ ⎜⎜ a ⎝ S

⎞ x 2 z + zy 2 ⎟⎟ dS = ∫ dS = a ∫ z dS a ⎠ S S

z a

dz 

dS = a n a dφ dz 

y φ

dφ 

x Fig. (1.30). Geometry of the surface of Example 1.12.

Referring to Fig. (1.30), in the cylindrical coordinate system dS = a dφ dz , and y = a sin φ . Substituting these quantities in the flux equation ψ above, yields π a

a

π ⎡ z2 ⎤ ψ = a ∫∫ z dφ dz = π a ⎢ ⎥ = a 4 ⎣ 2 ⎦0 2 0 0 2

2

1.13. THE DIVERGENCE OF A VECTOR Let A ( u , v , w ) be an arbitrary vector field flows in the direction from point P1 (u , v , w ) to P2 (u + Δu , v + Δv , w + Δw ) through the unit volume Δ
= ΔuΔvΔw in the generalized orthogonal coordinate system shown in Fig. (1.31). The divergence of A ( u , v , w ) at P1 (u , v , w ) is defined as

1 A ⋅ dS Δ
→0 Δ
∫ S

∇ ⋅ A = lim

(1.124)

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where A = a u Au + a v Av + a w Aw . Referring to Fig. (1.31), dS is the surface enclosing the volume Δ
. In the generalized coordinate system dS can be written as

dS = a u v t wt dvdw + a v wt u t dudw + a w u t v t dudv

Au ( u + Δu )

u

Differential volume Δ


Av ( v )  Δu 

Aw ( w ) 

(1.125)

P2

Aw ( w + Δw ) w

P1  Δv 

Δw 

Au ( u ) 

Av ( v + Δv )

v Fig. (1.31). The divergence of a vector.

Substituting A and dS into (1.124), yields

1 ∇ ⋅ A = lim Δ
→ 0 Δ
1 Δ
→ 0 Δ


+ lim

1 Δ
→ 0 Δ


+ lim

w + Δw v + Δv

∫ ∫ v w A dvdw t

t

u

w v w + Δw u + Δu

∫ ∫w u

t

∫ ∫u v

Aw dudv

t

(1.126)

Av dudw

w u v + Δv u + Δu

t

v

t

u

Au , Av , and A w can be considered constant between the points P1 and P2 . Hence, (1.126) becomes ∇ ⋅ A = lim

Δ
→ 0

+ lim

[v t ( u + Δu ) wt ( u + Δu ) Au ( u + Δu ) − v t ( u ) wt ( u ) Au ( u )] Δ
[ wt ( v + Δv )u t ( v + Δv ) Av ( v + Δv ) − wt ( v )u t ( v ) Av ( v )]

dvdw

(1.127) dudw Δ
[u ( w + Δw )v t ( w + Δw ) Aw ( w + Δw ) − u t ( w )v t ( w ) Aw ( w )] dudv + lim t Δ
→ 0 Δ
Δ
→ 0

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Using the definition of the differentiation, (1.127) simplifies to

∇⋅A =

∂ ( v t wt Au ) ∂ ( wt u t Av ) ∂ ( u t v t Aw ) ∂v∂w + ∂u∂w + ∂u∂v ∂
∂
∂


(1.128)

Substituting d
= u t v t wt dudvdw into (1.128), the divergence of a vector A in the generalized coordinate system is obtained as

∇⋅A =

1 ⎡ ∂ ( v t wt Au ) ∂ ( wt u t Av ) ∂ ( u t v t Aw ) ⎤ + + ⎥⎦ u t v t wt ⎢⎣ ∂u ∂v ∂w

(1.129)

Substituting the respective values for u t , v t , w t , u , v , w , a u = a r , a v , and a w from Table 1.2, the divergence in rectangular, cylindrical, and spherical coordinate system are obtained respectively as

∂Ax ∂A y ∂Az + + ∂x ∂y ∂z 1 ∂ ( ρAρ ) 1 ∂Aφ ∂Az ∇⋅A = + + ρ ∂ρ ρ ∂φ ∂z 2 1 ∂ (r Ar ) 1 ∂ ( sin θ Aθ ) 1 ∂Aφ ∇⋅A = 2 + + ∂r ∂θ r sin θ r sin θ ∂φ r ∇⋅A =

(1.130a) (1.130b) (1.130c)

Note the following important properties of the divergence of the vector field:

∇ ⋅ (A + B) = ∇ ⋅ A + ∇ ⋅ B ∇ × ( A × B ) = A (∇ ⋅ B ) − B (∇ ⋅ A ) ∇ ⋅ ( ΦA ) = Φ (∇ ⋅ A ) + (∇Φ ) ⋅ A ∇ ⋅ ( ∇Φ ) = ∇ 2 Φ

(1.131) (1.132) (1.133) (1.134)

where A and B are vectors fields and Φ is a scalar field. The operator ∇ 2 is called the Laplacian and it applies for scalar and vector quantities. In the generalized coordinate system ∇ 2 Φ = ∇ ⋅ ∇Φ can be obtained using (1.116) and (1.128) as

∇2Φ =

1 ⎡ ∂ ⎛ v t wt ∂Φ ⎞ ∂ ⎛ wt u t ∂Φ ⎞ ∂ ⎛ u t v t ∂Φ ⎞ ⎤ ⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟ (1.135) u t v t wt ⎢⎣ ∂u ⎝ u t ∂u ⎠ ∂v ⎝ v t ∂v ⎠ ∂w ⎝ wt ∂w ⎠ ⎥⎦

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Using (1.135) along with Table 1.2, ∇ 2 Φ in rectangular, cylindrical, and spherical coordinate systems is obtained respectively as

∇ 2Φ = ∇2Φ = ∇2Φ =

1 r2

∂ 2Φ

( ∂x ) 2

+

∂ 2Φ

( ∂y ) 2

+

∂ 2Φ

(1.136a)

( ∂z ) 2

1 ∂ ⎛ ∂Φ ⎞ 1 ∂ 2Φ ∂ 2Φ ρ + + ⎜ ⎟ ρ ∂ρ ⎝ ∂ρ ⎠ ρ 2 ( ∂φ ) 2 ( ∂z ) 2

(1.136b)

⎡ ∂ ⎛ 2 ∂Φ ⎞ 1 1 ∂ ⎛ ∂Φ ⎞ ∂ 2Φ ⎤ r sin θ + + ⎜ ⎟ ⎜ ⎟ ⎥ (1.136c) ⎢ ∂r ⎠ sin θ ∂θ ⎝ ∂θ ⎠ sin 2 θ ( ∂φ ) 2 ⎦ ⎣ ∂r ⎝

1.14. THE CURL OF A VECTOR FIELD For an arbitrary vector field A ( u , v , w ) in the generalized orthogonal coordinate system shown in Fig. (1.32), the curl or circulation in the direction of u is defined by

1 A ⋅ dl ∫ ΔS vw →0 ΔS vw l

Bu = lim

(1.137)

where A = a u Au + a v Av + a w Aw and l is a closed path enclosing the area ΔS u as shown in Fig. (1.32). The differential vector dl  along the path l is given by (1.64a).

u

Au ut

wt

,u , v, w -

Aw

w w

, u , v .v , w .w -

vt Av

v

Av , w .w -

Aw , v -

Aw , v .v Av , w -

Fig. (1.32). The curl of a vector.

v

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51

In the generalized coordinate system ΔS u is given by ΔS vw = v t wt ΔvΔw . Substituting A and dl into (1.137), yields 1 ΔS u → 0 ΔS u

Bu = lim

v + Δv ⎛ w + Δw ⎞ ⎜⎜ ∫ Av v t dv + ∫ Aw wt dw ⎟⎟ ⎝ w ⎠ v

(1.138)

This simplifies to Bu

1 ΔS u → 0 v w t t 1 = lim ΔS u → 0 v w t t = lim

⎡ − Av ( w + Δw )v t ( w + Δw ) + Av ( w ) v t ( w ) ⎤ ⎥⎦ ⎢⎣ Δw ⎡ Aw ( v + Δv ) wt ( v + Δv ) − Aw ( v ) wt ( v ) ⎤ ⎢⎣ ⎥⎦ Δv

(1.139)

Using the definition of the differentiation, (1.139) simplifies to

Bu =

1 ∂ ( wt Aw ) 1 ∂ ( v t Av ) − v t wt ∂v v t wt ∂w

(1.140a)

Following similar analysis as for B u , the curl in the directions of v and w can obtained as

1 ∂ ( u t Au ) 1 ∂ ( wt Aw ) − wt u t ∂w wt u t ∂u 1 ∂ ( v t Av ) 1 ∂ ( u t Au ) Bw = − u t vt ∂u u t vt ∂v

Bv =

(1.140b) (1.140c)

Hence, the curl of a vector field A in the generalized coordinate system becomes

1 v t wt 1 + av wt u t 1 + aw u t vt

∇ × A = au

⎡ ∂ ( wt Aw ) ∂ ( v t Av ) ⎤ − ⎢⎣ ∂v ∂w ⎥⎦ ⎡ ∂ ( u t Au ) ∂ ( wt Aw ) ⎤ ⎢⎣ ∂w − ⎥⎦ ∂u ⎡ ∂ ( v t Av ) ∂ ( u t Au ) ⎤ ⎢⎣ ∂u − ∂v ⎥⎦

(1.141a)

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52

Sameir M. Ali Hamed


This can be written in determinant form as

u t a u v t a v wt a w 1 ∇×A = ∂ ∂u ∂ ∂v ∂ ∂w u t v t wt u t Au v t Av wt Aw

(1.141b)

Substituting u t , v t , w t , u , v , w , a u = a r , a v , and a w from Table 1.2 for rectangular coordinate system into (1.141), the curl of A in rectangular coordinate system becomes ax ay az ∇ × A = ∂ ∂x ∂ ∂y ∂ ∂z Ax Ay Az ∂A y ⎞ ∂A ⎞ ⎛ ∂A ⎛ ∂A y ∂Ax ⎛ ∂A ⎟⎟ + a y ⎜ x − z ⎟ + a z ⎜⎜ = a x ⎜⎜ z − − ∂z ⎠ ∂x ⎠ ∂y ⎝ ∂z ⎝ ∂y ⎝ ∂x

(1.142a) ⎞ ⎟⎟ ⎠

Similarly, in cylindrical and spherical coordinate systems the curl of A is obtained respectively as

∇× A =

1

ρ

aρ ∂ ∂ρ Aρ

ρa φ az ∂ ∂φ ∂ ∂z ρAφ Az

⎛ 1 ∂A z ∂Aφ = a ρ ⎜⎜ − ∂z ⎝ ρ ∂φ

⎞ ⎟⎟ + a φ ⎠

⎛ ∂A ρ ∂A z ⎜⎜ − ∂ρ ⎝ ∂z

⎞ 1 ⎡ ∂ ( ρAφ ) ∂A ρ ⎤ ⎟⎟ + a z ⎢ − ⎥ ∂φ ⎦ ρ ⎣ ∂ρ ⎠ (1.142b)

and

r aθ ar 1 ∇×A = 2 ∂ ∂r ∂ ∂θ r sin θ Ar r Aθ

r sin θ a φ ∂ ∂φ

(1.142c)

r sin θ Aφ

Some important properties of the curl of the vector field are:

∇× (A + B) = ∇× A + ∇× B

(1.143)

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∇ × ( A × B ) = A (∇ ⋅ B ) − B (∇ ⋅ A ) ∇ × ( ΦA ) = Φ∇ × A + ∇Φ × A ∇ ⋅ (∇ × A ) = 0 ∇ × (∇Φ ) = 0

53

(1.144) (1.145) (1.146) (1.147)

where A and B are vector fields and Φ is a scalar field. Example 1.13 Given that A = r sin θ a r + r cosθ a θ + sin θ cosφ a φ , find ∇ ⋅ A and ∇ × A . Solution Since A is given in spherical coordinate system, then using (1.130c), we have 2 1 ∂ ( r Ar ) 1 ∂ ( sin θ Aθ ) 1 ∂Aφ ∇⋅A = 2 + + r sin θ r sin θ ∂φ ∂r ∂θ r 2 5 sin θ − 1 sin φ = − sin θ r

and from (1.142c) the curl is obtained as

ar r aθ r sin θ a φ 1 ∇×A = 2 ∂ ∂r ∂ ∂θ ∂ ∂φ r sin θ 2 r sin θ r cosθ r sin 2 θ cosφ 2 cosθ cosφ sin θ cosφ = ar − aθ + a φ cosθ r r 1.15. THE DIVERGENCE THEOREM The divergence theorem states that the total flux of a vector field through a closed surface equals to the volume integral of the divergence of the vector field within the volume enclosed by the surface. For a vector field A through a region of the volume
enclosed by a surface S as shown in Fig. (1.33), the divergence theorem in mathematical form is

∫ A ⋅ dS =
∫ ∇ ⋅ Ad
S

(1.148)

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To prove the divergence theorem, assume that the volume
is divided into N cells with the cell i = 1, 2, , N has a differential volume Δ
i . The flux crossing the boundaries of Δ
i is ∇ ⋅ AΔ
i . Hence, the total outwards flux from all differential volumes is

∫ ∇ ⋅ A d
= lim

N

Δ
i → 0




∑ ∇ ⋅ A Δ


i

(1.149)

i =1

Note that for the interior cells, the net flux crossing the boundaries of the cell is zero because the flux leaving each cell boundary is canceled by the opposite flux leaving the neighboring cell as illustrated in Fig. (1.33). The cells having a face on the surface S results in a nonzero flux. Hence, the right-hand side of (1.149) is due the flux crossing the faces on S of the adjacent cells to S . Assuming the number of cells having a face on S is K and the surface area of the kth cell is Δ S k , and if the unit normal to Δ S k is a n , then using (1.124), (1.149) becomes

∫ ∇ ⋅ Ad
= lim

ΔS k →0




K

∑A ⋅ a

n

ΔS k

(1.150)

k =1

Since Δ S k → 0 and Δ S k belongs to S for all k , it follows that the sum on the right-hand side of (1.150) changes to integral over S and the equation reduces to the divergence theorem, or

∫ ∇ ⋅ Ad
= ∫ A ⋅ a




Unit cell

S

n

dS = ∫ A ⋅ dS S

A a n .S k

.Y L

S

Net flux crossing the cell boundaries is zero Fig. (1.33). Illustration of the divergence theorem.

(1.151)

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1.16. STOKES’ THEOREM Stokes’ theorem states that the total flux of a curl of a vector field crossing an open surface equals to the line integral of the vector field along the closed path that bounds the open surface. For a vector field A over an open surface S bounded by the closed path l as shown in Fig. (1.34), Stokes’ theorem is stated mathematically as

∫ ∇ × A ⋅ dS = ∫ A ⋅ dl S

(1.152)

l

S ΔS i

A

l

ΔS k  Δl k

Fig. (1.34). Illustration of Stokes’ theorem.

To prove (1.152), let the surface S be divided into N infinitesimal cells and the area of the cell i = 1, 2, , N is ΔS i . Hence, the flux through each cell is

∇ × A ⋅ ΔS i , and the net flux crossing S is the sum of the individual cells, or

∫ ∇ × A ⋅ dS = lim

ΔS i → 0

S

N

∑ ∇ × A ⋅ ΔS

i

(1.153)

i =1

The net result of ∇ × A ⋅ ΔS i for the interior cells is zero due to the cancellation from the adjacent cells as illustrated in Fig. (1.34). The cells that having a side on the closed path l give a nonzero value. Hence, the right-hand side of (1.153) is due to the cells having a side on l only. Assuming the number of these cells is K , the surface area of the kth cell is Δ S k , and the length of the side of Δ S k on l be

Δl k , then using the definition of curl in (1.137), the curl on the right-hand side of (1.153) for each Δ S k can be replaced by A ⋅ Δl k Δ S k . Thus, (1.153) becomes

∫ ∇ × A ⋅ dS = lim S

Δl k → 0

K

∑ A ⋅ Δl k =1

k

(1.154)

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Since Δl k → 0 and Δl k belongs to l for all k , then (1.154) changes to integral along the closed path l , or

∫ ∇ × A ⋅ dS = ∫ A ⋅ dl S

(1.155)

l

The direction of dS and dl in (1.155) obeys the right-hand rule. According to the right-hand rule, if the fingers point in the direction of dl , the thumb will indicate the direction of dS . Using Stokes’ theorem stated in (1.152) or (1.155) with the definition of conservative vector field it can be shown for a conservative vector field A that ∇ × A = 0 and the conservative vector field A can be written in terms of a scalar potential Φ as A = −∇Φ

(1.156)

It follows that the line integral from point P to point Q in the vector field A can be obtained as Q

∫ A ⋅ dl l

Q

∂Φ ∂Φ ⎤ ⎡ ∂Φ = − ∫ ∇Φ ⋅ dl = − ∫ ⎢ dx + dy + dz ∂x ∂y ∂z ⎥⎦ P P⎣ Q

Q

dΦ ⎡ dx ∂Φ dy ∂Φ dz ∂Φ ⎤ = −∫⎢ + + dS = − ∫ dS ⎥ dS ∂x dS ∂y dS ∂z ⎦ dS P⎣ P

(1.157)

Q

= − ∫ dΦ = Φ ( P ) − Φ ( Q ) P

Example 1.14 Given that A = ra r , verify the divergence theorem over the closed hemisphere r = 1 , 0 ≤ θ ≤ π 2 , 0 ≤ φ ≤ 2π . Solution Given A = ra r over a closed hemisphere. The divergence theorem states

∫
∇ ⋅ Ad
= ∫ A ⋅ dS S

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In spherical coordinate system we have

dS = a r r 2 sin θ dθ dφ + a θ r sin θ dr dφ + a φ rdr dθ

d
= r 2 sin θ dr dθ dφ Since A = ra r , then using (1.19) and (1.130c) respectively, yields

A ⋅ dS = r sin θ dφ dθ ,

∇⋅A =

3

2 1 ∂ (r Ar ) =3 ∂r r2

Since r = 1 , 0 ≤ θ ≤ π 2 , 0 ≤ φ ≤ 2π , the right-hand side of the theorem is

∫ A ⋅ dS = S

2π π 2

∫ ∫ (r

3

sin θ

0 0

)

r =1

dθ dφ

= −2π [cosθ ]θ = 0 = 2π θ =π 2

The left-hand side is 2π π 2 1

∫
∇ ⋅ Dd
= 3 ∫ ∫ ∫ r 0

2

sin θ dr dθz dφ = −2π [r 3 ]r = 0 [cosθ ]θ = 0 r =1

θ =π 2

= 2π

0 0

Hence, both sides are equal which verifies the divergence theorem. Example 1.15 If F = xa x + ( x + y )a y + ( x + y + z )a z , verify Stokes’ theorem around the closed path x = a sin t , y = a cos t , z = a ( sin t + cos t ), 0 ≤ t ≤ 2 π . Solution Given that F = xa x + ( x + y )a y + ( x + y + z )a z and the closed path is x = a sin t , y = a cos t , z = a ( sin t + cos t ), 0 ≤ t ≤ 2 π . Stokes’ theorem states that

∫ ∇ × F ⋅ dS = ∫ ∇ × F ⋅ a S

S

n

dS = ∫ F ⋅ dl l

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Using (1.142a), we have ax ay az ∇ × F = ∂ ∂x ∂ ∂y ∂ ∂z = ax −ay +az (x + y) (x + y + z) x

Since x = a sin t , y = a cos t , z = a ( sin t + cos t ), then z = x + y and the function f ( x , y , z ) describing the surface containing this path can be obtained as

⇒ f ( x, y, z ) = x + y − z = 0

z = x+ y

The unit vector normal to the surface f ( x , y , z ) is

ax +ay −az ∇f  = ∇f 3 ax +ay −az 1 ⇒ ∇ × F ⋅ a n = (a x − a y + a z )⋅ =− 3 3 an =

The differential surface area on the xy plane is obtained from (1.123) as

dS xy =

∇f dxdy = a x + a y − a y dxdy = 3dxdy ∇f ⋅ a z

The path in the xy plane represents a circle x 2 + y 2 = a 2 . Hence, the left-hand side of the Stokes’ theorem becomes

∫ ∇ × F ⋅ dS = ∫ ∇ × F ⋅ a S

S xy

n

dS xy

=

2 2 + a+ a − y

⎛ 1 ⎞ ⎜− ⎟ 3dxdy ⎝ ⎠ 3 a2 −y2

∫ ∫

− a−

+a

= −2 ∫ a 2 − y 2 dy = − a 2 π −a

Since



dl = a x dx + a y dy + a z dz x = a sin t , y = a cos t , z = a ( sin t + cos t )



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Then

dl = a [a x cos t − a y sin t + a z ( cos t − sin t ) ]dt 

F = a [a x sin t + ( sin t + cos t )a y + 2 ( sin t + cos t )a z ] F ⋅ dl = a 2 (5 cos 2 t − 3 )dt =

a2 ( 5 cos 2 t − 1)dt  2

Hence, the right-hand side of the Stokes’ theorem becomes

a2 F ⋅ d l = ∫l 2

2π

∫ ( 5 cos 2 t − 1)dt = − a

2

π

0

Both sides of the Stokes’ theorem are equal, which verifies the theorem. SOLVED PROBLEMS Solved Problem 1.1 Show that elliptic cylindrical coordinates (ξ ,η , z ) defined by the equations

x = cosh ξ cosη  y = sinh ξ sinη  z = z  − ∞ ≤ ξ ≤ +∞  − π ≤ η ≤ +π  − ∞ ≤ z ≤ +∞  are orthogonal. Find the scale factors u t , v t , w t and ∇Φ . Solution Comparing to the generalized coordinate system, then u = ξ , v = η , and w = z . Given that x = cosh ξ cosη , y = sinh ξ sinη , and z = z . Thus,

dy dy dx dx dz dz = = cosh ξ sin η  = = sinh ξ cosη  = = 0 du dξ du dξ du dξ dy dy dx dx dz dz = = sinh ξ cosη  = = = − cosh ξ sinη  = 0 dv dη dv dη dv dη dx dx dz dz dy dy = = 0  = = 1 = = 0  dw dz dw dz dw dz

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Consequently, the tangent unit vectors of the elliptic cylindrical coordinate are

dy dx dz +ay +az = a x sinh ξ cosη + a y cosh ξ sin η du du du dy dx dz vt = ax +ay +az = − a x cosh ξ sin η + a y sinh ξ cosη dv dv dv dy dx dz wt = ax +ay +az = az dw dw dw ut = a x

Using the last three equations it is clear that u t ⋅ v t = v t ⋅ w t = w t ⋅ u t = 0 , and hence the elliptic cylindrical coordinates are orthogonal. The scale factors can be obtained using (1.63) as

u t = ( sinh ξ cosη ) + ( cosh ξ sinη ) = sinh 2 ξ + sin 2 η 2

2

v t = ( cosh ξ sinη ) + ( sinh ξ cosη ) = u t 2

2

wt = 1 Substituting u = ξ , v = η , w = z and obtained as

∇Φ = a ξ

u t , v t , w t into (1.117), then ∇Φ is

∂Φ + aη sinh 2 ξ + sin 2 η ∂ξ 1

∂Φ ∂Φ +az ∂z sinh 2 ξ + sin 2 η ∂η 1

Solved Problem 1.2 If r is the position vector of point ( x , y , z ) and r = r , show that for integer n ∇ ⋅ r n r = ( n + 3 )r n

Solution In rectangular coordinate system r = xa x + ya y + za z , then

r=

x2 + y2 + z2

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Hence, ∇ ⋅ r n r = ∇ ⋅ ( x 2 + y 2 + z 2 ) 2 ( xa x + ya y + za ) n n n ∂ ∂ ∂ = x (x 2 + y 2 + z 2 )2 + y (x 2 + y 2 + z 2 )2 + z (x 2 + y 2 + z 2 )2 ∂x ∂y ∂z n

[

] [

] [

]

Note that

[ [ [

] ] ]

n n−2 n ∂ ⎞ ⎛ nx 2 x ( x 2 + y 2 + z 2 ) 2 = nx 2 ( x 2 + y 2 + z 2 ) 2 + ( x 2 + y 2 + z 2 ) 2 = r n ⎜ 2 + 1 ⎟  ∂x ⎝ r ⎠ 2 n n−2 n ⎛ ny ⎞ ∂ y ( x 2 + y 2 + z 2 ) 2 = ny 2 ( x 2 + y 2 + z 2 ) 2 + ( x 2 + y 2 + z 2 ) 2 = r n ⎜⎜ 2 + 1 ⎟⎟  ∂y ⎝ r ⎠ 2 n n−2 n ∂ ⎛ nz ⎞ z ( x 2 + y 2 + z 2 ) 2 = nz 2 ( x 2 + y 2 + z 2 ) 2 + ( x 2 + y 2 + z 2 ) 2 = r n ⎜ 2 + 1 ⎟  ∂z ⎝ r ⎠

Substituting the last three equations into the right-hand side of the previous equation for ∇ ⋅ r n r , yields ⎛ ny 2 ⎞ ⎛ nx 2 ⎞ ⎛ nz 2 ⎞ ∇ ⋅ r n r = r n ⎜ 2 + 1 ⎟ + r n ⎜⎜ 2 + 1 ⎟⎟ + r n ⎜ 2 + 1 ⎟ ⎝ r ⎠ ⎝ r ⎠ ⎝ r ⎠

Since r 2 = x 2 + y 2 + z 2 , the above equation simplifies to ∇ ⋅ r nr = (n + 3) r n

Solved Problem 1.3 For the vector field F = ρ 2 cos 2 φ a ρ + z sin φ a φ , find the flux through the closed surface of the cylinder ρ = a , 0 ≤ φ ≤ 2π , 0 ≤ z ≤ b , where a and b are constants. Verify the divergence theorem. Solution The total flux through the surfaces of the cylinder shown in Fig. (1.35) is

ψ = ∫ F ⋅ dS = ∫ F ⋅ dS 1 + ∫ F ⋅ dS 2 + ∫ F ⋅ dS 3 S

S1

S2

S3

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where dS 1 = a z ρ dρ dφ , dS 2 = a ρ ρ dφ dz , and dS 3 = − a z ρ dρ dφ . Substituting

F , dS 1, dS 2 , and dS 3 into the last equation, yields

ψ = ∫ ( 0 ) ρ dρ dφ + S1

∫ [( ρ

2

S2

cos 2 φ a ρ + z sin φ a φ ) ⋅ a ρ ρ ]ρ = a dφ dz + ∫ ( 0 ) ρ dρ dφ S3

b 2π

3

a b ( cos 2φ + 1) dφ = π a 3 b ∫ 2 0

ψ = ∫ ∫ a 3 cos 2 φ dφ dz =

⇒

0 0

a z dS1 

2π

z

a

S1  b

S2

S3

a ρ dS 2 y

φ

− a z dS 3 

x Fig. (1.35). Geometry of Solved Problem 1.3.

where dS 1 = a z ρ dρ dφ , dS 2 = a ρ ρ dφ dz , and dS 3 = − a z ρ dρ dφ . Substituting

F , dS 1, dS 2 , and dS 3 into the last equation, yields

ψ = ∫ ( 0 ) ρ dρ dφ + S1

∫ [( ρ

S2

2

cos 2 φ a ρ + z sin φ a φ ) ⋅ a ρ ρ ]ρ = a dφ dz + ∫ ( 0 ) ρ dρ dφ S3

This reduces to b 2π

ψ = ∫ ∫ a 3 cos 2 φ dφ dz = 0 0

2π

a 3b ( cos 2φ + 1) dφ = π a 3 b ∫ 2 0

The divergence theorem states that

∫ F ⋅ dS =
∫ ∇ ⋅ Fd
S

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Since F = ρ 2 cos 2 φ a ρ + z sin φ a φ , then in cylindrical coordinate system

∇⋅F =

z cosφ 1 ∂ ( ρFρ ) 1 ∂Fφ ∂Fz + + = 3 ρ cos 2 φ + ∂z ρ ∂ρ ρ ∂φ ρ

Substituting ∇ ⋅ F and d
= ρ dρ dφ dz into the right-hand side of last equation, then b 2π a

∫ ∇ ⋅ Fd
= ∫




∫ ∫ (3 ρ

2

cos 2 φ + z cosφ )dρ dφ dz =π a 3 b

0 0 0

∫ ∇ ⋅ Fd
= ψ = ∫ F ⋅ dS

⇒




S

This verifies the divergence theorem. Solved Problem 1.4 In the triangle of Fig. (1.36a) c > a and c > b . Using vector algebra prove that

c 2 = a 2 + b 2 − 2 ab cos γ a b c = = sin α sin β sin γ

π −γ b

γ

α

a

b

γ

α

β

a β

c

c

(a)

(b)

Fig. (1.36). Geometry of Solved Problem 1.4.

Solution Let a , b , and c be vectors along the sides of lengths a , b , and c respectively as shown in Fig.(1.36b) such that a = a , b = b , and c = c . Thus, a+b =c

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Finding the dot product of both sides of the above equation by vector s a and b , yields a⋅a + b⋅a = c⋅a a⋅b + b⋅b = c⋅b

⇒ ⇒

a 2 + ab cos (π − γ ) = c ⋅ a ab cos (π − γ ) + b 2 = c ⋅ b

Adding the last equations, yields a 2 + b 2 + 2 ab cos (π − γ ) = c ⋅ a + c ⋅ b = c ⋅ ( a + b ) = c ⋅ c = c 2

or c 2 = a 2 + b 2 − 2 ab cosγ

Let a n be the unit vector perpendicular to the plane of the triangle, then a × b = a n a b sinγ ,

c × a = a n c a sin β ,

b × c = a n b c sinγ

Using the above relations it can be shown that a × b − c × a = a × ( b + c ) = a × a = a n a b sinγ − a n c a sinβ = 0 ⇒

ab sinγ = ac sinβ

Similarly, from c × a = a n c a sin β and b × c = a n b c sinγ it can be shown that ac sinβ = bc sinα

Thus, ab sinγ = ac sinβ = bc sinα

⇒

sinγ sinβ sinα = = c b a

or a b c = = sinα sinβ sinγ

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Solved Problem 1.5 Show that the vector field F = 2 xy 2 a x + (2 x 2 y + 6 yz 3 )a y + 9 y 2 z 2 a z is a conservative. Find the associated scalar potential and the work done by F to move an object from point ( 0 , − 1, 2 ) to (1, − 1,1) . Solution Given that F = 2 xy 2 a x + (2 x 2 y + 6 yz 3 )a y + 9 y 2 z 2 a z , then using (1.142a), we have ax ∇ × F = ∂ ∂x 2 xy 2

ay az ∂ ∂y ∂ ∂z = a x ( 0 ) − a y ( 0 ) + a z ( 0 ) = 0 2 3 (2 x y + 6 yz ) 9 y 2 z 2

Since , ∇ × F = 0 , then F is a conservative vector field. The associated a scalar potential Φ is related to F by F = −∇Φ . Hence,

Fx = 2 xy 2 = −

∂Φ ∂Φ ∂Φ , F y = 2 x 2 y + 6 yz 3 = − , Fz = 9 y 2 z 2 = − ∂x ∂z ∂y 2 2 ⇒ Φ = − x y + f x ( y, z )

Φ = − x 2 y 2 − 3 y 2 z 3 + f y ( x, z )

Φ = −3 y 2 z 3 + f z ( x , y ) Comparing the above three solutions, yields

f x ( y , z ) = −3 y 2 z 3 ,

f y ( x, z ) = c ,

f z ( x, y ) = − x 2 y 2

where c is a constant. Consequently, the required a scalar field is

Φ = −x2 y2 −3y2 z3 + c The work done W to move an object from point ( 0 , − 1, 2 ) to (1, − 1,1) by F can be obtained using (1.157) as

W = Φ (1, − 1,1) − Φ ( 0, − 1, 2 ) = −4 + 24 = 20

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Solved Problem 1.6 Verify the divergence theorem for the vector field D leaving the surfaces of the cube shown in Fig. (1.37), given that D = xy 2a x + y 3a y + y 2 za z .

z

dS xy = a z dxdy ( 0 , 0 ,1 )  D z  Dy dS xz = a y dxdz 

( 0 ,1, 0 )

(1, 0 , 0 ) x

Dx dS yz = a x dydz 

y

Fig. (1.37). The cube of Solved Problem 1.6.

Solution The divergence theorem states

∫ ∇ ⋅ Dd
= ∫ D ⋅ dS




S

Referring to Fig. (1.37), with D = xy 2a x + y 3a y + y 2 za z , the right-hand side is

S

0 0 1 1

(

x =1

+ ∫∫ − xy 2 0 0

1 1

1 1

1 1

2 ∫ D ⋅ dS = ∫∫ xy

dydz + ∫∫ y 3

)

0 0

y =1

1 1

x =0

dxdz + ∫∫ y 2 z 0 0

(

dydz + ∫∫ − y 3

[ ]

0 0

)

z =1

dxdy

1 1

y =0

[ ]

(

dxdz + ∫∫ − y 2 z

)

0 0

1 1 1 5 1 = y 3 0 [z ]10 + [x ]0 [z ]10 + [x ]10 y 3 10 = 3 3 3

The left-hand side is

∫ ∇ ⋅ Dd





1 1 1

= ∫∫∫ ∇ ⋅ [ xy 2 a x + y 3 a y + y 2 za z ]dxdydz 0 0 0 1 1 1

= ∫∫∫ 5 y 2 dxdydz = 0 0 0

5 1 3 1 1 5 [ x ]0 [ y ]0 [ z ]0 = 3 3

Hence, both sides are equal, which verifies the divergence theorem.

z =0

dxdy

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Solved Problem 1.7 Verify the Stokes theorem for the vector field H = − y 2 a x + xza y + z 2 a z over the surface S defined by the plane x + y + 2 z = 2 in the first octant of the coordinate system with a clockwise sense of direction along its boundary l . Solution Given that H = − y 2 a x + xza y + z 2 a z , and the surface S is bounded by the closed path l as shown in Fig. (1.38) . The Stokes’ theorem states that

∫ ∇ × H ⋅ dS = ∫ H ⋅ dl S

l

z The plane x + y + 2 z = 2 

( 0 , 0 ,1 ) 

l zy

l xz 

S

(0, 2, 0 )

(2, 0, 0 )

y

l yx

x Fig. (1.38). Geometry of Solved Problem 1.7.

Referring to Fig. (1.38), the path l has three segments l yx , l xz , and l zy , where

l xz = − a x dx + a z dz ,

l yx = a x dx − a y dy ,

l zy = a y dy − a z dz

Along the path l yx , x = y ; z = 0 ; H ⋅ dl yx = − y 2 dx = − x 2 dx . Hence

∫ H ⋅ dl

l yx

2

yx

= − ∫ x 2 dx = − 0

1 3 2 [x ]0 = − 8 3 3

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Along the path l xz , x = 2 z ; y = 0 ; H ⋅ dl xz = y 2 dx + z 2 dz = z 2 dz . Hence,

∫ H ⋅ dl

1

xz

l xz

= ∫ z 2 dz = − 0

1 3 1 1 [ z ]0 = 3 3

Along the path l zy , y = 2 z ; x = 0 ; H ⋅ dl zy = xzdy − z 2 dz = − z 2 dz . Hence, 1

2 ∫ H ⋅ dl xz = − ∫ z dz = −

l zy

0

1 3 1 [z ]0 = − 1 3 3

Thus, the left-hand side of the Stokes’ theorem is

∫ H ⋅ dl = ∫ H ⋅ dl l

l yx

yx

8 1 1 8 + ∫ H ⋅ dl xz + ∫ H ⋅ dl zy = − + − = − 3 3 3 3 l xz l zy

For the left-hand side of the Stokes’ theorem, we need to determine ∇ × H , then

ax ay az ∇ × H = ∂ ∂x ∂ ∂y ∂ ∂z = − x a x + ( z + 2 y )a z  − y2 xz z2 The equation of the surface is f = x + y + 2 z − 2 , then the unit normal to it is

an = ±

a x + a y + 2a z ∇f =± ∇f 6 

Referring to Fig. (1.38), since the direction of the path l is in the clockwise direction, then using the right-hand rule, the direction of the surface dS will be towards the origin or a n must be negative. Hence,

an = −

a x + a y + 2a z

6 ⎛ a + a y + 2a z ⎞ x − 4 y − 2 z ⎟= ⇒ ∇ × H ⋅ a n = −[− x a x + (z + 2 y )a z ]⋅ ⎜ x ⎜ ⎟ 6 6 ⎝ ⎠

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We have 2 z = x , and the projection of the unit area on the plane z = 0 , is

dS xy =

a x + a y + 2a z ∇f 6 dxdy = = ∇f ⋅ a z (a x + a y + 2 a z )⋅ a z 2

Thus

∫ ∇ × H ⋅ dS S

= ∫∫ ∇ × H ⋅ a n dS xy =−

4

2 x−2

∫∫ 6 0 0

[

2 6 dxdy 1 2 2 y = − ∫ ( 2 − x ) dx = ( x − 2 ) 2 3 0

= ∫ H ⋅ dl l

Hence, both sides are equal which verifies the Stokes’ theorem.

]

2

0

=−

8 3

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PROBLEMS 1.1.

For any two vectors A and B show that A⋅B ≤ A B A+B ≤ A + B

1.2.

1.3.

Find the dot product of A and B if (a) A = a x − 2a y + 5a z ,

B = 6a x − 2.5a y + a z .

(a) A = 2a x + 7a y − a z ,

B = 4a x − 2a y + 14a z .

(b) A = 5a ρ + aφ − 2a z ,

B = a ρ − 2aφ + 4a z .

Given

that

A = a x − 2a y + a z ,

B = 4a x − a y + 7a z ,

and

C = 2 a x − a y + a z , find (a) (b) (c) 1.4.

(A + B )⋅ C. (3C − B ) ⋅ A . ( A ⋅ C )C + B.

Given that A = a x + 2 a y + a z , B = 2 a x − a y + 2 a z , and C = a x + a y + a z , find (c) ( 2 A − B ) × C. (d) A normal vector in the plane containing (C ⋅ B )C + C and A . (e) A unit normal to the plane containing A and C .

1.5.

If A = a x + 4 a y + 2 a z , B = 5 a x − a y + a z , and C = 2 a x + 2 a y − a z , find (a) ( A × B ) × C (b) A × ( B × C )

1.6.

If the vectors A = a x + aa y + 2 a z , and B = ba x − a y + a z are coplanar, find the constant a and b .

1.7.

Show that

A × ( B × C ) + B × (C × A ) + C × ( A × B ) = 0

Vector Analysis


1.8.

If

the

Electromagnetics for Engineering Students

vectors

A = aa x + 2 a y + 8 a z ,

B = a x + ba y − 2 a z

71

and

C = 2 a x + a y + ca z are mutually orthogonal find the scalars a , b , and c. 1.9.

Given that A = aa x + 2 a y + 6 a z , B = a x + 2 a z and C = 2 a x + a y + ca z , if A + bB + C is parallel to the x -axis, find the constants a , b , and c .

1.10. Find the equation of the plane and a unit vector normal to the plane containing the points (1, 4 , 2 ), (5,1, −2 ) , ( 0 ,1, 4 ) . 1.11. Determine the unit vector normal to the surface x 2 + y 2 + z − 4 = 0 at the point (1,1, 2 ) . 1.12. Show that the vectors A = a x + 2 a y − 5 a z , B = a x + a y + 2 a z , and

C = a x + 4 a y − 19 a z are linearly dependent vectors. 1.13. The vectors A = a x + 2 a y + a z , B = 2 a x − a y + 2 a z , C = a x + a y + a z and D = a x + 6 a y satisfy aA + bB + cC + D = 0

where a , b and c are scalar constants. Find a , b and c . 1.14. Find the work done by a force F = yza x + zxa y + xya z in moving its point of application from ( 0 , 0 , 0 ) , ( 2 , 4 , 8 ) along (a) The straight line between the two points. (b) The curve x = t , y = t 2 , z = t 3 , 0 ≤ t ≤ 1 . 1.15. Show that parabolic cylindrical coordinates equations

x=

1 2 (u − v 2 ), 2

(u , v , z )

y = sinh ξ sin uv ,

z=z

are orthogonal. Find the scale factors u t , v t , w t and ∇Φ .

defined by the

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1.16. Find the net flux of the vector field F = ya x − xa y + a z through the first octant of the sphere x 2 + y 2 + z 2 = a 2 . 1.17. Determine the net flux of the vector field F = ρ a ρ + a φ + za z leaving a cylindrical closed surface ρ = 1 , 0 ≤ φ ≤ π , and 0 ≤ z ≤ 1 . 1.18. Determine the net flux of the vector field F = xza x + yza y + a z leaving a spherical closed surface x 2 + y 2 + z 2 = 25 . 1.19. Find the gradient of the following scalar fields at the specified points (a) Φ = ( x 2 + 2 y 2 + 4 z 2 ) at ( x = 1, y = 1, z = 0 ). −1

(b) Φ = ρ 2 sin φ + ρz 2 + 1, at ( ρ = 1,φ = 0, z = 0 ) . (c) Φ = r sin θ cosφ + sin 2 φ , at ( r = 1,θ = π 2,φ = 0 ). 1.20. Find the divergence of the following vector fields (a) F = cos xa x + sin ya y + z 2 a z . (b) F = (1 − x 2 )a x + sin ( yz )a y + e xyz a z . (c) F = ρz sin φa ρ + 3 ρ cosφ a φ + ρ cosφ sin φa z . (d) F = r 2 a r + sin θa φ . 1.21. Find the curl of the following vector fields (a) F = xyz 2 a x + x 2 yza y + xy 2 a z . (b) F = ρz 2 cosφ a ρ + z sin 2 φ a z . (c) F = r cosθ a r − ( sin θ r )a θ + 2 r 2 sin θ a φ . (d) F = r cosθ ar − (sin θ cosφ r )aθ 1.22. Given that F = 3 xyza x + 2 ya y − 4 za z , find

∇ (∇ ⋅ F ). Use the results to show that ∇ × ∇ × F = ∇ (∇ ⋅ F ) − ∇ 2 F

∇ ⋅ F , ∇ × F , ∇ 2 F , and

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1.23. Find the Laplacian of the following scalar fields at the specified points (a) Φ = x 2 y 2 z 2 + xz + xy at ( x = −1, y = 1, z = 2 ) . (b) Φ = ρ 2 sin φ + ρz , at ( ρ = 2 ,φ = π , z = 0 ). (c) Φ = e − r sin θ cosφ , at ( r = 1,θ = π 2 ,φ = π ). 1.24. Determine the net flux of the vector field F = 2 x 2 ya x + za y + ya z leaving the cube 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1, and 0 ≤ z ≤ 1 . Check the result with the divergence theorem. 1.25. Using F = 2 xya x + 3 xya y + ze x + y a z and the region bounded by the volume 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1, and 0 ≤ z ≤ 1 , Verify the divergence theorem. 1.26. Verify Stokes’ theorem for the vector field F = − y 2 a x + xa y + z 2 a z where the closed path is the intersection of the cylinder x 2 + y 2 = 1 with the plane y + z = 2 , oriented counterclockwise from above. 1.27. Verify Stokes’ theorem for the vector field F = cosθ a r by using a hemisphere defined by r = 1 , 0 ≤ θ ≤ π 2 , and 0 ≤ φ ≤ 2π . 1.28. Show for any scalar fields
and
that

∇ ( 
) =
∇ +  ∇
∇ ( 
) = (
∇ −  ∇
)
2 ,
≠ 0 1.29. Given that
= ye xy + z and
= xyz 2 + 1, find ∇ ( 


) and ∇( 
) .

1.30. If F = ∇f , prove that

∇f n = nf

n −1

F

Use the result to show that when f = r , where r is the magnitude of the position vector r = xa x + ya y + za z , then ∇r = r r

and

∇ (1 r ) = − r r 3

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1.31. Show for any scalar fields Φ and vector field F that

∇ ⋅ ( ΦF ) = Φ ∇ ⋅ F + F ⋅ ∇Φ ∇ × ( ΦF ) = Φ ∇ × F + F × (∇Φ ) 1.32. Given that Φ = 1 r and F = r cosθ a r + r sin θ a θ , find ∇ ⋅ ( ΦF ) and

∇ × ( ΦF ).

1.33. Given that Φ = ρz cosφ and F = ρ a ρ + za z , find ∇ ⋅ ( ΦF ) and ∇ × ( ΦF ) .

Φ = xy + yz + zx and F = yz 2 a x + x 2 za y + xy 2 a z , find ∇ ⋅ ( ΦF ) and ∇ × ( ΦF ).

1.34. Given that

1.35. If r is the position vector of point ( x , y , z ) and r = r , show that within a volume
enclosed by the surface S for integer n

∫r


n

d
=

1 r n r ⋅ dS ∫ (n + 3) S

1.36. Check whether the following vector fields are conservative or not (a) F = e x y 2 a x + e x y a y + 3 xza z .

(b) H = (1 ρ )a ρ .

(c) E = ( 2 cosθ a r + sin θ a θ

) r3 .

1.37. Given the vector field

(

)

(

)

F = axz 2 cosh x 2 + by 2 a x + cyz 2 cosh x 2 + by 2 a y

(

)

+ 2 z sinh x 2 + 2 y 2 a z If F is a conservative field, find the constants a , b , and c . 1.38. Given the vector field

F = e xyz (1 + xyz )a x + x 2 ze xyz a y + x 2 ye xyz a z

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Show that F is a conservative, then find the associated scalar potential. Use this scalar potential to find the work done by F in moving an object from point ( 0 , 0 , 0 ) to (1,1, 2 ) . 1.39. Show that the vector field F = y 2 a x + (2 xy + z 2 )a y + 2 yza z is a conservative. Find the work done by the vector F in moving a particle from point ( 0 , 0 , 0 ) to (1,1,1) . 1.40. For any scalar Φ and vector H show that (a) ∇ × ∇Φ = 0 . (b) ∇ ⋅ (∇ × H ) = 0 .

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CHAPTER 2

Electrostatic Fields Abstract: The field concept is fundamental to describe and understand the electrostatic phenomenon and its interaction with surrounding objects. Understanding the concepts and laws related to the electrostatic field is necessary before introducing the more advanced topics in electromagnetics. On the other hand, there is a wide range of applications in industry, medicine, and electronic devices that are based on electrostatics. This makes the subject of electrostatics an important topic as a prerequisite for other subjects or as a basic technology in many industrial applications. This chapter introduces the principles of electrostatic fields. The main topics that are covered in the chapter include Coulomb’s law and electrostatic force between the objects, the concept of electric field intensity, Gauss’s law and its applications, the concept of electric potential, the electrostatic energy and the energy stored in the electrostatic field, and the electric dipole. All topics of the chapter are supported by illustrative examples and figures that make the topics understandable easily. To enhance the student's problem solving skills, the chapter is provided by miscellaneous solved problems and numerous homework problems covering the topics of the chapter.

Keywords: Coulomb's law, electric charge, electric field, electric flux, electric flux density, electric dipole, electrostatic energy, electric potential, Gauss's law, permittivity. INTRODUCTION The field concept is fundamental to describe and understand the electromagnetic phenomenon. Electromagnetic fields possess energy and momentum and interact with surrounding objects like particles of matter. The electrostatic field is an essential topic within the broad electromagnetics field theory. On the other hand, there is a wide range of applications in industry, medicine, and electronic devices that are based on electrostatics. Understanding the concepts and laws related to the electrostatic field is necessary before introducing the more advanced topics in electromagnetics. That is why we begin our discussion on engineering electromagnetics with electrostatic fields. In this chapter, the concept of the electrostatic field is introduced. The chapter begins with a brief discussion on the sources of the electrostatic fields. The main topics that are covered include Coulomb's law, the electrostatic force, the electric field intensity, Gauss’s law, the concept of the electric potential, the electrostatic energy, and the electric dipole. Students may refer also to some excellent textbooks [16-53] for more details treatment on these topics. Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

Electrostatic Fields

2.1.

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SOURCES OF ELECTROSTATIC FIELDS

Electric charge constitutes the source of all electromagnetic fields. The unit of measurement of charge is the Coulomb (C). The first known observations on the effects of charge back to before 600 BC, when the Greek philosopher Thales observed that, rubbing a piece of amber could enable the amber to attract bits of a straw. Although the nature of the charge is unknown, the understanding of its properties and effects has been developed since Thales observation up to date. The main properties of the electric charge can be summarized as follows [19] and [41]: 1. 2. 3. 4.

Charge is a fundamental property of matter, as is mass. There are two kinds of charges, positive or negative. Charges of opposite kind attract and charge of the same kind repel. Charge is quantized. There exists a smallest quantity of charge that can be associated with matter. The larger amounts occur in integral multiples of this smallest quantity. 5. Charge may be static or in movement. The movement of charges produces an electric current. The electric current I is the rate of change of charge Q with respect to time t , or I = dQ dt . The current I is constant when the rate of change dQ dt is constant. A time-varying current results when the rate of change dQ dt is a function of time. 6. Charge is the source of electromagnetic fields. Energy and momentum propagate away from the charge YLD the field. Static charges produce an electrostatic field, while constant currents produce a static magnetic field. Time-varying electromagnetic fields are produced by time-varying currents. In engineering electromagnetics, we are interested in the electromagnetic effects that can be predicted by classical techniques assuming continuous sources of charge and current densities. In microscopic scale, the charge is not of a continuous nature. Hence, charge and its effects are well described only by quantum theory. However, in macroscopic scale when the physical scale of the charged source is much larger than the individual charged particle and includes a vast number of charges, the source can be viewed as a continuous charge distribution. This macroscopic distribution of charge is useful in many situations in engineering electromagnetics, such as the determination of voltages and currents in lumped circuits, torques exerted by electrical machines, and electromagnetic fields radiated by antennas [19].

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For the consideration of electromagnetic study, the source of the charge is assumed to take one of the following forms: 1. Point charge: A charge Q is assumed concentrated at an isolated point in space. The concept of a point charge is useful in the study of the electrostatic field. 2. Linear distribution of charge along a line l ′ with a charge density ρl C m . The total amount of charge QT in a segment of length l ≤ l ′ along the line is

QT = ∫ dQ = ∫ ρl dl ′ l

l

(2.1a)

3. Surface distribution of charge on a surface S ′ with a surface charge density 2 ρ s C m and the total charge QT on a surface S ≤ S ′ is

QT = ∫ dQ = ∫ ρ s dS ′ S

S

(2.1b)

4. Volume distribution of charge within a volume v′ with a volume charge 3

density ρv C m . In this case, the total charge QT within a volume v ≤ v′ is

QT = ∫ dQ = ∫ ρv dv′ v

v

(2.1c)

Example 2.1 A dielectric cube 0 ≤ x ≤ a , 0 ≤ y ≤ a , 0 ≤ z ≤ a has a volume charge density ρ v = ρ 0 x a C m3 , where ρ0 is a constant. Find the total charge in the cube. Solution It is appropriate to use the rectangular coordinate system for this problem. Hence, the differential volume is dv′ = dx′dy′dz′ . Using (2.1c), the cube is bounded by 0 ≤ x′ ≤ a , 0 ≤ y′ ≤ a , 0 ≤ z′ ≤ a , then ρ v = ρ 0 x′ a C m3 . The total charge is a a a

QT = ∫ ρv dv′ = ∫∫∫ ρ0 v

0 0 0

x′ 1 dx′dy′dz′ = ρ0 a3 C a 2

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COULOMB’S LAW

In 1785, a French army Engineer Colonel Charles Augustine De Coulomb determined experimentally the force between two charged objects. Based on the results of the experiments Coulomb stated that the force between two very small objects separated in a vacuum or free space by a distance that is larger compared to their size is proportional to the charge on each object and inversely proportional to the square of the distance between them. This statement is known as Coulomb’s law. Coulomb’s law can be formulated mathematically by considering the charged objects as two point charges Q1 and Q2 at points P1 ( x1, y1, z1 ) and P2 ( x2 , y2 , z2 ) as shown in Fig. (2.1). The force exerted on Q1 due to the presence of Q2 is

F12 = K

Q1Q2 R12

2

a12 = K Q1Q2

R12 R12

3

= K Q1Q2

r1 − r2 r1 − r2

3

N

(2.2)

where K is a constant. r1 and r2 are the position vectors of Q1 and Q2 respectively as shown in Fig. (2.1). R12 = r1 − r2 is the vector in the direction from Q2 to Q1 , and a12 = R12 R12 is the unit vector in the direction of R12 . The charges Q1 and Q2 in (2.2) may be positive or negative in sign. Hence, a positive signed charge is substituted as a positive quantity, while a negative signed charge as a negative quantity in (2.2). In the rectangular coordinate system

r1 = a x x1 + a y y1 + a z z1

(2.3)

r2 = a x x2 + a y y2 + a z z2

(2.4)

The force on Q2 due to Q1 is F21 = −F12 . The constant K depends on the properties of the medium in which the charges are placed. It has the value K = 1 4π ε , where ε is known as the permittivity of the medium. For free space

ε = ε0 =

1 × 10−9 = 8.85 × 10 −12 F/m 36π

(2.5)

Substituting K = 1 4π ε into (2.2), the force exerted on Q1 due to Q2 becomes

F12 =

Q1Q2 a12 Q1Q2 r1 − r2 = 4π ε R 12 3 4π ε r1 − r2 3

N

(2.6)

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If Q1 and Q2 are opposite in sign, the force between them is an attractive force, and if they are similar in sign, the force between them is a repulsive force. F12 P1 (x1, y1, z1)

Q1 R12 = r1 - r2

r1

a12 Q2

Origin

P2 (x2, y2, z2)

r2

Fig. (2.1). Illustration of the electrostatic force between two charges and the position vectors of the charges.

2.3.

FORCE ON A CHARGE DUE TO SYSTEM OF POINT CHARGES

Consider a point charge q at a point ( x, y, z ) . If there are N point charges Q1 , Q2 , , QN at points ( x1, y1, z1 ) , ( x2 , y2 , z2 ) , …, ( xN , yN , z N ) respectively, the resultant force exerted on q is the vector sum of the forces due to the individual point charges Q1, Q2 ,, QN on q . Let Fn be the force on q due to a charge Qn at ( xn , yn , zn ) , where n = 1, 2, , N . Hence, from (2.6), Fn can be written as

Fn =

1 qQn 4πε R n 2

(2.7)

a Rn

where R n = r − rn and a Rn = (r − rn ) R n . In the rectangular coordinate

r = ax x + a y y + az z

(2.8)

rn = a x xn + a y yn + a z zn

(2.9)

The resultant force on q due to the system of charges is N

F = F1 + F2 +  + FN = ∑ Fn = n =1

q

N

Qn

∑R 4π ε n =1

2 n

a Rn

(2.10)

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Example 2.2 Two point charges Q1 = +2n C and Q2 = +9n C are placed in free space such that Q1 on the x axis at a distance 3 cm from the origin, while Q2 on the y axis at 4 cm from the origin. Find the force on each charge. Solution Given that Q1 = +2n C at P1 (3,0,0) cm and Q2 = +9n C at P2 (0,4,0) cm, then

r1 − r2 = a x ( x1 − x2 ) + a y ( y1 − y2 ) + a z ( z1 − z2 ) = 0.03a x − 0.04a y

R12 = r1 − r2 = 0.03a x − 0.04a y = 0.032 + 0.042 = 0.05 m a12 =

r1 − r2 0.03a x − 0.04a y = = 0.6a x − 0.8a y R12 0.05

The force on Q1 is

Q1Q2 a12 2 × 10 −9 × 9 × 10 −9 ⎛ 0.6a x − 0.8a y ⎞ ⎜ ⎟⎟ = F12 = 4π ε R12 2 4π × 8.85 × 10 −12 ⎜⎝ 0.05 2 ⎠ = 6.48 × 10 − 4 (3a x − 4a y ) N The force on Q2 is F21 = −F12 = −6.48 × 10−4 (3a x − 4a y ) N . Example 2.3 Two point charges Q1 = +60 n C and Q2 = −30 n C are placed in free space where Q1 at P1 (1,0,1) and Q2 at P2 (0,1,1) . If a third charge Q3 = +3 2n C is placed at the origin, find the force on Q3 . The units of the coordinate system are in cm. Solution The total force on Q3 is the resultant of the forces F31 and F32 due to Q1 and Q2 respectively, where

F31 =

1 Q1Q3 a31, 4π ε R 31 2

F32 =

1 Q2Q3 a32 4π ε R 32 2

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Since Q1 = +60 n C is at P1 (1,0,1) and Q3 = +3 2n C is at P3 (0,0,0) , then

r3 − r1 = a x ( x3 − x1 ) + a y ( y3 − y1 ) + a z ( z3 − z1 ) = −(a x + a z )

R 31 = r3 − r1 = (a x + a z ) = 2 cm r − r (a + a z ) a + az ⇒ a31 = 3 1 = x =− x R 31 R 31 2 Thus,

F31 = −

3 2 × 10 −9 × 60 × 10 −9 ⎛ a x + a z ⎞ −4 ⎟ = −81 × 10 (a x + a y ) N ⎜ 4π × 8.85 × 10 −12 ⎝ 2 2 × 10 − 4 ⎠

Following the same analysis, the force on Q3 due to Q2 is obtained as F32 = −

3 2 × 10 −9 × (− 30 )× 10 −9 ⎛ a y + a z ⎞ ⎟ = 40.5 × 10 − 4 (a y + a z ) N ⎜⎜ −12 −4 ⎟ 4π × 8.85 × 10 ⎝ 2 2 × 10 ⎠

The total force on Q3 is F = F31 + F32 = −40.5 × 10−4 (2a x + a y − a z ) N .

Example 2.4 Two point charges Q1 and Q2 are placed in free space. A third charge Q3 is placed at a point that divides the line between Q1 and Q2 in the ratio 1:3, find the ratio of Q1 to Q2 so that the force on Q3 is zero. Solution Let d is the distance between Q1 and Q2 , and Q1 be at P1 (0,0, z) , then Q2 will be at P2 (0,0, z + d ) and Q3 at P3 (0,0, z + d 4) . The total force on Q3 is the resultant of the forces F31 and F32 due to Q1 and Q2 respectively, where

F31 =

1 Q3Q1 1 Q3Q2 a , F32 = a32 2 31 4π ε R 31 4π ε R 32 2

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where R 31 and a31 are obtained as

r3 − r1 = a x ( x3 − x1 ) + a y ( y3 − y1 ) + a z ( z3 − z1 ) = a z d 4

R31 = r3 − r1 = d 4 r −r ⇒ a31 = 3 1 = a z R31 Substituting R31 = d 4 and a31 = a z in F31 above, yields

F31 =

1 Q3Q1 az 4πε (d 4)2

Similarly, the force on Q3 due to Q2 is obtained as

F32 =

1 Q3Q2 az 4πε (3d 4)2

Since the total force on Q3 is zero, then

F = F31 + F32 = a z

Q ⎞ 1 4Q3 ⎛ Q + 2⎟=0 2 ⎜ 1 9 ⎠ πε d ⎝

Q1 to Q2 that makes zero force on Q3 is Q1 Q2 = −1 9. The minus sign indicates that Q1 and Q2 are opposite in sign.

From this equation, the ratio of

2.4.

ELECTRIC FIELD INTENSITY

2.4.1. Electric Field due to a Point Charge Consider a point charge Q at the origin. According to Coulomb’s law, if a charge q is placed in the vicinity of the charge Q , then q will be under the influence of an electrostatic force field originating from Q . Using (2.6), this force is given by

F=

1 qQ ar 4π ε r 2

(2.11)

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where r is the distance between the two charges and a r is the unit vector along r . If q is considered as a positive test charge, then the direction of the force F is determined by the sign of Q . If Q is positive, q will be under the influence of repulsive force away from Q , and if Q is negative q will be under the influence of attractive force towards Q . The magnitude of the force on q depends on the value of Q . This indicates that Q has an electrostatic influence on any charge q placed in the region surrounding it. This effect per unit positive charge can be expressed as

F 1 Q = ar q 4πε r 2

(2.12)

When q is removed, or q → 0 , the electrostatic influence originating from Q is known as the electric field intensity E of the charge Q . The electric field intensity is a vector field and given mathematically by

E = lim q →0

F 1 Q = ar q 4πε r 2

(2.13)

The direction of the electric field produced by Q at any point is the direction of the force on a positive test charge at that point. In (2.13), Q is assumed at the origin. If Q is located at any general point ( x ′ , y ′ , z ′ ) as shown in Fig. (2.2), (2.13) can be expressed in terms of this point and the point ( x, y , z ) where the electric field to be measured. The point ( x ′, y ′, z ′) is referred to as source point, while ( x, y, z ) as a field point. E P1 (x1, y1, z1) Field Point

R12 = r1 - r2

r1

Origin

aR

r2

Source Point Q P2 (x2, y2, z2)

Fig. (2.2). The electrostatic field due to charge and the position vectors of the source and field points.

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Referring to Fig. (2.2) the position vectors of the source and field points are r′ and r respectively. Therefore, the electric field E at ( x, y, z ) due to charge Q at ( x ′, y ′, z ′) can be expressed as

E=

1

Q

4π ε R

2

aR =

Q r − r′ 4π ε r − r ′ 3

(2.14)

Vm

The unit of measurement of the electric field intensity is (V m). The electric field intensity is represented by lines known as streamlines or lines of force. The density of these lines in space is proportional to the magnitude of the electric field intensity and the direction of these lines indicates the direction of the electric field. The streamlines are always originating from positive charges and entering negative charges, as illustrated in Fig. (2.3).

+Q

-Q

(a)

(b)

Fig. (2.3). Streamlines of electric field intensity. (a) Positive charge. (b) Negative charge.

2.4.2. Electric Field due to a Discrete System of Point Charges Equation (2.14) gives the electric field produced by a single point charge Q . If there are N point charges Q1 , Q2 , , QN at the points (x1, y1, z1 ), (x2 , y2 , z2 ), …, (xN , yN , z N ) respectively, the total electric field intensity at a field point (x, y, z ) is the vector sum of the electric fields of individual point charges Q1, Q2 ,, QN at this point. The electric field En due to a charge Qn at n = 1, 2, , N , can be obtained from (2.14) as

En =

1

Qn

4πε R n 2

a Rn

(xn , yn , zn ),

where

(2.15)

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where | R n |=| r − rn | and a Rn = (r − rn ) / | R n | . r and rn are the position vectors of point (x, y, z ) and (xn , yn , zn ) respectively. Superimposing the field of all individual charges, then the total electric field E due to the system of charges at the point (x, y, z ) is N

E = E1 + E 2 +  + E N = ∑ En = n =1

1

N

Qn

∑R 4π ε n =1

2

a Rn

(2.16)

n

Example 2.5 A point charge Q = +10n C is placed in free space at the point P1 (2,1,2) cm. Find the electric field intensity at the point P2 (4,2,0) cm. Solution The charge Q = +10n C is at P1 (2,1, 2) . It is required to find the electric field at P2 (4, 2, 0) , then P1 is the source point, and P2 is the field point. The electric field is given by (2.14), where R and a R are obtained as follows

r − r′ = a x ( x − x′) + a y ( y − y′) + a z ( z − z′) = 2a x + a y − 2a z

R = r − r′ = 4 + 1 + 4 = 3 cm ⇒ a R =

r − r′ 2a x + a y − 2a z = R 3

Using (2.14), the electric field at P2 (4, 2, 0) is

E=

1

Q

4π ε R

a = 2 R

⎛ 2a x + a y − 2a z ⎞ 10 × 10 −9 ⎜ ⎟ −9 ⎜ 4π × (1 36π ) × 10 ⎝ 27 × 10 − 4 ⎟⎠ .

= 6.67a x + 3.33a y − 6.67a z V m Example 2.6 Four point charge Q1 , Q2 , Q3 , and Q4 are placed in free space at the points P1 (1,1, 0) , P2 (−1,1, 0) , P3 (−1, − 1, 0) , and P4 (1, − 1, 0) respectively, which represent

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vertices of a square. If Q1 = Q3 = +1μ C and Q2 = Q4 = −1μ C , and the units of the coordinate system are in meter, find the electric field (a) At the center of the square. (b) On the axis of the square at a height 1m above the center of the square. Solution The geometry of the problem is shown in Fig. (2.4). The center of the square is at the origin (0,0,0) . (a) Electric field due to Q1 , Q2 , Q3 , and Q4 at the point (0,0,0) is

E=

1

4

Qn

∑R 4π ε n =1

2

a Rn

n

r − rn = a x ( x − xn ) + a y ( y − yn ) + a z ( z − zn ) R = r − r ⇒ a = (r − r ) R , n = 1, 2,, 4 n

n

Rn

n

n

For Q1 at P1 (1,1,0), we have

r − r1 = a x (0 − 1) + a y (0 − 1) = −(a x + a y )

R1 = a x + a y = 2 m ⇒ a R1 = − (a x + a y )

2

Similarly, for Q2 , Q3 , and Q4 , we have R 2 = R 3 = R 4 = 2 m ⇒ a R 2 = (a x − a y )

2 , a R 3 = (a x + a y )

2 , a R 4 = − (a x − a y )

2

Then the electric field at the center of the square (at the origin) becomes

36π × 10 −6 ⎛ − a x − a y a x − a y a x + a y − a x + a y ⎞ ⎜ ⎟=0 V m E= − + − 4π × 10 −9 ⎝⎜ 2 2 2 2 2 2 2 2 ⎠⎟ This result is expected, because the field of Q1 cancels that of Q3 and the field of Q2 cancels that of Q4 . Therefore, the total electric field is zero.

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(b) The electric field on the axis of the square at a height 1m above the center is

E=

4

1

Qn

∑R 4π ε n =1

2

a Rn

n

For Q1 at P1 (1,1,0), we have

r − r1 = a x (0 − 1) + a y (0 − 1) + a z (1 − 0) = −(a x + a y + a z )

R1 = r − r1 = 3 m ⇒ a R1 = − (a x + a y + a z )

3

Similarly for Q2 , Q3 , and Q4 , we have R 2 = R 3 = R 4 = 3 m

(

⇒ aR2 = ax − a y + az

)

3,

a R 3 = (a x + a y + a z )

a R 4 = (− a x + a y + a z )

3,

3

The electric field on the axis of the square at a height 1m above the center of the square becomes

E=

⎛ − ax − a y + az ax − a y + az ax + a y + az − ax + a y + az ⎞ 10 6 ⎜ ⎟⎟ − + − 4π × (1 36π ) ⎜⎝ 3 3 3 3 3 3 3 3 ⎠

= 1.732 a z k V m z (0,0,1) P2(-1,1,0) Q2 = -1μC

P3(-1,-1,0) Q3 = +1μC (0,0,0)

y Q1 = +1μC P1(1,1,0)

Q4 = -1μC P4(1,-1,0)

x Fig. (2.4). Geometry of the problem in Example 2.5.

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2.4.3. Electric Field due to a Continuous Distribution of Charge The electric field intensity at any field point ( x, y, z ) due to a continuous distribution of charges can be determined as follows: 1. Choose the appropriate coordinate system. 2. Choose an infinitesimal unit charge ΔQ in the region of charge distribution at an arbitrary source point ( x′, y′, z′) . 3. Find the electric field intensity ΔE at ( x, y, z ) due to the charge ΔQ at ( x′, y′, z′) . When ΔQ → 0 , then the electric field can be written as dE =

dQ r − r′ dQ a R = 3 4πε r − r′ 4πε R 2

(2.17)

where r′ and r are the position vectors of the source and field point respectively. R = r − r ′ and a R = R / | R |. 4. The total electric field due to the continuous charge distribution is obtained by integrating both sides of (2.17). The integration on the right-hand side is taken over the entire region containing the charge distribution. Therefore, the total electric field due to the continuous charge distribution becomes E=

1

4πε ∫

Q

dQ

r − r′ 1 = 3 4πε r − r′

∫

Q

dQ

aR R

2

(2.18)

2.4.3.1. Electric Field due to a Linear Distribution of Charge Consider an amount of charge is distributed along a line l with a density ρl C m as shown in Fig. (2.5). The amount of charge along a unit length dl ′ is dQ = ρ l dl ′ ; hence the total electric field due to the charge along l at the field point is E=

1

4πε ∫ l

ρl dl ′ R

2

aR

(2.19)

where R = r − r′ and a R = R / | R |. r′ and r are the position vectors of the source and field point respectively.

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Sameir M. Ali Hamed

Field point (x, y, z)

Source point (x’, y’, z’) dQ = ρl dl’

R

r

r′

l

y x Fig. (2.5). Electric field intensity due to a linear distribution of charge.

2.4.3.2. Electric Field due to a Surface Distribution of Charge For a surface charge distribution on a surface S with charge density ρl C m as shown in Fig. (2.6), the total charge dQ per unit surface area dS ′ is dQ = ρ s dS ′ . Substituting dQ into (2.18), the total electric field at the field point due to the surface distribution of charge on S becomes E=

1

4πε ∫∫

ρ s dS ′

S

z Surface charge density ρs C/m2

R

2

(2.20)

aR

Source point (x’,y’,z’) dQ = ρs dS’ R

Field point ( )

S r r'

y x Fig. (2.6). Electric field intensity due to a surface distribution of charge.

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2.4.3.3. Electric Field due to a Volume Distribution of Charge The geometry for an arbitrary distribution of charge within a volume v with 3

charge density ρ v C m is shown Fig. (2.7). In this case, charge dQ per unit volume dv′ is dQ = ρ s dv′ and the total electric field intensity at the field point due to the total charge within the volume v is obtained as E=

1

4πε ∫∫∫ V

ρ v d v′ R

2

aR

(2.21)

Fig. (2.7). Electric field intensity due to a volume distribution of charge.

Example 2.7 Find the electric field intensity produced by a uniform charge density ρl C m along a line of length 2 L at a distance d from the axis of the line, and then find the electric field when the line is infinite in length. Solution Let the line coincides with z-axis and symmetrical about the origin as shown in Fig. (2.8). The cylindrical coordinate is appropriate for this problem. Thus, the source and field points are (0,0, z′) and (d ,φ ,0) respectively. Referring to Fig. (2.8), it can be shown that

r = da ρ , r′ = z′a z ⇒ R = r − r′ = da ρ − z′a z

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⇒ aR =

Sameir M. Ali Hamed

R da ρ − z ′a z = R d 2 + z′2

The electric field at distance d from the axis of the line is

E=

1 4πε

∫

ρl dl ′ R

l

2

aR =

L

1 4πε

ρl dz′ d 2 + z ′2

∫(

−L

)

3 2

∫ (d

2

(da

ρ

− z′a z )

This integral can be written in the form

ρ d E= l 4πε

L

dz′ 2 d + z′2

∫(

−L

)

3 2

ρ aρ − l 4πε

L

−L

z′dz′ + z′2

)

3 2

az

The integration of the second term on the right-hand side is zero and the electric field reduces to E=

ρl d 4πε

L

∫ (d

−L

dz′ 2 + z′2

)

3 2

aρ

Using the substitution z′ = d tan θ the above integral simplifies to

E = aρ

tan −1 ( dL )

ρl 4πε d

∫ cosθ dθ 0

= aρ

ρl

4πε d ρl = aρ 2πε d

θ = tan −1 ( dL )

sin θ θ = 0 L

L2 + d 2

Note that the electric field is in the radial direction. If the line is infinite in length, then L → ∞ . The electric field, in this case, can be obtained by letting L → ∞ in the expression of E that is obtained for the line of length 2 L . Thus, E due to the infinite line charge at distance d becomes E = aρ

⎛ L lim ⎜ ⎜ → ∞ L 2πε d 2 2 ⎝ L +d ρl

⎞ ⎟ = ρl a . ⎟ 2πε d ρ ⎠

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z (0, 0, L) Source point (0, 0, z′)

dQ = ρl dz′

r'

R

y r (d,φ, 0) Field point

x

Linear charge density ρl C/m

(0, 0, -L)

Fig. (2.8). Geometry of the problem in Example 2.7.

Example 2.8 Find the electric field intensity produced by a uniform surface charge density 2 ρ s C m over a square plate of the side 2 L on its axis at a distance d from its center, and then find the electric field at the same point when the plate is infinitely extended. Solution Let the plate coincides with the xy plane and symmetrical about the origin as shown in Fig. (2.9). The geometry of the problem is suitable for the rectangular coordinate system. With reference to Fig. (2.9), the source point (x′, y′,0) is on the xy plane and the field point is at the point (0,0, d ), then

r = daz , r′ = x′a x + y′a y ⇒ R = r − r′ = −(x′a x + y′a y − da z ) x′a + y′a y − da z R ⇒ aR = =− x R d 2 + x′2 + y′2 The electric field due to the charge on the plate at a distance d is

E=

1

ρ s dS ′

2 4π ε ∫∫ R S

aR = −

1

L L

ρ s dx ′dy ′

4π ε −∫L−∫L d 2 + x ′ 2 + y ′ 2

(

)

3 2

(x′a

x

+ y ′a y − da z )

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This integral can be written in the form L L L L (x′a x + y′a y ) dx′dy′ ρs d ρs d dx′dy′ E = az − 3 3 4π ε −∫L−∫L (d 2 + x′2 + y′2 )2 4π ε −∫L−∫L (d 2 + x′2 + y′2 )2

The integral in the second term of the right-hand side is zero, then

E = az

L L dx′dy′ ρs d ∫ ∫ 4π ε − L− L d 2 + x′2 + y′2

(

)

3 2

2 2 2 The integral with respect to y ′ can be evaluated first by letting d + x′ = α and using the substitution y′ = α tan θ , as

L

ρ d E = a z s ∫ dx′ 2π ε − L

tan −1

∫ 0

( αL )

ρ s d dx′ θ = tan −1 ( L ) sin θ θ = 0 α 2 ∫ 2π ε − L α L dx′ ρ s Ld = az ∫ 2π ε − L (d 2 + x′2 ) L2 + d 2 + x′2

cosθ dθ α2

L

= az

2

2

2

The integral with respect to x′ can be evaluated by letting d + L = β and using the substitution x′ = β sinh θ . Doing this, the integral reduces to

E = az

ρs Ld πε

⎛ sinh −1⎜ ⎝

∫ 0

⎞ L ⎟ d 2 + L2 ⎠

(d

sech 2 θ dθ 2

+ L2 tanh 2 θ ⎛ θ =sinh −1⎜ ⎝

ρ ⎛ L tanh θ ⎞ = a z s tan −1 ⎜ πε d ⎟⎠ θ =0 ⎝

L

) ⎞ ⎟

d 2 + L2 ⎠

Substituting the limits of the integral with simplification, E due to the surface charge on the plate at a distance d above its center is obtained as

E = az

ρs L2 ⎛ ⎞ tan −1⎜ ⎟ 2 2 πε ⎝ d d + 2L ⎠

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z Field point (0, 0, d) Surface charge density ρs C/m2

r R

(0,L,0)

r'

x

y

Source point dQ = ρs dS′ = ρs dx′dy′

(L,0,0)

Fig. (2.9). Geometry of the problem in Example 2.8.

Note that the electric field is normal to the plate. If the plate is infinitely extended, then L → ∞ . The electric field, in this case, is obtained by letting L → ∞ in the above expression for E . Thus, ⎡ −1 ⎛ ρs L2 ⎜ E = az lim ⎢tan ⎜ 2 2 π ε L →∞ ⎢⎣ ⎝ d d + 2L

⎞⎤ ρ ⎟⎟⎥ = s a z , ⎠⎥⎦ 2 ε

z>0

Note that for an infinitely extended plate the resulted electric field is independent of the distance of the field point from the plate. Similarly, the electric field intensity in the region z < 0 can be obtained as

E=−

ρs az , 2ε

z a from its center, if the charge density is

3q 4π a 3 3q (b) ρ v = , 0 ≤ r′ ≤ a . 4π a 2 r ′ (a) ρ v =

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Solution Let the center of the sphere coincides with the origin of the coordinate system. Using the spherical coordinate, the source point will be (r′,θ ′, φ′), where 0 ≤ r ′ ≤ 0 . Since the sphere is symmetric in θ and φ directions, the resulted electric field at any field point (r ,θ , φ ) depends on the distance r only. Therefore, (r,θ , φ ) can be chosen on z -axis at a distance r from the origin. With reference to Fig. (2.10), we have

r = r a z , r ′ = a x r ′ sin θ ′ cos φ ′ + a y r ′ sin θ ′ sin φ ′ + a z r ′ cosθ ′ ⇒ R = r − r ′ = a x r ′ sin θ ′ cos φ ′ + a y r ′ sin θ ′ sin φ ′ + a z (r − r ′ cosθ ′) R a x r ′ sin θ ′ cos φ ′ + a y r ′ sin θ ′ sin φ ′ + a z (r − r ′ cosθ ′) ⇒ aR = = R r ′ 2 + d 2 − 2r ′d cosθ Substituting dv′ = r′2 sin θ ′dr′dθ ′dφ ′ , R , and a R into (2.21), yields

E=

2π π a

1 4π ε

ρv r′2 sin θ ′dr ′dθ ′dφ ′

∫∫∫ (r′

)

3

+ r 2 − 2r′r cosθ ′ 2 × a x r′ sin θ ′ cosφ ′ + a y r′ sin θ ′ sin φ ′ + a z (r − r′ cosθ ′)

0 0 0

2

[

]

The above integral with respect to φ ′ in the direction of a x and a y is zero. In the direction of a z the integral becomes π a 1 ρv r ′2 sin θ ′dr ′dθ ′ E = az 2 2 2 ε ∫∫ 0 0 r ′ + r − 2r ′r cosθ ′

(

)

3 2

(r − r′ cosθ ′)

The above integral depends on ρ v . There are two cases for ρ v as follows: (a) When ρ v = 3q (4π a 3 ) , then π a 3q r ′2 sin θ ′dr ′dθ ′ E = az 2 2 8π ε a 3 ∫∫ 0 0 r ′ + r − 2r ′r cosθ ′

(

)

3 2

(r − r′ cosθ ′)

Electrostatic Fields

Electromagnetics for Engineering Students Field point (r,θ,φ)

Source point dQ = ρv dv′ = ρv r2

z

R

97

(r′,θ′,φ′) Volume charge density ρv C/m3

r

r'

(0,a,0)

y

v

x Fig. (2.10). Geometry of the problem in Example 2.9.

The integral can be written in the form E = az

(

)

π a ⎡π a 3q r ′2 sin θ ′dr ′dθ ′ r ′2 r 2 − r ′2 sin θ ′dr ′dθ ′ ⎤ + ⎢ 1 3 ⎥ 16π ε a 3r ⎢⎣ ∫0 ∫0 r ′2 + r 2 − 2r ′r cosθ ′ 2 ∫0 ∫0 r ′2 + r 2 − 2r ′r cosθ ′ 2 ⎥⎦

(

)

(

)

Using the substitution u = cosθ ′ , then du = − sin θ ′dθ ′ and the limits of the integrals become from u = 1 to u = −1 . Hence, a

E = az

(

3q r ′dr ′ r ′2 + r 2 − 2r ′r u 16π ε a 3r 2 ∫0

3q − az 16π ε a 3r 2

a

(

)

r ′ r 2 − r ′ 2 dr ′

∫ (r′

1 2

u = −1 u = +1

u = −1

)

1

+ r − 2r ′r u 2 a 3q q = az r ′ 2 dr ′ = a z 3 2 ∫ 4π ε a d 0 4π ε r 2 0

2

)

2

u = +1

(b) When ρ v = 3q (4π a 2 r ′) the integral becomes E = az

(

)

π a ⎡π a 3q r ′ sin θ ′dr ′dθ ′ r ′ r 2 − r ′2 sin θ ′dr ′dθ ′ ⎤ + ⎢ 1 3 ⎥ 16π ε a 2 r ⎢⎣ ∫0 ∫0 r ′2 + r 2 − 2r ′r cosθ ′ 2 ∫0 ∫0 r ′2 + r 2 − 2r ′r cosθ ′ 2 ⎥⎦

(

)

(

)

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Using the substitution u = cos θ ′ , then du = − sin θ ′dθ ′ and the limits of the integrals become from u = 1 to u = −1 as in (a) above, then a

(

3q dr ′ r ′2 + r 2 − 2r ′r u E = az 2 2 ∫ 16π ε a r 0 3q − az 16π ε a 2 r 2

a

∫ (r′ 0

(r 2

2

)

− r ′ 2 dr ′

+ r 2 − 2r ′r u

)

1 2

u = −1 u = +1

u = −1

)

1 2

u = +1

Substituting the limits and simplifying yields a

E = az 2.5.

3q 3q 3q = az r′dr′ = a z 2 2 ∫ 2 4π ε a r 0 8π ε r 8π ε r 2

ELECTRIC FLUX AND ELECTRIC FLUX DENSITY

Consider two concentric conducting spheres as shown in Fig. (2.11). The space between the sphere is a dielectric of permittivity ε . When the outer sphere is charged with a positive charge + Q , a negative charge − Q will be induced in the inner sphere. The charge in the inner sphere is directly proportional to the charge on the outer sphere. Michael Faraday carried out a similar experiment in 1837 and he concluded that there is a form of a displacement takes place from the charged sphere to the sphere of the induced charges. This displacement is independent of the dielectric medium in the space between the spheres. This displacement is termed electric flux ψ e . The electric flux is directly proportional to the amount of the charge Q . The proportionality constant does not depend on the medium between the spheres and its value is found to be equal to unity, hence.

ψe = Q

(2.22)

The electric flux is measured in Coulomb (C) . The electric flux is represented by lines of electric field originating from the charge + Q on the outer sphere in the radial direction and ending at the charge − Q on the inner sphere as shown in Fig. (2.11). If the charge + Q is uniformly distributed on the surface of the sphere, then the flux crossing a spherical surface of radius r in the space between the two

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99

spheres is ψ e = Q . The distribution of this flux on a surface S that is perpendicular to the direction of electric field is known as the electric flux density D , which is a vector quantity. The direction of D is in the same direction of the electric field intensity E . In the case of the two conducting spheres of Fig. (2.11), the electric field is always perpendicular to the spherical surface S of radius r , then the electric flux density can be expressed as

D=

ψe S

ar =

Q ar 4π r 2

(2.23)

D is a vector field measured in Coulomb per meter squared (C m 2 ) . The electric field E on the spherical surface of radius r shown in Fig. (2.11), is

E=

Q ar 4π ε r 2

(2.24)

Comparing (2.23) and (2.24), the relation between E and D is obtained as

D = εE

(2.25)

where ε is the permittivity of the medium as introduced in Section 2.2. ε gives a quantitative measure of the polarization ability of materials that occurs when a material is interacted with an external electric field as will be shown in Chapter 3.

Spherical surface of radius r

ε

Conducting sphere of radius b and charge + Q

Conducting sphere of radius a and charge - Q

Fig. (2.11). Illustration the concept of electric flux.

2.6.

GAUSS’S LAW FOR ELECTRIC FLUX

Gauss’ law states that the total electric flux crossing a closed surface is equal to the total charge enclosed in that surface. To explain this statement, consider a

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total charge QT is enclosed in an arbitrary closed surface S as illustrated in Fig. (2.12). Δψ e is the electric flux due to QT crossing an infinitesimal unit surface area ΔS . The unit surface area ΔS is a vector quantity in the direction of the unit vector n which is directed normally out of the surface S as shown in Fig. (2.12). If the direction of the flux density D makes an angle θ to the direction of ΔS = nΔS , the flux Δψ e crossing the unit surface area ΔS can be expressed as

Δψ e = D ΔS cosθ . This can be written in vector form as

Δψ e = D • ΔS = D • nΔS

QT

(2.26)

ΔS = n ΔS

n

θ

Closed surface S

D Fig. (2.12). The electric flux density D through a closed surface S enclosing a total charge QT .

The total flux can be obtained be summing the flux at all points on the surface S . As ΔS → 0 , the sum becomes an integral over S . Hence

ψ e = ∫∫ D • dS

(2.27)

S

If the total enclosed charge in S is QT , then using (2.22) and (2.27), Gauss’s law is expressed as

ψ e = ∫∫ D • dS = QT

(2.28)

S

The relation between the total charge QT and the volume charge density in the region enclosed by S is also given by (2.1c). Substituting QT from (2.1c) into (2.28), and using the Divergence theorem, yields

QT = ∫∫ D • dS = ∫ ∇ • D dv′ = ∫ ρv dv′ S

v

v

(2.29)

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It follows from (2.29) that

∇ • D = ρv

(2.30)

Using (2.25), (2.30) can be written as

∇•E=

ρv ε

(2.31)

Based on Gauss’s law, (2.1a) and (2.1b), the relation between the linear and surface charge densities and the electric flux density can be written as

∫∫ D S

2.7.

•

dS = ∫ ρl dl′ ,

∫∫ D

l

•

S

dS = ∫∫ ρ s dS ′

(2.32)

S

APPLICATIONS OF GAUSS’S LAW

Gauss’s law is used to find the electric field intensity produced by charges enclosed in closed surfaces. When applying Gauss’s law we follow the steps: 1. A suitable closed surface enclosing the charge is chosen. This surface is known as a Gaussian surface. 2. The unit vector in the direction of the differential area dS and the direction of electric flux density through dS are determined. 3. The total enclosed charge QT in the Gaussian surface is determined. If the enclosed charges are discrete system of charges Q1, Q2 ,, QN , then

QT =

n= N

∑Q

n

(2.33)

n =1

If the enclosed charge is continuously distributed along a line of charge density ρ l C m , on a surface of charge density ρ s C m 2 , or within a volume of charge density ρ v C m 3 , then the total enclosed charge QT is given respectively by (2.1a) - (2.1c). 4. Application of Gauss’s law

∫∫ D S

•

dS = QT

(2.34)

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2.7.1. Electric Field due to a Point Charge The electric field intensity produced by a point charge Q at a distance r from it can be obtained using Gauss's law as follows: 1. It is suitable to choose a sphere as a Gaussian surface in this problem. The radius of the sphere is r and point charge Q is at its center. 2. Referring to Fig. (2.13), the differential surface area in spherical coordinates is dS = a r r 2 sin θ dθ dφ , where a r is the unit vector in the radial direction. E and D are also in the radial direction, or D = a r ε E . 3. Finding the total enclosed charge. The total charge in this case is QT = Q . 4. Applying the Gauss’s law. Substituting dS and D into (2.34), yields

∫∫ D

•

dS =

S

2π π

∫∫ ε Er

2

sin θ dθ dφ = Q

(2.35)

0 0

Evaluating the integral, then solving for E and using E = D ε = ar E , yields

E=

Q ar 4π ε r 2

(2.36)

This is the electric field intensity produced by a point charge Q at a distance r from it, which is obtained previously in (2.13). z

dS = ar r2 sinθ ar

θ r Q

y φ S

x Fig. (2.13). The electric field due to a point charge using Gauss’s law.

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2.7.2. Electric field due to an Infinite Line of Charge The electric field produce by an infinite line charge of a uniform charge density ρ l C m at a distance d from it can be obtained using Gauss’s law as follows: 1. In this problem, it is suitable to choose a cylindrical surface of radius d as a Gaussian surface. Let the axis of the Gaussian surface enclosing a length l of the infinite line of charge such that the line charge coincides with the axis of the Gauss’s surface. In this case, the closed surface is composed of following three surfaces as shown in Fig. (2.14): (1) The top base of the cylinder S1 . (2) The bottom base of the cylinder S2 . (3) The cylindrical surface S3 . 2. Referring to Fig. (2.14), the differential surfaces for S1 , S 2 , and S 3 are dS1 = a z ρ dρ dφ , dS 2 = −a z ρ dρ dφ , and dS 3 = a ρ r dφ dz respectively. The direction of the electric field and the electric flux density for the line charge is in the radial direction only, then D = a ρ ε E . 3. Finding the total charge. In this problem the total charge enclosed is l

QT = ∫ ρ l dz = ρ l l

(2.37)

0

4. Applying Gauss's law.

∫∫ D S

•

dS = ∫∫ D • dS1 + ∫∫ D • dS 2 + ∫∫ D • dS3 = ρl l S1

S2

(2.38)

S 31

Substituting dS1 , dS2 , dS 3 , and D into (2.38), putting in mind that D • dS1 = 0 and D • dS2 = 0 , the remaining integral can be obtained as l 2π

∫ ∫ ε Ed dφ dz = 2π ε d lE = ρ l l

(2.39)

0 0

Solving for E and using E = D ε = a ρ E , the electric field produced by an infinite line charge of charge density ρl C m at a radial distance d from it is

E=

ρl aρ 2π ε d

(2.40)

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Which is the same result obtained in Example 2.6. dS1 = az ρ dρ dφ

z S1

d

S3

l

dS3 = aρ d dφ dz S2

φ

y dS2 = -az ρ dρ dφ Unifrom linear charge density ρl C/m

x

Fig. (2.14). The electric field due to a line charge using Gauss’s law.

2.7.3. Electric Field due to a Uniform Distribution of Surface Charge The electric field intensity above an infinite conducting plate of a uniform surface 2 charge density ρ s C m can be obtained using Gauss’s law as follows: 1. In this problem, a closed cylinder of radius r can be chosen as a Gauss surface. The Gaussian surface is composed of following three surfaces as shown in Fig. (2.15): (1) The top base of the cylinder S1 . (2) The bottom base of the cylinder S 2 . (3) The cylindrical surface S 3 . 2. Assuming that the axis of Gaussian surface coincides with the z-axis, then the differential surfaces for S1 , S 2 , and S 3 become dS1 = a z ρ dρ dφ

dS 2 = −a z ρ dρ dφ , and dS 3 = a ρ ρ dφ dz respectively. The directions of E and D are normal to the plate. Therefore, the flux density through S1 , S 2 , and S 3 are D = a z ε E , D = −a z ε E , and D = 0 respectively. 3. Finding the total enclosed charge in the Gaussian surface. Referring to Fig. (2.15), the total enclosed charge inside the Gaussian surface is

QT =

2π r

∫ ∫ρ 0 0

s

ρ dρ dφ = ρ s π r 2

(2.41)

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z

Surface charge

105

dS1 = az ρ dρ dφ

density ρs C/m2

D = az D S1

r

D=0 dS3 = aρ d dφ dz

Conducting plate S3 S2

y

φ D = -az D dS2 = -az ρ dρ dφ

x

Fig. (2.15). The electric field due to a conducting plate of a uniform charge density using Gauss’s law.

4. Applying Gauss’s law yields

∫∫ D S

•

dS = ∫∫ D • dS1 + ∫∫ D • dS 2 + ∫∫ D • dS3 = ρ s π r 2 S1

S2

(2.42a)

S 31

Substituting dS1 , dS2 , dS 3 , and D into (2,42a), with D • dS3 = 0 , it reduces to 2π r

2π r

0 0

0 0

∫∫ ε Eρ dρ dφ + ∫∫ ε Eρ dρ dφ

= 2 ε E π r 2 = ρ sπ r 2

(2.42b)

Solving (2.42b) for E and using E = D ε = a z E , the electric field intensity due to 2

a uniform surface charge ρ s C m on an infinite conducting plate is

E = az

ρ s ⎧+ 1, z > 0

⎨ 2 ε ⎩− 1, z < 0

This result is similar to that is obtained in Example 2.8.

(2.43)

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2.7.4. Electric Field due to a Uniform Distribution of Volume Charge 3

Consider a sphere of radius a with a uniform volume charge density ρ v C m as shown in Fig. (2.16). The electric field intensity at a distance r from the center of the sphere can be obtained by applying Gauss’s law as follows: 1. The Gaussian surface is chosen as a sphere of radius r . 2. The differential area in given by dS = ar r 2 sin θ dθ dφ . The directions of E and D are also in the radial direction, or D = a r ε E . dV = r2 sinθ drdθ dφ

z

dS = ar r2 sinθ dθ dφ

r a

dV

θ

φ

y

Volume charege density ρv C/m3 V

x Fig. (2.16). The electric field due to a sphere of a uniform volume charge density using Gauss’s law.

3. Finding the total charge inside the Gaussian surface. There are two cases for the Gaussian surface: (a) r ≥ a . (b) r ≤ a . Therefore, the total charge is

⎧ 2π π a 4 2 3 ⎪ ∫∫∫ ρv r sin θ dr dθ dφ = ρv π a 3 ⎪ QT = ⎨ 20π π0 0r 4 ⎪ ρv r 2 sin θ dr dθ dφ = ρv π r 3 ∫∫∫ ⎪⎩ 0 0 0 3

r≥a (2.44)

r≤a

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4. Applying Gauss’s law with dS = ar r 2 sin θ dθ dφ , Substituting QT from (2.44), we get 2π π

⎧ a3 4 2 Er d d θ θ φ ρ π = ε sin ⎨ 3 v ∫∫ 3 ⎩r 0 0

107

D = ar ε E , and

r≥a

(2.45)

r≤a

Solving (2.45) for E , and using E = D ε = ar E , the electric field intensity due to 3

a charged sphere of radius a with a uniform charge density ρ v C m can be expressed as

E = ar 2.8.

2 ρv a ⎧(a r ) r ≥ a

⎨ 3 ε ⎩r a

(2.46)

r≤a

ELECTROSTATIC POTENTIAL

Consider a positive charge q in an electric field of intensity E as shown in Fig. (2.17). The work done to move the charge from a point A that is defined by the vector rA to a point B defined by the vector rB along the path l is given by A

B

WAB = ∫ F cosθ dl = − ∫ F • dl B

A

(2.47)

A

l′

q

l

dl

qEdlcosθ

θ

E

rA r

qE B

Origin

rB

Fig. (2.17). The work done to move the charge q from a point A to B in an electric field E .

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where F = F is the magnitude of the force exerting on the positive charge q. Using (2.12) and (2.13), the force on a charge q placed in an electric field E is

F = qE

(2.48)

Substituting F from (2.48) into (2.47), the work done, or the potential energy, to move the positive charge q from point A to point B along the path l becomes B

WAB = − q ∫ E • dl

(2.49)

A

The work done to move the charge along closed path from A to B , then from B to A must be zero, or B

A

− q ∫ E • d l − q ∫ E • dl ′ = 0 A

(2.50)

B

Consequently, B

B

A

A

WAB = −q ∫ E • dl = −q ∫ E • dl′

(2.51)

It is evident from (2.51) that the work done along the path l is the same as the work done along the path l ′ to move the charge between the points A and B . This means that the work does not depend on the path between A and B along which the charge is moved. It depends only on the coordinates of the points A and B . The electric potential at a certain point is defined as the work done to move a unit positive charge from infinity to the point. The electric potential is measured in Volts (V ). In (2.49), if point A is at infinity, then the work done per unit positive charge q is the electric potential of point B . Mathematically, the electric potential of point B is expressed as B

VB =

WAB = − ∫ E • dl q ∞

(2.52)

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Similarly, the electric potential at a point C is the work done per unit positive charge q from infinity (point A ) to point C , which is C

W VC = AC = − ∫ E • dl q ∞

(2.53)

The potential difference between B and C is the work done per unit positive charge from B to C . This can be explained as follows

WAB q

B

C

∞

∞

B

= − ∫ E • dl = − ∫ E • dl − ∫ E • dl A C

B

∞

∞

= − ∫ E • dl + ∫ E • d l

(2.54)

= VC − VB = VBC Consequently, the potential difference between two points B and C in an electric field E is given by C

VBC = VC − VB = − ∫ E • dl

(2.55)

B

The differential element dl along the path is given by

dl = a x dx + a y dy + a z dz dl = a ρ dρ + aφ ρ dρ + a z dz dl = a r dr + aθ rd θ + aφ r sin θ dφ

In rectangular coordinate

(2.56a)

In cylindrical coordinate

(2.56b)

In spherical coordinate

(2.56c)

Example 2.10 Find the work done to move a 10 μC charge in the electric field E = −8xya x − 4 x2a y + a z V m

From the point (1, 8, 5) m to (2,18, 6) m along the path y = 3x 2 + z , z = x + 4 .

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Solution Let the points (1, 8, 5) m and (2,18, 6) m be A and B respectively. The work done to move a charge q = 10 μC from A to B in E is given by (2.49). In rectangular coordinate dl = a x dx + a y dy + a z dz , then B

WAB = WAB = −q ∫ E • dl = −q A

(2,18, 6 )

(− 8xya ∫ ( )

x

1, 8, 5

⇒ W AB = 8q

(2,18, 6 )

)

− 4 x 2a y + a z ⋅ (a x dx + a y dy + a z dz ) (2,18, 6 )

∫ xydx + 4q ( ∫ x)

(1, 8, 5 )

2

dy − q

1, 8, 5

(2,18, 6 )

∫ dz

(1, 8, 5 )

The work is done along the path y = 3x 2 + z , z = x + 4 . Therefore, all above three integrals can be expressed in terms of x only as follows y = 3x 2 + z ⇒ dy = 6 xdx + dz z = x + 4 , ⇒ dz = dx , ⇒ y = 3 x 2 + x + 4

Substituting y , dy and dz in the above expression for WAB , yields 2

2

W AB = 8 q ∫ (3 x + x + 4 x )dx + 4 q ∫ (6 x + x 3

2

3

1

2

2

1

)dx − q ∫ dx 1

Evaluating the integrals and substituting q = 10 μC , yields

W AB

3 1 = q ⎡⎢ 8 ⎛⎜ x 4 + x 3 + 2 x 2 3 ⎣ ⎝4

⎞ ⎛6 4 1 3 ⎟ + 4⎜ x + x 3 ⎠ ⎝4

x=2

⎤ ⎞ = 2.55 m J ⎟ − x⎥ ⎠ ⎦ x =1

Alternative Solution Since the work done to move q from A to B is independent of the path between A (1, 8, 5) m and B (2,18, 6) m, any other path can be taken. Let this path be the straight line between A and B . This path may be any two of the following

x − xB =

x A − xB (z − zB ) z A − zB

⇒

x= z−4

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y − yB = z − zB =

y A − yB ( x − xB ) x A − xB

z A − zB ( y − yB ) y A − yB

⇒

y = 10 x − 2

⇒

10 z = y + 42

111

The work done to move q from A to B is B

W AB = − q ∫ E • dl = − q A

( 2 ,18 , 6 )

∫ (−) 8 xya

x

(1,8 , 5

− 4 x 2 a y + a z )⋅ (a x dx + a y dy + a z dz )

( 2 ,18 , 6 )

⇒

( 2 ,18 , 6 )

( 2 ,18 , 6 )

∫ xydx + 4 q ( ∫ x) dy − q ( ∫ dz)

W AB = 8 q

2

(1,8 , 5 )

1,8 , 5

1,8 ,5

From the straight line path equations between A and B we have

y = 10 x − 2 ⇒ dy = 10 dx x= z−4 ⇒ dx = dz Substituting these into the expression of W AB with q = 10 μC , yields 2

W AB = 10 × 10 − 6

∫ (120 x

2

− 16 x − 1 )dx = 2.55 m J

1

2.8.1. Electrostatic Potential due to a Point Charge For a charge Q at the origin of the coordinate system as shown in Fig. (2.18), the electric field intensity E at a radial distance r in spherical coordinate system is

E=

Q ar 4π ε r 2

(2.57)

The potential difference between the points A and B in Fig. (2.18) is B

B

Q a r • dl 4π ε r 2 A

VAB = − ∫ E • dl = − ∫ A

(2.58)

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dl = rdr ar+rdθ aθ+rsinθ dφ aφ

z

A (rA,θA,φA)

ar E (r,θ,φ)

θ

r dθ B (rB,θB,φB)

r dr

Q

x

y

φ

r sinθ dφ

Fig. (2.18). The electric potential due to a point charge.

Substituting dl from (2.56c) into (2.58), VAB becomes (rB ,θ B ,φ B )

VAB = −

Q

∫ 4π ε r θ φ )

(rA ,

A,

2

dr =

A

Q 4π ε

⎛1 1⎞ ⎜⎜ − ⎟⎟ ⎝ rB rA ⎠

(2.59)

It is often assumed that the potential at infinity is zero. Therefore, if A is removed to infinity, or rA → ∞, then the result is the potential of point B , or

VB =

Q

1 4π ε rB

(2.60a)

Since B is an arbitrary point, then it represents any point at a distance rB = r from Q . Thus, the electric potential due to charge Q at a distance r from it becomes

V (r ) =

Q 4π ε r

(2.60b)

Equation (2.60b) gives the electric potential at a radial distance r for a point charge Q at the origin. If Q is placed at a general point defined by a position vector r ′ , then the electric potential at a point defined by a position vector r , is

V (r ) =

Q 4π ε r − r′

(2.61)

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2.8.2. Electrostatic Potential due to System of Charges Consider charges Q1 , Q2 , , QN at the points (x1, y1, z1 ), (x2 , y2 , z2 ), …, (xN , yN , zN ) respectively. The electric potential due to these charges at a point (x, y, z ) is the algebraic sum of the electric potential of individual charges at this point. The electric potential Vn (x, y, z ) due to any charge Qn where n = 1, 2,  , N , is obtained by letting Q = Qn and r′ = rn in (2.61), then

Vn (r ) =

Qn 4π ε r − rn

(2.62)

The electrical potential due to all charges at the field point is

V (r ) =

1

N

∑ 4π ε n =1

Qn r − rn

(2.63)

For a continuous charge distribution, the potential V ( r ) due to a differential charge is obtained first, then the total V ( r ) due to the charge distribution is determined by integrating over the whole charge distribution. Accordingly, V ( r ) produced by a linear distribution of charge with density ρl C m is obtained as

V (r ) =

ρl dl ′

1

4π ε ∫ r − r′

(2.64a)

l

where r ′ is the position vector of the source point (x′, y′, z′) and r is position vector of the field point (x, y, z ). Similarly, for a surface charge distribution with charge density ρs C m2 and a volume charge distribution with charge density 3 ρv C m , the electric potential can be expressed by replacing ρl dl ′ with ρ s dS ′ and ρ v dv′ respectively in (2.64a), as

V (r ) =

1

S

V (r ) =

ρ s dS ′

(2.64b)

ρv dv′

(2.64c)

4πε ∫∫ r − r′ 1 4πε

∫∫∫ r − r′ v

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Equations (2.64a) – (2.64c) give the electric potential with respect to a reference point at infinity, assuming that the electric potential at infinity is zero. If the reference point is (xR , yR , zR ) and the potential at this reference point is V R , then the electric potentials with respect to the point (xR , yR , zR ) becomes V (r ) − VR . If the electric potential is the same at all points on a certain surface, this surface is referred to as an equipotential surface. There is no potential difference between any two points on the equipotential surface. Example 2.11 Three point charges Q1 , Q2 , and Q3 are placed respectively in free space at the points P1 (1,0, 0), P2 (− 1,0,0) , and P3 (0,1,0), which represent vertices of a triangle. If the units of the coordinate system are in meters and Q1 = Q2 = −Q3 = +90 2n C , find the electric potential with respect to a zero reference potential at the point P(0,0,1) . Solution Given that Q1 = +90 2n C at P1 (1,0, 0), Q2 = +90 2n C at P2 (− 1,0,0), and Q3 = −90 2n C at P3 (0,1,0). It is required to find the electric potential at P(0,0,1) . From given data, we can write

r − r1 =

(x − x1 )2 + ( y − y1 )2 + (z − z1 )2

= 2

r − r2 =

(x − x2 )2 + ( y − y 2 )2 + (z − z 2 )2

= 2

r − r3 =

(x − x3 )2 + ( y − y3 )2 + (z − z 3 )2

= 2

The electric potential at P(0,0,1) produced by the charges in free space can be obtained using (2.63) as

V=

1 4π ε 0

N =3

∑ n =1

Qn r − rn

90 2 × 10 −9 ⎛⎜ 1 1 1 ⎞⎟ + − ⎟ 4π × (1 36π ) × 10 −9 ⎜⎝ 2 2 2⎠ = 810 V =

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115

Example 2.12 A charged thin-circular loop with a linear charge density ρ l C m is placed such that its axis coincides with the z-axis and its center is at the origin. If the radius of the loop is a , find the electric potential produced by the loop (a) On its axis at a distance d from the origin. (b) At the center of the loop. Solution (a) Referring to Fig. (2.19), the position vectors for the source r ′ and field r points and r − r′ can be obtained as 2 2 r = d a z , r ′ = a a ρ , r − r′ = d + a

The differential length along the loop is dl ′ = adφ ′ . z

Linear charge density ρl C/m

Field point (0,0,d) r

y r' dl′ = a dφ′ Source point (a,φ,0)

x Fig. (2.19). Geometry of the charged thin-circular loop of Example 2.12.

Substituting r − r′ and dl ′ = adφ ′ into (2.64a), yields

V=

ρl

a

4πε 0

d +a

2

2π 2

∫ dφ′ 0

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Evaluating the integral then the electric potential on the axis of the loop at distance d from the origin is

V=

ρl a 2 ε0 d 2 + a2

(b) At the center of the loop d = 0 , then

V=

ρl 2 ε0

Example 2.13 The charge density along a line of length 2 L is uniform and equal to ρl C m . Find the potential difference due to the line between two points at distances ra and rb from the axis of the line. Assume that rb > ra . Find the potential difference between the points when the line is infinite in length. Solution Referring to Fig. (2.20), the position vectors for the source point r ′ , field point r and r − r′ are

r = ra z , r ′ = z ′a ρ , r − r′ = r 2 + z′2 The differential length along the line is dl ′ = dz′ . Substituting r − r′ and dl ′ = dz′ into (2.64a), yields V (r ) =

ρl 4π ε 0

L

∫

−L

ρl dz′ r + z′ 2

2

=

2 ⎛L ⎞ ρl ρl −1 ⎛ L ⎞ ⎛L⎞ sinh ⎜ ⎟ = ln⎜ + ⎜ ⎟ + 1 ⎟ ⎟ 2π ε 0 ⎝ r ⎠ 2π ε 0 ⎜⎝ r ⎝r⎠ ⎠

The potential at distance ra is obtained be letting r = ra , as

V a = V ( ra ) =

(

ρl 2 ln L ra + ( L ra ) + 1 2π ε 0

)

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117

Similarly, the potential at distance rb is obtained be letting r = rb , as

Vb = V ( rb ) =

(

ρl 2 ln L rb + ( L rb ) + 1 2π ε 0

)

Consequently, the potential difference between the points at distances ra and rb is

V ab

⎛ L + L2 + r 2 ρ l ⎡ ⎛ ra ⎞ b ⎜ ⎢ ln ⎜ ⎟ − ln = Vb − V a = ⎜ L + L2 + r 2 2π ε 0 ⎢ ⎝ rb ⎠ ⎝ ⎣ a

⎞⎤ ⎟⎥ ⎟ ⎠ ⎥⎦

When the line is infinite in length, then L → ∞ , the V ab becomes

Vab =

⎡ ⎛r ⎞ ⎛ L + L2 + r 2 ρl b lim ⎢ln⎜⎜ a ⎟⎟ − ln⎜ 2 2 → ∞ L ⎜ 2π ε 0 ⎢ ⎝ rb ⎠ ⎝ L + L + ra ⎣

⎞⎤ ⎟⎥ = ρl ln⎛⎜ ra ⎞⎟ ⎟ 2π ε 0 ⎜⎝ rb ⎟⎠ ⎠⎥⎦

z (0, 0, L) Source point (0, 0, z′)

dl′ = dz′

r'

y

r

(r, φ, 0) Field point

x (0, 0, -L)

Linear charge density ρl

Fig. (2.20). Geometry of the infinite line charge of Example 2.13.

Alternative Solution The electric field intensity due to line charge of length 2 L and charge density at a radial distance ρ from Example 2.7 is given

E = aρ

ρl 2π ε 0 ρ

L L2 + ρ 2

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Then the potential difference between the points at distances ra and rb from the axis of line can be obtained using B

Vab

= − ∫ E • dl A rb

ρl 2π ε 0 ρ ra

= −∫

L L +ρ 2

2

a ρ • dl

In cylindrical coordinate dl = dρ a ρ + ρdφ aφ + dz a z . Thus, Vab becomes rb

Vab

L

ρl 2π ε 0 ρ ra

= −∫ =−

ρl L 2π ε 0

L2 + ρ 2 dρ

rb

∫ρ

a ρ • (dρ a ρ + ρdφ aφ + dz a z )

L2 + ρ 2

ra

Using the substitution ρ = L tan(2 α), we get Vab = −

(

ρl 2π ε 0

⎡ tan(α )⎤ sec2α d α ρ ∫α tanα = 2π lε0 ln⎢⎣ tan(αba )⎥⎦ a αb

)

)

(

where tan(αa ) = ra L + L2 + ra 2 and tan(αb ) = rb L + L2 + rb 2 . Substituting tan (α a ) and tan (αb ) into the above expression of Vab , yields

⎡ ⎛ ⎜ L + L2 + rb 2 ρl ⎢ ⎛ ra ⎞ Vab = ln⎜ ⎟ − ln⎜ 2π ε 0 ⎢ ⎜⎝ rb ⎟⎠ ⎜ L + L2 + r 2 ⎢⎣ a ⎝

⎞⎤ ⎟⎥ ⎟⎥ ⎟⎥ ⎠⎦

For a line of infinite length L → ∞ , the potential difference reduces to

Vab =

⎛r ⎞ ln⎜⎜ a ⎟⎟ 2πε 0 ⎝ rb ⎠ ρl

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Example 2.14 A sphere of radius a has a uniform volume charge density ρv C m3 . Find the electric potential due to the sphere at a distance r from its center, if (a) r > a . (b) r < a . Solution The electric vector potential is given by (2.64c). Referring to Fig. (2.21), in spherical coordinate the differential volume is dv′ = r ′ 2 sin θ ′ dr ′ dθ ′ dφ ′ . The position vectors of the source r ′ and observation r points are

r = ra z

r′ = a x r′ sin θ ′ cosφ ′ + a y r′ sin θ ′ sin φ ′ + a z r′ cosθ ′

(

⇒ r − r′ = r′2 + r 2 − 2r′r cosθ ′ z

)

1 2

Source point (r′,θ′,φ′) dv′ = r′ 2 sinθ′ dr′dθ

Field point (r,θ,φ)

r - r′ r

θ a

r ′

φ Volume charge densityVρv C/m3

x Fig. (2.21). Geometry of the sphere in Example 2.14 for the case r > a .

y

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(a) If r > a , then 1 4π ε 0

V=

2π π a

ρ v r ′2 sin θ ′dr ′dθ ′dφ ′

∫ ∫ ∫ (r′ 0 0 0

2

+ r 2 − 2r ′r cosθ ′)2 1

=−

1 2 ε0

π a

∫ ∫ (r′ 0 0

ρ v r ′2 sin θ ′dr ′dθ ′ 2

+ r 2 − 2r ′r cosθ ′)2 1

Using the substitution u = cosθ ′ , then du = − sin θ ′dθ ′ and the limits of the integrals become from u = 1 to u = −1 . Evaluating the resulted integrals, yields a

V

(

ρv r ′dr ′ r ′2 + r 2 − 2r ′r u ∫ 2r ε 0 0 a ρv ρv a3 2 ′ ′ = = r d r ε 0 r ∫0 3ε0 r =

)

1 2

u = −1 u = +1

(b) If r < a , the potential is due to the charge within a sphere of radius r in addition to the potential due to the charge enclosed in the region r < r ′ < a as illustrated in Fig. (2.22), then

V=

r a ρ v r ′ 2 dr ′ ρ v r ′ 2 dr ′ sin θ ′dθ ′dφ ′ ⎡ +∫ 1 1 ∫0 4π ε 0 ⎢⎢ ∫0 ′ 2 2 ′ 2 2 2 2 r ( r ′ + r − 2 r ′r cosθ ′ ) ⎣ (r + r − 2 r r cosθ ′ )

2π π

∫ 0

⎤ ⎥ ⎥⎦

Integration with respect to φ ′ yields

V=

a ⎡r ρ v r ′ 2 dr ′ ρ v r ′ 2 dr ′ ′ ′ sin d θ θ + ⎢ 1 1 ∫ ∫ 2 ε 0 ∫0 ⎢⎣ 0 (r ′ 2 + r 2 − 2 r ′r cosθ ′ ) 2 r (r ′ 2 + r 2 − 2 r ′r cosθ ′ ) 2

1

π

⎤ ⎥ ⎥⎦

Using the substitution u = cos θ ′ , then du = − sin θ ′dθ ′ and the limits of integrals become from u = 1 to u = −1 . Evaluating the resulted integrals with respect u , yields r

V=

a

1 u = −1 1 u = −1 ρv ρv 2 2 2 2 2 2 ′ ′ ′ ′ ′ ′ ′ ′ + − + + − r d r r r r r u r d r ( r r 2 r r u ) ( 2 ) ∫ ∫ = + u = +1 1 u 2ε 0 r r 2ε 0 r 0

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z

Observation point (r, θ, φ)

121

Source point (r′, θ′, φ′) dv′ = r′ 2 sinθ′ dr′dθ ′dφ′

r′ r a

θ

r ′

φ

y

V

Volume charge density ρv C/m3

x Fig. (2.22). Geometry of the sphere in Example 2.14 for the case r < a .

In the first term on the right-hand side r > r ′ , then the quantity between the brackets is (r ′ + r ) − (r − r ′) = 2r ′ , while in the second term r ′ > r , then the quantity between the brackets is (r ′ + r ) − (r ′ − r ) = 2r . Thus, the electric potential reduces to r

V=

a

(

ρv ρ ρ ρ r′2 dr′ + v ∫ r′dr′ = v r 2 + v a 2 − r 2 ∫ ε0 r 0 ε0 r 3ε0 2 ε0

)

2 ρv a 2 ⎡ ⎛ r ⎞ ⎤ ⇒ V= ⎢3 − ⎜ ⎟ ⎥ 6 ε 0 ⎣⎢ ⎝ a ⎠ ⎦⎥

Consequently, the potential due to the sphere of radius a with a uniform volume charge density ρv C m3 at any distance r , where 0 ≤ r ≤ ∞ is ⎧ ρ a 2 ⎡ ⎛ r ⎞2 ⎤ ⎪ v ⎢3 − ⎜ ⎟ ⎥, 0 ≤ r ≤ a ⎪ 6ε a V = ⎨ 0 ⎢⎣ ⎝ ⎠ ⎥⎦ 3 ⎪ ρv a , a≤r≤∞ ⎪3ε r ⎩ 0

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2.9.

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ELECTRIC FIELD AS A GRADIENT OF THE POTENTIAL

It has been shown that in (2.50), the work done to move a charge in a closed path is zero. Consequently, we can write

∫E

•

dl = 0

(2.65)

l

where the integral is along a closed path l . Using Stokes’ theorem, the line integral in (2.65) is changed to a surface integral as follows

∫∫ ∇ × E

•

dS = 0

(2.66)

S

where S is the surface enclosed by the closed path l . (2.66) implies that ∇×E = 0

(2.67)

This equation reveals that the electrostatic field is a conservative field or irrotational field. Assuming that the electric field intensity is

E = a x Ex + a y E y + a z Ez

(2.68)

the electric potential can be obtained from the electric field in (2.68) at any point (x, y, z ) using (2.55), and (2.56a) as V ( x, y , z ) = −

(x, y, z )

∫E

∞

•

dl = −

(x, y, z )

(x, y, z )

∫ E dx − ∫ E x

∞

y

dy −

∞

(x, y, z )

∫ E dz z

(2.69)

∞

Finding partial differentiation for both sides of (2.69), yields

∂V (x, y, z ) = − E x ( x, y , z ) ∂x ∂V (x, y, z ) = − E y (x, y, z ) ∂y ∂V (x, y, z ) = − E z ( x, y , z ) ∂z

(2.70a) (2.70b) (2.70c)

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Note that the components of the electric field vector are the negative of the gradient of the electric potential in the direction of these components. Substituting E x , E y , and Ez from (2.70) into (2.68), yields

E = −a x

∂V (x, y, z ) ∂V (x, y, z ) ∂V (x, y, z ) − ay − az ∂x ∂y ∂z

(2.71a)

Comparing (2.71a) with (1.118), yields E = −∇V

(2.71b)

The gradient in rectangular, cylindrical, and spherical coordinate systems is given respectively by (1.118) – (1.120). Note that: satisfies (2.67), since ∇ × ∇ = 0 . This indicates that the electrostatic field is a conservative field. 2. The maximum rate of change of the potential V is the magnitude of its gradient ∇V . This implies that the maximum rate of change of V is equal to E . To show this, consider the change of V in the direction of the differential element dl along the path l . The unit vector au in the direction of dl is

1.

E = −∇V

au =

dl a x dx + a y dy + a z dz dx dy dz = = ax + ay + az dl dl dl dl dl

(2.72)

where dl = dl . Carrying out the dot product of both sides of (2.71a) and (2.72) leads to

∇V ⋅ a u = ∇V ⋅

dl ∂V dx ∂V dy ∂V dz dV = + + = dl ∂x dl ∂y dl ∂z dl dl

(2.73)

If the angle between the direction of ∇V and au is θ ; hence

dV = ∇V cosθ dl

(2.74)

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It is clear from (2.74) the rate of change dV dl is maximum when θ = 0 , or

dV dl

= ∇V = E

(2.75)

max

3. The electric field E is always normal to the equipotential surface in the direction of the lower potential as will be shown in the following analysis. Using E = −∇V , (2.73) can be written as

∇V ⋅ a u = −E ⋅ a u =

dV dl

(2.76)

If V is constant along the path l , then dV dl = 0 and (2.76) becomes

E ⋅ au = 0

(2.77)

This implies E is perpendicular to au under the condition that dV dl = 0. Since au is the unit vector along the path l , then E is always normal to the path l at the point where V is constant. Any path on an equipotential surface satisfies dV dl = 0 . Therefore, the electric field E is always normal to the equipotential surfaces in the direction of the lower potential. Example 2.15 The potential at any point (r ,θ , φ ) in spherical coordinate system is given by

V (r,θ ,φ ) = 50(2 + r + 3r sin θ sin φ ) V Find the electric field at the point (r = 1m,θ = 90 ,φ = 0 ) . Solution In spherical coordinate system, the electric field is

E = −∇V = −a r

∂V 1 ∂V 1 ∂V − aθ − aφ ∂r r ∂θ r sin θ ∂φ

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Thus,

E = −50 ⎡ar (1 + 3sin θ sin φ ) + aθ 3 cosθ sin φ + aφ 3sin φ ⎤ V m ⎢⎣ ⎥⎦ At the point (r = 1m,θ = 90 ,φ = 0 )

E = −50 ⎡ar (1 + 3sin 90 sin 0 ) + aθ 3 cos 90 sin 0 + aφ 3sin 0 ⎤ = −50 ar V m ⎢⎣ ⎥⎦ 2.10. ELECTROSTATIC ENERGY The electrostatic energy of a system of point charges Q1 , Q2 , , QN placed at points defined by position vectors r1, r2 ,, rN , respectively, is the work done to move these charge from infinity to these point. The electric potential produced due to the presence of charge Qn at a point defined by a vector r can be obtained using (2.62), as

Vn (r ) =

Qn , 4πε r − rn 1

n = 1, 2, N

(2.78)

The work done by the charge Qn to move any other charge Qi , (n ≠ i ) from infinity to the point defined by the vector ri is

Wni

= QiVn (ri ) 1 QiQn = 4πε ri − rn

(2.79)

The work done by all charges n = 1, 2, N to move any other charge Qi , (n ≠ i ) from infinity to the point defined by the vector ri is N

N

n =1 n ≠i

n =1 n ≠i

∑Wni = Qi ∑Vn (ri ) = QiVi

(2.80)

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Evaluating the work Wni done by all charges Qn , n = 1, 2, N in (2.80) to move each charge Qi , where i = 1, 2,, N and (n ≠ i ) , from infinity to the point, we get the system of equations

W21 + W31 +  + W N 1 = Q1 [V2 (r1 ) + V3 (r1 ) +  + V N (r1 )] = Q1V1

W12 + W32 +  + W N 2 = Q2 [V1 (r2 ) + V3 (r2 ) +  + V N (r2 )] = Q2V 

(2.81)

W1N + W2 N +  + W( N −1)N = Q N [V1 (rN ) + V2 (rN ) +  + V N −1 (rN )] = Q N V N Summing both sides of the equations, with Win = Wni , yields

2W12

+ 2W13 + 2W23

+ 2W14 + 2W24

+  + 2W1N +  + 2W1N  + 2W( N −1)N

(2.82) N

N

n =1 n≠i

i =1

= 2WT = ∑Wni = ∑ QiVi

Consequently, the total energy of the charges becomes WT =

1 N 1 N W = QiVi ∑ ni 2 ∑ 2 n =1 i =1

(2.83)

n ≠i

2.10.1. Electrostatic Energy due to a Continuous Distribution of Charge Consider a total charge q is continuously distributed with a linear density ρ l . If q is divided into N unit charges each of amount Qi , where i = 1, 2, N , then Qi = Δqi (r ) = ρ l (ri )Δli

(2.84)

Substituting Qi from (2.84) into (2.83), the approximated work done to move the charges Qi , i = 1, 2, N from infinity to their current positions becomes

WT ≈

1 N ∑ ρ l (ri )ΔliVi 2 i =1

(2.85)

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Note that the original continuously distributed charge is approximated by dividing the total charge to discrete charge units located at points defined by the unit vectors ri . Therefore, (2.85) gives an approximate value of the work. The accurate value is obtained by letting the length of the unit charges approaching zero, or Δli → 0 , so that the effect of all charge distribution will be taken into account and the summation changes to integration along the line of charge. Consequently, the accurate work or energy is

WT =

1 ρ l (r ′)V (r ′)dl ′ 2 ∫l

(2.86a)

Similar expressions for surface charge distribution with density ρ s and volume charge distribution with density ρ v can be obtained respectively, as

1 ρ S (r ′)V (r ′)dS ′ 2 ∫S 1 WT = ∫ ρ v (r ′)V (r ′) dv ′ 2v

WT =

(2.86b) (2.86c)

2.10.2. Energy Stored in an Electrostatic Field The energy expression can be written in terms of the electric flux density D by substituting the volume charge density ρv from (2.30) into (2.86c), as

1 V (r ′)∇ • D dv′ 2 ∫v

(2.87)

∇ • DV = V ∇ • D + D • ∇V

(2.88a)

WT = Using vector identities, we have

Using (2.71b), − ∇V can be replaced by E in (2.88), then

V ∇ • D = ∇ • DV − D • ∇V = ∇ • DV + D • E

(2.88b)

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Substituting the quantity V ∇ • D from (2.88b) into (2.87), the energy becomes

WT =

1 1 D • E dv′ + ∫ ∇ • DV (r ′) dv′ ∫ 2v 2v

(2.89a)

Changing the volume integral in the second term of the right-hand side, then

WT =

1 1 D • E dv′ + ∫ V (r ′)D • dS ′ ∫ 2v 2S

(2.89b)

The second term on the right-hand side of equation (2.89b) can be written in terms of surface charge density ρ s . Hence,

WT =

1 1 D • E dv′ + ∫ V (r ′) ρ s dS ′ ∫ 2v 2S

(2.89c)

If the medium is charge-free, or ρ s = 0 , the second term in (2.89c) vanishes and the electrostatic energy stored in E within a volume v is obtained from (2.89c) as

WT =

ε 1 D • E dv ′ = 0 ∫ 2v 2

∫E

2

dv ′ =

v

1 2ε0

∫D

2

dv ′

(2.90)

v

Consequently, the electrostatic energy per unit volume is

1 ε 1 2 Wv = D • E = 0 E = D 2 2 2 ε0

2

(2.91)

If the surface charge density of the medium is ρ s ≠ 0 and the electric field intensity in the medium is zero, or E = 0 , then (2.89c) reduces to

WT =

1 V (r′) ρ s dS ′ 2 ∫S

(2.92)

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Example 2.16 The electric field intensity in cylindrical coordinate system is given by

E=

100

ρ

aφ V m

in the region bounded by 1 < ρ < 4 , 0 < φ < 3π 2 , 0 < z < 6 . (a) Find the total energy stored in the region. (b) Show that the surfaces φ = 0 and φ = 3π 2 are equipotential surfaces. (c) Find the potential difference between the surfaces at φ = 0 and φ = 3π 2 . Solution (a) In the cylindrical coordinate the differential volume dv ′ = ρ ′ dρ ′dφ ′dz ′ . Replacing ρ by ρ ′ in E and using (2.90), the total energy is obtained as

ε WT = 0 2

ε0 ∫v E dv′ = 2 2

6 3π 2 4

∫∫ 0

0

2

10 4 ε 0 100 ′ ′ ′ ′ (ln ρ ′)14 (φ )30π 2 (z )60 = ρ d ρ d φ d z ∫1 ρ ′ 2

This simplifies to

WT =

3π 10 4 × 10 −9 × 6 ln 4 = 1.155 μ J 2 2 × 36π

In cylindrical coordinate system, the potential is

V (ρ , φ , z ) = −∫ E • dl = −∫ Eρ dρ − ∫ Eφ ρ dφ − ∫ Ez dz Since Eρ = Ez = 0 , then

V (ρ ,φ , z ) = −100∫

1

ρ

ρ dφ = −100∫ dφ = −100φ + C

where C is a constant. This constant is the reference potential.

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(b) At φ = 0 the potential is V (ρ ,φ = 0, z ) = C which is constant. Similarly, at φ = 3π 2 V (ρ ,φ = 3π 2 , z ) = −150π + C , which is also constant. Hence, the surfaces φ = 0 and φ = 3π 2 are equipotential surfaces. (c) The potential difference between the surfaces at φ = 0 and φ = 3π 2 is V (ρ , φ = 0, z ) − V (ρ , φ = 3π 2 , z ) = C − (− 150π + C ) = 150π V 2.11. THE ELECTRIC DIPOLE The system of two charges + Q and − Q separated by a small distance d is known as an electric dipole. Note that the charges are equal in magnitude and different in sign. Consider the electric dipole shown in Fig. (2.23). The axis of the electric dipole is along the z-axis and it is symmetrical about the origin of the coordinate. The electric potential of the electric dipole at a point (r ,θ ,φ ) can be obtained using (2.60b) as

V=

Q ⎛1 1⎞ ⎜ − ⎟ 4π ε ⎜⎝ r1 r2 ⎟⎠

(2.93)

For the electric dipole r >> d , then referring to Fig. (2.23), r1 , r2 , θ1 , and θ 2 can be approximated as follows

d 2

d 2

θ ≈ θ1 ≈ θ 2 , r1 ≈ r − cosθ , r2 ≈ r + cosθ

(2.94)

Substituting r1 , r2 , θ1 , and θ 2 into (2.93), yields

V≈ 2

2

2

Qd cosθ 1 4π ε r 2 + d 2 cos 2 θ 4 2

(2.95)

2

Since r >> d cos θ 4 , then d cos θ 4 can be ignored and the potential of the electric dipole in terms of r and θ becomes

V=

Qd cosθ 4πε r 2

(2.96)

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The product of the charge of the electric dipole Q and the distance between the charges d is known as the dipole moment. The dipole moment of the electric dipole p is a vector quantity with its unit vector along the line joining the two charges in the direction of the positive charge. Therefore, the dipole moment of the electric dipole of Fig. (2.23) is

p = a z Qd

(2.97) (r,θ,φ) Field point

z r1

θ1 r

+Q

r2

θ d

(d/2) cosθ

θ2

x

y

-Q

Fig. (2.23). The electric dipole.

Since az ⋅ ar = cosθ , then

p ⋅ ar = Qd cosθ

(2.98)

From (2.96), (2.98) the potential can be written in the form

V=

p ⋅ ar 4π ε r 2

(2.99)

If the center of the dipole is at a general point describe by the position vector r ′ and the observation point is at a point described by the position vector r , (2.99) can be generalized as

V=

p ⋅ (r − r′) 3 4π ε r − r′

(2.100)

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The electric field E of the electric dipole can be obtained from (2.96) using the relation (2.71b). In spherical coordinate system

E = −∇V = −a r

∂V 1 ∂V 1 ∂V − aθ − aφ ∂r r ∂θ r sin θ ∂φ

(2.101)

Substituting V from (2.96) into (1.101), taking into consideration that there is no variation of V in the direction of φ , the electric field intensity of an electric dipole of dipole moment Qd at a point (r ,θ , φ ), where is r the distance from the center of the dipole and θ is the angle between the axis of the dipole and the direction of r , is obtained as

E=

Qd (ar 2 cosθ + aθ sin θ ) 4π ε r 3

(2.102)

Example 2.17 A dipole having a moment p = 10 a x − 3 a y + 5 a z n C . m is located at (2, 4,1) m in free space. Find the potential at (5,1, 0) m . Solution The position vectors of the source point (2, 4,1) m and the field point ( x, y, z ) are

r′ = 2 a x + 4 a y + a z , r = x a x + y a y + z a z Substituting p , r , and r′ into (2.100), the potential due to dipole at ( x, y, z ) is

(10 a x − 3a y + 5 a z ) ⋅ [( x − 2) a x + ( y − 4) a y + ( z − 1) a z ] 10 −9 V= × 3 4π × (1 36π )× 10 −9 [( x − 2)2 + ( y − 4)2 + ( z − 1)2 ]2 ⇒ V (x, y, z ) =

9[10 ( x − 2) a x − 3 ( y − 4) a y + 5 ( z − 1) a z ]

[( x − 2) 2 + ( y − 4) 2 + ( z − 1) 2 ]

3 2

V

The potential at the point (5,1, 0) is

V (5,1, 0 ) = 9 ×

[10 × (5 − 2) − 3 × (1 − 4) + 5 × (1 − 4)] = 3.695 V [(5 − 2)2 + (1 − 42 + (1 − 42 )] 3 2

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SOLVED PROBLEMS Solved Problem 2.1 Find the total charge inside the volume 0 ≤ ρ ≤ 2 , 0 ≤ φ ≤ π 2, 0 ≤ z ≤ 3 , if the charge density is ρv = 2 ρ 2 z 2 sin 2φ C m3 . Solution Using (2.29), with dv′ = ρ ′dρ ′dφ ′dz′ and ρ v = 2 ρ ′2 z′2 sin 2φ ′ C m3 , then the total charge contained in the region 0 ≤ ρ ≤ 2 cm , 0 ≤ φ ≤ π 2, 0 ≤ z ≤ 3 cm is

= ∫ ρv dv′ =

QT

v

0.03 π 2

0.02

∫ ∫ ∫ 2 ρ ′ z′ 3

0

0

2

sin 2φ ′dρ ′dφ ′dz′

0

0.02 0.03 1 π 2 = − ρ ′4 z′3 cos 2φ ′ 0 = 2.16 ρ C 0 0 12

Solved Problem 2.2 Three point charges Q1 , Q2 , and Q3 are arranged along a straight line in free space, such that Q2 divides the distance between Q1 and Q3 in the ratio 1:4. Determine the condition that the charges to satisfy such they are at equilibrium. Solution Let the distance between Q1 and Q2 is d . Since Q2 divides the distance between Q1 and Q3 in the ratio 1:4, then the distance between Q1 and Q3 will be 4d . Assuming the charges are along the x-axis as illustrated in Fig. (2.24), then the position vectors of the charges are

r1 = x a x , r2 = (x + d )a x , r3 = (x + 5d )a x The force on Q1 due to Q2 is obtained using (2.6) as

F12 =

Q1Q2 r1 − r2 Q Q 1 ⎡ x a − (x + d ) a x ⎤ Q1Q2 = 1 2 2 ×⎢ x = − a x 2 ⎥ 4π ε 0 r1 − r2 4π ε 0 d ⎣ d 4π ε 0 d 2 ⎦

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(x, 0, 0)

(x+d, 0, 0)

Q1

Q2

Sameir M. Ali Hamed

(x+5d, 0, 0) Q3

x

Fig. (2.24). Geometry of the Solved Problem 2.2.

The force on Q1 due to Q3 can be obtained following the above steps as

F13 = −

1 Q1Q3 ax 4π ε 0 25d 2

The total force F1 on Q1 is the resultant of the forces F21 due Q2 and F31 due to Q3 , then

F1 = F21 + F31 = −a x

Q ⎞ Q1 ⎛ 1 Q1Q2 1 Q1Q3 − ax = −a x Q + 3⎟ 2 2 ⎜ 2 2 4π ε 0 d 4π ε 0 25d 4π ε 0 d ⎝ 25 ⎠

Similarly, the forces F2 and F3 on Q2 and Q3 respectively can be obtained as

Q2 ⎛ Q ⎞ 1 Q1Q2 1 Q2 Q3 − ax = ax Q − 2⎟ 2 2 2 ⎜ 1 4π ε 0 d 4π ε 0 16d 4π ε 0 d ⎝ 16 ⎠ Q3 ⎛ Q1 Q2 ⎞ 1 Q1Q3 1 Q2 Q3 + ax = ax F3 = F31 + F32 = a x ⎟ ⎜ + 2 2 4π ε 0 25d 4π ε 0 16d 4π ε 0 d 2 ⎝ 25 16 ⎠ F2 = F21 + F23 = a x

Since the charges are at equilibrium, then F1 = F2 = F3 = 0 . Consequently,

F1 = 0 ⇒ Q2 + Q3 25 = 0 , ⇒ Q3 = −25Q2 F2 = 0 ⇒ − Q1 + Q3 16 = 0 , ⇒ Q3 = 16Q1 Therefore, the condition that the charges to satisfy such they are at equilibrium is

16 Q1 = −25 Q2 = Q3 Note that Q1 and Q3 are similar in sign, while Q2 is opposite in sign to them.

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Solved Problem 2.3 Six identical charges Q = 50 μ C each, are placed at the vertices of a regular hexagon of sides length a = 3 m . Find the electrostatic force on q = 25 μ C charge located d = 4 m above the center of the hexagon. Solution Let the hexagon plane lies in the xy plane with its center at the origin as shown in Fig. (2.25). Then the angles φn are given by

φn = 2π

(n − 1) = π (n − 1) , 6

n = 1, 2, , 6

3

z Fn (0, 0, d)

q

Qn

(acosφn, acosφn, 0)

Q a

Q

φn

Qn = Q n = 1, 2, …, 6

Q

a

x

Q1 = Q

Q

y

Fig. (2.25). Geometry of Solved Problem 2.3.

Note that for a regular hexagon the diagonal length is twice the side length, then the charge Q n is at the point (a cosφn , a sin φn , 0), and q is at (0, 0, d ). The force

Fn on q due to any charge Q n is Fn =

1 qQn an , 4π ε 0 R n 2

n = 1, 2,, 6

The position vector of the charge Qn is rn and that of q is r , where

rn = a x a cos φ n + a y a sin φ n , r = d a z

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Since we are interested in the force on q due to Qn , then

R n = r − rn = −a x a cosφn − a y a sin φn + d a z − a x a cosφn − a y a sin φn + d a z R an = n = Rn a2 + d 2 Substituting Qn = Q , R n and a n in the above expression of Fn , yields

Fn =

qQ (− a x a cos φn − a y a sin φn + d a z ) 3 4π ε 0 (a 2 + d 2 )2

Fn is the force due to any charge Qn on q . The total force F is superposition of the force due to all charges correspond to n = 1, 2, , 6 , or 6

F = ∑ Fn = n =1

qQ 1 3 4π ε 0 (a 2 + d 2 )2

6

∑ (− a a cos φ x

n =1

n

− a y a sin φn + d a z )

This can be written as

F = az

6 qQ 6d 3 − ∑ (a x a cos φn + a y a sin φn ) 4π ε 0 (a 2 + d 2 )2 n =1

Let the summation in the second term of the right-hand be S , then 6 ⎧ ⎡π ⎤ ⎡π ⎤⎫ S = ∑ ⎨a x a cos⎢ (n − 1)⎥ + a y a sin ⎢ (n − 1)⎥ ⎬ ⎣3 ⎦ ⎣3 ⎦⎭ n =1 ⎩

Expanding the right-hand side, yields

π 2π 4π 5π ⎞ ⎛ S = a x a ⎜1 + cos + cos + cos π + cos + cos ⎟ 3 3 3 3 ⎠ ⎝ π 5π ⎞ 4π 2π ⎛ + a y a ⎜ 0 + sin + sin + sin π + sin + sin ⎟ 3 ⎠ 3 3 3 ⎝

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Using cos ( x + π ) = − cos x and sin ( x + π ) = − sin x , and simplifying ⇒ S = 0 . Consequently, the force on q reduces to

F = az

qQ 6d 3 4π ε 0 (a 2 + d 2 )2

Substituting Q , q , a , and d , then

25 × 10 −6 × 50 × 10 −6 6 × 4 F = az N 3 = 2.6 a z 4π × (1 36π ) × 10 − 9 (32 + 42 )2 Solved Problem 2.4 Find the force on a charge q at the focus of a parabola y 2 = 4ax due to a linear charge distribution along the parabola from point ( a, 2a ) to point (a, − 2a) with a uniform density ρl C m . Solution The source point is any point ( x′, y′) on the parabola. The field point is the focus of the parabola, or (a, 0) . Hence

r′ = x′a x + y′a y , r = aa x ⇒ R = r − r′ = (a − x′) a x − y′a y R (a − x′) a x − y′a y ⇒ aR = = (a − x′) 2 + y′2 R Since ( x′, y′) satisfying the given equation of the parabola, then y′2 = 4ax′ . Substituting a R and R into (2.18), and using F = qE , the force on the charge q is obtained as F=

(a − x′) a x − y′a y q ρl dl ′ 3 ∫ 4πε l (a − x′) 2 + y′2 2

[

]

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where dl ′ = (dx′) 2 + (dy′) 2 = (1 + (dx′ dy′) dy′ 2

Since y′2 = 4ax′ ⇒ x′ = y′2 4a ⇒ dx′ dy′ = y′ 2a ⇒ dl′ = (1 + ( y′ 2a )2 dy′. Substituting these quantities in the above expression of F and simplifying, yields F = ax

q ρl 4π ε a 2

2a

∫

− 2a

1 − ( y′ 2a )

2

[1 + ( y′ 2a ) ]

5 2 2

y′ − a y

q ρl 4π ε a 2

2a

∫

− 2a

y′dy′

[1 + ( y′ 2a ) ]

5 2 2

It is clear that the integral in the direction of a y is zero. Using the substitution

y′ 2a = tan u , then dy′ = 2a sec2 u and F becomes +π 4

q ρl (1 − tan 2 u ) sec2 u du F = ax 5 2π ε a −π∫ 4 1 + tan 2 u 2 +π 4 q ρl = ax (1 − 2 sin 2 u ) cos u du ∫ 2π ε a −π 4

[

]

+π 4

⇒ F = ax

2q ρ l q ρl ⎡ 2 ⎤ = ax sin u − sin 3 u ⎥ ⎢ 2π ε a ⎣ 3 3π ε a ⎦ −π 4

Solved Problem 2.5 Two infinite sheets of charge, each with a uniform surface charge density ρ s = (1 4π ) n C m 2 , are located at z = 0 cm and z = 8 cm in free space. Determine the electric field at all regions. Solution Let the sheet at the plane z = 0 cm be sheet 1, and the that at the z = 8 cm be sheet 2 as illustrated in Fig. (2.26). The electric field in the + z direction and − z direction due to sheet 1, from (2.43) are

E1+ = +a z

ρs , z>0 2 ε0

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E1− = −a z

ρs , z8 2 ε0 ρ E2− = −a z s , z < 8 2 ε0

E2+ = +a z

Therefore, the electric field in the region z > 8 is

E+ = E1+ + E2+ = a z

ρs ρ ρ + az s = az s 2 ε0 2 ε0 ε0

The electric field in the region 0 < z < 8 is

E0 = E1+ + E−2 = a z

ρs ρ − az s = 0 2 ε0 2 ε0

The electric field in the region z < 0 is

E− = E1− + E−2 = −a z

ρs ρ ρ − a z s = −a z s 2 ε0 2 ε0 ε0

z E1

+

E 2+ Sheet 2 ρs C/m

E2

-

E1

2

+

z = 8 cm Sheet 1 ρs C/m2

E 2-

E 1-

x Fig. (2.26). Geometry of Solved Problem 2.5.

y

139

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Substituting ρ s = (1 4π ) n C m 2 into the above expressions and simplifying, the electric field at all regions in V m becomes

⎧+ 1, z > 8 ⎪ 0< z a .

(

)

2.30. The spherical region, 0 < r < 10 cm , contains a uniform volume charge density 4 μ C m3 . (a) Find the total charge enclosed in the region 0 < r < 10 cm . (b) Find the electric flux density in the region 0 < r < 10 cm . (c) The non-uniform charge density ρv = − 3 (r 3 + 0.001) n C m3 exists in the region 10 cm < r < a . Find a so that the total charge in the region 0 < r < a is zero.

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2.31. The electric flux density is given by

D = 3 y 2a x + 3x 2 y a y + 5a z C m 2 Find the total charge enclosed in the region 0 < x < 2 , 0 < y < 2 , 0 < z < 2 using (a) Gauss’s law. (b) Divergence theorem.

(

)

2.32. A non-uniform surface charge density ρ 1 + ρ 2 n C m 2 lies in the plane 0 ≤ ρ ≤ 5 , 0 ≤ φ ≤ 2π , z = 2 . Find (a) The total electric flux density leaving the region enclosed by 0 ≤ ρ ≤ 5 , 0 ≤ φ ≤ 2π , z = 2 . (b) The electric flux density crossing the plane z = 0 . 2.33. A line charge of uniform charge density 5 n C m lies along the z-axis, a surface charge of 1 π n C m 2 on the plane x = 6 cm , and a point charge of 3n C at (3,1, 2) cm , all in free space. Find (a) the electric field at (4, 3, 0) . (b) the potential at (4, 3, 0) cm if the potential at (0,1, 0) cm is zero. 2.34. The circular ring in Fig. (2.30) has surface charge density ρl = K ρ C m 2 where K is a constant and ρ is the radial distance in the cylindrical coordinate system. Find the Potential and electric field intensity at the point (0, 0, c). z (0, 0, c) Surface charge density ρl C/m2 a

ρ

y b

x Fig. (2.30). Problem 2.34.

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2.35. Three point charges Q1 = +10 2n C , Q2 = −5 5n C , and Q3 = +10 5n C are placed respectively in free space at the points P1 (1,0, 0) m, P2 (0,2,0) m, and P3 (0,0,2) m. Find the electric potential at the point P (1,2,2) with respect to a zero reference potential. 2.36. Two identical point charges each carrying a charge Q , are located at the points (0, 0, a) and (0, 0, − a ) in free space. Show that the potential at any point on the surface of the sphere r = a with respect to a zero reference potential can be expressed as

V=

⎛ 1 + cosθ Q 1 − sin θ ⎜ + 4π ε 0 a sin θ ⎝⎜ 2 + cosθ 2 − sin θ

⎞ ⎟ , θ ≠ 0, π ⎟ ⎠

2.37. Two-line charge L1 and L2 carry a uniform distribution of charges with charge density ρl C m each. L1 extends from (0, 0, a ) to (0, 0, L + a) and L2 from (0, 0, − a) to (0, 0, − L − a ), where a and L are positive real quantities. Find the potential difference between the points (xa , ya , 0) and

(xb , yb , 0) .

2.38. An infinitely long cylinder of radius a is filled with a charge of uniform volume density ρ v C m 3 . Assuming the potential on the axis of the cylinder is V0 , determine potential at any point within the cylinder. 2.39. An infinitely long cylinder of radius a is filled with a charge of volume density varies with radial distance ρ from the axis of the cylinder as ρ v = ρ 0 ρ a C m 3 , where ρ 0 is a constant. Assuming the potential on the axis of the cylinder is V0 , determine potential at any point within the cylinder. 2.40. A charged thin-circular loop of radius 6 cm carries a uniform linear charge density 16 n C m . Find the electric potential produced by the loop (a) On its axis at a distance 8 cm from the origin. (b) At the center of the loop.

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2.41. The volume charge density of a spherical cloud varies with radial distance from the cloud center as ρv = ρ0 r a C m3 , where a is the radius of the cloud and ρ 0 is a constant. Determine the electric field intensity and potential at all points. 2.42. Four identical charges of 4 n C each are located at the vertices of a square of 5 ×10−4 m on a side in free space. Find (a) The potential at the center of the square (b) The work required to move one of the charges to the center of the square. 2.43. For the system of charges of Problem 2.11, find the work required to move the charge q from its position to the center of the polygon. 2.44. Find the work done to move an 8 μC charge in the electric field E = 3x 2 y a x + (2 x − z )a y V m

From the point (2,1,1) m to (4, 3,1) m along the path: (a) A straight line y = x − 1, z = 1 . (b) A curve 6 y = x 2 + 2 , z = 1 . 2.45. The potential at any point (x, y, z ) in rectangular coordinate is given by

V (x, y, z ) = 2x + yz − xy V Find (a) the electric field at any point. (b) the work done to move a 10 nC from the point (10, − 10,10) cm to the origin. 2.46. The potential at any point (ρ , φ , z ) in cylindrical coordinate is given by

V (ρ , φ , z ) = 10 (1 + ρ ) z 2 cos φ V Find (a) The electric field at any point. (b) The energy stored in the region bounded by 0 ≤ ρ ≤ 5 cm , 0 ≤ φ ≤ π 2, 0 ≤ z ≤ 10 cm . 2.47. An electric dipole having a moment p = 2 a x − a z n C. m is located at (−10, 20,10) cm in free space. Find the potential at (10,10,10) cm.

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CHAPTER 3

Conducting and Dielectric Materials Abstract: This chapter focuses on the conducting and dielectric materials and their properties under static fields condition. Conducting materials conduct electric current efficiently, while dielectrics possess high insulation capabilities, in addition to its ability to store electric energy. Conducting and dielectric materials are essential in all electrical and electronics systems and equipment. The ability of the material to conduct the electric current is called the conductivity of the material. On the other hand, the interaction between a dielectric material and electrostatic field leads to the formation of dipole moments in the atoms of the material, which is known as polarization. This polarization ability of a material is quantitatively described by a constant known as the permittivity of the material. The permittivity of a dielectric material relative to that of free space is known as the relative permittivity or dielectric constant of the material. Both the conductivity and dielectric constant depend on the intrinsic properties of the material. This chapter presents a detailed derivation for the conductivity and permittivity under static field conduction. In addition to the properties of the conducting and dielectric materials, the chapter discusses the concept of conservation of charge and relaxation time; the boundary conditions between different media; the resistance; the capacitance; the stored energy in a capacitor. The topics of the chapter are supported by numerous illustrative examples and figures in addition to solved problems and homework problems at the end of the chapter.

Keywords: Capacitance, conduction current, conductivity, conductor, convection current, dielectric breakdown, dielectric constant, dielectric strength, dielectric, displacement current, electric flux, mobility, perfect electric conductor, polarization, relaxation time, resistance, resistivity. INTRODUCTION The behavior of electromagnetic fields in a certain medium is determined mainly by the electric and magnetic properties of the medium. This chapter focuses on the electric properties of materials and related concepts, while Chapter 6 deals with the magnetic materials and their properties. The electric properties are related to the ability of transferring and storing the electric energy. Materials that possess the ability of transferring the electric energy in the form of a flow of free charges under the influence of an applied electric field are known as conducting materials. On the other hand, the materials that cannot permit the flow of charges and possess the ability of storing the electric energy in the form of the electric field are known as dielectric materials. Therefore, conducting materials conduct electric current efficiently, while Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

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dielectrics possess high insulation capabilities, in addition to its ability to store electric energy. These properties make the conducting and dielectric materials essential in all electrical and electronics systems and equipment. In addition to the properties of the conducting and dielectric materials, the chapter discusses some related topics including, the conservation of charge and relaxation time; the boundary conditions between different media; the resistance; the capacitance; the stored energy in a capacitor. The student may refer to [19, 24, 34, 54-58] for more details on the aforementioned topics. 3.1.

CONDUCTION CURRENT AND CONDUCTIVITY

Electric conduction is the property that is related to the ability of flow of electrons in the materials under the influence of an applied electric field. Based on this property the materials are divided into conducting and dielectric materials. Materials possess this property are known as electric conductors. When an electric field is applied between the terminals of an electric conductor, the free electrons in the conductor experience a force in the direction opposite to the applied electric field. Therefore, a flow of electrons takes place in the conductor in the direction opposite to the applied electric field. The rate of flow of these free electrons with respect to time is referred to as conduction current. The movement of electrons in dielectric materials results in another kind of currents. In vacuum tubes, the movement of electrons from cathode to anode is accomplished by convection and the resulted current is known as convection current. If the charges are under the influence of a time-varying electric field in a dielectric material, a displacement of charge takes place producing what is called a displacement current. In all cases, the current is the rate of change of charge with respect to time. In this chapter, we will focus on the conduction current. Consider an electric conductor of cross-section area S and length l in the x direction as shown in Fig. (3.1). When a voltage source is connected between the terminals of the conductor as illustrated in Fig. (3.1), the electric field in the conductor will be

E = a x Ex

(3.1)

Assuming the number of electrons per unit volume is N , hence the charge density is

ρv = eN

(3.2)

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where e is a charge of the electron. The total charge that travels a distance Δx in a period of time Δt is

Δq = ρv SΔx = eNSΔx

(3.3)

.x

8xi Jx

S

l x = l B

Ex

x = 0 A

I V Fig. (3.1). Flow of the conduction current.

Dividing both sides of (3.3) by Δt and assuming that the average velocity of the total charge in the x direction is υ x , yields

dq Δq Δx = lim = eNS lim = eNυ x S Δt → 0 Δt dt Δt →0 Δt

(3.4)

The rate of change of charge with respect to time is the electric current, or

Δq dq = Δt → 0 Δt dt

I = lim

(3.5)

The unit of measurement of current is the Ampere ( A ), which is equivalent to Coulomb per second ( C s ). The volume current density J is the current per unit normal surface area. J is a vector quantity in the same direction as the electric field E that causes the flow of current. The relation between the current and current density is

I = ∫ J ⋅ dS S

(3.6)

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Using (3.5), the current density through the conductor of Fig. (3.1) becomes

J = ax

I 1 dq = ax S S dt

(3.7)

Using (3.2) and (3.4), the current density in (3.7) can be written in terms of υ x as

J = ax J x = ax

1 dq = a x eNυ x = a x ρv υ x S dt

(3.8)

In the absence of the external electric field E , the free electrons in the conductor are in a continuous random motion in the crystal structure of the conductor. This is because the atoms in the lattice structure of the conductor have a thermal energy that makes the atoms vibrate about their equilibrium positions. This vibration causes collisions between the free electron and these atoms. These collisions in addition to the presence of distortions, voids, and impurities in the structure of the conductor, make the free electrons scatter randomly from one atom to the other. The random motion of free electrons does not result in a displacement in the total charge and hence no net current is produced in the absence of the external electric field. When the electric field E is applied, each free electron experiences a force

Fi = eE = a xeEx ,

i = 1, 2,, N

(3.9)

The electron acquires an average speed υ xi in the direction of x due to this force. Since Fi = m dυ xi dt , then from (3.9) we can be write

dυ xi eE x = dt m

(3.10)

where m is the mass of the electron. Assuming that the initial speed of electron immediately after each collision is
xi at instant t i , then the speed at t is

υ xi =
xi +

eEx (t − ti ) m

(3.11)

The average speed of the total charge in the conductor is the average speed of the electrons in the unit volume ( N electrons), or

1 N 1 N 1 N eE x (t − ti ) υ x = ∑υ xi = ∑
xi + ∑ N i =1 N i =1 N i =1 m

(3.12)

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The sum in the first term of (3.12) is the average speed of the total charge after the collision. The initial speed of the electron
xi is random and it may be in any direction with equal probabilities. Therefore, the average value of
xi is zero, and the average speed of the total charge becomes

υx =

eE x N (t − ti ) eτ = Ex ∑ m i =1 N m

where N

(t − ti )

i =1

N

τ=∑

(3.13a)

(3.13b)

τ is called the mean scattering time of the electron or the relaxation time. Equation (3.13a) reveals that the velocity of the free electrons is directly proportional to the applied electric field. The constant of proportionality depends on the properties of the material and known as the mobility of the material. The mobility is given by

μ=

eτ m

(3.14)

From (3.13a) and (3.14), the average velocity, which is known also as the drift velocity, becomes

υ x = μ Ex

(3.15)

Substituting υ x from (3.15) into (3.8), then the relation between the current density and the applied electric field is obtained as

J x = eN μ Ex

(3.16)

The quantity eN μ measures the ability of the material to conduct the electric current and known as conductivity of the material. Thus, the conductivity σ is a function of intrinsic properties of the material and temperature. Using (3.14) and (3.16), the conductivity can be expressed as

σ = eN μ =

e2 N τ m

(3.17)

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The conductivity is measured in Siemens per meter ( S/m ). From (3.16) the relation between the current density and the electric field in vector form becomes J = σE

(3.18)

This relation is known as Ohm’s law in point form. Example 3.1 The current density through a wire of radius a and parallel to the z-axis is given by

J = az

400 ρ A/m2 π a a−ρ

If a = 1.6 mm , find the total current through the wire. Solution In cylindrical coordinate system dS = az ρ dρ dφ . The boundaries of the crosssection of the wire are 0 ≤ ρ ≤ 0.0016 , 0 ≤ φ ≤ 2π , then I = ∫ J • dS = S

400 πa

⇒ I

2π 0.0016

∫ ∫ 0

0

ρ a−ρ

a z • a z ρ dρ dφ = −

[

800 a

0.0016

∫ 0

2 ⎛ ⎞ ⎜ ρ + a − a ⎟ dρ ⎜ a−ρ⎟ ⎝ ⎠

]

ρ =a 800 2 ρ 2 + aρ + a 2 ln(a − ρ ) ρ = 0 a = −800× 0.0016 × [3 2 + ln(0.0016)] = 6.32 A

=−

Example 3.2 The electric potential in a region − 1 ≤ x ≤ 1 , − 1 ≤ y ≤ 1, − 1 ≤ z ≤ 1 of permittivity ε = 2 ε 0 , is given by

V (x, y, z ) = x 2 y 2 z V (a) Find the charge density in the region. (b) If the charge is moving with a velocity 104 y a y m s , find the current flowing through the surface 0 ≤ x ≤ 1 , z ≤ 0.5 , y = 1 .

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Solution The charge density in the region can be obtained from ∇ ⋅ E = ρv ε , where

⎛ ∂V ∂V ∂V ⎞ ⎟ + ay + az E = −∇V = − ⎜⎜ a x ∂y ∂z ⎟⎠ ⎝ ∂x ⇒ E = − 2 xy 2 z a x + 2 x2 yz a y + x2 y 2a z V m

(

⇒ ∇⋅E =

ρv

)

∂ ∂ ∂ 2 xy 2 z − 2 x 2 yz − 2x 2 y 2 ∂x ∂y ∂z ρ v = −4ε 0 z (y 2 + x 2 ) C m3

=−

ε ⇒

(

)

(

)

(

)

For υ = 104 y a y m s

(

)

J = ρv υ = −4 × 104 ε0 yz x 2 + y 2 a y A m2

(-1,-1,1)

z (-1,1,1)

Charge density ρv C/m3 S

y (-1,1, 1) (1,1,-1)

dS =ay dx dz (1,1, 1)

x Fig. (3.2). Geometry of the problem of Example 3.2.

The charge is enclosed in the cube shown in Fig. (3.2), and the current flows through the surface 0 ≤ x ≤ 1 , z ≤ 0.5 , y = 1 , is

I = ∫ J • dS = −4 × 10 4 ε 0 S

0.5

1

0.5 1

∫∫ z (x

−1 0

2

)

+ 1 a y • a y dx dz

5 ⎡ 1 ⎤ ⎡1 ⎤ ⇒ I = −4 × 10 4 ε 0 ⎢ z 2 ⎥ ⎢ x3 + x⎥ = − × 10 4 × 8.85 × 10 −12 = −73.75 nA 6 ⎣ 2 ⎦ −1 ⎣ 3 ⎦0

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PROPERTIES OF PERFECT CONDUCTORS

Materials of high conductivity, or σ → ∞ are called perfect conductors. Inside an isolated perfect conductor from any source, the static electric field E is always zero. This obvious, since for non-zero current density ( J ≠ 0) , the electric field in the conductor E → 0 as σ → ∞ , which is clear from (3.18). Consequently, based on Gauss’s law, the volume charge density ρ v inside the perfect conductors is zero, since there is no static field. If an amount of charge is placed inside a perfect isolated conductor, it is immediately redistributed itself on the surface of the conductor such that the electric field inside the conductors remains zero. The perfect conductor is characterized by the following properties: 1. The electric field inside and on the surface of the perfect conductor is zero E = 0. 2. The total charge of the conductor is distributed on its surface so that the electric field in the conductor and on its surface remains always zero. 3. The electrical field is always normal to the surface of the perfect conductor, and there is no electric field component tangential to its surface. Accordingly, the electric flux density D is normal to the surface and equal to the total charge on the surface of the conductor. 4. The electric potential inside and on the surface of the conductor is constant since E = −∇V = 0 . Thus, the surface of the conductor is equipotential. Medium of permittivity ε En S

En B

Et C

A Δy

Δx

Δh S′

Surface charge density ρs

D E=0 Perfect conductor (σ = ∞) Fig. (3.3). Boundary conditions between a perfect conductor and a dielectric medium.

These properties are explained in the following analysis. Consider a perfect conductor in a medium of permittivity ε as shown in Fig. (3.3). The tangential

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and normal components of the electric field E and electric flux density D to the surface of the conductor are respectively E t , D t , E n , and D n where

E = Et + E n ,

Et = Et ,

En = E n

(3.19)

D = Dt + D n ,

Dt = Dt ,

Dn = D n

(3.20)

The electric potential along a closed path is zero, then

∫E

•

dl = 0

(3.21)

Applying (3.21) along the closed rectangular loop ABCDA of sides Δx and Δy at the boundary shown in Fig. (3.3), assuming Δy → 0 , and taking into consideration that the electric field interior to the conductor is zero, and the fields along the paths AD and BC are opposite in directions, then B

∫E A

•

C

D

A

B

C

D

dl + ∫ E • dl + ∫ E • dl + ∫ E • dl Δy Δy (0) + Δx(0) + Δy (0) + Δy En = 0 = ΔxEt − En − 2 2 2 2

(3.22)

This reduces to Et = 0

(3.23)

The magnitude of the normal component En can be obtained by applying Gauss's law to the surface of the conductor. Choosing a circular cylinder of small height Δh → 0 , the base of the surface area S , and the cylindrical surface area S ′ shown in Fig. (3.3), as a Gaussian surface, then

∫εE

•

dS = ε En S + S ′ ε En + S (0) = S ρ s

(3.24)

S

where ρ s is the surface charge density on the surface of the conductor. Since Δh → 0 , then S ′ → 0 , and (3.24) reduces to

En = ρ s ε

(3.25)

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Consequently, the boundary conditions on the surface of a perfect electric conductor immersed in a medium of permittivity ε , become

Dt = Et = 0 Dn = ε En = ρ s 3.3.

(3.26) (3.27)

RESISTANCE

When a conductor of a finite conductivity is connected between terminals of a voltage source, a non-zero electric field E ≠ 0 will result in the conductor and the flow of electrons take place as illustrated in Fig. (3.1). The flow of electrons inside the conductor suffers a resistance. This resistance depends on the properties and dimensions of the conductor in addition to the conductor temperature. Appling Ohm’s law on the conductor in Fig. (3.1), then J x = σ Ex

(3.28)

Using (3.6), the current in the conductor is

I = ∫ J • dS = J x S

(3.29)

s

The potential difference V between the terminals of the conductor A and B can be obtained from (2.55) as A

0

B

l

V = VBA = VA − VB = −∫ E • dl = −∫ E • dl = Exl

(3.30)

Eliminating J x and E x from (3.29) and (3.30), yields

V=

l I σS

(3.31)

This equation indicates that the potential difference between the terminals of the conductor is directly proportional to the current through the conductor. The proportionality constant is known as the resistance R of the conductor. It is clear that from (3.31), this constant is

R=

l σS

(3.32)

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The unit of resistance is Ohm ( Ω ). The resistance is directly proportional to the length of the conductor and inversely with its cross section area and conductivity. The resistance is often expressed in terms of a parameter called resistivity ρ , which is the inverse of the conductivity, or

ρ=

1 σ

(3.33)

The unit of resistivity is Ω. m . The temperature affects the dimensions and intrinsic properties of the conductor. Therefore, the resistance changes with temperature. The variation of the resistance with temperature makes Ohm’s law valid only under the condition of a constant temperature. The resistance of some conductors changes abruptly to zero at certain temperature as Aluminum. This property is known as superconductivity. For a non-uniform dimension conductor, the resistance between any two points A and B in the conductor can be obtained using the general formula A

V R = = − ∫ E • dl σ ∫ E • dS I B s

(3.34)

The conductance G is the inverse of the resistance and can be expressed as

G=

1 = − σ ∫ E • dS R s

A

∫E

•

dl

(3.35)

B

The conductance is measured in Siemens ( S ). Example 3.3 The conductor shown in Fig. (3.4) have a conductivity σ . Find the resistance between the surfaces (a) A and D. (b) B and E. (c) C and F. Solution The resistance of the differential shaded element shown in Fig. (3.4) is

ΔR =

1 Δl σ ΔS

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z

Sameir M. Ali Hamed

D

x

C

a E y

B

A

a

x

c

a E

C

dr

F (a)

B

A

r

y

b

Length of the arc is r a.

D

z

F

(b)

Fig. (3.4). Geometry of the problem of Example 3.3.

(a) For the resistance between A and D, Δl = ρ α and ΔS = c Δρ . Hence,

ΔRAD =

1 Δl 1 ρ α = σ ΔS σ c Δρ

ΔGAD =

σ c Δρ ρα

As Δρ → 0 , then

dG AD =

σ c Δρ ρα

b

c 1 c ⇒ G AD = σ ∫ dρ = σ ln(b a) αaρ α 1 α ⇒ R AD = = G AD σ c ln(b a)

(b) Referring to Fig. (3.4), between B and E Δl = Δρ and ΔS = c ρ α . Hence,

ΔRBE =

1 Δl 1 Δρ = σ ΔS σ c ρ α

As Δρ → 0 , then

dRBE = b

⇒

RBE =

1 dρ σ cρα

1 1 1 1 ln(b a) dρ = ∫ σ cα a ρ σcα

c

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(c) Referring to Fig. (3.4), between C and F Δl = c and ΔS = α ρ dρ , hence

ΔRCF =

1 Δl 1 c = σ ΔS σ α ρ Δρ

As Δρ → 0 , then

dRCF = 3.4.

b 1 c 1 2c 1 σα σα ⇒ = ρ dρ = (b 2 − a 2 ) ⇒ RCF = 2 ∫ σ α ρ dρ σα b − a 2 dRCF c a 2c

DIELECTRIC MATERIALS

3.4.1. Polarization Atom is electrically neutral, or the total charge in the atom is zero, since the positive charge (protons) and negative charge (electrons) in the atom are equal, and in average, the center of mass for both negative and positive charge coincides with the center of the atom. When a material is placed in an external electric field E , the center of mass of electrons, which are much less in weight compared to protons, displaces by a distance d from the center of the atom in a direction opposite to that of E as illustrated in Fig. (3.5). The presence of equal amount of negative charge and positive charge with small displacement between their center of mass resembles an electric dipole. For an atom with n electrons, the total amount of positive and negative charge is q = ne . Therefore, the resulted dipole moment is

p = a x ned = a x qd

(3.36)

The formation of dipole moments in the atoms of a dielectric material in the presence of an external field is known as electronic polarization. A similar process occurs in a molecule composed of positively and negatively charged ions. In this case, the process of formation of dipole moments in the molecules of a material is known as ionic polarization. A mechanism of polarization, known as orientational or dipolar polarization, occurs in materials that possess randomly oriented permanent dipoles. When these materials are placed in an external electric field, the randomly oriented dipoles tend to align themselves in the direction of the external field. A fourth mechanism of polarization, known as interfacial polarization, takes place when there is an accumulation of charges at the interface between two materials or between two regions within a material. All

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mechanisms of polarizations mentioned above lead to the formation of an average dipole moment p per atom or molecules. E

E=0

Center of mass of positive and negative charges

Center of -ve charges Nucleus p -q + +q d Center of +ve charges

+ Nucleus

Negative charge x Fig. (3.5). Formation of dipole moments in the atom.

The process of atom polarization is illustrated in Fig. (3.5), where the center of mass of negative (-ve) charges is displaced by small distance from the center of mass of positive (+ve) charges when an external field E is applied. The presence of the electric field makes the negatively charged electrons under the influence of two forces, the force Fr due to positive charges in the nucleus of the atom, and the force Fs due to the electric field. Fr attracts the electrons towards the nucleus, while Fs is opposite to E and tends to move the electrons away from the nucleus. Fr is proportional to the displacement d with proportionality constant κ , or

Fr = −κ d a x

(3.37)

The minus sign indicates that Fr is towards the nucleus. Fs is given by

Fs = neE = neExa x

(3.38)

At equilibrium, the total force on the charges is zero, hence

Fr + Fs = −κd a x + neExa x = 0

(3.39)

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Solving (3.39) for d , then

d=

ne

(3.40)

Ex

κ

Substituting d from (3.40) in (3.36), the dipole moment induced in the atom due to the applied electric field E x is

p = ax

n 2e 2

κ

Ex =

n 2e 2

κ

E = ξE

(3.41)

This equation reveals that the dipole moment is proportional to the applied electric field E . The constant of proportionality ξ is known as the electronic polarizability. It is clear that from (3.41)

ξ=

n 2e 2

(3.42)

κ

Consider a volume Δv of the polarized material by the electric field E as shown in Fig. (3.6). Assuming the induced electric dipole moments in the volume Δv are p i , where i = 1, 2,, N , then the total dipole moment in Δv becomes N

Δp v = ∑ p i

(3.43)

i =1

Dn

p1

+

p2

pi

E

+

+

n

Dn

pN

+

Dpn

Dpn E

Fig. (3.6). The total dipole moment in a differential volume element Δv of a dielectric material.

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The polarization vector P in a material is defined as the total dipole moments induced in the material per unit volume. Mathematically, the polarization vector is

P=

Δp v 1 N = ∑ pi Δv Δv i =1

(3.44)

When Δv → 0 the polarization becomes

Δp v dp v 1 N = = lim ∑ pi Δv → 0 Δv dv Δv → 0 Δv i =1

P = lim

(3.45)

The dipole moment p given by (3.41), is equivalent to the average of the dipole moments induced in all atoms in the volume Δv , or

p=

1 N ∑ pi N i =1

(3.46)

Using (3.36), (3.45) and (3.46) along with (3.41), the polarization vector becomes

P=N

1 NΔv ∑ pi = Np = a x Nqd = N ξ E NΔv i =1

(3.47)

This relates the polarization to the applied external electric field. In the absence of the electric field, the polarization is zero. Equation in (3.47) can be written in terms of a quantity χ e as

P = χ eε 0 E

(3.48)

where χ e = N ξ ε 0 is referred to as the electric susceptibility of the dielectric. ε 0 is the permittivity of free space and its value is given in (2.5). Example 3.4

χ e = 0.52 , and it contains 2.71 × 10 atoms m at a certain temperature. If the applied electric field is E = a x1 kV m , find the polarization, the electronic polarizability and the effective dipole length. (The number of electron per atom in Argon is n = 10 ). The

electric 28

susceptibility 3

of

Argon

is

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Solution

χ e = 0.52 , a number of atoms per unit volume N = 2.71 × 1028 atoms m3 , and the electric field E = a x1 kV m . From (3.48), the polarization assuming the relation is linear is

Given

that

P = χ eε 0 E = 0.52 × 8.85 × 10 −12 × 103 = 4.61 n C m 2 From (3.47) and (3.48), the electronic polarizability of Argon is

ξ=

χ eε 0 0.52 × 8.85 × 10 −12 = = 1.7 × 10 − 40 F⋅ m2 22 2.71 × 10 N

From (3.40) and (3.42), the effective dipole length can be obtained as

d=

ξ Ex 1.7 × 10 −40 × 103 = = 1.06 × 10 −18 m ne 10 × 1.6 × 10 −19

3.4.2. Relation between the Polarization Vector and the Potential The electric potential at a point of a position vector r , of a single dipole moment p v located at a point of a position vector r′ , can be written based on (2.100) as

V (r ) =

p v ⋅ (r − r′) 3 4π ε r − r′

(3.49)

This relation can be used to determine the electric potential due to the polarization vector P at any point surrounding a polarized dielectric. When the material shown in Fig. (3.7) is immersed in an electric field, a dipole moments will be induced in the atoms of the material resulting in a polarization vector P . According to (3.45), the polarization vector P of the material is

P=

dp v dv′

(3.50)

where dp v is the dipole moment of the differential volume dv′ shown in Fig. (3.7).

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Source point (x `, y `, z)`

v S

r-r`

dv`

dpv

r`

Field point (x, y, z) r

y x Fig. (3.7). The electric potential of a polarized dielectric material.

The differential electric potential dV due to dp v at a field point ( x, y, z ) can be obtained using (3.49) and (3.50) as

dV =

dp v ⋅ (r − r′) P ⋅ (r − r′) dv′ = 3 3 4π ε r − r′ 4π ε r − r′

(3.51)

Integrating both sides of (3.51), the electric potential V due to a polarized material of a polarization vector P is obtained as

V=

1

4π ε ∫ v

P ⋅ (r − r′) dv′ 3 r − r′

(3.52)

Taking the differentiations in the gradient operator with respect to r′ , we have

⎛ 1 ⎞ r − r′ ⎟= ∇⎜⎜ 3 ⎟ ⎝ r − r′ ⎠ r − r′

(3.53)

Using (3.53), (3.52) can be written as

V=

⎛ 1 ⎞ ⎜ ⎟ ′ ⋅ ∇ P ⎜ r − r′ ⎟dv 4π ε ∫v ⎝ ⎠ 1

(3.54)

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Applying the vector identity ∇ ⋅ (ΦA) = Φ∇ ⋅ A + A ⋅ ∇Φ , with Φ = 1 r − r′ and A = P , (3.54) becomes

V=

1

P

∇⋅P

1

dv′ − dv′ ∇⋅ 4π ε ∫ r − r′ 4π ε ∫ r − r′ v

(3.55)

v

The differentiations in the gradient operator in (3.53) – (3.55) are with respect to r′ . The surface enclosing the volume v is S and the differential surface area of the source is dS′ = an dS ′ where a n is the normal unit vector to the surface. Applying the divergence theorem to the first term of (3.55), yields

V=

P ⋅ an 1 ∇⋅P ′ d S dv′ − 4π ε ∫S r − r′ 4π ε ∫v r − r′ 1

(3.56)

It is clear that the first term is the potential due to a surface charge density equivalent to P ⋅ a n C m 2 , while the second term is the potential due to a volume charge density equivalent to − ∇ ⋅ P C m3 . Let Pn be the component of P that is normal to the surface S , ρ sb is the equivalent surface charge density, and ρ vb is the equivalent volume charge density of the polarized material, then

ρ sb = Pn = P ⋅ an

(3.57)

ρvb = −∇ ⋅ P

(3.58)

ρ sb is the surface charge density of the surface polarization charge Qb that is induced as the result of the polarization process. These charges are also known as the bound charges because they are bound to the atoms or molecules of the dielectric and cannot move freely within the dielectric. Example 3.5 A dielectric disk of radius a and thickness t as shown in Fig. (3.8) is polarized with a dipole moment per unit volume P = a z Pz . Find the electric potential at any point along the disk axis.

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z (0, 0, z)

(ρ′, φ′, z′)

a t

y dv′ = ρ′ dρ′ dφ′ dz′

φ′

x

Fig. (3.8). Geometry of the problem of Example 3.5.

Solution The electric potential using (3.52) is given by

V=

1

4π ε ∫ v

P ⋅ (r − r′) dv ′ 3 r − r′

where

r′ = a x ρ ′ cosφ ′ + a y ρ ′ sin φ ′ + a z z′ , r = az z , r − r′ = a x ρ ′ cosφ ′ + a y ρ ′ sin φ ′ + a z ( z − z′)

r − r′ = ρ ′2 + ( z − z′)2 P = a z Pz ⇒ P ⋅ (r − r′) = ( z − z′) dv′ = ρ ′dρ ′dφ ′dz′ Hence,

Pz d ( z − z′) dv′ 4π ε ∫v ρ ′2 + ( z − z′) 2

]

t Pz ⎧⎪ ( z − z′) = − ∫⎨ 2 ε 0 ⎪⎩ a 2 + ( z − z′) 2

]

V=

[

[

3 2

1 2

=

t 2π a ρ ′( z − z′) dρ ′dφ ′dz′ Pz 3 4π ε ∫0 ∫∫ ρ ′2 + ( z − z′) 2 2 0 0

⎫⎪ − 1⎬ dz′ ⎪⎭

[

]

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Simplifying, the electric potential along the z-axis due to the polarized disk is t ( z − z′) P ⎧⎪ V = − z ∫⎨ 2 ε 0 ⎪⎩ a 2 + ( z − z′) 2

=

(

[

]

1 2

⎫⎪ P − 1⎬ dz′ = − z 2ε ⎪⎭

Pz t + a 2 + t 2 − a 2 + ( z − t )2 2ε

(a

2

+ ( z − z′) − z′ 2

)

z′=t z′=0

)

3.4.3. Permittivity and Relative Permittivity It has been shown in the previous section that, when a dielectric is placed in an electric field, the atoms of the dielectric become polarized. This polarization depends on the electric field and can be described mathematically by the polarization vector as in (3.48). When the electric field is removed, the atoms of the dielectric return to their original state and the polarization becomes zero. A charge Qb of surface charge density ρ sb is induced on the surface of the material as the result of polarization. To explain this, consider the two metallic plates shown in Fig. (3.9). When an electric field is applied between the plates, such that the medium between them is free space, free charges + QT and QT will be induced on the plates. When a dielectric material is inserted in the space between the plates in the presence of the electric field E , the atoms of the material will be polarized as illustrated in Fig. (3.9), and more charges will be transferred from the source to the plates to neutralize the bound charges Qb . This process increases the free charges on the plates from QT to Q f . The produced dipoles as the result of the polarization process are in the direction such that the total polarization vector P is in the same direction as E . Referring to Fig. (3.9), it is clear that the total charge in the dielectric is zero since each positive charge neutralizes the adjacent negative charge of the same amount. The positive charges at the right boundary between the dielectric and the right plate do not neutralize any negative dipole charge. Hence, a total charge + Qb will be induced on the surface of the right plate due to the process of polarization as illustrated in Fig. (3.9). Similarly, a charge − Qb will be induced on the surface of the left plate. Since the applied electric field is the same before and after the insertion of the dielectric material between the plates, the total charge before and after the insertion of the dielectric must be equal. The total charge before the insertion of the dielectric is QT and after the insertion of the dielectric is Qb plus the free charge Q f . Thus,

Q f + Qb = QT

(3.59)

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V

V

Conducting plates of surface area S

-Qb

+ +

Free space

Dielectric

+Qb

+ - + - +

…-

+ - +

+ - + - +

…-

+ - +

+ - + - +

…-

+ - +

+ - + - +

…-

+ - +

+ - + - +

…-

+ - +

P +

E

+ +

d

+QT

-QT

+Qf

-Qf

E x

x

Fig. (3.9). Polarization of a dielectric material in the space between two parallel conducting plates.

Note that both QT and Q f are free charges. Let the volume of the dielectric material is v and enclosed by the surface S . If ρ vb is the equivalent volume charge density due to the polarization of the material, then the bound charge can be obtained as Qb = ∫ ρvb dv

(3.60a)

v

The volume charge density is given in terms of polarization P by (3.58) as

ρvb = −∇ ⋅ P

(3.60b)

Substituting ρ vb from (3.60b) into (3.60a) and using the divergence theorem, yields Qb = −∫ ∇ ⋅ Pdv = −∫∫ P ⋅ dS v

S

(3.61)

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The relation between the applied electric field E and the total charge QT when the space between the plates is free space is governed by Gauss’s law as QT = ∫∫ ε 0 E ⋅ dS

(3.62)

S

Using (3.59), (3.61) and (3.62) the total free charge on the surface S in the presence of the dielectric material is obtained as

Q f = QT − Qb = ∫∫ (ε 0 E + P ) ⋅ dS

(3.63)

S

According to Gauss’s law, the total charge is related to the electric flux density D by

Q f = ∫∫ D ⋅ dS

(3.64)

S

Comparing (3.63) and (3.64), yields

D = ε0 E + P

(3.65)

Substituting P from (3.48) into (3.65), the relation between the electric field and electric flux density in a dielectric material is obtained as

D = (1 + χ e )ε 0 E

(3.66)

The quantity ( 1+ χ e ) is a dimensionless ratio and known as the relative permittivity ε r of the dielectric material or the dielectric constant. Then

εr = 1+ χ e

(3.67)

Consequently, the relation between E and D in a dielectric material is

D = εrε0 E

(3.68)

It follows that the permittivity of the material is

ε = εrε0

(3.69)

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Using (3.47), (3.48), and (3.67) the relative permittivity can be written in terms of the electronic polarizability ξ and the number of atoms per unit volume N as

εr = 1+

ξN ε0

(3.70)

For time-varying fields, the permittivity is a complex quantity that depends on the frequency. The frequency-dependent permittivity is discussed in Chapter 6. The relative permittivity of vacuum is 1.0 since its permittivity is ε 0 . Table 3.1 lists the relative permittivity for some dielectric materials at 20° C. Table 3.1. Relative permittivity for some dielectric materials at 20°° C. Material Air (1 atm) Epoxy glass Fused quartz Mica Paper Paraffin Polystyrene Porcelain Rubber Transformer oil Teflon Water

Relative Permittivity 1.00059 5.2 3.8 7.0 3.7 2.0 – 3.0 2.5 – 2.6 6.0 – 8.0 3.0 – 4.0 4.5 2.1 80.4

3.4.4. Linear, Isotropic and Homogeneous Media Electrical media may be linear, isotropic, or homogeneous according to the properties of susceptibility of the media χ e and the relation between the polarization vector P and the applied electric field E . 3.4.4.1. Linear Media In most dielectrics used in practical applications, χ e is constant quantity, hence the relation between P and E is linear. Such dielectrics are said to be linear media. Dielectrics in which the susceptibility is a function of applied electric field are said to be non-linear media.

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3.4.4.2. Isotropic Media Isotropic media are those in which P and E are parallel. The media in which P and E are not parallel are said to be anisotropic. The susceptibility and permittivity of anisotropic media depend on the directionLH their values are not the same in all directions. Therefore, the polarization in the x , y , and z directions is related to the applied electric field components respectively by

Px = ε 0 χ xx Ex + ε 0 χ xy E y + ε 0 χ xz Ez

(3.71a)

Py = ε 0 χ yx Ex + ε 0 χ yy E y + ε 0 χ yz Ez

(3.71b)

Pz = ε 0 χ zx Ex + ε 0 χ zy E y + ε 0 χ zz Ez

(3.71c)

Then the relation between the electric field E and electric flux density D in the anisotropic materials can be written in a matrix form as

⎡ Dx ⎤ ⎡ε xx ⎢ D ⎥ = ⎢ε ⎢ y ⎥ ⎢ yx ⎢⎣ Dz ⎥⎦ ⎢⎣ε zx

ε xy ε yy ε zy

ε xz ⎤ ⎡ Ex ⎤ ⎥ ε yz ⎥ ⎢⎢ E y ⎥⎥ ε zz ⎥⎦ ⎢⎣ Ez ⎥⎦

(3.72)

For isotropic materials ε xx = ε yy = ε zz = ε and all other elements in (3.72) are equal to zero. 3.4.4.3. Homogeneous Media When the permittivity of the medium is independent of the position in it, then the medium is said to be a homogeneous, while that in which the permittivity is a function of the position is an inhomogeneous medium. One of the typical examples of inhomogeneous media is the ionosphere. The ionosphere is an ionized layer surrounding the earth between 90 and 1000 km above the earth's surface. At high frequencies the permittivity of the ionosphere varies with electron density which is a function of the height above the surface of the earth. The media that are isotropic, linear, and homogeneous are known as simple media. Most of the materials of practical interest are simple when the applied electric field is within certain limits. At large values of an applied electric field, many of these linear materials become non-linear. In the conductors, the applied electric field E is related to the current density J by Ohm’s law J = σ E . For practical conductors under normal operation conditions, σ is independent of E , E and J are in the same direction, and σ has the same value at all points within

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the conductor. Therefore, these conductors are, linear, isotropic, and homogeneous. 3.5.

CONSERVATION OF CHARGE AND RELAXATION TIME

Consider a charge with a volume charge density ρ v is enclosed within a volume v bounded by a surface S . The rate of decrease of the charge within v is equal the current passing through the surface S . The total charge is given by

Q = ∫ ρv dυ

(3.73)

υ

Hence, the current through the surface S is

I = ∫ J • dS = − S

∂Q ∂ = − ∫ ρv dυ ∂t ∂t υ

(3.74)

∂ ρv dυ ∂t υ∫

(3.75)

Using the divergence theorem

∫J S

•

dS = ∫ ∇ • J dυ = − υ

Consequently, the law of conservation of charge or the continuity equation is obtained from (3.75) as

∇•J =−

∂ ρv ∂t

(3.76)

As mentioned before, if an initial charge is placed in an isolated perfect conductor in a homogeneous dielectric, they move towards the boundaries of the conductor due to resulted coulomb forces and accumulate on the surface of the conductor. These charges cannot leave the conductor because it is surrounded by a dielectric medium. The accumulation of charges on the surface of the conductor continues until all the interior charges are removed and the interior of the conductor become electrically neutral. Hence, since the net charge is zero within the interior region of the conductor, then the electric field according to Gauss’s law is zero inside the conductor. The time required to remove about 63% of the charge in the material to its surface is known as the relaxation time. The same process can occur in the dielectrics with finite conductivities.

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Consider a linear, isotropic, and homogeneous material with finite conductivity σ and permittivity ε . A charge of a volume charge density ρ0 is placed in the material at an instant t = 0 s . Recalling (2.31), then

ρv ε

(3.77)

1 ∂ ρv σ ∂t

(3.78)

∇•E= Replacing J in (3.76) by σ E , yields

∇•E=− From (3.77) and (3.78), we have

1 ∂ ρv σ 1 =− =− ρ v ∂t ε τ

(3.79)

where τ = ε σ is the relaxation time. Letting t = t ′ in (3.79), and integrating both sides in the range t ≥ t ′ ≥ 0 with respect to time, and ρ0 ≥ ρv ≥ 0 with respect to the charge density, then ρ0

t

1 σ ∫0 ρv d ρv = −∫0 ε dt′

(3.80)

Evaluating the integrals in (3.80), the variation of the charge density with time in the material, can be obtained as

ρv (t ) = ρ0 e

σ − t ε

= ρ0 e−t τ

(3.81)

The relaxation time τ = ε σ is the time required to decrease the volume charge density ρ v to 1 e of its initial value, or 37% of the initial value. The relaxation time of metallic conductors is extremely short while it is very long for dielectrics. The relaxation time for copper, distilled water and fused quartz are: For copper ε = ε 0 and σ = 5.7 × 107 S m ; hence

τcopper =

ε 8.85 × 10−12 = = 1.53 × 10−19 s σ 5.7 × 107

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For distilled water ε = 80 ε0 and σ = 10−4 S m ; hence

τ water =

ε 80 × 8.85 × 10−12 = = 7.08 × 10 − 6 s −4 σ 10

For fused quartz ε = 3.8 ε 0 , σ = 10−17 S m ; hence

τquartz =

ε 3.8 × 8.85 × 10 −12 = = 33.63 × 105 s −17 σ 10

Note that the relaxation time for copper is nearly zero and this result can be generalized for all conducting materials. This explains why there is no free charge inside the conductors. For fused quartz, the relaxation time is about 39 days, which means that the initial charges remain inside the dielectrics. 3.6.

BOUNDARY CONDITIONS FOR ELECTROSTATIC FIELDS

3.6.1. Interface between Two dielectrics Consider a static electric field crossing the boundary between a medium of relative permittivity ε r1 , which will be referred to as medium 1, to a second medium of relative permittivity ε r 2 , which will be referred to as medium 2. The respective electric fields in medium 1 and medium 2 are E1 and E2 , where

E1 = E t1 + E n1

E1 = E1 ,

Et1 = Et1 ,

(3.82a)

En1 = E n1

E 2 = Et 2 + E n2

E2 = E 2 ,

Et 2 = E t 2 ,

(3.82b) (3.82c)

En 2 = E n 2

(3.82d)

Let the angles between the direction of E1 and E2 and the normal to the interface are θ1 and θ 2 respectively as illustrated in Fig. (3.10), then the tangential and normal electric field components to the boundary in medium 1 and medium 2 are

Et1 = E1 sin θ1 , En1 = E1 cosθ1 Et 2 = E2 sin θ 2 , En 2 = E2 cosθ2

(3.83a) (3.83b)

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Applying (2.55) along the closed rectangular small loop ABCDA of sides Δx and Δy at the boundary shown in Fig. (3.10), yields B

C

D

A

A

B

C

D

∫ Edl + ∫ Edl + ∫ Edl + ∫ Edl = − ΔxEt1 +

Δy Δy En1 + En 2 + ΔxEt 2 2 2 Δy Δy − En1 − En 2 = 0 2 2

(3.84)

On the surface of the interface Δy → 0 , hence ( Et 2 − Et1 )Δx = 0 ,which leads to

Et 2 = Et1 En1

E1

01

Dy

Dn1 = e0er1 En1

Dh

C

D

E2

02

Dt1 S

Dh

Dx

Et2

Medium 1 er1

S

Et1

B A

(3.85)

`

En2

Medium 2 er2

Dt2 Dn2 = e0er2 En2

Fig. (3.10). Boundary conditions at the interface between two dielectric media.

This means that the tangential component of an electric field that is crossing the interface between the two media is contusions across the interface. Applying Gauss’s law, the boundary conditions for the normal components of the electric field can be obtained. The Gaussian surface is the small circular cylinder shown in Fig. (3.10). The height of the cylinder is Δh → 0 , the area of its base is S and that of the curved surface is S ′ . If the enclosed charge is QT , then

∫D

•

dS = QT

(3.86a)

S

∫D

dS − ∫ Dn 2dS − ∫ Dt1dS ′ + ∫ Dt1dS ′ − ∫ Dt 2dS ′ + ∫ Dt 2dS ′ = QT (3.86b)

n1

S

S

S′

S′

S′

S′

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Since Δh → 0 , then S ′ → 0 and the Gaussian surface diminishes to a surface of area S . Therefore, (3.86b) reduces to

Dn1S − Dn 2 S = QT

(3.87)

Dividing both sides of (3.87) by the surface area S , yields

Dn1 − Dn 2 =

QT = ρs S

(3.88)

where ρ s C m 2 is the density of the free charge on the boundary between the two media. Since the two media are perfect dielectrics, then the charge on the boundary between them is zero. Putting ρ s = 0 in (3.88), the boundary conditions for normal components is obtained as

Dn1 = Dn 2

(3.89)

This means that the normal component of an electric flux density that is crossing the interface between two media is continuous across the interface. In terms of the electric field intensity the boundary conditions for normal components becomes

ε r1 En1 = ε r 2 En 2

(3.90)

Equation (3.85) can be written in terms of electric flux density as

ε r 2 Dt1 = ε r1 Dt 2

(3.91)

In vector form, (3.85) and (3.90) may be expressed as

an × (E2 − E1 ) = 0 an • (ε r 2 E2 − ε r1 E1 ) = 0

(3.92a) (3.92b)

where a n is the unit vector in the normal direction to the interface. The boundary conditions for the static electric flux density are

an × (ε r1 D2 − ε r 2 D1 ) = 0 an • (D2 − D1 ) = 0

(3.93a) (3.93b)

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The boundary conditions in (3.92) and (3.93) can be used to express the electric field in medium 2 in terms of that of medium 1. Let θ1 and θ 2 be the angles between the direction of E1 and E2 and the normal to the interface respectively, then from Fig. (3.10), we have in medium 1

Et1 = E1 sin θ1 , En1 = E1 cosθ1

(3.94)

Et 2 = E2 sin θ 2 , En 2 = E2 cosθ2

(3.95)

In medium 2

where E1 = E1 and E2 = E 2 . Applying the boundary conditions given by (3.90) at the interface, yields

En1 E1 cosθ1 ε r 2 = = En 2 E2 cosθ 2 ε r1

(3.96)

εr 2 E2 cosθ2 = εr1 E1 cosθ1

(3.97)

Hence

Similarly, applying (3.85), yields

Et1 = E1 sin θ1 = Et 2 = E2 sin θ2

(3.98)

E2 sin θ2 = E1 sin θ1

(3.99)

Hence,

Squaring (3.98) and (3.99), and then adding the results, the magnitude of the electric field in medium 2 in terms of that in medium 1 can be obtained as

[

]

E2 = E1 1 + (ε r1 ε r 2 ) − 1 cos2 θ1 2

(3.100)

Similarly, applying the boundary conditions (3.89) and (3.91), the magnitude of the electric flux density in medium 2 in terms of that in medium 1 is obtained as

[

]

D2 = D1 1 + (ε r 2 ε r1 ) − 1 sin 2 θ1 2

(3.101)

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Dividing (3.97) by (3.99), yields

ε r1 tanθ2 = ε r 2 tanθ1

(3.102)

This equation is known as the refraction law of electric field at the interface between two perfect dielectrics. Example 3.6 The electric field E1 = a x + a y + a z V m is incident on the interface between medium 1 which occupies the space z > 0 with ε r1 = 2 and medium 2 which occupies the space z < 0 with ε r 2 = 1.5. Find the electric field in medium 2. Solution It is clear that the interface is the xy plane and then the normal and tangential electric field components in medium 1 are

E n1 = a z ,

E t1 = a x + a y

The boundary conditions are

ε r1 E n1 = ε r 2 E n 2,

E t1 = E t 2

Applying the boundary conditions, we have

E t 2 = E t1 = a x + a y

En 2 = (ε r1 ε r 2 )En1 = a z (ε r1 ε r 2 ) The electric field in medium 2 becomes E2 = Et 2 + En 2 = a x + a y + (ε r1 ε r 2 )a z V m

3.6.2. Interface between Two Dielectrics of Finite Conductivities A dielectric of a finite conductivity σ , is known as a lossy dielectric. Consider the interface between two lossy dielectric media shown in Fig. (3.11). Let σ1 and σ 2 be the conductivities of medium 1 and medium 2 respectively. Assuming that there exist electric field E1 and E2 in medium 1 and medium 2 respectively, then current densities J1 and J 2 will be induced in the media due to the presence of

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E1 and E 2 as illustrated in Fig. (3.11). E1 and E 2 are given in terms of tangential and normal components to the interface as in (3.82a) and (3.82b). Similarly, J 1 and J 2 can be written as

J 1 = J t1 + J n1 ,

J1 = J1 ,

J t1 = J t1 ,

J n1 = J n1

(3.103a)

J 2 = J t 2 + J n2 ,

J2 = J2 ,

Jt2 = Jt2 ,

J n2 = J n2

(3.103b)

Using the continuity equation given by (3.76), we have

∫J

•

dS = −

S

∂ ρv dυ = 0 ∂t υ∫

(3.104)

Then

J n1 = J n 2

(3.105)

Using J = σ E , then (3.105) can be written as

σ1 E n1 = σ 2 E n 2 ,

σ1 En1 = σ 2 En 2

(3.106)

We know that from (3.92a) the tangential electric field across the interface between the media is continuous, which can be written as

J t1 J t 2 , = σ1 σ 2 En

C Dx

Et2 D

J2 = s2 E2

02 E2

En2

(3.107)

Dn1 = e0er1 En1

Medium 1 er, , s1

S

Et1

B

A Dy

E1

01

J1 = s1

σ1 ≠ 0 and σ2 ≠ 0

Dh Dt S

Dh

`

Dt

Medium 2

Dn2 = e0er2 En2

Fig. (3.11). Boundary conditions at the interface between two lossy dielectric media.

er2 , s2

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The surface charge density satisfies (3.88), or

ε r1 En1 − ε r 2 En 2 = ρ s

(3.108)

From (3.106) and (3.108) the surface charge density on the boundary between two media with finite conductivities assuming a static charge, is obtained as

ρs =

σ 2 ε r1 − σ 1 ε r 2 σ ε −σ ε En1 = 2 r1 1 r 2 E n 2 σ2 σ1

(3.109)

If the charge is time-varying, then (3.109) becomes invalid and the relation between volume charge density ρ v and the current density is given by

∫J

•

dS = −

S

∂ ρv dv ∂t ∫v

(3.110)

Example 3.7 In Fig. (3.12), let ε 2 = 2 ε1 = 4 ε 0 and σ 2 = 2 σ1 = 6 × 10 −4 S m . The electric field in medium 1 is E1 = 90a x − 60a z V m . Find the (a) Electric field in medium 2. (b) The current density in each medium. (c) Surface charge density on the boundary between the media. zz J1 = σ1 E 1 J2 = σσ2 2E2

Medium 1 ε 1, σ 1 Medium 2 ε 2, σ 2

x

Fig. (3.12). Geometry of the problem in Example 3.7.

Solution Given that ε 2 = 2 ε1 = 4 ε 0 and σ 2 = 2 σ1 = 6 × 10 −4 S m . In medium 1

E1 = 90a x − 60a z V m ⇒ Et1 = 90a x , ⇒ En1 = −60a z

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The current density across the boundary is continuous, then

J n1 = J n 2

⇒ σ1 En1 = σ 2 En 2

⇒ En 2 =

1 σ1 E n1 = E z1 = −30a z 2 σ2

The tangential components satisfy

Et1 = Et 2

⇒

Et 2 = Et1 = 90a x

(a) The electric field in medium 2 becomes

E1 = Et 2 + En 2 = 90a x − 30a z V m (b) The current densities in medium 1 and medium 2 are

J1 = σ1 E1 == 3 × 10 −4 × (90a x − 60a z ) = 27a x − 18a z m A m J 2 = σ 2 E1 = 6 × 10 −4 × (90a x − 30a z ) = 54a x − 18a z m A m

(c) Since ε r 2 = 2 ε r1 and σ 2 = 2 σ1 , then the surface charge density is

ρs = 3.7.

σ 2ε r1 − σ1ε r 2 2 σ ε − 2 σ1ε r1 En1 = 1 r1 En1 = 0 C m 2 σ2 σ2

CAPACITANCE

3.7.1. Parallel-Plates Capacitor Consider a dielectric of a relative permittivity ε r is inserted in the space between two metallic plates as shown in Fig. (3.13). If a voltage source is connected between the terminals of the plates, a charge + Q will induce on the surface of the plate connected to positive pole of the voltage source, and a charge − Q on the plate connected to the negative pole. This system of charges produces an electric field E in the space between the plates. It is assumed that the width of each the metallic plate is much greater than the separation d between them. Hence, the electric field in the space between the plates is uniform except at edges of the plate where the edge fringing occurs as shown in Fig. (3.14).

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Vο

Metallic plates each of area S + + + + + + +Q +

εr E -Q

+ + + x=d V = Vο

x=0 V=0

x Fig. (3.13). Parallel-Plates Capacitor.

V = Vο

x x=d

+Q + + + + + + + + + +

Edge Fringing

E

x=0 -Q V=0 Fig. (3.14). The electric field between the parallel-plates capacitor and the edge fringing.

The surfaces area of each plate is S and their planes are perpendicular to the x axis as shown in Fig. (3.14). Since the width of the plates is much larger than the separation d , then S is large and the field is uniform between plates. The surface

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charge density on the surface of the plates is ρ −s = − Q S on the plate at the potential V0 and ρ +s = + Q S on the plate at zero volts. Since S is considered large, the electric field between the plates can be obtained as

ρ +s ρ −s Q Q Q Ε = −a x + ax = −a x − ax = −a x 2ε 2ε 2ε S 2ε S εS

(3.111)

The potential difference between the two plates is V0 , then d

Qd Q a x ⋅ a x dx = = V0 εS εS 0

(3.112)

Q εS = V0 d

(3.113)

V = − ∫ E ⋅ dl = ∫ l

It follows from this equation that

The quantity on the right-hand side of (3.113) is a constant that depends on the properties and dimensions of the region between the two plates. Therefore, the ratio between the induced charge Q and the applied voltage Q between the plates is a constant quantity. This constant quantity is known as the capacitor or condenser. The unit of measurement of the capacitance is the Farad ( F ), which is equivalent to Coulomb per Volt ( C V ). According to (3.113), the capacitance of two parallel plates each of a surface area S , separated by a distance d and the region between them is filled with a dielectric of relative permittivity ε r is

C=

ε 0ε r S d

(3.114)

Using the definition of the capacitance, a general formula of the capacitance between any arbitrary conductors as those shown in Fig. (3.15) can be obtained. Let VAB , Q , ε r are a potential difference between the conductors, the induced charge on their surfaces and the relative permittivity of the space between them. If the conductor of the positive charge + Q is at a potential VA and the other one is at VB . Applying Gauss’s law by choosing the surface S shown in Fig. (3.15) as a

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Gaussian surface, the total charge Q on each conductor can be obtained. The potential difference VAB can be obtained by integrating the electric field in the space between the conductors E along the path l . Hence, substituting these Q and VAB in (3.113), a general formula of the capacitance between two conductors can be written as

Q C= = VAB

ε 0 ε r ∫ E • dS S

(3.115)

A

∫E

•

dl

B

Surface S

+

+ +

+ +Q + +

Conductor at VB

er

+ + VA + +

E 1

-Q VB

Conductor at VA Fig. (3.15). The capacitance between two arbitrary conductors.

If the dielectric of the capacitor has an infinite conductivity σ (a lossy dielectric), then a leakage current will flow through the capacitor when connected to a voltage source. The resistance to this current is known as the leakage resistance of the capacitor. The leakage resistance and corresponding conductance of the capacitor can be obtained from (3.34) and (3.35) respectively. Multiplying both sides of (3.34) and (3.115) and using (3.79), the relation between the leakage resistance and conductance of the capacitor to its capacitance is obtained as

RC =

C ε 0ε r ε = = =τ σ σ G

(3.116)

This indicates that the leakage resistance of a capacitor multiplied by its capacitance is equal to the relaxation time τ of the dielectric of the capacitor.

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3.7.2. Spherical Capacitor The spherical capacitor is composed of two conducting concentric spheres of radii a and b separated by a dielectric of permittivity ε as shown in Fig. (3.16). Assume that the outer sphere is charged such that the total charge on its surface is + Q , then a charge − Q will be induced on the surface of the inner sphere of radius a . To obtain the capacitor of the spherical capacitor the electric field E between the spheres must be determined. Let the spheres concentric at the origin. The electric field can be obtained by applying Gauss’s law. Choosing a sphere of radius r shown in Fig. (3.16) as a Gaussian surface, and applying Gauss's law, yields

− ε ∫∫ E ⋅ dS = − ε E (4πr 2 ) = −Q

(3.117)

S

From this equation, the electric field is obtained as

E = ar

Q , 4π ε r 2

a≤r ≤b

(3.118)

The capacitance can be obtained by substituting E from (3.118) into (3.115), then

Q a • a r dS ε Q (4π r 2 ) 2 r 2 r 4 ε π 4π ε C = AS = bS = 4bπ ε r = − [1 r ] rr == ba Q Q ∫B E • dl ∫a 4π ε r 2 ar • ar dr ∫a 4π ε r 2 dr ε ∫ E • dS

ε∫

(3.119a)

This simplifies to

C = 4π ε

ab b−a

(3.119b)

The capacitance CS for the inner sphere as an isolated sphere can be obtained by letting the radius of the outer sphere approaching infinity. Doing this, yield

⎛ ab ⎞ CS = 4π ε lim ⎜ ⎟ = 4π ε a b →∞ b − a ⎝ ⎠

(3.120)

Which is the capacitance of an isolated sphere of the radius a in a medium of permittivity ε .

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Conducting sphere of radius b and charge +Q

ε Spherical surface of radius r

Conducting sphere of radius a and charge -Q

Fig. (3.16). Spherical capacitor.

3.7.3. Cylindrical Capacitor The cylindrical capacitor is composed of a two coaxial cylindrical conductors separated by a dielectric material as shown in Fig. (3.17). The radii of the inner and outer conductors are a and b respectively. The length of the cylindrical capacitor is l . Assume that the outer is charged such that the total charge density on its surface is + ρ s , then a charge of a density − ρ s will be induced on the surface of the inner conductor. dl = aρ dρ

z ρ

a b

ε

-ρ s

l

+ρs

E

dS = aρ ρ dφ dz

y x Fig. (3.17). The cylindrical capacitor.

φ

dφ

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To obtain the capacitance we need to determine the electric field E in the region between the conductors. Referring to Fig. (3.17), let the Gaussian surface be the cylindrical surface of radius ρ , and applying Gauss’s law, we have

QT = ∫∫ ε E ⋅ dS = ε ES = ε E (2πρ l ) = ρ s (2π a l )

(3.121)

S

It follows that the electric field in the region between the conductors is

E = aρ

ρs a , ε ρ

a≤ρ ≤b

(3.122)

Neglecting the fringing at the ends of the conductors and referring to Fig. (3.17), we have d l = a ρ dρ , and dS = ρ dφ dz . Hence, using (3.115), the capacitance of the cylindrical capacitor is obtained as l 2π

l 2π ρs a ∫s E ⋅ dS ∫0 ∫0 ε ρ a ρ ⋅ a ρ ρ dφ dz ∫0 ∫0 dρ dz 2π ε l (3.123a) =ε b = C=ε =ε b ln( b a ) a 1 ρ d E l ⋅ s ∫l ∫a ε ρ a ρ ⋅ a ρ dρ ∫a ρ dρ

The capacitance of the cylindrical capacitor per unit length is

Cl =

C 2π ε = l ln(b a)

(3.123b)

3.7.4. The Energy Stored in a Capacitor The energy stored a capacitor can be determined using one of the following methods: 1. By finding the work done by the applied electric potential V to produce the charges + Q and − Q on the conducting surfaces of the capacitor. 2. By finding the stored energy in the dielectric filling the space between the conducting surfaces due to the presence of the electric field E .

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3.7.4.1. Parallel-Plates Capacitor In the first method, the work required to produce a charge of a surface charge density ρ +s = + Q S on the plate of potential VA and a surface density

ρ −s = − Q S on the plate of potential VB , where VA > VB , is obtained as

WT =

1 1 Q 1 Q V (r′) ρ s dS ′ = + ∫ VA dS ′ − ∫ VB dS ′ ∫ 2S S 2S S 2S

(3.124a)

1Q (VA − VB )S = 1 QV 2S 2

(3.124b)

Which reduces to

WT = +

In the second method, the medium between the plates is charge free, then the stored energy can be obtained using (2.90) as

WT =

ε0 ε0 2 ε0 V 2 1 1 2 ′ E d v = E Sd = Sd = CV 2 = QV 2 ∫ 2υ 2 2 d 2 2

(3.124c)

3.7.4.2. Spherical Capacitor The stored energy in the spherical capacitor can be obtained following the same analysis as in the case of the parallel-plates capacitor, as

Q2 ⎛ 1 1 ⎞ Q2 1 1 WT = = CV 2 = QV ⎜ − ⎟= 8π ε ⎝ a b ⎠ 2C 2 2

(3.125)

where V is the potential difference between the spheres and Q the charge on each sphere. 3.7.4.3. Cylindrical Capacitor Using a similar analysis as for parallel-plates and spherical capacitors, the stored energy in the cylindrical capacitor can be obtained as

Q2 Q2 1 1 WT = = CV 2 = QV ln(b a ) = 4π ε l 2C 2 2

(3.126)

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where V is the potential difference between the inner and outer conductors and Q the total charge on each conductor. Example 3.8 A spherical capacitor of radii a = 8 cm and b = 18 cm . The potential difference between the inner and outer spheres is 100 V and the space between them is filled with a dielectric of relative permittivity ε r = 2.56 . Find the (a) Capacitance of the capacitor. (b) Total charge on each sphere. (c) Stored energy in the capacitor. Solution (a) Given that a = 8 cm , b = 18 cm , and ε r = 2.56 , the capacitance from (3.119b) is

10 −9 × 2.56 8 × 10−2 × 18 × 10 −2 ab = 4π × × = 40.96 ρ F C = 4π ε 0ε r 36π 18 × 10− 2 − 8 × 10 − 2 b−a (b) The total charge on each sphere is

Q = CV = 40.96 × 10 −12 × 100 = 4.096 n C (c) The stored energy is

W= 3.8.

CV 2 4.096 × 10−12 × 1002 = = 20.48 n J 2 2

DIELECTRIC STRENGTH

The dielectric materials have been widely used in electrical engineering systems and equipment. Their insulation behavior makes them ideal for isolation media between conductors at different voltages to prevent the ionization of air and hence current flashover between conductors. The dielectric material can withstand certain limits of the applied voltage and hence the electric field within it cannot be increased beyond a certain limit EB . If the voltage between the terminals of the dielectric is increased such that the electric field in the dielectric exceeds EB , a

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conduction current will flow between the electrodes which leads to a dielectric breakdown. In solid dielectrics, the current does not obey Ohm’s law at strong electric fields. In this case, the current density increases almost exponentially with the electric field and at a certain field E B , the current density jumps to a very high magnitude at which the breakdown occurs. The occurrence of dielectric breakdown in solid dielectrics leads to a permanent damage in the dielectric, while in gaseous and many liquid dielectrics the breakdown mechanisms is different than that of solid dielectrics and the material restores its original behavior if the voltage causing the breakdown is removed. The limit EB is known as the dielectric strength, which is defined as the maximum field that can be applied to an insulating medium without causing dielectric breakdown. The dielectric strength for practical dielectrics at room temperature and atmospheric pressure are shown Table 3.2. Table 3.2. The dielectric strength for practical dielectrics at room temperature and atmospheric pressure. Dielectric Sulfur hexafluoride SF6 gas Polybutene Transformer oil Transformer oil Agip ITE 360 Porcelain Glass-bonded mica Natural rubber Polystyrene Distilled water Air (1 atm)

Dielectric Strength in kV/mm 8.5 13.8 110.7 9-12.6 35-160 14-15.7 100-225 19.7 65-70 3

SOLVED PROBLEMS Solved Problem 3.1 The current density in a certain region is given in spherical coordinate system by

J = 3r 2 cosθ a r − r 2 sin θ aθ A/m 2 Find the current passing across the surface 0 ≤ r ≤ 2 m , 0 ≤ φ ≤ 2π , θ = π 6 .

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Solution The surface of the cone 0 ≤ r ≤ 2 m , 0 ≤ φ ≤ 2π , θ = π 6 is shown in Fig. (3.18). In spherical coordinate system dS = aθ r 2 sin (π 6 )dr dφ . The current passes through the surface of the cone 0 ≤ r ≤ 2 sin(π 6), 0 ≤ φ ≤ 2π with the current density

J = 3r 2 cos(π 6)a r − r 2 sin (π 6)aθ A/m 2 Then the current through the surface of the cone

I

= ∫ J • dS =

S 2π 2 sin (π 6 )

∫ ∫ (3r 0

2

cos(π 6 )a r − r 2 sin (π 6)aθ

)

•

aθ r 2 sin (π 6 )dr dφ

0

⇒ I =−

1 4

2π 1

∫∫ r

3

dr dφ = −2π A

0 0

ar

z π /6

dS = r2 sin(π /6) dr dφ

S

aθ dS

φ

y

x Fig. (3.18). Geometry of Solved Problem 3.1.

Solved Problem 3.2 Find the resistance of the sector of a spherical cap of conductivity σ , with the cross section shown in Fig. (3.19), between the spherical surfaces A and B.

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z

z

dr

x

2α

A

a

y

α

x

B

a

y

A

B

r

b

b

(b)

(a) Fig. (3.19). Geometry of Solved Problem 3.2.

Solution The resistance of the differential spherical cap of thickness dr at a radius r shown in Fig. (3.19b) is

dR =

1 dr σ S (r )

where S (r ) is surface area of the spherical cap and can be obtained as

S (r ) =

2π α

∫∫ r

2

sin θ dθ dφ = 2π r 2 (1 − cosα )

0 0

Hence, the resistance between the surfaces A and B is b

R=

b

1 dr 1 1 dr ⎛1 1⎞ = = ⎜ − ⎟ 2 ∫ ∫ σ a S (r ) 2π σ (1 − cosα ) a r 2π σ (1 − cosα ) ⎝ a b ⎠

Solved Problem 3.3 The space of a parallel-plates capacitor is filled with two dielectrics as shown in Fig. (3.20). S1 is the plate area corresponding to the dielectric of permittivity ε1 and S2 is that corresponding to the dielectric of permittivity ε 2 . Find the (a) Electric field in each dielectric. (b) Total charge.

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(c) Energy stored in each dielectric. (d) Capacitance of the capacitor. Solution (a) The tangential field at the boundary between the dielectric are equal, then

E1 = E2 = Vo d V = Vο

+Q S2

S1

ε2

ε1

d

E1

E2 -Q

V=0

Fig. (3.20). Geometry of Solved Problem 3.3.

(a) The charge on the surface S1 and S 2 are

ε1 S1 Vo d ε S Q2 = D2 S2 = ε 2 E2 S2 = 2 2 Vo d Q1 = D1S1 = ε1 E1S1 =

Then the total charge

⎛ε S ε S ⎞ Q = Q1 + Q2 = V ⎜ 1 1 + 2 2 ⎟ d ⎠ ⎝ d (b) The stored energy in each dielectric is obtained as

1 1 ε1 S1 2 2 ε1 E1 × S1d = Vo 2 2 d 1 1 ε 2 S2 2 2 W2 = ε 2 E2 × S2 d = Vo 2 2 d W1 =

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Then the total energy in the capacitor is

Vo ⎛ ε1 S1 + ε 2 S 2 ⎞ ⎜ ⎟ d 2 ⎝ ⎠ 2

W = W1 + W2 = (c) The capacitance can be obtained as

C=

Q ε1 S1 ε 2 S 2 ε1 S1 + ε 2 S 2 = + = Vo d d d

Solved Problem 3.4 The space of a parallel-plates capacitor is filled with two dielectrics of permittivity ε1 and ε 2 , and thickness d1 and d 2 respectively as shown in Fig. (3.21). The surface area of each plate is S . Find the (a) (b) (c) (d)

Electric field in each dielectric. Total charge. Energy stored in each dielectric. Capacitance of the capacitor.

Solution Let the dielectric of permittivity ε1 be medium 1 and that of permittivity ε 2 be medium 2 . Referring to Fig. (3.21), E1 , E2 are the electric fields in medium 1 and medium 2 respectively. V1 and V2 are potential differences across medium 1 and medium 2 respectively. (a) The electric fields in the two dielectrics satisfy

∫D

•

dS = Q = ε1 E1S = ε 2 E2 S

S

⇒ ε1 E1 = ε 2 E2 Since the fields in the dielectrics is uniform, then

E1 =

V1 , d1

E2 =

V2 d2

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V = Vο S +Q +

d1

ε1

E1

d2

ε2

E2

V1 + V - 2

-Q V=0 Fig. (3.21). Geometry of Solved Problem 3.4.

The total electric potential across the capacitor is Vo , then

ε1 d 2 = E1 (ε1 d 2 + ε 2 d1 ) ε2 ε1 Vo E2 = (ε1 d 2 + ε 2 d1 )

Vo = V1 + V2 = E1d1 + E2 d 2 = E1d1 + E1 ⇒ E1 =

ε 2 Vo , (ε1 d 2 + ε 2 d1 )

(b) The charge on S is

Q = ε1 E1S =

ε1ε 2 S V (ε1 d2 + ε 2 d1 ) o

(c) The energy in each dielectric is

1 1 ε1 ε 2 Sd1 2 2 ε1 E1 × Sd1 = Vo 2 2 (ε1 d 2 + ε 2 d1 ) 1 1 ε1 ε 2 Sd 2 2 2 W2 = ε 2 E2 × Sd 2 = Vo 2 2 (ε1 d 2 + ε 2 d1 ) W1 =

Then the total energy in the capacitor is

W = W1 + W2 =

1 ε1ε 2 (d1 + d 2 ) S 2 Vo 2 (ε1 d 2 + ε 2 d1 )

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(d) The capacitance is

C=

Q ε1ε 2 S = Vo (ε1 d 2 + ε 2 d1 )

Solved Problem 3.5 The space between the plates of parallel-plates capacitor is filled with two dielectrics of finite conductivities σ1 and σ 2 permittivity ε1 and ε 2 respectively as shown in Fig. (3.22). Find the (a) Electric field in each dielectric. (b) Conduction current through the capacitor. (c) Charge density on the boundary between the two dielectrics. Solution Let the medium of permittivity ε1 be medium 1 and that of permittivity ε1 be medium 2 . Referring to Fig. (3.22), E1 , E2 are the electric fields in medium 1 and medium 2 respectively. J 1 and J 2 are current densities in medium 1 and medium 2 respectively. V = Vο S

d1

ε 1, σ 1

E1

J1

d2

ε 2, σ 2

E2

J2

+ V - 1 + V - 2 V=0

Fig. (3.22). Geometry of Solved Problem 3.5.

(a) The current densities in the dielectrics are equal, then

J = J1 = J 2 = σ1 E1 = σ2 E2 The electric field is uniform in the two dielectrics, then

E1 = V1 d1 ,

E2 = V2 d2

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The total electric potential is Vo , then

Vo = V1 + V2 = E1d1 + E2d2 = E1d1 + E1 σ1 d2 σ2 = E1 (σ1 d2 + σ2 d1 ) σ2 σ2 σ1 ⇒ E1 = Vo , E2 = Vo σ 1 d 2 + σ 2 d1 σ1 d 2 + σ 2 d1 (b) The current density is

J = J1 = J 2 = σ1 E1 = σ 2 E2 =

σ1σ 2 Vo σ1 d 2 + σ 2 d1

(c) Applying the boundary conditions at the interface between the dielectrics, with an = a x , then the charge density on the interface is obtained as

ρ s = a n ⋅ (ε 2 E2 − ε1 Ε1 ) = a x ⋅ (a x ε 2 E2 − a x ε1 E1 ) =

ε 2σ1 − ε1σ 2 Vo σ1 d 2 + σ 2 d1

Solved Problem 3.6 A long coaxial cable is filled with a dielectric of permittivity ε = 2 ε 0 and conductivity σ = 3 × 10−3 S m . The radii of the inner and outer conductors of the cable are a = 0.5 mm and b = 3 mm respectively. If the potential difference between the inner and outer conductors is V = 10 V, find (a) Conductance and capacitance per unit length of the cable. (b) Current density. (c) Dissipated energy per unit length of the cable. Solution Referring to Fig. (3.23), assume the surface charge density on the inner conductor is ρ s , then chosen the cylindrical surface of radius ρ and applying Gauss’s law, we have

QT = ∫∫ ε E ⋅ dS = ε ES = ε E (2πρ l ) = ρs (2π a l ) S

⇒ E = aρ

ρs a ε ρ

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(a) Referring to Fig. (3.23), d l = a ρ dρ , and dS = ρ dφ dz . Then the conductance per unit length ( l = 1 m ) is 1 2π

1 2π ρs a ∫s E ⋅ dS ∫0 ∫0 ε ρ a ρ ⋅ a ρ ρ dφ dz ∫0 ∫0 dρ dz 2π σ G=σ =σ b = =σ b ln(b a) a 1 ρ d E l ⋅ s ∫l ∫a ε ρ a ρ ⋅ a ρ dρ ∫a ρ dρ

2π σ 2π × 3 × 10 −3 6π × 10 −3 G= = = = 10.52 m S ln(b a) ln(3 0.5) ln(6)

⇒

The capacitance of a length l = 1 m of the cable can be obtained using (3.116) as

G σ = C ε

⇒

C=

2π ε 2π × 2 × (1 36π ) × 10 −9 10 −9 = = = 62 ρ F ln(b a) ln(3 0.5) 9 ln(6)

(b) The potential difference between the inner and outer conductors is b

V = −∫ E ⋅ dl = ∫ l

a

a

ρ

a ρ ⋅ a ρ dρ =

ρs a ln(b a) 2 ε0

⇒ ρs =

2 ε0 V a ln(b a )

The current density is

J = σ E = aρ

σ V 3 × 10 −3 10 16.7 ρs σ a = aρ = aρ = m A m2 ε ρ ρ ln(b a) ρ ln(6) ρ

(c) Referring to Fig. (3.23), dv = ρ dρ dφ dz . The dissipated energy per unit length of the cable can be obtained as 2

⎡ V ⎤ 1 2π b 1 1 2 Wd = ∫ J ⋅ E dv = ∫ J dv = σ ⎢ dρ dφ dz A m2 ⎥ ∫ ∫∫ σv ⎣ ln(b a) ⎦ 0 0 a ρ v 2π σ 2 ⇒ Wd = V = GV 2 = 10.52 × 10 − 3 × 10 2 = 1.05 J ln(b a)

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dl = aρ dρ

z ρ

a b

(ε, σ)

l E

dS = aρ ρ dφ dz

y φ

x

dφ

Fig. (3.23). Geometry of Solved Problem 3.6.

Problem 3.7 Two conductors are immersed in a material of conductivity σ = 10 −4 Ω m and permittivity ε = 80 ε 0 . If the resistance between the conductors is 10 5 Ω , find an expression for the capacitance between the conductors and compute its value. Solution Let the potential difference between the conductors be V , then the resistance is A

E • dl 1 ∫B 1 V ε V εV R= = = = σ ∫ E • dS σ ES σ ε ES σ Q s

The capacitor can be obtained from the above result as C=

Q V

ε Rσ 80 × 8.85 × 10 −12 = = 70.8 ρ F 10 5 × 10 − 4 =

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Solved Problem 3.8 The half of the space between two conducting concentric spheres radii is filled with a dielectric of permittivity ε1 and the other half with a different dielectric of permittivity ε 2 is shown in Fig. (3.24). If the total charge on the inner and outer spheres are + Q and − Q respectively, find the electric field in each dielectric and the capacitance between the spheres. z S a

b

ε2

r

V

a

ε1

y

+Q

at

an

E1

E2

x

-Q

Fig. (3.24). Geometry of Solved Problem 3.8.

Solution Referring to Fig. (3.24), and assuming the center of the spheres at the origin of the coordinate system, then both E1 and E2 are in the radial direction or

E1 = ar E1 ,

E2 = ar E2

(a) Applying the boundary conditions on the interface between the two dielectrics, then

an ⋅ (ε 2 E2 − ε1 E1 ) = 0 ,

at ⋅ (E2 − E1 ) = 0

Since a n and a r are perpendicular, and a r are parallel, then E2 = E1 . From Gauss’s law, we have

∫D S

•

dS = Q = ε1 E1S + ε 2 E2 S = (ε1 + ε 2 )E1S

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Then the electric fields in the dielectrics at a distance r from the center are

E1 = E2 =

Q Q ar = ar S (ε1 + ε 2 ) 2π (ε1 + ε 2 ) r 2

(b) The flux density in the media are

D1 = ε1 E1 =

ε1 Q ar , 2π (ε1 + ε 2 ) r 2

D2 = ε 2 E2 =

ε2 Q ar 2π (ε1 + ε 2 ) r 2

The capacitance between the spheres is

C=

Q = VAB

Q

=

A

Q b

1 Q ∫B E • dl 2π (ε1 + ε 2 ) ∫a r 2 dr ab ⇒ C = 2π (ε1 + ε 2 ) (b − a)

Solved Problem 3.9 The inner and outer radii of two concentric spheres are a and b respectively. The space between the spheres is filled with a dielectric of permittivity ε and conductivity σ as shown in Fig. (3.25). At the instant t = 0 a charge qο is placed on the outer sphere. Find the (a) Electric current in the dielectric (b) Dissipated energy in the dielectric and show that it is equivalent to the decrease in the electrostatic energy. Solution At t = 0 , the electric field in the dielectric is given by

E=

qο ar 4π ε r 2

At any subsequent instant t , the charge is q (t ) and E becomes

E(t ) =

q(t ) ar 4π ε r 2

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(a) The current through the surface S in Fig. (3.25) is

I = ∫ J • dS = − S

∂q ∂t

Using J = σ E , we have

σq σ ∫ E • dS = 4π ε S

2π π

∂q

∫ ∫ sin θ dθ dφ = − ∂t 0 0

Which simplifies to qο

σq ∂q =− ε ∂t

⇒

t

1 σ ∫0 q dq = −∫0 ε dt

⇒

q(t ) = qοe

z b E

r

S

q (t)

y

a ε, σ

x Fig. (3.25). Geometry of Solved Problem 3.9.

The current through S becomes

∂q σ qο − ε t I =− = e ε ∂t σ

σ − t ε

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(b) The dissipated energy through S is 2

⎡ qο ⎤ 1 − 2εσ t W (t ) = σ E = σ ⎢ ⎥ 4e ⎣ 4π ε ⎦ r 2

The total dissipated energy in the region a ≤ r ≤ b during the time 0 ≤ t ≤ ∞ is ∞

2

⎡ q ⎤ 1 − 2σt WT = ∫ dt ∫ σ ⎢ ο ⎥ 4 e ε 4π r 2 dr 4π ε ⎦ r 0 a ⎣ b

2

⇒

b

2

1 q ⎡1 1⎤ q WT = ο ∫ 2 dr = ο ⎢ − ⎥ 8π ε a r 8π ε ⎣ a b ⎦

The electrostatic energy at the beginning of the charge flow at t = 0 is obtained as a

WE =

2

1 1 q qοV = qο ∫ E • dl = − ο 2 2 b 8π ε

a

1

∫r

2

ar • ar dr

b

2

⇒

WE =

qο ⎡ 1 1 ⎤ − 8π ε ⎢⎣ a b ⎥⎦

Which is equal to dissipated energy, or WT = WE . Solved Problem 3.10 One plate of a parallel-plates capacitor is placed on the plane x = 0 and charged with a charge + Q , while the other is at x = d and charged with a charge − Q . The permittivity between the plates of the capacitor varies with x as ε = ε 0 (a + x d ). If the area of each plate is S and a > 1 is a constant, find the (a) Capacitance of the capacitor. (b) Polarization charges at each plate. (c) Polarization volume charge density in the medium between the plates.

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Solution Using (3.111), and assuming the charge density on the plates is ρ s , the electric field between the two plates is

E = ax

ρs ρs ρ d = ax = ax s ε ε 0 (a + x d ) ε 0 (ad + x )

(a) The capacitance can be obtained from (3.115) as

ρs a x • a x dS x=d ε ε0 S S S C= A = d = d ln(ad + x ) x = 0 d ρs E a • dl • a dx x x ∫B ε 0 ∫0 (ad + x ) ε0 S ⇒ C= d ln(1 + 1/ a ) ε ∫ E • dS

ε∫

(b) Using (3.65), the polarization P is related to E by

d ⎞ ⎛ ⇒ P = (ε − ε 0 )E = a x ρ s ⎜1 − ⎟ ⎝ ad + x ⎠ From (3.61) the surface polarization charge (bound charge) is

D = ε E = ε0 E + P

d ⎞ d ⎞ ⎛ ⎛ Qb = − ∫∫ P ⋅ dS = − ∫∫ ρ s ⎜1 − ⎟ a x ⋅ a x dS = − ρ s ⎜1 − ⎟S ⎝ ad + x ⎠ ⎝ ad + x ⎠ S S On the plate at x = 0 , the charge is + Q , ρ s = +QS ,

d ⎞ ⎛ Qb x = 0 = − ρ s S ⎜1 − ⎟ ⎝ ad + x ⎠

x =0

⎛ a −1⎞ ⎛ 1⎞ = − ρ s S ⎜1 − ⎟ = − ⎜ ⎟Q ⎝ a ⎠ ⎝ a⎠

On the plate at x = d , the charge is − Q , and from (3.57), ρ s = Pn = −QS , then

d ⎞ ⎛ Qb x = d = − ρ s S ⎜1 − ⎟ ⎝ ad + x ⎠

x=d

⎛ a ⎞ ⎛ a ⎞ = − ρs S ⎜ ⎟Q ⎟ = +⎜ ⎝ a + 1⎠ ⎝ a +1⎠

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(c) The polarization charge density in the medium between the plates can be obtained using (3.58) as

ρvb = −∇ ⋅ P = −

−1 Q d ⎞⎤ ∂ ⎡ ⎛ . ρ s ⎜1 − ⎟⎥ = ⎢ ∂x ⎣ ⎝ ad + x ⎠⎦ (ad + x )2 S

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PROBLEMS 3.1.

The current density through a square cross section wire of side a = 2 mm with its axis along the z-axis, is given by

⎛ x2 + y 2 ⎞ ⎟⎟ A/m2 J = a z 600 ⎜⎜1 + 4 ⎠ ⎝ Find the total current through the wire. 3.2.

The current density through a circular cross section wire of radius a and along the z-axis is given by

J = az

2

π

(1 − e ) A/m −ρ a

2

If a = 2.5 mm , find the total current through the wire. 3.3.

The current density passing across the spherical surface 0 ≤ r ≤ 2 m , 0 ≤ φ ≤ 2π , − π 4 ≤ θ ≤ + π 4, is given by

J = 3 cosφ cosθ aθ + 2 sin θ aφ A/m2 Find the current passing across the surface 3.4.

The conductivity of a circular cross section wire of radius a and length l varies along its length as

σ=

1 Sm σ 0 (1 + x l )

where σ 0 is a constant. Find the resistance of the wire. 3.5.

A conductor circular cross section is constructed of aluminum in the region 0 ≤ ρ < 1.5 mm , and copper in the region 1.5 ≤ ρ < 2 mm . If the current in the conductor is 10 A , find in the aluminum and copper, the

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(a) Current densities. (b) Electric fields. 3.6.

A disk of thickness c and radius b has a central hole of radius a , if the conductivity of the disk is σ , show that the resistance between the hole and the rim of the disk is given by

R= 3.7.

Show that the resistance of truncated cone shown in Fig. (3.26) between the faces A and B is given by

1 R= σπ 3.8.

ln(b a ) 2π σ c

(b − a) 2 + c 2 ab

A small spherical charged body of the total charge − 1.9 n C is located at the center of the spherical dielectric body of the radius 10 μ m and relative permittivity ε r = 3. Find the (a) The electric field and polarization vectors at all points. (b) The potential at all points. (c) The surface polarization charge densities.

3.9.

A dielectric disk of radius a and thickness t is placed with its axis along z-axis and centered at the origin. The disk is polarized with a dipole moment per unit volume P = a z P . Find the surface polarization charge density and the electric field intensity along the disk axis.

3.10. In Fig. (3.27), σ1 = σ2 = 0 . The electric field in medium 1 is given by

E1 = 30a x − 20a y − 40a z V m Find the (a) Electric field in medium 2. (b) Angle that the electric field in medium 2 makes with normal. (c) Energy density in each region.

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3.11. In Fig. (3.27), let ε1 = 2.56 ε 0 , ε 2 = 5.8 ε 0 , σ1 = 5 × 10 −66 S m , and

σ2 = 3 × 10−36 S m . The electric field in medium 1 is E1 = 200a x + 150a y − 80a z V m Find the (a) (b) (c) (d)

Electric field in medium 2. The current density in each medium. Surface charge density on the boundary between the media. Dissipated energy in each medium.

3.12. A parallel plates capacitor has a capacitance 15μ F and plate separation 2 mm . The charge density on the plates is 90 μ C . Find the area of the plates and the electric field between them. 2b B

c

zz

Medium 1 ε 1, σ 1

σ σ2

A

Medium 2 ε 2, σ 2

x

2a Fig. (3.26). Problem 3.7.

Fig. (3.27). Problems 3.10 and 3.11.

3.13. Find the capacitance of the spherical capacitor of two conducting concentric spheres of radii a and b , assuming the space between them is filled with a dielectric of permittivity ε . If the dielectric has a small conductivity σ , determine the conductance of the capacitor. 3.14. The space between the plates of parallel-plates capacitor is filled with three dielectrics as shown in Fig. (3.28). Find the (a) (b) (c) (d)

Electric field in each dielectric. Total charge. Energy stored in each dielectric. Capacitance of the capacitor.

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Assume that the area of the plates is much large compared to the separation between them. 3.15. The space between the plates of parallel-plates capacitor shown in Fig. (3.29) is filled with two dielectrics of conductivities σ1 = 0 and σ2 = 0 permittivity ε1 = 2 ε 0 and ε 2 = 3 ε 0 . If V0 = 50 V , S1 = 4S2 = 80 cm 2 and d = 4 mm . Find the (e) (f) (g) (h)

Electric field in each dielectric. Total charge. Energy stored in each dielectric. Capacitance of the capacitor.

3.16. The surface area of each plate of a parallel-plates capacitor is S = 200 cm 2 , and the distance between the plates is d = 3 mm . When a voltage source of 200 V is applied between the plates, the capacitor charges up, and with a charge of the amount 52 n C . Find the relative permittivity of the dielectric between the plates. 3.17. The space between the plates of parallel-plates capacitor shown in Fig. (3.29) is filled with two dielectrics of finite conductivities σ1 and σ 2 and permittivity ε1 and ε 2 respectively. Find an expression for the (a) Electric field in each dielectric. (b) Conduction current through the capacitor. (c) Capacitance of the capacitor. V = Vο

V = Vο S d1

ε1

d2

ε2

d3

ε3

S2

S1

d

ε1

ε2

σ1

σ2 V=0

V=0

Fig. (3.28). Problems 3.14, 3.18, and 3.19.

Fig. (3.29). Problems 3.15 and 3.17.

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3.18. The space between the plates of parallel-plates capacitor is filled with three dielectrics as shown in Fig. (3.28). Find an expression for the (a) (b) (c) (d)

Electric field in each dielectric. Total charge. Energy stored in each dielectric. Capacitance of the capacitor.

Assume that the area of the plates is much large compared to the separation between them. 3.19. In the capacitor shown in Fig. (3.28), if V0 = 100 V , ε1 = ε 2 2 = ε 3 = ε 0 , S1 = 100 cm 2 and d1 = d 2 4 = d3 = 1 mm . Find the (a) (b) (c) (d) (e)

Electric field in each dielectric. The polarization vector in each dielectric. Free charge and polarization charges on all boundary surfaces. Energy stored in each dielectric. Capacitance of the capacitor.

Assume that the area of the plates is much large compared to the separation between them. 3.20. One plate of parallel plates capacitor is placed on the plane x = 0 and charged with a charge + Q while the other is at x = d and charged with a charge − Q . The permittivity between the plates of a parallel plates capacitor varies as ε = ε 0 (1 + x d ) . If the area of each plate is S , find the 2

(a) Capacitance of the capacitor. (b) Polarization charges at each plate. (c) Polarization charge density in the medium between the plates. 3.21. The space between the inner and outer conductor of a coaxial cable is filled with two dielectrics as shown in Fig. (3.30). The potential difference between the inner and outer conductors is V . If σ1 = σ2 = 0 , find the (a) (b) (c) (d)

Electric field in each dielectric. Total charge on the inner and outer conductors. Energy stored in each dielectric. Capacitance per unit length of the coaxial cable.

Conducting and Dielectric Materials

Electromagnetics for Engineering Students

a ε2 σ2

ε1

c

b

V

ε1 σ1

Fig. (3.30). Problems 3.21, 3.23, and 3.24.

σ1

a ε2 V σ2

219

b

Fig. (3.31). Problems 3.22, and 3.25.

3.22. The space between the inner and outer conductor of a coaxial cable is filled with two dielectrics as shown in Fig. (3.31). The potential difference between the inner and outer conductors is V . If σ1 = σ2 = 0 , find the (a) (b) (c) (d)

Electric field in each dielectric. Total charge on the inner and outer conductors. Energy stored in each dielectric. Capacitance per unit length of the coaxial cable.

3.23. In the coaxial cable of Problem 3.22, if 3 ε1 = 2 ε 2 = 7 ε 0 , a = 0.5 mm , b = 3 mm and c = 4 mm , find the capacitance of the coaxial cable per unit length. 3.24. For the coaxial of Problem 3.21, if σ1 ≠ 0 and σ2 ≠ 0 , find the (a) (b) (c) (d) 3.25.

Current density in each dielectric. Total charge on the boundary between the dielectrics. Dissipated energy in each dielectric. Conductance per unit length of the coaxial cable.

For the coaxial of Problem 3.22, if σ1 ≠ 0 and σ2 ≠ 0 , find the (a) (b) (c) (d)

Current density in each dielectric. Total charge on the boundary between the dielectrics. Dissipated energy in each dielectric. Conductance per unit length of the coaxial cable.

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3.26. A spherical capacitor of an inner sphere of the radius a and an outer sphere of radius b . The space between the spheres is filled with a lossy dielectric of permittivity ε and conductivity σ . Find the conductance of the capacitor. 3.27. Show that the capacitance of the spherical capacitor shown in Fig. (3.32) is given by

C=

4π ε1ε 2ε 3 abcd ε 2ε 3 bd (c − a ) + ε1ε 3 ad (b − c ) + ε1ε 2 ac (d − b )

ε2

c

ε1

σ1

a

a ε3

ε1

c

b V

ε2 V σ2

b

d Fig. (3.32). Problems 3.27.

Fig. (3.33). Problems 3.28.

3.28. In the spherical capacitor shown in Fig. (3.33), if ε1 = 2.5 ε 2 = 2.5 ε 0 ,

σ1 = 2 σ 2 = 10 −6 S m , a = 0.5 mm , b = 2 mm and c = 3 mm , find the capacitance of the capacitor. 3.29. A cylindrical capacitor with inner radius a and outer radius b , is filled with an inhomogeneous dielectric having ε = b ε 0 ρ . Determine the capacitance per unit length of the capacitor. 3.30. Show that the half of the energy inside a coaxial cable with a homogeneous dielectric, of inner conductor radius a and outer conductor radius b , is contained within a cylinder a < ρ < ab .




Electromagnetics for Engineering Students, 2017, 221-303

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CHAPTER 4

Electrostatic Boundary-Value Problems Abstract: This chapter deals with more advanced techniques than those discussed in Chapter 2 to solve electrostatic problems. The techniques presented in the chapter include the method of images, the multi-pole expansion, analytical methods, and some numerical methods, which may be invoked as an alternative techniques or when the analytical solution for a particular problem is not available. The method of images and multi-pole expansion techniques are applied to analyze the problems involving charges near a perfectly electric conducting or dielectric interfaces. For the media that are subject to certain boundary conditions, the electrostatic problem is modeled by Poisson's or Laplace's equation then solved analytically or numerically. Analytical solutions for the Laplace's equation in different coordinate systems are presented in details. The method of moments and finite difference method are presented as examples of numerical techniques for the solution of Poisson's or Laplace's equation. The discussions of the topics in the chapter are supported by illustrative examples, figures, solved problems, and computer programs in addition to homework problems at the end of the chapter.

Keywords: Boundary conditions, Dirichlet conditions, finite difference method, electrostatic potential, Laplace's equation, method of images, method of moments, multi-pole expansion, Neumann conditions, Poisson's equation, uniqueness. INTRODUCTION The problem of electrostatic fields and potentials due to point charges, the discrete system of charges, and continuous distribution of charges in homogeneous dielectrics is discussed in Chapter 2. When the medium is not homogeneous or there is a conducting or dielectric object near the charges or the medium is a charge-free and subject to certain boundary conditions, more advanced concepts and techniques are needed to solve the electrostatic problem. In this chapter, some of these techniques such as the method of images, the multi-pole expansion [59, 60], in addition to the analytical and numerical method will be presented. The method of images and multi-pole expansion techniques are suitable for the problems involving charges near a perfectly electric conducting or dielectric interfaces. Analytical solutions are applied for the problems of charge-free media and subject to certain boundary conditions. Numerical methods such as available in [61-69] are invoked as an alternative techniques or when the analytical solution fails. The Moment Method [61, 62] and the Finite Difference Method [63, 64] are applied to solve electrostatic problems as examples of numerical techniques. Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

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4.1.

THE METHOD OF IMAGES

Sameir M. Ali Hamed


In the method of images, the original problem to be solved is replaced by an equivalent one that gives the same solution [52]. The method of images is suitable for the electrostatic problems involve charges near perfect electric conducting or dielectric objects as will be shown in the following analysis. 4.1.1. A Point Charge above a Conducting Ground Plane The concept of the method of images can be illustrated by considering a point charge + q at height d above an infinitely extended perfect electric conducting plane as shown in Fig. (4.1a). Assuming the surface of the conducting plane coincides with the z = 0 plane, and the point charge is at ( x1, y1, z1 = d ) . Based on the properties of the electric field and electric potential in a conductor, we have: 1. 2. 3. 4.

The electric field E on the surface of the plane satisfies that E = 0 . The potential V on the surface of the conducting grounded plane is V = 0 . A charge − q induces on the plane due to the presence of the charge + q . The lines of force of the electric fields start from the positive charge + q and end at the negative charges on the surface of the plane as shown in Fig. (4.1a). +q

+q

d -

-

-

-

-

-

-

Perfect electric conducting plane -q

(a)

(b)

Fig. (4.1). A point charge in front of a conducting grounded plane. (a) Original problem. (b) Equivalent problem.

The problem of a point charge + q at a height d above the plane shown in Fig. (4.1a) can be replaced by an equivalent problem shown in Fig. (4.1b), which gives a solution identical to that can be obtained from the original problem. The new problem is obtained by replacing the conducting plane by an image point charge q′ located at a point (x2 , y2 , − z2 ), such that it gives the same solution as

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the original one. The boundary conditions can be enforced to determine q′ and its location. The total electric field at any point (x, y, z ) is the vector sum of the field E1 = a1E1 due to the charge + q and the field E2 = a2 E2 due to the image charge q′ . The total electric field E at a point (x0 , y0 , 0) on the surface of the plane is E=

1 ⎡ q(r1 − r0 ) q′(r2 − r0 ) ⎤ + ⎢ 3 3 ⎥ 4π ε ⎢⎣ r1 − r0 2 r2 − r0 2 ⎥⎦

(4.1)

The electrostatic potential on the conducting surface is

V=

1 ⎛ q q′ ⎜⎜ + 4π ε ⎝ r1 − r0 r2 − r0

⎞ ⎟⎟ ⎠

(4.2)

where r1 = a x ( x1 − x0 ) + a y ( y1 − y0 ) + a z d , r2 = a x ( x2 − x0 ) + a y ( y2 − y0 ) − a z z 2 and

r0 = a x x0 + a y y0 . Using E1 = a1E1 and E2 = a2 E2 , (4.1) can be written as

E=

⎞ 1 ⎛⎜ q q′ ⎟ + a a 1 2 2 2 ⎟ 4π ε ⎝⎜ r1 − r0 r2 − r0 ⎠

(4.3)

The boundary conditions are that the tangential electric field Et = 0 and V = 0 on the surface of the plane. Substituting Et = 0 and V = 0 at z = 0 in (4.2) and (4.3), yields

q − q′ = 2 2 r1 − r0 r2 − r0

(4.4)

q − q′ = r1 − r0 r2 − r0

(4.5)

Dividing both sides of (4.4) by the corresponding sides of (4.5), then r1 − r0 = r2 − r0 , which implies that from (4.5) q′ = − q , and x2 = x1, y2 = y1 and z2 = − z1 = −d . Consequently, the equivalent problem is to replace the

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conducting plane by a charge q′ = − q at (x1 , y1 , − d ) as shown in Fig. (4.1b). The equivalent problem can be used to obtain the electrostatic potential V and electric field E above the plane or in the region z > 0 . Let the charge be on the z-axis at a height d above the conducting plane, then x2 = x1 = 0 , y2 = y1 = 0 , and z2 = − z1 = −d . Thus, E and V are obtained respectively as

q ⎧⎪ a x x + a y y + a z ( z − d ) a x x + a y y + a z ( z + d )⎫⎪ − ⎨ 3 3 ⎬, 2 2 2 2 4π ε ⎪ x 2 + y 2 + ( z − d )2 2 x + y + (z + d ) ⎪⎭ ⎩ ⎫ q ⎧ 1 1 V= − ⎨ 2 ⎬, 2 2 4π ε ⎩ x + y 2 + (z − d ) x 2 + y 2 + (z + d ) ⎭ E=

[

]

[

]

z≥0

(4.6)

z≥0

(4.7)

4.1.2. A Point Charge in Front of a Conducting Grounded Sphere Consider a point charge + q in front of a conducting grounded sphere of radius a and at a distance f from its center. Let the center of the sphere be at the origin of the coordinates and the charge + q at A (0, 0, f ) as shown in Fig. (4.2). It is required to find the electrostatic field and potential due to the charge and the sphere. The problem can be simplified by using the method of images. In this case, the sphere is replaced by an image charge q′ at some point A′ , a distance zo from the center of the sphere. Since the sphere is grounded, then the potential on its surface is zero. Setting the potential at any point P on the surface of the sphere to zero, we get

V= where r1 =

1 ⎛ q q′ ⎞ ⎜ + ⎟=0 4π ε ⎝⎜ r1 r2 ⎟⎠

(4.8)

f 2 + a 2 − 2af cosθ and r2 = zo 2 + a 2 − 2azo cosθ . Hence, r2 = r1

zo + a 2 − 2 zo a cosθ − q′ = q f 2 + a 2 − 2 fa cosθ 2

(4.9)

Equation (4.9) holds for 0 ≤ θ ≤ π on the surface of the sphere r = a . At P′ , θ = 0 and (4.9) reduces to

(a − z o ) ( f − a ) = − q′ q

(4.10)

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y P r1

a r2 θ q′ O

P′′

A′

q P′

z

A

zo f

(a) P (r, θ)

y

r

θ O

a2/f

q

q A′ ′

A

z

f

(b) Fig. (4.2). Point charge in front of a conducting grounded sphere. (a) original problem. (b) The equivalent problem.

Similarly, at P′′ , θ = π , then

(a + z o ) ( f + a ) = − q′ q

(4.11)

From (4.10) and (4.11), it can be shown that the location of the image q ′ is

zo = a 2 f

(4.12)

Substituting zo from (4.12) into (4.11), the charge of the image is obtained as

q ′ = − aq f

(4.13)

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The original problem is equivalent to a point charge q at A (0, 0, f ) and a point charge q ′ = − aq f at A′ (0, 0, a 2 f ). Hence, the potential at any point (r,θ ) is

q ⎡ 1 1 ⎢ − V ( r ,θ ) = 2 4π ε ⎢ r 2 + f 2 − 2 rf cosθ a 2 + ( r a ) f 2 − 2 rf cosθ ⎣

⎤ ⎥ (4.14) ⎥⎦

The electric field can be obtained using E = −∇V . 4.1.3. A Point Charge in Front of a Conducting Isolated Sphere Consider a point charge + q in front of a conducting isolated sphere of radius a and at a distance f from its center as shown in Fig. (4.3). It is required to find the electrostatic field and potential due to the charge and sphere. Since the sphere is conducting and isolated, then the sphere is equipotential surface and the total charge inside the sphere is zero. y P r

a r′ θ q′ O

P′′

A′

q A

P′

z

z f (a) P (r, θ)

y r q′′ O

θ a2/f

q

q′ A′

A

z

f (b)

Fig. (4.3). A charge in front of a conducting sphere. (a) Original problem. (b) The equivalent problem.

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The charge + q will induce an image charge q′ = −(a f ) q inside the sphere at a distance x from its center. Since the total charge inside the sphere is zero, then there must be a second image of charge q′′ such that q′ + q′′ = 0 , or q′′ = +(a f )q. The charge q′′ is located at the center of the sphere. Therefore, the problem of the charge + q in front of a conducting isolated sphere can be replaced by a charge + q at a point A ( 0 , 0 , f ) , charge q ′ = − ( a f ) q at A ′ (0, 0, a 2 f ) , and charge q ′′ = + ( a f ) q at the origin O as shown in Fig. (4.3b). The electrostatic potential at any point P ( x , y , z ) of the original problem in Fig. (4.3a), is equivalent to the resultant potential of the point charges + q , q ′ , and q ′′ at P ; thus ⎡ ⎢ q ⎢ a f a f + − V ( x, y , z ) = a x ⎢ 2 2 2 2 4π ε ⎢ x + y + z ⎛ a2 ⎞ ⎜⎜ z − ⎟⎟ + y 2 + x 2 ⎢ f ⎠ ⎝ ⎣

⎤ ⎥ ⎥ 1 ⎥ (z − f )2 + y 2 + x 2 ⎥ ⎥ ⎦

(4.15) 4.1.4. A Conducting Sphere in a Uniform Electric Field In the problem of the isolated conducting sphere, in the previous section, if the point charge + q is moved to a point far away from the sphere such that f >> a , then the field from the charge + q can be considered as a plane wave of a uniform field E 0 in the vicinity of the sphere. Furthermore, the distance a 2 f between the image charges q′ = −(a f ) q and q′′ = +(a f ) q becomes very small. Under this situation q′ and q′′ represents an electric dipole with a dipole moment

(a

f 2 )q . The electric field expressions for this dipole can easily be obtained from (2.102) by replacing the dipole moment by (a 3 f 2 )q . Hence, the electric field in the polar coordinates at any point (r,θ ) due to the charges q′ and q′′ becomes 3

Ed

≈ E q ′ + E q ′′ q a3 1 (a r 2 cosθ + aθ sin θ ) ≈ 4π ε f 2 r 3

(4.16)

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The total field E near the sphere is the sum of the uniform field E0 and the field E d due to the electric dipole, then

E ≈ E0 + Ed = E0 +

q a3 1 (ar 2 cosθ + aθ sin θ ) 4π ε f 2 r 3

(4.17)

the uniform field E0 at any point (r,θ ) can be written as

E0 = E0 (a r cosθ − aθ sin θ )

(4.18)

From (4.17) and (4.18) the total field becomes

⎛ q a3 1 + E0 E ≈ a r cosθ ⎜⎜ 2 3 2 ε π f r ⎝

⎞ ⎛ q a3 1 ⎟⎟ − a θ sin θ ⎜⎜ − E0 2 3 4 ε π f r ⎠ ⎝

⎞ ⎟⎟ ⎠

(4.19)

On the surface of the sphere r = a , the tangential electric field Eθ = 0 ; hence

E0 =

q

1 4π ε f 2

(4.20)

Using (4.19) and (4.20), the electric field of a conducting sphere in a uniform electric field can be written in terms of E0 as 3 3 E ≈ ar E0 cosθ ⎡2 (a r ) + 1⎤ − aθ E0 sin θ ⎡(a r ) − 1⎤ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦

(4.21)

On the surface of the sphere r = a , the tangential components vanishes and the radial component reduces to Er ≈ 3E0 cosθ . When θ = π 2 , then E = 0 , and the plane θ = π 2 is called the plane of charge neutralization. At this plane, the charge changes sign from positive to negative as illustrated in Fig. (4.4). Assuming V0 is a reference potential, the electrostatic potential of a conducting sphere in a uniform electric field can be obtained easily from (4.21) as

V = − ∫ E ⋅ a r dr

[

]

[

]

= − E0 cosθ ∫ 2(a r ) + 1 dr = (a r ) − 1 rE0 cosθ + V0 3

3

(4.22)

Electrostatic Boundary-Value Problems


Electromagnetics for Engineering Students

θ = π/2

-

E

- + + -

-

-

+

229

+ + + + + +

θ


z

Fig. (4.4). A conducting sphere in a uniform electric field.

4.1.5. A Point Charge in Front of a Dielectric Planar Interface Assume that a dielectric of relative permittivity ε1 fills a half of the space and the second half is filled with a different dielectric of relative permittivity ε 2 . Let the dielectric of relative permittivity ε1 be referred to as medium 1 and the second dielectric as medium 2, and the boundary between the dielectrics is coincided with the plane z = 0 . Consider a point charge + q placed at the point P (x1 , y1 , d ) in medium 1 as shown in Fig. (4.5). The problem is to find the electric field and potential in both dielectrics using the method of images. The boundary conditions can be imposed to determine the induced charge images and their locations for both dielectrics. The boundary conditions are 1. The electric flux density at the interface must be continuous,LH the normal components of the flux Dn1 in medium 1 and Dn 2 in medium 2, satisfy Dn1 = Dn 2 at z = 0 . 2. The electrostatic potential at the boundary is continuous. There are two different equivalent problems that may replace the original problem depending on the location of the observation point. q

z Dn1 Dn2

Medium 1 d

B1 B2 Medium 2

Fig. (4.5). A point charge in front of the interface between two dielectrics.

y

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4.1.5.1. Equivalent Problem for Medium 1 (Field Point in the Region z > 0 ) In this case, medium 2 can be replaced by a dielectric of relative permittivity ε1 and image of charge q′ as shown in Fig. (4.6). Therefore, the problem reduces to two charges + q at the point P (x1 , y1 , d ) and q′ at P′ (x1 , y1 , − d ) in a homogeneous dielectric of relative permittivity ε1 . The electric flux D q in medium 1 at any point (x, y, z ) due to the charge + q can be obtained as

q a x ( x − x1 ) + a y ( y − y1 ) + a z ( z − d ) 3 4π ( x − x )2 + ( y − y )2 + ( z − d )2 2 1 1

Dq = ε 0ε1 E =

[

]

(4.23)

On the interface z = 0 , the electric flux due to the point charge + q is

Dq

z =0

=−

q a x (x − x1 ) + a y ( y − y1 ) + a z d 3 4π (x − x )2 + ( y − y )2 + d 2 2 1 1

[

]

(4.24)

The normal component of the flux density in medium 1 due to q using (4.24) is Dnq = −

q 4π

[(x − x )

d

2

1

+ ( y − y1 ) + d 2

q

z

(4.25)

]

3 2

2

Medium 1

Dn1

d

D′n1

d

ε1

y

q′ Fig. (4.6). The equivalent problem in medium 1 of the point charge of Fig. (4.5).

Similarly, the normal component of the flux density in medium 1 due to q′ is ′ = Dnq

q′ 4π

[(x − x )

2

1

d + ( y − y1 ) + d 2

2

]

3 2

(4.26)

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Adding (4.25) and (4.26), the total normal flux Dn1 in medium 1 at the interface becomes Dn1 (z = 0) =

(q′ − q ) 4π

d

[(x − x )

+ ( y − y1 ) + d 2

2

2

1

]

3 2

(4.27)

Similarly, the electrostatic potential V1 due to the charge + q and its image of charge q ′ at the interface (z = 0) can be obtained as

V1 (z = 0) =

(q′ + q ) 4π ε 0ε1

1

(x − x1 ) + ( y − y1 )2 + d 2 2

(4.28)

4.1.5.2. Equivalent Problem for Medium 2 (Field Point in the Region z < 0 ) In this case, medium 1 can be replaced by a dielectric of relative permittivity ε 2 with a charge q′′ . Therefore, the problem reduces to a charge q′′ at the point P (x1, y1, d ) in a homogeneous dielectric of relative permittivity ε 2 as shown in Fig. (4.7). Following the same analysis as in the case of medium 1, the total flux Dn 2 due to q′′ in medium 2 at normal to the interface (z = 0) can be obtained Dn 2 (z = 0) = −

q′′ 4π

[(x − x )

d

2

1

+ ( y − y1 ) + d 2 2

]

3 2

(4.29)

The electrostatic potential V2 at the interface is obtained as

V2 =

q′′ 4π ε 0ε 2

1

(4.30)

(x − x1 ) + ( y − y1 )2 + d 2 2

q′

z

d Dnq′

ε2 Medium 2

Fig. (4.7). The equivalent problem in medium 2 of the point charge of Fig. (4.5).

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Enforcing the boundary conditions at z = 0 , Dnq = Dnq′ , and V1 = V2 then q′ − q = −q′′ and (q′ + q) ε1 = q′′ ε 2 . Solving for q′ and q′′ we obtain

⎛ ε −ε ⎞ q′ = ⎜ 1 2 ⎟q , ⎝ ε1 + ε 2 ⎠

⎛ 2ε1ε 2 ⎞ q ′′ = ⎜ ⎟q ⎝ ε1 + ε 2 ⎠

(4.31)

The total electric field at any point (x, y, z ) in the region z > 0 is the superposition of the fields from q and q′ , then E=

q a x ( x − x1 ) + a y ( y − y1 ) + a z ( z − d ) 3 4π ε 0ε1 ( x − x )2 + ( y − y )2 + ( z − d )2 2 1 1

[

]

q ⎛ ε1 − ε 2 ⎞ a x ( x − x1 ) + a y ( y − y1 ) + a z ( z + d ) + ⎜ ⎟ 3 4π ε 0ε1 ⎜⎝ ε1 + ε 2 ⎟⎠ ( x − x )2 + ( y − y )2 + ( z + d )2 2

[

1

1

,

z≥0

(4.32)

]

The potential at any point (x, y, z ) is V=

q 4π ε 0ε1

1

(x − x1 ) + ( y − y1 )2 + (z − d )2 2

q ⎛ ε1 − ε 2 ⎞ ⎜ ⎟ + 4π ε 0ε1 ⎜⎝ ε1 + ε 2 ⎟⎠

,

z≥0

(4.33)

1

(x − x1 )2 + ( y − y1 )2 + (z + d )2

The electrostatic field E and potential V at any point (x, y, z ) in the region z < 0 are due to the charge q′′ and can be obtained, respectively as

E=

q ⎛ ε1 ⎞ a x (x − x1 ) + a y ( y − y1 ) + a z (z − d ) ⎜ ⎟ z≤0 3 , 2π ε 0 ⎜⎝ ε1 + ε 2 ⎟⎠ (x − x )2 + ( y − y )2 + (z − d )2 2

[

1

1

]

(4.34)

and

V=

q ⎛ ε1 ⎞ ⎜ ⎟ 2π ε 0 ⎜⎝ ε1 + ε 2 ⎟⎠

1

(x − x1 ) + ( y − y1 )2 + (z − d )2 2

, z≤0

(4.35)

The lines of force for the electric field intensity are shown in Fig. (4.8) for ε1 < ε 2 .

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+q Medium 1 ε1 ε2 Medium 2 Fig. (4.8). Lines of electric field for a point charge near the interface between two dielectrics of dielectric constants ε1 and ε2, where ε1 < ε 2 .

4.2. MULT-POLE EXPANSION Electrostatic potential due to a point charge or system of charges can be expressed in terms of infinite series that is known as a multi-pole expansion. To drive a general formula, assume a point charge q at a point (r′,θ ′φ′) as shown in Fig. (4.9), then the potential at any observation point (r,θ ,φ ) is

V=

q 4π ε R

(4.36)

Where R = r − r′ is the distance between the source and field points. In the spherical coordinate system

r = a x r sin θ cosφ + a y r sin θ sin φ + a z r cosθ r ′ = a x r ′ sin θ ′ cosφ ′ + a y r ′ sin θ ′ sin φ ′ + a z r ′ cosθ ′

(4.37b)

R = r − r′ = r 2 + r′2 − 2rr′[sin θ sin θ ′ cos(φ − φ ′) + cosθ cosθ ′]

(4.38)

z

Source point ( r′, θ′′, φ′) q

R

r′

Field point (r, θ, φ)

r x

y

Fig. (4.9). Coordinates of the field point and source point of a point charge.

(4.37a)

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1 R can be written as

1 +∞ = ∑ Ψn e − jn(φ −φ ′ ) R n=−∞

(4.39)

⎧ 1 ⎛ r′ ⎞k ⎪ ⎜ ⎟ , r > r′ +∞ ( k − n )! n ⎪r r n Ψn = ∑ Pk (cosθ )Pk (cosθ ′)⎨ ⎝ ⎠ k k = n (k + n )! ⎪ 1 ⎛⎜ r ⎞⎟ , r < r ′ ⎪⎩ r ′ ⎝ r ′ ⎠

(4.40)

where

Pkn (⋅) is associated Legendre functions of degree k and order n. Assuming the charge is at a point along the z-axis, then the source point is (r′ = z′,θ ′ = 0,φ′). Substituting r ′ = z′ and θ ′ = 0 in (4.40), taking into account that

⎧1, n = 0 Pkn (1) = ⎨ ⎩0, n ≠ 0

(4.41)

⎧(z′ r ) , r > z′ 1 1 +∞ = ∑ Pk (cosθ ) ⎨ k +1 R r k =0 ⎩(r z′) , r < r ′

(4.42)

Then (4.39) reduces to k

From (4.36) and (4.42) the potential due to the charge q at any point on the surface of a sphere of radius r can be expressed in the general form

V (r ,θ ) =

⎧ ( z′ r ) , r > z′ q +∞ Pk (cosθ )⎨ ∑ k +1 4π ε r k = 0 ⎩(r z′) , r < r′ k

(4.43)

The solution in the regions r > z′ represents the case when q is inside the sphere while the solution in the region r < z′ represents the case of q outside the sphere as shown in Fig. (4.10). For the case r > z′ , we can write 2 ⎤ q ⎡ z′ 1 ⎛ z′ ⎞ 2 V ( r ,θ ) = ⎢1 + cosθ + ⎜ ⎟ (3 cos θ − 1) + ⎥ r 2⎝ r ⎠ 4π ε r ⎣ ⎦

(4.44)

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z

z

Source point (0,0, z′) q

Field point (r > z′, θ, φ)

R

Source point (0,0, z′) q

r

r′

Field point (r < z′, θ, φ) R

r′

r

y

x

235

x

(a)

y

(b)

Fig. (4.10). The regions of the field point of a point charge. (a) r > z′. (b) r < z′.

The first term in (4.44) is the monopole term, the second term represents a dipole term, the third term is the quadrupole term and so on. It is clear that the second term represents the potential of an electric dipole with a dipole moment qz ′ . Example 4.1 Consider a charge + q in front of a conducting grounded sphere of the radius a at a distance f > a . Find the potential at any point r > a . Solution Since the sphere is conducting and grounded, then an image of charge q ′ = − aq f will be induced at a distance z o = a 2 f from the center of the sphere. The total potential outside the sphere is the algebraic sum of potentials due to the charge q and its image q ′ , then

V = Vq + Vq ′ Solution in the region a < r < f In this region, the charge q is outside the sphere or r < f , then the potential can be obtained using (4.42) by replacing z ′ by f to obtain

q V q ( r ,θ ) = 4π ε f

+∞

∑(r k =0

f ) Pk ( cosθ ), k

r< f

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The image is inside the sphere, then r > z o and its potential can be written as +∞

V q ′ ( r ,θ ) = ∑ k =0

Ck r k +1

Pk ( cosθ ),

r > zo

where Ck is a constant to be determined. The total potential can be obtained by substituting V q and V q ′ into the above expression of V , then +∞ C ⎡ q ( r f ) k + k +k1 V ( r ,θ ) = ∑ ⎢ r k =0 ⎣ 4 π ε f

⎤ ⎥ Pk ( cosθ ) ⎦

Since the sphere is grounded and conducting, then on the surface of the sphere r = a and V = 0 . Imposing this boundary condition on the last equation, yields +∞ C ⎡ q ( a f ) k + k k+1 0 = ∑⎢ a k =0 ⎣ 4 π ε f

⎤ ⎥ Pk ( cosθ ) ⎦

Since Pk (cosθ ) ≠ 0 , then

Ck = −

(a f )q 4π ε

a k (a f )

k

Substituting Ck into the expression of Vq′ , then Vq ′ (r ,θ )

k

a q +∞ ⎛ a2 ⎞ =− ∑ ⎜ ⎟ Pk (cosθ ) 4π ε rf k = 0 ⎜⎝ rf ⎟⎠ , k q′ + ∞ ⎛ zo ⎞ = ∑ ⎜ ⎟ Pk (cosθ ) 4π ε r k = 0 ⎝ r ⎠

r > zo

Alternative V q ′ can be obtained by replacing z ′ by z o = a 2 f and q by

q ′ = − ( a f ) q in (4.42) for the region r > z ′ . Substituting Ck in the above expression of V , we get the total potential V in the region a < r < f as V ( r ,θ ) =

q 4π ε rf

+∞

1 ∑ ( rf ) (r k =0

k

2 k +1

− a 2 k +1 ) Pk ( cosθ ),

a f In this region, the charge q is inside the sphere or r > f , then the potential Vq can be obtained using (4.42) by replacing z′ by f to obtain

V q ( r ,θ ) =

q +∞ ( f r ) k Pk ( cosθ ) ∑ 4π ε r k =0

r> f

Similarly, the potential Vq′ for the image can be obtained by replacing z′ by a f 2

and q by q′ = −(a f ) q in (4.42) for the region r > z′ , then Vq ′ (r ,θ ) = −

(a f ) q + ∞ ⎛ a 2 f ⎞ 4π ε r

∑ ⎜⎜⎝ k =0

r

k

⎟⎟ Pk (cosθ ), ⎠

r > a2 f

The total solution in the region r > f is the sum of V q and V q ′ , then

V (r , θ )

(a f ) q +∞ ⎛⎜ a 2 f ⎞⎟ P (cosθ ) q +∞ k ( ) ( ) = − f r P cos θ ∑ ∑ k k 4π ε r k =0 4π ε r k =0 ⎜⎝ r ⎟⎠ +∞ q 1 = f 2 k +1 − a 2 k +1 Pk (cos θ ), r> f ∑ k 4π ε rf k =0 (rf ) k

(

)

Combining, the total solution in the whole region outside the sphere (r > a ) can be expressed as q V (r ,θ ) = 4π ε rf

4.3.

⎧(r 2 k +1 − a 2 k +1 ), a < r < f 1 Pk (cosθ ) ⎨ 2 k +1 ∑ k r> f − a 2 k +1 ), k =0 (rf ) ⎩( f +∞

POISSON’S AND LAPLACE’S EQUATIONS

Assume that ρ v is the charge density per unit volume in a medium. If the volume of the medium is v and enclosed by a surface S . Using Gauss’s law, we have

∫∫ ε E ⋅ dS = ∫ ρ S

v

v

dv

(4.45)

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where E is the electrostatic field and ε is the permittivity of the medium. The surface integral in (4.45) can be written as a volume integral to obtain

ε ∫ ∇ ⋅ Edv = ∫ ρv dv V

(4.46)

V

Comparing both sides of (4.46), yields

∇⋅E =

ρv ε

(4.47)

Substituting E = −∇V in (4.47), we get the following differential equation

∇ 2V = −

ρv ε

(4.48)

Equation (4.48) is known as Poisson's equation and can be solved analytically or numerically to determine the electrostatic potential in a region subject to certain boundary conditions. Consider an arbitrarily shaped region R bounded by a closed surface S , occupied by a charge distribution with volume charge density ρ v . The following boundary conditions lead to a unique solution of Poisson's equation in the region R : 1. Dirichlet boundary conditions: the electrostatic potential V is given on the closed surface S enclosing the region R 2. Neumann boundary conditions: the normal derivative of the electrostatic potential ∂V ∂n is given on the closed surface S enclosing the region R 3. Mixed boundary conditions: Dirichlet boundary conditions are given on a portion of the closed surface S , and Neumann boundary conditions are given for the remainder of S . In many problems concern the solution of Poisson's equation, the region where the potential distribution to be determined is a charge-free,LH the charge density ρ v is zero. Hence, for such problems we can substitute ρv = 0 in (4.48) to get the following simple form for Poisson's equation ∇ 2V = 0

(4.49)

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This special case of Poisson's equation is known as Laplace’s equation. The solution of Poisson's equation in (4.48) can be viewed as the potential distribution due to a volume charge density ρ v in a region of volume v and defined by the position vector r , which is given by the integral formula (2.64c). Thus, the solution of Poisson's equation may be written in integral form as

V (r ) =

1 ρv (r′)dv′ 4π ε ∫∫∫ r − r′ v

(4.50)

where r′ and r are the position vectors of the source and field points respectively. 4.4.

UNIQUENESS OF THE SOLUTION

It can be shown that Poisson's and Laplace’s equations give a unique solution in a particular region for a given boundary conditions. Assume that there are two solutions V1 and V2 ≠ V1 satisfying Poisson's equation in a region R of a volume v and enclosed by surfaces S1 , S2 , and S3 such that S = S1 ∪ S2 ∪ S3 and

S1 ∩ S2 ∩ S3 = ∅ as shown in Fig. (4.11). Assume also that V1 and V2 satisfy the same boundary conditions imposed on S . The volume charge density in R is ρ v  then

∇ 2V1 = −

ρv , ε

and

∇ 2V2 = −

ρv ε

(4.51)

Letting V12 = V2 − V1, then from (4.51) we can write

∇2 (V2 − V1 ) = ∇2V12 = 0

(4.52) S = S 1∪ S 2∪ S 3 S 1∩ S 2∩ S 3 = ∅

Region of volume v S2 S1

R

ρv

S3

Fig. (4.11). A region R of a volume v with volume charge density ρ v .

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Which means that V12 satisfies Laplace’s equation. Since ∇ 2V12 = 0 in the region R , then we have either V12 = 0 or ∂V12 ∂n = 0 on the surface S , where ∂V12 ∂n = ∇V12 ⋅ an and a n is normal to the surface S . Green’s theorem states that

∂Φ

∫∫∫(Φ ∇ Φ + ∇Φ ⋅ ∇Φ)dv = ∫∫ Φ ∂n dS 2

v

(4.53)

S

Letting Φ = V12 and using (4.53), taking into consideration that on S either V12 = 0 (for Dirichlet boundary conditions) or ∂V12 ∂n = 0 (for Neumann boundary conditions), the right-hand side of (4.53) vanishes and it reduces to

∫∫∫ ∇V

12

2

dv = 0

(4.54)

v

Since ∇V12 is positive, (4.54) is true only when ∇V12 = 0 and hence V12 is a constant in the region R . As the consequence, we can conclude that: 2

1. For Dirichlet boundary conditions, V12 = 0 on S . Therefore, the constant must be zero or V12 = V2 − V1 = 0 in the region R . Hence, V2 = V1 which contradict the initial assumption that V2 ≠ V1 and the solution is unique. 2. For Neumann boundary conditions, ∂V12 ∂n = 0 on S . Therefore, the constant must be zero and hence V2 = V1 and the solution is unique. 3. For mixed boundary conditions, V12 = 0 on a part of S . Therefore, as above, V2 = V1 in R and the solution is unique. Note that for charge-free region the electrostatic potential V satisfies Laplace’s equation, or ∇ 2V = 0 and its solution has the following properties: 1. There can never be a maximum or minimum in the potential in a charge-free region because the second derivative of V is always zero 2. The value of the potential at the center of a line, circle, or sphere is equal to the mean value of the potentials at their boundaries respectively.

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SOLUTION OF LAPLACE’S EQUATION IN ONE DIMENSION

4.5.1. Rectangular Coordinates System In rectangular coordinate system Laplace’s equation is given by

∂ 2V ∂ 2V ∂ 2V + + =0 ∂x 2 ∂y 2 ∂z 2

(4.55)

Based on the geometry and the boundary conditions of the problem, there are three cases for the one-dimensional solution of the Laplace’s equation in rectangular coordinates. These are: 1) variation of V with x only, 2) variation of V with y only, and 3) variation of V with z only. Let us consider the solution for the case when the potential varies with z only, then ∂ 2V ∂x 2 = ∂ 2V ∂y 2 = 0 , and (4.55) reduces to

∂ 2V =0 ∂z 2

(4.56)

The solution of (4.56) is obtained by direct integration with respect to z twice. Doing this, yields

V (z ) = K1 z + K2

(4.57)

where K1 and K 2 are constants. These constants can be determined by applying the boundary conditions of the problem. Following the same analysis, similar expressions as in (4.57) can be obtained for the cases when the potential varies with x only or y only as shown in Table 4.1. 4.5.2. Cylindrical Coordinate System In cylindrical coordinate system Laplace’s equation is given by

1 ∂ ⎛ ∂V ⎞ 1 ∂ 2V ∂ 2V + =0 ⎜ρ ⎟+ ρ ∂ρ ⎝ ∂ρ ⎠ ρ 2 ∂φ 2 ∂z 2

(4.58)

The one-dimensional solution of (4.58) may be along the ρ , φ , or z direction depending on the variation of V along these directions. The variation of V depends on the geometry of the problem and the boundary conditions. The case of

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variation of the potential with z has been considered in the previous subsection. Here we will consider the cases of V varies with ρ only or φ only. 4.5.2.1. Case 1: V Varies with ρ only In this case, ∂ 2V ∂φ 2 = ∂ 2V ∂z 2 = 0 and (4.58) reduces to

1 ∂ ⎛ ∂V ⎞ ⎜ρ ⎟=0 ρ ∂ρ ⎝ ∂ρ ⎠

(4.59)

Multiplying both sides of (4.59) by ρ and integrate with respect to ρ , yields

ρ

∂V = K1 ∂ρ

(4.60)

where K1 is a constant. Divide both sides of (4.60) by ρ and integrate again with respect to ρ , the solution is obtained as

V (ρ ) = K1 ln ρ + K2

(4.61)

where K 2 is a constant. The constants K1 and K 2 can be obtained from the boundary conditions of the problem. 4.5.2.2. Case 2: V Varies with φ only In this case, ∂V ∂ρ = ∂ 2V ∂ρ 2 = ∂ 2V ∂z 2 = 0 , then from (4.58)

1 ∂ 2V =0 ρ ∂φ 2

(4.62)

Multiplying both sides of (4.62) by ρ and integrate with respect to φ twice, the solution becomes

V (φ ) = K1φ + K2

(4.63)

where K1 and K 2 are constants that can be determined from the boundary conditions of the problem.

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4.5.3. Spherical Coordinate System Laplace’s equation in spherical coordinate systems is given by

1 ∂ ⎛ 2 ∂V ⎞ 1 1 ∂ 2V ∂ ⎛ ∂V ⎞ =0 ⎟+ ⎜r ⎜ sin θ ⎟+ r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ∂θ ⎝ ∂θ ⎠ r 2 sin 2 θ ∂φ 2

(4.64)

The case of variation of V with φ has been considered in the previous subsection. The other two cases are as follows. 4.5.3.1. Case 1: V Varies with r only In this case, (4.64) reduces to

1 ∂ ⎛ 2 ∂V ⎞ ⎜r ⎟=0 r 2 ∂r ⎝ ∂r ⎠

(4.65)

Multiplying both sides of (4.65) by r and integrate twice with respect to r , yield

V (r ) = −

K1 + K2 r

(4.66)

where K1 and K 2 are constants. 4.5.3.2. Case 2: V Varies with θ only In this case, (4.64) reduces to

∂ ⎛ ∂V ⎞ 1 ⎜ sin θ ⎟=0 ∂θ ⎠ r sin θ ∂θ ⎝ 2

(4.67)

Multiplying both sides of (4.67) by r 2 sin θ and integrating with respect to θ , then dividing both sides by sin θ and performing the integration again with respect to θ , yields

K1 K1 K1 cos 2 (θ 2) V (θ ) = ∫ dθ = ∫ dθ = dθ sin θ 2 sin (θ 2) cos(θ 2) 2 ∫ 2 tan (θ 2) = K1 ln[tan (θ 2)] + K 2

(4.68)

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where K1 and K 2 are constants to be determined. Solutions for one-dimensional Laplace’s equation are summarized in Table 4.1. Table 4.1. Solutions for Laplace’s equation in one dimension. Rectangular Coordinates

Cylindrical Coordinates

V varies with

V varies with ρ only

V varies with y only

V varies with

φ

only

V varies with

z

only

x only V (x) = K1x + K2

V (ρ ) = K1 ln ρ + K2

V ( y ) = K1 y + K 2

V varies with

z

V (φ ) = K1φ + K2

only

V (z ) = K1 z + K 2

V (z ) = K1 z + K 2

Spherical Coordinates V varies with

r

only

V (r ) = − K1 r + K 2

V varies with θ only



V (θ ) = K1 ln[tan(θ 2)] + K 2 V varies with

φ

only

V (φ ) = K1φ + K2

Example 4.2 An infinitely long conducting plate of width w at potential V = 0 volt is extended along x-axis such that its plane coincides with the plane z = 0 . If a second similar plate at potential V = V0 volts is placed parallel to the first one and at a height d 0

(4.75)

where Ak and Bk are constants. Similarly, the solution of (4.74b) is

α=0 ⎧C y + D0 , Y (y) = ⎨ 0 ⎩Ck sinh (α y ) + Dk cosh (α y ), α > 0

(4.76)

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where Ck and Dk are constants. Substituting (4.75) and (4.76) into (4.71), the solution becomes

α=0 ⎧( A x + B0 )(C0 y + D0 ), Vk ( x, y ) = ⎨ 0 αy −α y ⎩( Ak sinα x + Bk cosα x )(Ck e + Dk e ), α > 0

(4.77)

For γ 2 = + α2 , the sign of α 2 changes from (+) to (-) in (4.74a) and from (-) to (+) in (4.74b), and the solution in this case becomes

α=0 ⎧( A x + B )(C y + D0 ), Vk (x, y ) = ⎨ 0 α x 0 −0α x ⎩(Ak e + Bk e )(C sinα y + D cosα y ), α > 0

(4.78)

The complete solution for the potential is the superposition of all solutions Vk , then V (x, y ) = ( A0 x + B0 )(C0 y + D0 )

(

)

∞ ⎧ ( A sinα x + Bk cosα x ) Ck eα y + Dk e−α y , γ 2 = − α 2 + ∑⎨ k α x + Bk e− α x (C sinα y + D cosα y ), γ 2 = + α 2 k =1 ⎩ Ak e

(

)

(4.79)

Similar solutions for electrostatic potential V can be obtained for the cases: 1. When V varies with x and z while constant with y . In this case, V (x, z ) = X (x) Z (z ), where X and Z are functions of x and z respectively. 2. When V

varies with y and z while constant with x . Then, V (x, z ) = Y ( y ) Z (z ) , Y and Z are functions of y and z respectively.

All possible solutions for Laplace’s equation in two-dimensional rectangular coordinates system are listed in Table 4.2. Important points regarding the boundary conditions in the rectangular System: Let u stand for x , y or z , and V ′ = ∂V ∂u then: 1. If − ∞ ≤ u ≤ ∞ , then V does not vary with u . 2. If 0 ≤ u ≤ ∞ , then the coefficient of the factor e α u is zero. 3. If − ∞ ≤ u ≤ 0 , then the coefficient of the factor e − α u is zero.

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4. If V = 0 when u = 0 , then the coefficient of cos α u is zero. 5. If V ′ = 0 when u = 0 , then the coefficient of sin α u is zero. 6. If V = 0 when u = a , and V = 0 when u = −a , then the coefficient of cosα u is zero, and α = kπ a , k = 0,1, 2, 7. If V = 0 when u = 0 , and V = 0 when u = 0 , then α = kπ a , k = 0,1, 2,. Table 4.2. Solutions of two-dimensional Laplace’s equation in the rectangular coordinate system. Variation of V

Differential Equations

V varies

X ′′ − γ 2 X = 0

with x and y

V varies

with x and z

V varies with y

and z

Y ′′ + γ 2 Y = 0

X ′′ − γ 2 X = 0 Z ′′ + γ 2 Z = 0

Solution

V (x, y ) = ( A0 x + B0 )(C0 y + D0 ) +

⎧( Ak sinα x + Bk cosα x ) (Ck e α y + Dk e −α y ), γ 2 = − α 2 αx + Bk e −α x ) (Ck sinα y + Dk cosα y ) , γ 2 = + α 2 k k =1 ∞

∑ ⎨⎩(A e

Vk (x, z ) = ( A0 x + B0 )(C0 z + D0 ) +

∞

∑ k =1

Y ′′ − γ 2 Y = 0 Z ′′ + γ 2 Z = 0

⎧( Ak sinα x + Bk cosα x ) (Ck e α z + Dk e −α z ), γ 2 = − α 2 ⎨ αx 2 2 −α x ⎩(Ak e + Bk e )(Ck sinα z + Dk cosα z ) , γ = + α

Vk ( y, z ) = ( A0 y + B0 )(C0 z + D0 ) +

∞

∑ k =1

⎧( Ak sinα y + Bk cosα y ) (Ck e α z + Dk e −α z ), γ 2 = − α 2 ⎨ αy 2 2 −α y ⎩(Ak e + Bk e )(Ck sinα z + Dk cosα z ) , γ = + α

The constants Ak , Bk , Ck , and Dk in (4.79) are determined by applying the boundary conditions. Example 4.5 Find the electrostatic potential inside a rectangle 0 < x < a , 0 < y < b shown in Fig. (4.15). The problem satisfies the Dirichlet boundary conditions as illustrated in Fig. (4.15).

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Solution The region is infinitely extended in the z-direction or − ∞ < z < ∞ , then V varies with x and y only. Since 0 < x < a and 0 < y < b , the solution may be periodic with x or y , depending on the value of α . y

V4

b V1 = 0

z V2 = V0 a

V3 = 0

x

Fig. (4.15). Geometry of the problem in Example 4.5.

The solution is

V (x, y ) = ( A0 x + B0 )(C0 y + D0 )

⎧( Ak sinα x + Bk cosα x )(Ck eα y + Dk e− α y ), γ 2 = − α 2 + ∑⎨ αx + Bk e− α x )(Ck sinα y + Dk cosα y ), γ 2 = + α 2 k =1 ⎩(Ak e ∞

The required solution must satisfy all boundary conditions. let γ 2 = − α2 , then ∞

V (x, y ) = ( A0 x + B0 )(C0 y + D0 ) + ∑ (Ak eα x + Bk e − α x )(Ck sinα y + Dk cosα y ), k =1

The boundary conditions of the problem are

V (x, y = 0) = 0 , V (x, y = b) = 0 V (x = 0, y ) = 0 , V (x = a, y ) = V0 Applying the boundary conditions, then

V (x = 0, y ) = 0 ⇒

Ak = − Bk , B0 = 0 ⇒

(A e k

αy

+ Bk e − α y ) = 2 Ak sinhα y

V (x, y = 0) = 0 , ⇒ D0 = Dk = 0 V ( x, y = b ) = 0 , ⇒ C 0 = 0,

sinα b = 0 ,

⇒ α = kπ b , k = 1, 2,

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Substituting these constants, then the solution becomes ∞

V ( x , y ) = ∑ Fk sinh ( kπ x b )sin ( kπ y b ) k =1

where Fk = Ak Ck . The boundary condition V (x = a, y ) = V0 , implies that ∞

∞

k =1

k =1

f ( y ) = V 0 = ∑ Fk sinh ( kπ a b )sin ( kπ y b ) = ∑ Fk′ sin ( kπ y b ) The solution at x = a represents a trigonometric Fourier series expansion for the function f ( y ) = V0 with period b , then the Fourier coefficients Fk′ are

Fk sinh ( kπ a b ) = Fk′ =

b

2 f ( y )sin ( kπ y b ) dy b ∫0

2V b 2V a cos ( kπ y b ) ⇒ Fk′ = 0 ∫ sin ( kπ y b ) dy = − 0 b 0 b kπ

⎧⎪0, 1 ⇒ Fk = ⎨ 4V0 sinh(kπ a b ) ⎪ ⎩ kπ

y =b y =0

⎧0, ⎪ = ⎨ 4V 0 ⎪⎩ kπ

k even k odd

k even k odd

Hence, the solution becomes

4V0 sinh[(2k − 1)π x b] ⎡ (2k − 1)π y ⎤ sin ⎢ ⎥⎦ b ⎣ k =1 π (2k − 1) sinh[(2k − 1)π a b] ∞

V (x, y ) = ∑

The second solution for γ 2 = + α2 is given by ∞

V ( x, y ) = ( A0 x + B0 )(C0 y + D0 ) + ∑ ( Ak sinα x + Bk cosα x )(Ck eα y + Dk e − α y ) k =1

Imposing the boundary conditions on this solution, we obtain B0 = Bk = 0 , Ck = − Dk , and C0 = D0 = 0 , then ∞

V (x, y ) = ∑ Fk sin (α x )sinh (α y ) k =1

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The boundary condition V (x, y = b) = 0 , gives Fk sin (α x )sinh (α b ) = 0 which implies eα b (e2 α b − 1) = 0 . This means that, for V (x, y = b) = 0 to be satisfied then α = 0 and the solution vanishes. Since the solution must satisfy all boundary conditions, the solution for γ 2 = + α2 does not exist. Example 4.6 Find the electrostatic potential inside a rectangle 0 < x < a , 0 < y < b , if it satisfies Neumann boundary conditions shown in Fig. (4.16). y ∂V /∂x = 0

b V = sinπx

z

∂V /∂x = 0 V=0

a

x

Fig. (4.16). Geometry of the problem in Example 4.6.

Solution Since − ∞ < z < ∞ , then V varies with x and y only, and the solution is 

V (x, y ) = ( A0 x + B0 )(C0 y + D0 )

(

)

⎧( Ak sinα x + Bk cosα x ) Ck eα y + Dk e− α y , γ 2 = − α 2 + ∑⎨ αx + Bk e− α x (Ck sinα y + Dk cosα y ), γ 2 = + α 2 k =1 ⎩ Ak e ∞

(

)

The boundary conditions are

V (x, y = 0) = 0 , V (x, y = b) = sin(π x), ∂ ∂x [V ( x = 0, y )] = 0 , ∂ ∂x [V (x = a, y )] = 0 The required solution must satisfy all boundary conditions. V (x, y = b) = sin(π x) cannot be satisfied for the case γ 2 = + α2 , then the general solution is

( )

∞

V x, y = ( A0 x + B0 ) (C0 y + D0 ) + ∑ ( Ak sinα x + Bk cosα x) (Ck eα y + Dk e − α y ) k =1

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The boundary conditions of the problem are

V (x, y = 0) = 0 , V (x, y = b) = sin(π x) ,

∂ ∂ V (x = 0, y ) = 0 , V ( x = a, y ) = 0 ∂x ∂x

To apply all boundary conditions, V must be differentiated with respect to x as ∞ ∂ V (x, y ) = x ( A0C0 y + A0 D0 ) + ∑ ( Ak k cosα x − Bk k sinα x )(Ck eα y + Dk e − α y ) ∂x k =1

Hence,

∂ V (x = 0, y ) = 0 , ⇒ A0 = Ak = 0 ∂x Substituting A0 = Ak = 0 and applying ∂ ∂x [V (x = a, y )] = 0 for α > 0 , yields

∂ V ( x = a , y ) = 0 = − B k α sin ( α a )[C sinh ( α y ) + D k cosh ( α y )] ∂x For the solution to exist, Bk ≠ 0 , then sin(α a ) = 0 , and the constant α can be obtained as α = kπ a , k = 0,1, 2,. Consequently, the solution becomes ∞

V (x, y ) = B0 (C0 y + D0 ) + ∑ Bk cosα x (Ck eα y + Dk e − α y ) k =1

Applying the boundary condition V (x, y = 0) = 0 , then for k = 0 , B0 D0 = 0 . For k > 0 :

B k ( C k + D k )cos ( kπ x a ) = 0 , ⇒

Bk (C k + Dk ) = 0

Since Bk ≠ 0 , and B0 ≠ 0 then D0 = 0 , Dk = Ck , and Ck (e ky − e − ky ) = 2Ck sinh ky. The quantities B0C0 and 2 Bk Ck can be replaced by F0 = B0 C0 and Fk = 2Bk Ck respectively. Hence, the complete solution is ∞

V ( x , y ) = F0 y + ∑ Fk sinh ( kπ y a ) cos ( kπ x a ) k =1

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Applying the boundary condition V (x, y = b) = sin(πx), yields ∞

V ( x , y = b ) = f ( x ) = sin (πx ) = F0 b + ∑ Fk sinh ( kπ b a ) cos ( kπ x a ) k =1

The solution at y = b represents a trigonometric Fourier series expansion for the function f (x) = sin(πx ) with a period a . We can write f (x ) in the form ∞

f ( x ) = F0′ + ∑ Fk′ cos ( kπ x a ) k =1

where

F0′ = F0 b ,

Fk′ = Fk sinh ( kπ b a )

The Fourier coefficients F0′ and Fk′ can be obtained from a

a

F0′ =

1 f ( x ) dx , a ∫0

Fk′ =

2 f ( x ) cos ( kπ x a ) dx a ∫0

The solution in terms of F0′ and Fk′ is ∞

V n ( x , y ) = F0′ y b + ∑ Fk′ k =1

sinh ( kπ y a ) cos ( kπ x a ) sinh ( kπ b a )

Since f (x) = sin(πx ), then F0′ =

a cos (πa ) − 1 1 sin (π x ) dx = ∫ a0 πa

a

Fk′ =

a

2 1 ⎧ ⎡ k ⎤ ⎡ k ⎤⎫ sin (πx ) cos ( kπ x a ) dx = ∫ ⎨ sin ⎢ ⎛⎜ + 1 ⎞⎟ πx ⎥ − sin ⎢ ⎛⎜ − 1 ⎞⎟ πx ⎥ ⎬ dx ∫ a0 a 0 ⎩ ⎣⎝ a ⎠ ⎦ ⎣⎝ a ⎠ ⎦ ⎭ 1 ⎫ ⎧ cos[(k + a )π ] cos[(k − a )π ]⎫ ⎧ 1 ⇒ Fk′ = ⎨ − − ⎬−⎨ (k − a )π ⎭ ⎩ (k + a )π (k − a )π ⎬⎭ ⎩ (k + a )π

=

k 2a ⎡ (− 1) cos aπ − 1⎤ ⎥ π ⎢⎣ k 2 − a2 ⎦

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Then the complete solution for the electrostatic potential in the region becomes k cos πa − 1 ⎤ 2 a ∞ ⎡ ( − 1) cos aπ − 1 ⎤ sinh ( kπ y a ) cos ( kπ x a ) + y V ( x , y ) = ⎡⎢ ∑⎢ k 2 − a2 ⎥ π k =1 ⎣ sinh ( kπ b a ) ⎣ πa ⎥⎦ ⎦

4.6.2. Cylindrical Coordinate System 4.6.2.1. V Varies with ρ and z The Laplace’s equation in cylindrical coordinates is given by

1 ∂ ⎛ ∂V ⎞ 1 ∂ 2V ∂ 2V =0 + ⎜ρ ⎟+ ρ ∂ρ ⎝ ∂ρ ⎠ ρ 2 ∂φ 2 ∂z 2

(4.80)

Assuming that the potential varies in the direction of ρ and z and constant with φ ,then ∂ 2V ∂φ 2 = 0 and (4.80) reduces to

1 ∂ ⎛ ∂V ⎞ ∂ 2V ∂ 2V 1 ∂V ∂ 2V = + + =0 ⎜ρ ⎟+ ρ ∂ρ ⎝ ∂ρ ⎠ ∂z 2 ∂ρ 2 ρ ∂ρ ∂z 2

(4.81)

Using separation of variables method, the potential V can be expressed as a multiplication of two independent functions Ρ and Z as follows

V (ρ , z ) = Ρ (ρ ) Z (z )

(4.82)

Substituting V from (4.82) into (4.81) and dividing both sides of (4.81) by ΡZ , we get

1 ∂ 2Ρ 1 1 ∂Ρ 1 ∂ 2 Z + + =0 Ρ ∂ρ 2 ρ Ρ ∂ρ Z ∂z 2

(4.83)

Since Ρ and Z are independent functions, then

1 ∂ 2Ρ 1 1 ∂Ρ 1 ∂2Z + =− = γ2 2 2 Ρ ∂ρ Z ∂z ρ Ρ ∂ρ where γ is a constant.

(4.84)

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Let Ρ′′ = ∂ 2 Ρ ∂ρ 2 , Ρ′ = ∂Ρ ∂ρ and Z ′′ = ∂ 2 Z ∂z 2 , then from (4.84) with γ 2 = − α2 ( α is real and positive), we can write

Ρ′′ +

1

ρ

Ρ′ + α 2 Ρ = 0

Z ′ − α2 Z = 0

(4.85a) (4.85b)

Bessel’s function equation is given by [see equation (A.1) in Appendix A]

Ρ′′ +

⎛ ν2 ⎞ Ρ′ + ⎜⎜ α 2 − 2 ⎟⎟ Ρ = 0 ρ ρ ⎠ ⎝ 1

(4.86)

The solution of (4.86) is in the form

Ρ = AJν (α ρ ) + BYν (α ρ )

(4.87)

where J v (α ρ ) and Yv (α ρ ) are Bessel functions of the first and second kind of order ν respectively. When ν = 0 and α > 0 , (4.87) reduces to

Ρ = AJ 0 (α ρ ) + BY0 (α ρ )

(4.88)

For α = 0 the solution is obtained directly from (4.85a) as Ρ = A0 + B 0 ln ρ . Thus, the kth solution is

α=0 ⎧ A + B 0 ln ρ , Ρk ( ρ ) = ⎨ 0 ⎩ A k J 0 ( α ρ ) + B k Y 0 ( α ρ ), α > 0

(4.89)

where Ak and Bk , are constants. The solution of (4.85b) is similar to that obtained for Z in the case of rectangular coordinate system, then

α=0 ⎧(C + D z ), Z (z ) = ⎨ 0 α z 0 − α z ⎩Ck e + Dk e , α ≠ 0 where Ck and Dk are constants.

(4.90)

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From (4.82), (4.89), and (4.90) the kth solution of the Laplace’s equation for ρ and z dependence potential can be written as

α=0 ⎧( A + B0 ln ρ )(C0 + D0 z ), Vk (ρ , z ) = ⎨ 0 αz −α z ⎩[Ak J 0 (α ρ ) + BkY0 (α ρ )] Ck e + Dk e , α > 0

(

)

(4.91)

For the case γ 2 = α 2 , equations (4.85a) and (4.85b) change to

Ρ′′ +

1

ρ

Ρ′ − α 2 Ρ = 0

Z ′ + α2 Z = 0

(4.92a) (4.92b)

Modified Bessel’s function equation is given by

Ρ′′ +

⎛ 2 ν2 ⎞ ′ Ρ − ⎜⎜ α + 2 ⎟⎟ Ρ = 0 ρ ρ ⎠ ⎝ 1

(4.93)

The solution of this equation is

Ρ = AIν (α ρ ) + BKν (α ρ )

(4.94)

where Iν (α ρ ) and Kν (α ρ ) are modified Bessel functions of the first and second kind of order ν respectively. When v = 0 , the solution of (4.93) is

Ρ = AI 0 (α ρ ) + BK0 (α ρ )

(4.95)

The solution for z dependence factor is similar to the rectangular coordinate case. Hence, the kth solution becomes

α=0 ⎧( A + B0 ln ρ )(C0 z + D0 ), Vk (ρ , z ) = ⎨ 0 ⎩[Ak I 0 (α ρ ) + Bk K 0 (α ρ )](Ck sinα z + Dk cosα z ), α > 0

(4.96)

The general solution for (4.81) is ∞

V (ρ , z ) = ∑Vk (ρ , z ) k =0

(4.97)

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When V varies with ρ and z , there are two possible solutions for Vk (ρ , z ) based on the values of γ 2 = ± α2 . These solutions are given by (4.91) and (4.96) and the selection between them is determined by the geometry of the region in which Vk (ρ , z ) to be obtained. If the region is finite in the z direction, then the solution is periodic with z , and it will be in the form of equation (4.96). For regions of infinite dimensions in the z direction the solution will be in the form given by (4.91). Both solutions depend on Bessel’s functions and modified Bessel’s functions J 0 (α ρ ), Y0 (α ρ ) , I 0 (α ρ ) and K0 (α ρ ) . These functions are tabulated and can be computed simply using many programming packages. 4.6.2.2. V Varies with ρ and φ In this case, the two-dimensional Laplace’s equation is

∂ 2V 1 ∂V 1 ∂ 2V + + =0 ∂ρ 2 ρ ∂ρ ρ 2 ∂φ 2

(4.98)

Let V (ρ ,φ ) = Ρ (ρ )Φ (φ ) and substituting in (4.98), we can separate it into

ρ 2Ρ′′ + ρΡ′ − γ 2 Ρ = 0 Φ′′ + γ 2 Φ = 0

(4.99a) (4.99b)

where γ is a constant, Ρ′′ = ∂ 2Ρ ∂r 2 , Ρ′ = ∂Ρ ∂r , and Φ′′ = ∂ 2Φ ∂θ 2 . Since Φ(φ ) = Φ (φ + 2π ), the solution of (4.99b) must be in terms of trigonometric functions. Thus, γ 2 is greater than zero and the solution for γ 2 < 0 is not acceptable. Let γ 2 = k 2 , where k is a positive integer, then (4.99a) represents Cauchy-Euler equation and its solution for k ≠ 0 is in the form Aρ k + Bρ − k . Hence, the solution of (4.99a) for k ≠ 0 is

Ρk (ρ ) = Ak ρ k + Bk ρ − k

(4.100)

where Ak and Bk are constants. The solution for k = 0 is obtained by letting k = 0 in (4.100) andperforming the integration with respect to ρ twice, to get

Ρ0 (ρ ) = A0 + B0 ln ρ

(4.101)

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The solution for (4.99b) can be obtained as

k =0 ⎧C φ + D0 , Φ k (φ ) = ⎨ 0 ⎩Ck sin kφ + Dk cos kφ , k > 0

(4.102)

Using (4.100), (4.101), and (4.102), along with V (ρ ,φ ) = Ρ (ρ )Φ (φ ), the kth solution for the potential is

k =0 ⎧( A + B ln ρ )(C0 + D0φ ), Vk (ρ ,φ ) = ⎨ 0 k 0 −k ⎩(Ak ρ + Bk ρ )(Ck sin kφ + Dk cos kφ ), k > 0

(4.103)

The case of variation of V with φ and z while constant with ρ is not of practical interest and will be ignored. Table 4.3, summarizes the solutions of twodimensional Laplace’s equation in cylindrical coordinates for most practical cases. The relation between Bessel’s functions and modified Bessel’s functions are

I 0 (α ρ ) = J 0 ( j α ρ )

K 0 (α ρ ) = j

(4.104a)

π [J 0 ( j α ρ ) + jY0 ( j α ρ )] 2

(4.104b)

( )

The behavior of the functions J 0 (α ρ ), Y0 (α ρ ) , I 0 (α ρ ) and K0 α ρ in the limits of ρ → 0 and ρ → ∞ are as follows 1. As ρ → 0 , J 0 (α ρ ) → 1, I 0 (α ρ ) → 1 , Y0 (α ρ ) → −∞ , and K0 (α ρ ) → +∞ . 2. As ρ → ∞ , J 0 α ρ → 0 , I 0 (α ρ ) → ∞ , Y0 (α ρ ) → 0 , and K0 (α ρ ) → 0 .

( )

Note the following points when solving Laplace’s equation for cylinder of radius a and length l along z -axis assuming the solution exists and finite everywhere: 1. For finite l , the solution is periodic in z (function of sinα z and cosα z ), while for infinitely extended cylinders (l → ±∞), the solution will be in terms of e α z and e − α z  2. For the solution in the range 0 ≤ ρ ≤ a : It is known that as ρ → 0 , Y0 (α ρ ) → −∞ , K0 (α ρ ) → +∞ , and ρ −k → ∞ . Since the solution is assumed finite at ρ = 0 , then the coefficients of Y0 (α ρ ) , K0 (α ρ ) and ρ − k vanish.

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3. For the solution in the range ρ > a : As ρ → ∞ , I 0 (α ρ ) → ∞ , and ρ k → ∞ , then the coefficients of I 0 (α ρ ) and ρ vanish. 4. V is continuous on the surface of the cylinder. 5. For dielectric cylinders, the flux density on the surface of the cylinder is continuous. 6. For infinitely long conducting cylinders V is constant on the surface of the cylinder. k

Table 4.3. Solutions of two-dimensional Laplace’s equation in cylindrical coordinate system. 9DULDWLRQRI 9

V varies with ρ and z

V varies with ρ and φ

'LIIHUHQWLDO (TXDWLRQV

6ROXWLRQ V (ρ , z ) = ( A0 + B0 ln ρ )(C0 z + D0 )

Ρ′′ +

1

Ρ′ − γ Ρ = 0 2

Z ′′ρ+ γ 2 Z = 0

ρ 2 Ρ′′ + ρΡ′ − γ 2 Ρ = 0 Φ′′ + γ 2 Φ = 0

⎧[Ak I 0 (α ρ ) + Bk K 0 (α ρ )](Ck sinα z + Dk cosα z ), γ 2 = − α 2

∞

+

∑ ⎨⎩[A I (α ρ ) + B K (α ρ )](C sinα z + D cosα z ), k =1

k 0

k

0

k

k

γ2 = + α2

V (ρ ,φ ) = ( A0 + B0 ln ρ )(C0φ + D0 ) ∞

+

∑ (A ρ k

k

+ Bk ρ −k )(Ck sin kφ + Dk cos kφ )

k =1



Example 4.7 Find the potential inside a long tunnel of the cross section shown in Fig. (4.17). The boundary conductions are: V = 0 on the plane plate, and V = V0 on the curved surface. The curved surface is a semi-circle of radius a . y

V = V0 a x Plate

V=0

Fig. 4.17: Geometry of the problem in Example 4.7.

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Solution Assume the tunnel is a long z-axis. Since the tunnel is long then − ∞ < z < ∞ and the potential does not vary with z and the solution in cylindrical coordinates is ∞

Vk (ρ ,φ ) = ( A0 + B0 ln ρ )(C0φ + D0 ) + ∑ (Ak ρ k + Bk ρ − k )(Ck sin kφ + Dk cos kφ ) k =1

On the plate V = 0 , then V (ρ ,φ = 0) = V (ρ ,φ = π ) = 0 , ⇒

Dk = 0

It is assumed the solution exist, then C ≠ 0 . Since k ≠ 0 , then the solution for k = 0 does not exist or A0 = B0 = C0 = D0 = 0 . Applying the boundary condition V (ρ = 0, φ ) = 0 , then

ρ → 0 , Ck ( Ak ρ k + Bk ρ −k ) = 0 , ⇒ Bk = 0 Hence, the solution reduces to

Vk (ρ , φ ) = Ak Ck ρ − k sin kφ = Fk ρ k sin kφ where Ak Ck is replaced by Fk . For ρ = 0 , and φ ≠ 0, π

Fn sin kφ = 0 ,

⇒ k = 0,1, 2 ,

Since the solution exists, then the case for k = 0 is not included and k = 1, 2 , . Thus, the complete solution becomes ∞

V (ρ , φ ) = ∑ Fk ρ k sin kφ k =1

V (ρ = a, φ ) = V0 in the interval 0 < φ < π , then ∞

V0 = ∑ Fk a k sin kφ , k =1

0 0 (5.13)

z a illustrated in Fig. (5.9a) and b respectively. The total current enclosed by loop in each region can be obtained as

⎧( ρ a )2 I , 0 ≤ ρ ≤ a I T = ∫ J ⋅ dS = ⎨ S ρ >a ⎩I ,

the the as the

(5.14)

3. Since the current flowing in the z direction, the magnetic field H is in the φ direction, or H = a φ H φ . 4. Application of Ampere’s law in each region, yields

⎧2π 2 ⎪ ∫ H φ a φ ⋅ a φ ρ dφ = ( ρ a ) I , 0 ≤ ρ ≤ a ⎪0 ∫lH ⋅ dl = ⎨ 2 π ⎪ H a ⋅ a ρ dφ = I , ρ >a ⎪⎩ ∫0 φ φ φ

Fig. (5.9). Application of Ampere’s law for an infinitely long current carrying conductor.

(5.15)

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From (5.15), the magnetic field at a radial distance ρ from the axis of an infinite current carrying conductor in the z direction is obtained as

Iρ ⎧ ⎪⎪ a φ 2 π a 2 , 0 ≤ ρ ≤ a H=⎨ ⎪a φ I , ρ > a ⎪⎩ 2 π ρ

(5.16)

5.2.4. The Magnetic Field inside a Long Solenoid In practical applications, the solenoid is considered long when its length l and radius a satisfy l >> a . Consider the long solenoid of the circular cross section shown in Fig. (5.10) with N turns and current I . The magnetic field inside long solenoids is considered approximately constant provided that the turns N are closely wound. Let the axis of the solenoid coincides with the z-axis, then the surface current density on its surface will be K . The magnetic field intensity inside the solenoid can be obtained using Ampere’s law as follows: 1. A rectangle ABCDA of dimensions w × h shown in Fig. (5.10b) is chosen as the amperian loop. 2. Determining the total current enclosed by the closed loop, which is I T = wK y , where K y = NI l . 3. Since the current flowing in the y direction and for the symmetry of the problem, the magnetic field components in the y and x directions are zero. Also, the magnetic field outside the solenoid is zero. Therefore, the magnetic field has only z - directed component, that is H = a z H z . 4. Application of Ampere’s law, yields B

C

D

A

A

B

C

D

wK y = −∫ (0) a z ⋅ a z dz − ∫ (0) a x ⋅ a x dx + ∫ ( H za z ⋅ a z dz + ∫ (0) a x ⋅ a x dx (5.17) Simplifying (5.17), then H z = K w = NI l . Thus, the approximate magnetic field inside the solenoid is obtained as

H = azHz = az

NI l

(5.18)

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x w l

K = a yK y

A

B

D

C

h a

H

z N

I

z

I

(a)

(b)

Fig. (5.10). Application of Ampere’s law for a long solenoid.

5.2.5. The Magnetic Field inside a Toroid Consider a toroid with N turns wound closely and current I as shown in Fig. (5.11). The inner and outer radii of the toroid are a and b . The cross section of the toroid may be circular or rectangular. y

y

N

H b

I

a

x

I

(a)

ρ

x

(b)

Fig. (5.11): Application of Ampere’s law for a toroid.

The magnetic field intensity inside the toroid can be obtained using Ampere’s law as follows: 1. A circle of the radius a < ρ < b is chosen as an amperian loop as shown Fig. (5.11b). From the geometry of the differential length along the loop is dl = a φ ρ dφ . 2. The total current enclosed by the amperian loop is I T = NI .

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3. It is clear that from Fig. (5.11b) the magnetic field is H = − a φ H φ . 4. Application of Ampere’s law, yields 2π

∫ H ⋅ dl = − ∫ a φ H φ ⋅ a φ ρ dφ = − NI l

(5.19)

0

Simplifying (5.19), yields H φ = NI 2 πρ . Thus,

H = −aφ

NI , 2πρ

a c ) is H = 0 . Thus,

⎧ ⎪aφ ⎪ ⎪⎪a H=⎨ φ ⎪ ⎪aφ ⎪ ⎪⎩0,

ρI , 2π a 2

0≤ρ ≤a

I

a≤ρ ≤b , 2π ρ I ⎛ c2 − ρ 2 ⎞ ⎜ ⎟, b ≤ ρ ≤ c 2π ρ ⎜⎝ c 2 − b 2 ⎟⎠ ρ >c

5.2.6. Ampere’s Law in Point Form Ampere’s law can be written in point form by converting the line integral to surface integral using Stokes' theorem. Consider a current density J crossing a surface S , hence Ampere’s law in integral form is

∫ H ⋅ dl = ∫ J ⋅ dS l

S

(5.21)

where l is a closed loop enclosing the current J . Converting the line integral in the left hand side of (5.21) to a surface integral, yields

∫ ∇ × H ⋅ dS = ∫ J ⋅ dS S

S

(5.22)

Ampere’s law in point form is obtained by equating the integrands in both sides of (5.22). Thus, ∇×H = J

(5.23)

Accordingly, the circulation of the magnetic field is equal to the current density J , which means that loops of the magnetic field are caused by the current density. Example 5.3 If the magnetic field in a conducting medium is given by

H=

1 (2ar sin θ cosφ + aθ cosθ cosφ − aφ sin φ ) A m r

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Find the (a) Current density. (b) Total current in the direction aθ through the surface θ = 45 , 0 ≤ φ ≤ π , 0 ≤ r ≤ 2 m. Solution (a) Using (5.23) and (1.142c), the current density can be obtained as

J=

a r ⎡ ∂ (Hφ sin θ ) ∂Hθ ⎤ aθ ⎡ 1 ∂H r ∂(rHφ )⎤ aφ ⎡ ∂(rHθ ) ∂H r ⎤ − + − + − r sin θ ⎢⎣ ∂θ ⎥⎦ ∂r ⎥⎦ r ⎢⎣ ∂r ∂φ ⎥⎦ r ⎢⎣ sin θ ∂φ ∂θ

Substituting the H components in the above equation and simplifying, we get

J=

2 (ar sin φ cotθ − aθ sin φ − aφ cosθ cosφ ) A m r2

(b) The differential surface element in the direction of θ is dS = aθ r sin θ drdφ as illustrated in Fig. (1.17). Hence, the total current in the direction aθ through the surface θ = 45 , 0 ≤ φ ≤ π , 0 ≤ r ≤ 2 m is obtained as π 2

IT = ∫ J ⋅ aθ dS = −2∫∫ r sin φ sin 45 drdφ = − S

0 0

5.3.

1 2 r 2

r =2 r =0

cosφ

φ =π φ =0

=4 2 A

MAGNETIC FLUX AND MAGNETIC FLUX DENSITY

We know that from Chapter 2, the electric flux density D is related to the electric field intensity E by D = ε 0 E , where ε 0 is permittivity of free space. A magnetic quantity B , which is similar to D can be defined in terms H as

B = μ0 H

(5.24)

The quantity B is the magnetic flux density and it is measured in Weber per meter squared ( Wb m 2 ). The constant μ 0 is known as the permeability of free space. and its value is

μ 0 = 4 π × 10 −7 H m

(5.25)

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The magnetic flux ψ m passing through a surface S as illustrated in Fig. (5.13), is related to the magnetic flux density B by

ψ m = ∫ B ⋅ dS

(5.26)

S

The magnetic flux is measured in Weber ( Wb ). ψm

B

S Fig. (5.13). Magnetic flux and magnetic flux density.

Over a closed surface S shown in Fig. (5.14), the magnetic flux enters the surface S through the area S 1 with magnetic flux density B 1 and leaves it through the area S 2 with magnetic flux density B 2 . Therefore, the integral of the magnetic flux density over the closed surface S is

∫∫ B ⋅ dS = ∫∫ B

1

⋅ dS 1 + ∫∫ B 2 ⋅ dS 2 = − ψ m + ψ m = 0

(5.27)

1 2

 

 

S

S

S

Flux entering

Flux leaving

S2

S1

dS1 B

S

dS2 B2

Fig. 5.14. Gauss’s law of magnetostatic fields.

Equation (5.27) is similar to Gauss’s law of electrostatics for a source-free closed surface S , which is given by

∫∫ D ⋅ dS = 0 S

(5.28)

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Based on (5.27) and (5.28), Gauss’s law of magnetostatic can be stated as

∫∫ B ⋅ dS = 0

(5.29)

S

Example 5.4 Find the magnetic flux density and the magnetic flux per unit length in the region between the inner and outer conductors of the coaxial cable in Example 5.2. Solution Using the results of Example 5.2, the magnetic flux in the region between the conductors is obtained as follows

H = aφ

I 2πρ

⇒

B = μ 0 H = aφ

μ0 I 2πρ

Web m2 , a ≤ ρ ≤ b

Using (5.26), the magnetic flux is obtained as

ψm = ∫ B ⋅ dS = ∫∫ S

S

μ0 I aφ ⋅ dS 2πρ

Referring to Fig. (5.15), dS = aφ dρ dz , hence the magnetic flux per unit length is 1 b

μ0 I μ I aφ ⋅ aφ dρ dz = 0 2πρ 2π 0 a

ψm = ∫∫

y

1 b

1

∫∫ ρ dρ dz = 0 a

μ0 I ln(b a ) Web 2π

dS = aφ dρ dz l=1m B

z

a b

dS

aφ

x

dρ

dz

Fig. (5.15). The differential surface for the coaxial cable of Example 5.4.

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MAGNETOSTATIC POTENTIALS

5.4.1. Magnetic Scalar Potential For a source free region, the current density J is zero, hence by substituting J = 0 in Ampere’s law that is given by (5.23), we obtain ∇×H = 0

(5.30)

It is known from vector properties that the curl of the gradient of a scalar quantity is zero. Based on (5.30) we can define a scalar quantity V m such that H = −∇V m , which satisfies (5.30) as follows

∇ × (∇V m ) = ∇ × H = 0

(5.31)

V m is know as the magnetostatic scalar potential and related to the magnetic field H by

H = −∇V m ,

(J = 0)

(5.32)

This is valid for source-free regions only or when J = 0 . Note that (5.32) is similar to the relation between the electric field and electric scalar potential E = −∇V . Assuming there exists a magnetic charge q m , the magnetic scalar potential may be defined in the same manner as the electric potential. Therefore, the magnetic scalar potential at a point may be defined as the work done in bringing a unit positive magnetic charge from infinity to the point. The force acting on a magnetic charge in a magnetic field H is given by F = q mH . Accordingly, the work done to move the charge from infinity to the point A along a path l is A

A

W = − ∫ F ⋅ dl = − q m ∫ H ⋅ dl ∞

(5.33)

∞

The magnetic scalar potential of point A becomes A

W V = m = − ∫ H ⋅ dl q ∞ m A

(5.34)

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where the differential element dl is along the path l . If l is a closed path, then

V m = −∫ H ⋅ dl = IT

(5.35)

l

Where IT is the total enclosed current by the path l , which is positive when H and dl in the same direction and negative when they are in opposite directions. The unit of the magnetic scalar potential is Ampere ( A ). As in the case of scalar electrostatic potential, (5.35) can be written in the differential form as H = −∇V m

(5.36)

Based on our assumption that a magnetic charge exists, then the expression of Gauss’s law in (5.29) for a volume v bounded by a closed surface S and containing a magnetic charge with a volume charge density ρ vm , changes to

∫∫ B ⋅ dS = ∫ ρ

m v

dv

(5.37)

v

S

Using the divergence theorem, (5.37) can be written as

∇ ⋅ B = ρ , or m v

ρvm ∇⋅H = μ0

(5.38)

From (5.36) and (5.38), we can write Poisson's equation for the scalar magnetic potential as

∇ 2V m = − ρ vm μ 0

(5.39)

∇ 2V m = 0

(5.40)

When ρ vm = 0 , (5.39) reduces to

5.4.2. Magnetic Vector Potential We know from the vector properties that, the divergence of a vector results in a scalar quantity, and the curl of a vector results in another vector. Let A be a vector quantity, then P = ∇×A

(5.41)

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The resulted vector P is normal to both A and ∇ . Since P and ∇ are perpendicular to each other, then their dot product must be zero, or

∇ ⋅ P = ∇ ⋅ (∇ × A) = 0

(5.42)

∇ ⋅ B = ∇ ⋅ (∇ × A ) = 0

(5.43)

Replacing P by B , then

Consequently, we can define a vector A such that the magnetic flux density B can be expressed as a curl of the vector A as B = ∇×A

(5.44)

In terms of the magnetic field intensity, (5.44) can be written as

H=

1 ∇×A μ

(5.45)

The vector A is known as the magnetic vector potential that results from electric current sources. The definition of A in (5.45) does not a unique value. For example, consider A and A ′ given respectively by

A = a x Ax ( x, y, z ) + a y Ay ( x, y, z ) + a z Az ( x, y, z )

(5.46a)

A′ = a x Ax′ ( x) + a y A′y ( y ) + a z Az′ ( z )

(5.46b)

and

It is clear that both A and A + A ′ satisfy (5.45), or ∇ × ( A + A′) = ∇ × A + ∇ × A′ = ∇ × A

(5.47)

To define A uniquely we need to define its divergence beside its curl. For static fields the divergent of A is set to zero, or ∇⋅A = 0

(5.48)

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Equation (5.48), which is known as a Coulomb gauge. This equation with (5.45) yield a unique vector field A . The magnetic vector potential A in magnetic fields plays the role of electric scalar potential V in electric fields. The electric field E can be obtained once V is known. Similarly, the magnetic field H can be obtained from A using (5.45). Taking the curl for both sides of (5.45) and using the vector identity A × B × C = ( A ⋅ C )B − ( A ⋅ B )C , yields

∇×H =

1 1 ∇ × ∇ × A = ∇(∇ ⋅ A ) − ∇2 A μ μ

(5.49)

Comparing (5.23), (5.49), we get

∇ 2 A − ∇(∇ ⋅ A ) = − μ 0 J

(5.50)

Substituting ∇ ⋅ A = 0 from (5.48) into (5.50), yields

∇2A = − μ0 J

(5.51)

Equation (5.51) is the Poisson's equation for the magnetic vector potential, which is similar in form to the Poisson's equation for the electric scalar potential given by

∇ 2V = −

ρv ε0

(5.52)

We know that from Chapter 4, the solution of (5.52) is

V (r ) =

1 4π ε 0

ρv ∫v r − r′ dv′

(5.53)

By similarity between (5.51) and (5.52), we can immediately write the solution of (5.51) as A (r ) =

μ0 4π

J (r ′ ) ∫v R dv ′

(5.54)

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where R = r − r ′ . r ′ and r are the position vectors of the source and field points respectively as illustrated in Fig. (5.16). In the rectangular coordinate system

Source Point (x`, y`, z`) I d1`

r′ = a x x′ + a y y′ + a z z′

(5.55a)

r = ax x + a y y + az z

(5.55b)

z r - r` l

r`

Field Point

s

r

r - r`

(x, y, z) Field Point

r

r`

y

y

x

z

Source Point (x`, y`, z`) K dS`

(x, y, z)

x (b)

(a) z Source Point r - r`

(x`, y`, z`)

v

J dv`

(x, y, z)

r

r`

Field Point

y x (c) Fig. (5.16). Rectangular coordinates system for the vector potential. (a) linear current element. (b) Surface current element. (c) Volume current element.

For a surface, current distribution K and a linear current I , (5.54) becomes μ 0 K (r ′ ) dS ′ 4 π ∫S R μ I (r′) A(r ) = 0 ∫ dl′ 4π l R

A (r ) =

(5.56) (5.57)

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Using (5.26) along with (5.44), the relation between the magnetic flux ψ m through an area S enclosed by a closed loop l , and the vector potential A , can be obtained as

ψ m = ∫∫ B ⋅ dS = ∫ ∇ × B ⋅ dl = ∫ A ⋅ dl S

l

(5.58)

l

Example 5.5 A thin-wire of length L is placed along the z-axis. If the wire carries a current I, find the vector potential and magnetic field produced by the wire on the plane z = 0 . Find the approximate vector potential and magnetic fields when L >> ρ . Solution Let the conductor be along z-axis and symmetrical about the origin as shown in Fig. (5.17), then the position vectors of the field point r and source point r′ are r = a x ρ cosφ + a y ρ sin φ + a z z , r′ = a z z′ ⇒ R = r − r′ =

(z − z′)2 + ρ 2

The current is along z-axis, then dl′ = a z dz . From (5.57), the vector potential is

A = az

μ0 I z dz′ 4π ∫z R

Substituting R and I z = I , the vector potential at any point (ρ ,φ , z ) becomes

μ I Az = 0 4π

+L 2

1

∫ ( z − z′)

2

−L 2

+ρ

μ I ⎡ (z − L 2) + = − 0 ln ⎢ 4π ⎣⎢ ( z + L 2) +

2

dz′ = −

[

μ0 I ln z − z′ + 4π

(z − z′)2 + ρ 2 ]− L 2

+L 2

(z − L 2)2 + ρ 2 ⎤ ⎥ (z + L 2)2 + ρ 2 ⎦⎥

On the plane z = 0 , Az becomes

(

2 μ 0 I ⎡ − L 2 + (L 2) + ρ 2 ⎤ μ 0 I ⎡ L 2 + Az = − ln ⎢ ln⎢ ⎥= 4π ⎣⎢ L 2 + (L 2)2 + ρ 2 ⎦⎥ 2π ⎣⎢

(L 2)2 + ρ 2 )⎤ ρ

⎥ ⎦⎥

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(ρ , φ , z )

z R + L2

r

(0, 0, z′)

r′

y

φ

x

−L2

Fig. (5.17). Vector potential for a finite length thin-wire current-carrying conductor.

For a long wire L >> ρ , Az reduces to

Az =

μ 0 I ⎡⎛ ln ⎜ L 2 + 2π ⎢⎣⎝

(L 2)2 + ρ 2 ⎞⎟ ⎠

ρ⎤ ≈ ⎥⎦

μ0 I ⎛ L ⎞ ln⎜ ⎟ 2π ⎜⎝ ρ ⎟⎠

Since Aρ = Aφ = 0 , using (1.142b) the magnetic field in cylindrical coordinate is

H=

1 1 ∂Az 1 ∂Az ∇ × A = aρ − aφ μ0 μ 0 ∂φ μ 0 ∂ρ

Since Az does not depend on φ , ∂Az ∂φ = 0 , and H on the plane z = 0 becomes

H = −aφ

1 ∂Az I ∂ ⎡⎛ ln ⎜ L 2 + = −aφ μ 0 ∂ρ 2π ∂ρ ⎢⎣⎝

⎡ I ⎢ = −aφ 2π ⎢ ⎢ ⎣

(L 2)2 + ρ 2 ⎞⎟ ⎠

ρ⎤ ⎥⎦

⎤ 1⎥ I − ⎥ = aφ 2πρ (L 2)2 + ρ 2 ⎜⎛ L 2 + (L 2)2 + ρ 2 ⎞⎟ ρ ⎥ ⎝ ⎠ ⎦

ρ

L

(L 2)2 + ρ 2

For long wire L >> ρ , then (ρ L ) → 0 and H reduces to 2

H = aφ

I 2πρ

This result can be obtained using Ampere’s law as discussed in Section 5.2.1.

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Example 5.6 If the vector potential in free space due to an electric current distribution is 2

2

A = x ya x + xy a y − 4 xyza z

Wb m

Find the magnetic flux density at (1, 2, 0) and the flux through the surface z = 1 , 0 ≤ x ≤ 1, −1 ≤ y ≤ 4 . Solution Given that A = x2 ya x + xy 2a y − 4 xyza z , then the magnetic field is

ax a y az ∂ ∂ ∂ B = μ0 H = ∇ × A = ∂z ∂x ∂y 2 2 x y xy − 4 xyz = −4 xza x + a y 4 yz + ( y 2 − x 2 ) a z At the point (1, 2, 0) ⇒ B = 3a z Web m2 . With reference to Fig. (5.18), the flux through the surface z = 1 , 0 ≤ x ≤ 1 , − 1 ≤ y ≤ 4 is 4 1

ψ = ∫∫ B ⋅ dS = ∫∫ [− 4 xza x + a y 4 yz + (y 2 − x 2 )a z ]⋅ a z dxdy −1 0

S

4 1

⇒ ψ =

∫∫ (y

−1 0

2

)

− x 2 dxdy =

∫ [y x − x 3] 4

2

3

x =1

4

x = 0 dy =

−1

∫ (y

2

−1

y y=4

D

C

dS = az dxdy

y = -1

x A

B x=1

Fig. (5.18). The differential surface for the problem of Example 5.6.

)

− 1 3 dy = 20 Wb

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Alternatively, the flux can be obtained by finding the line integral of A along the loop ABCDA, then

ψ = ∫ A ⋅ dl = ∫ A( y = −1) ⋅ a x dx + ∫ A(x = 1) ⋅ a y dy l

AB

BC

+ ∫ A( y = 4) ⋅ a x dx + ∫ A(x = 0) ⋅ a y dy CD

⇒ ψ

DA

1

0

4

−1

0

1

−1

4

= (−1) ∫ x 2 dx + (4)∫ x 2 dx + (1) ∫ y 2 dy − (0) ∫ y 2 dy 1 4 65 =− − + = 20 Wb 3 3 3

5.5.

BIOT–SAVART’S LAW

The vector potential A at a point defined by the position vector r , can be determined for a given current density J at a point defined by the position vector r ′ within a volume v , from (5.54) as

A (r ) =

μ0 4π

∫

v

J (r ′ ) dv ′ r − r′

(5.59)

Once the vector potential is known, the magnetic field H can be obtained from (5.45). Substituting A from (5.59) into (5.45), yields

H=

J(r′) 1 1 dv′ ∇ × A(r ) = ∇× ∫ r − r′ μ0 4π v

(5.60)

Taking into account that r − r′ is scalar quantity, then using the vector identity ∇ ×ψA = ∇ψ × A + ψ∇ × A , (5.60) can be written as H=

1 4π

⎧ ⎛

∫ ⎨⎩ ∇ ⎝⎜⎜ v

1 r − r′

⎫ ⎞ 1 ⎟⎟ × J ( r ′ ) + ∇ × J ( r ′ ) ⎬ dv ′ r − r′ ⎠ ⎭

(5.61)

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Note that in (5.61), the operator ∇ is with respect to r only, and hence r ′ is constant with respect it. Thus,

⎛ 1 ⎞ r − r′ ⎟⎟ = − ∇⎜⎜ , 3 r − r′ ⎝ r − r′ ⎠

∇ × J(r′) = 0

(5.62)

Consequently, (5.61) becomes H=

1 4π

⎛

∫ ∇ ⎜⎜⎝ v

1 r − r′

⎞ 1 ⎟⎟ × J ( r ′ ) dv ′ = 4π ⎠

r − r′

∫ dv ′J (r ′ ) × r − r ′ v

3

(5.63)

According to (5.63), the magnetic field dH at a point described by the position vector r , due to a current density J at a point described by the position vector r ′ can be obtained as

dH =

1 r − r′ dv′J (r ′)× 3 4π r − r′

(5.64)

Equation (5.64) is known as Biot – Savart’s law. For a surface current density element KdS ′ and a linear current element Idl ′ , we have dv′J = KdS ′ = Idl′

(5.65)

Replacing dv ′J by KdS ′ and Idl ′ in (5.64), Biot – Savart’s law for surface current distribution and linear current is obtained as

dH =

1 r − r′ dS ′K × 3 4π r − r′

(5.66)

1 r − r′ Idl′ × 3 4π r − r′

(5.67)

and

dH =

The geometry of the source and field points when applying Biot – Savart law for a linear current source is shown in Fig. (5.19).

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z

333

(x′, y′, z′) Source point

dl

r - r′ r′

l

dH (x, y, z) Field point r

I

y

x Fig. (5.19). Biot – Savart law for a linear current source.

Example 5.7 Find the magnetic field strength produced by a thin-wire of length L carrying a current I at a radial distance ρ from wire. Assume the wire is along the z-axis and extends from (0,0, d − L) to (0,0, d ). Then find the magnetic field when the wire extends (a) from − ∞ to + ∞ , (b) from 0 to + ∞ . Solution Let the wire be along the z-axis extends from (0, 0, d − L) to (0, 0, d ) as shown in Fig. (5.20). Using cylindrical coordinate system, the source point is at ( 0, 0, z ′ ) and the field point at ( ρ , φ , 0 ). Referring to Fig.(5.20), we have

r = ρ a ρ , r ′ = z ′a z

Idl = a z Idz′ ,

⇒ r − r′ = a ρ ρ − a z z ′

Using Biot – Savart’s law, the magnetic field due to the wire is

H=

1 Idl × (r − r′) 4π ∫l r − r′ 3

Substituting r − r′ and Idl = a z Idz′ into Biot – Savart’s law, yields

I H= 4π

dz′

d

∫ (ρ

d −2L

2

+ z ′2

′ 3 a z × (ρ a ρ − z a z ) =

)

2

Iρ 4π

dz′

d

∫ (ρ

d −2L

2

+ z′2

)

3 2

aφ

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z

z (0, 0, d) Source point (0,0,z′)

Idl = az I dz′ d r - r′

r'

x

(0, 0, d-L)

ρ

y

r

L- d

(ρ,φ, 0) Field point

I

(a)

θ2 θ1

Field point

(b)

Fig. (5.20). Geometry of the problem in Example 5.7.

Using the substitution z ′ = ρ tan θ , the above integral becomes H = aφ = aφ

I

θ 2 = tan −1

4π ρ θ I 4π ρ

() d

ρ

∫ cosθ dθ

−1 1 = tan

( ) d −L

ρ

(sin θ 2 − sin θ1 )

Referring to Fig. (5.2b), we have

sin θ 2 =

d d + ρ2 2

and

sin θ1 =

d −L

(d − L)2 + ρ 2

Hence, the field at a radial distance ρ from the wire becomes

H = aφ

I ⎡ d + ⎢ 2 4π ρ ⎣ d + ρ 2

⎤ ⎥ (L − d ) + ρ ⎦ L−d 2

2

If the wire is symmetrical about the origin, then d = L 2 and H reduces to

H = aφ

I

L

4π ρ

(L 2)2 + ρ 2

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(a) When the wire extends from − ∞ to + ∞ , L → ∞ . Thus, the magnetic field at a distance ρ from the wire reduces to

H = aφ

⎡ lim ⎢ 4π ρ L →∞ ⎢⎣ I

L2

(L 2 ) + ρ 2 2

+

⎤ I = a ⎥ φ 2π ρ (L 2) + ρ 2 ⎥⎦ L2 2

(b) When the wire extends from 0 to + ∞ , then d = L and L → ∞ . Thus, the magnetic field, in this case, reduces to

H = aφ

⎤ ⎡ L I lim ⎢ 2 = aφ ⎥ 2 4π ρ L →∞ ⎣ L + ρ ⎦ 4π ρ I

Example 5.8 A thin circular loop of radius a carries a constant current I . Assume that the loop lies in the xy plane with its center coincides with the origin, find the magnetic field intensity on the axis of the loop at a distance z above its center. Solution The geometry of the loop is shown in Fig. (5.21). From the figure, we have

r = a z z , r′ = a x a cosφ ′ + a y a sin φ ′ − a x a cosφ ′ − a y a sin φ ′ + a z z r − r′ = ⇒ 3 3 r − r′ (z 2 + a 2 ) 2 Idl′ = aφ Iadφ ′ = Iadφ ′(− a x sin φ ′ + a y cosφ ′) Using (5.67), then dH =

( a z cosφ ′ + a y z sin φ ′ + a z a ) 1 1 r − r′ = Idl′ × Iadφ ′ x 3 3 4π 4π r − r′ (z 2 + a 2 ) 2

Integrating both sides, the magnetic field on the axis of the loop is obtained as

H=

2π

Ia

4π (z 2 + a 2 ) 2 3

∫ ( a z cosφ ′ + a z sin φ ′ + a a ) dφ ′ = a x

0

y

z

Ia2 z

3

2 ( z 2 + a2 ) 2

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z Field point (0, 0, z) r φ′

I

x

y r'

Idl′ = aφ a Idφ′ Source point (a, φ′, 0)

Fig. (5.21). Geometry of the thin circular current loop of Example 5.8.

Example 5.9 Find the magnetic field intensity produced by a uniform surface current density K = a x K x A m over a square plate of side 2 L on its axis at a distance z from its center, then find the magnetic field at the same point when the plate is infinitely extended. Solution Let the plate coincides with the xy plane and symmetrical about the origin as depicted in Fig. (5.22). The geometry of the problem is suitable for the rectangular coordinate system. With reference to Fig. (5.22), the source point (x′, y′,0) is on the xy plane and the field point is at the point (0,0, z ) , then

r = zaz , r′ = x′a x + y′a y , r − r′ = −(x′a x + y′a y − za z ) r − r′ = z 2 + x′2 + y′2 z Field point (0, 0, z) r

r - r′

K = ax (L, 0, 0)

x Fig. (5.22). Geometry of the problem in Example 5.9.

r

(0, L, 0)

Source point K dS′ = K dx′dy′

y

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The electric field due to the charge on the plate at a distance z above its center is

H=

⇒

H=−

1 4π

L L

∫ ∫ (z

− L− L

1 4π

r − r′

∫∫ dS ′K × r − r′

3

S

K x dx′dy′ ′ ′ 3 a x × (x a x + y a y − za z ) 2 + x′2 + y′2 ) 2

This integral can be written in the form L L Kx dx′dy′ ′ Η=− 3 (za y + y a z ) 4π −∫L−∫L (z 2 + x′2 + y′2 ) 2

The integral in the direction of a z is zero, then L L Kx z dx′dy′ H = ay 3 4π −∫L−∫L (z 2 + x′2 + y′2 ) 2

This integral can be evaluated following the same procedure as in Example 5.7, as

H = −a y

⎛ L2 tan −1 ⎜⎜ 2 2 π ⎝ z z + 2L

Kx

⎞ ⎟⎟ ⎠

If the plate is infinitely extended, then L → ∞ . The magnetic field, in this case, is obtained by letting L → ∞ . Doing this, H due to a uniform surface current density K = a x K x A m on an infinitely extended plate is obtained as

⎡ −1 ⎛ L2 lim tan ⎜⎜ H = −a y 2 2 π L →∞ ⎢⎣ ⎝ z z + 2L Kx

⎞⎤ K ⎟⎟⎥ = −a y x , 2 ⎠⎦

z>0

Similarly, the magnetic field intensity in the region z < 0 can be obtained as

H = ay

Kx az , 2

z0

The force on a differential element Idl = Idza z along the wire due to B is

μ K ⎞ μ IK ⎛ dF = Idl × B = Idza z × ⎜ − a y 0 x ⎟ = a x 0 x dz 2 ⎠ 2 ⎝ The total force experienced by the wire is obtained by integrating dF along the wire, as

μ IK F = ax 0 x 2

1.25 L .

∫ dz = a L

x

μ 0 IK x L 8

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z (0, 0, 1.25L) Idl = az Idz

B = -ay Kx/2 (0, 0, L)

I

K = ax Kx

y x Fig. (5.26). Geometry of the problem in Example 5.13.

5.6.3. The Force between Two Current Elements Consider two filamentary conductors 1 and 2, which are arbitrarily positioned as shown in Fig. (5.27) in free space. Conductor 1 carries a constant current I 1 and conductor 2 carries a constant current I 2 , with their lengths are l 1 and l 2 respectively. Let the differential current element I1d l1 on conductor 1 at a point described by the position vector r1 and I 2d l 2 on conductor 2 at a point described by the position vector r1 . The magnetic flux density produced by current element I1d l1 at the element I 2d l 2 can be obtained using Biot – Savart’s law as

dB1 = μ 0 dH1 =

μ 0 I1 dl1 × (r2 − r1 ) 3 4π r2 − r1

(5.77)

The magnetic force d ( dF12 ) on the current element I 2d l 2 due to B 1 is obtained by replacing Id l by I 2d l 2 and B by B 1in (5.77). Hence,

d (dF12 ) =

μ0 dl × (r2 − r1 ) I1 I 2 dl 2 × 1 3 4π r2 − r1

(5.78)

If conductor 1 and conductor 2 are in a shape of closed loops as shown in Fig. (5.28), then the force on loop 2 due to the current of loop 1 can be obtained by integrating the right-hand side of (5.78) around each loop. Thus,

F12 =

μ0 dl × dl1 × (r2 − r1 ) I1 I 2 ∫ ∫ 2 3 4π r2 − r1 l 2 l1

(5.79)

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Fig. (5.27). Force between two current elements.

Following the same procedure, it can be shown that the force on loop 1 due to the current of loop 2 is F21 = −F12 . Loop 1

Loop 2

I1 r2 – r1

I1dl1 l1

I2dl2 r2

r1

l2

Origin Fig. (5.28). Force between two current loops.

Using the vector identity A × B × C = ( A ⋅ C )B − ( A ⋅ B )C , (5.79) can be written as

F12 =

⎡ d l ⋅ (r − r ) ⎤ μ0 (r − r ) I1I 2 ⎢ ∫ ∫ 2 2 3 1 d l1 − ∫ ∫ 2 1 3 d l 2 ⋅ d l1 ⎥ 4π r − r1 l 2 l1 2 ⎣⎢l 2 l1 r2 − r1 ⎦⎥

(5.80)

Substituting ∇(1 r2 − r1 ) = − (r2 − r1 ) r2 − r1 3 into the first term of (5.80), yields

F12 =

⎡ ⎤ μ0 (r − r ) I1I 2 ⎢ ∫ d l1 ∫ d l 2 ⋅ ∇(1 r2 − r1 ) − ∫ ∫ 2 1 3 d l 2 ⋅ d l1 ⎥ 4π r − r1 l2 l 2 l1 2 ⎣⎢l1 ⎦⎥

(5.81)

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The line integral around l 2 in the first term of (5.81) can be converted to a surface integral by applying Stokes’ theorem. Hence,

F12 =

⎡ ⎤ μ0 (r − r ) I1I 2 ⎢ ∫ d l1 ∫ ∇ × ∇(1 r2 − r1 ) dS 2 − ∫ ∫ 2 13 d l 2 ⋅ d l1 ⎥ (5.82) 4π S2 l 2 l1 r2 − r1 ⎣⎢l1 ⎦⎥

Where S 2 is the area enclosed by l2 . Since ∇ × ∇(1 r2 − r1 ) = 0 , then (5.82) reduces to

F12 = −

r −r μ0 I1I 2 ∫ ∫ 2 13 d l 2 ⋅ d l1 4π l 2 l1 r2 − r1

(5.83)

Example 5.14 Two differential current elements I1d l1 = 2a x A⋅ m and I 2d l 2 = 5a z A⋅ m are located at (0, 0, 3) m and (1, 0, 0) m respectively. Find the force on (a) I 2d l 2 due to I1d l1 . (b) I1d l1 due to I 2d l 2 . Solution

I1d l1 and I 2d l 2 and the position vectors of their locations are given by

I1d l1 = 25a x , I 2d l 2 = 20πaz r1 = 3a z , r2 = 4a x (a) Using (5.78), the force on I 2dl 2 due to I1dl1 can be obtained as d ( dF12 )

⇒

μ0 dl × (r2 − r1 ) I1I 2 dl 2 × 1 3 4π r2 − r1 μ a × (4a x − 3a z ) = 0 500a z × x 4 (16 + 9)3 2

=

d (dF12 ) = −3 μ0 a x N

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(b) Similarly, the force on I1dl1 due to I 2dl 2 is

d (dF21 )

d l × (r1 − r2 ) μ0 I1I 2 d l1 × 2 3 4π r1 − r2 a × (− 4a x + 3a z ) μ = 0 500a x × z 4 (16 + 9)3 2 =

⇒ d (dF21 ) = −4 μ 0 a z N Note that for current elements, d (dF12 ) is not equal to d (dF21 ) . Example 5.15 Find the force per unit length between two infinitely long parallel conductors. The conductors are separated by a distance d , and carry currents I1 and I 2 . Solution Let the conductors as shown in Fig. (5.29), where conductor 1 carries the current I1 and conductor 2 carries I 2 . The magnetic force d ( dF12 ) on the current element I 2 dl 2 due to I1dl1 is given by (5.78) as

d (dF12 ) =

dl × (r − r ) μ0 I1I 2 dl 2 × 1 2 3 1 4π r2 − r1

From Fig. (5.29), we have

I1dl1 = az I1dz1 , I 2dl 2 = a z I 2dz2 ,

r1 = a z z1 , r2 = a y d + a z z2

Substituting these quantities into the above expression for d (dF12 ) , yields

μ d (dF12 ) = 0 I1I 2 dz 2a z × 4π =

dz1a z × ⎡a y d + a z ( z2 − z1 )⎤ ⎢⎣ ⎥⎦ 3 ⎡d 2 + ( z − z ) 2 ⎤ 2 1 ⎢⎣ ⎥⎦ dz1dz 2

μ0 d I1I 2 3 4π ⎡d 2 + ( z − z ) 2 ⎤ 2 1 ⎢⎣ ⎥⎦

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z (0, 0, z1)

I1dl1 I2dl2 (0, d, z2)

r1 r1

y I2

I1

x

d

Conductor 1

Conductor 2

Fig. (5.29). Force between two long current conductors.

The total force on the current element I 2 dl 2 due to conductor 1 is obtained by integrating the last equation along the entire length of the conductor; thus

dF12 = −a y

+∞

μ0 d I1I 2 dz 2 ∫ 4π −∞

[d

dz1 2

]

3 2 2

+ (z2 − z1 )

Using the substitution z2 − z1 = d tanθ , the above integral reduces to +π 2

μ I I dl dF12 = −a y 0 1 2 2 4π d

−

∫ cosθ dθ = −a π 2

y

μ 0 I1I 2 dl2 2π d

The force per unit length on conductor 2 due to the current of conductor 1 is obtained by integrating dF12 along a unit length of conductor. Hence, 1

F12 = −a y

μ 0 I1I 2 μ II dl2 = −a y 0 1 2 ∫ 2π d 0 2π d

Nm

Similarly, the force per unit length on conductor 1 due to conductor 2 is

F21 = +a y

μ 0 I1I 2 2π d

Nm

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The minus sign in F12 indicates that the force on conductor 2 is towards conductor 1, while the plus sign in F21 indicates that the force on conductor 1 is towards conductor 2, which means the conductors attract each other. Example 5.16 A rectangular loop of current I and dimensions l × w is situated on the yz plane with its center at a distance d from the y-axis. Find the force on the rectangular loop due to an infinitely long wire carrying current I ′ along the y-axis as shown in Fig. (5.30). l

z

z

B1 I

A

I

F1

F4

w

B4 D

d I′

F3

I

B

B2

F2 C

B3 I′

y

y

Fig. (5.30). Geometry of the problem in Example 2-6.

Solution H and B due to I ′ at a distance z from the y-axis on the yz plane are

H = ax

I′ , 2πz

B = ax

μ0 I ′ 2πz

Referring to Fig. (5.30), sides AB and CD are at distances z = d + w 2 and z = d − w 2 from the y-axis respectively. Thus,

B1 = a x

μ0 I ′ 1 , 2π d + w 2

B3 = a x

μ0 I ′ 1 2π d − w 2

Using (5.75), the force on the current loop is

F = ∫ Idl × B = F1 + F2 + F3 + F4 l

= ∫ Idya y × B1 − ∫ Idya z × B2 − ∫ Idya y × B3 + ∫ Idya z × B 4 AB

BC

CD

DA

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It is clear that the force on side BC is equal and opposite to that on side DA, or ( F2 = −F4 ), then the net force in the y-direction is zero. Thus,

F

= ∫ Idya y × B1 + ∫ Idya y × B3 AB

CD

μ Il ⎡ I′ I′ ⎤ = 0 ⎢− a y × a x + a y × ax ⎥ 2π ⎣ d+w 2 d − w 2⎦

⇒ 5.7.

F = az

μ 0 I ′Il ⎛ 1 1 ⎜⎜ − 2π ⎝ d + w 2 d − w

⎞ μ I ′I ⎛ 2w l ⎞ ⎟⎟ = −a z 0 ⎜ 2 ⎟ N π ⎝ 4d − w2 ⎠ 2⎠

MAGNETOSTATIC ENERGY

It has been shown that in Section 2.10.1, the electrostatic energy for a continuous distribution of electric charges with a volume charge density ρ v (r′) at a point defined by the position vector r′ is given by

WT =

1 ρv (r′)V (r′)dv′ 2 ∫v

(5.84)

Where r′ is the position vector defining a point within the volume v , and V (r′) is the electric potential at that point. Assuming that the magnetic charge exists and using the concept of magnetic scalar potential that is introduced in Section 5.4.1, the magnetostatic energy for a continuously distributed magnetic charge with a volume charge density ρvm (r′) can be written in a similar manner as in (2.86c).

Assuming the magnetic scalar potential at a point defined by r′ is V m (r′) , then the total magnetostatic energy of the charge distribution is

WT =

1 m ρv (r′)V m (r′)dv ∫ 2v

(5.85)

The energy expression in (5.85) can be written in terms of the magnetic flux density B by substituting the volume charge density ρ vm from (5.38) into (5.85), as

WT =

1 m V (r′)∇ ⋅ B dv′ 2 ∫v

(5.86)

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Using vector identity ∇ ⋅ Aψ = ψ ∇ ⋅ A + A ⋅ ∇ψ , we have

V m ∇ ⋅ B = ∇ ⋅ BV m − B ⋅ ∇V m

(5.87)

Using (5.32), − ∇V m can be replaced by H in (5.87), then

V m ∇ ⋅ B = ∇ ⋅ BV m + B ⋅ H

(5.88)

Substituting the quantity V m ∇ ⋅ B from (5.88) into (5.86), the energy becomes

WT =

1 1 B ⋅ H dv′ + ∫ ∇ ⋅ B V m (r′) dv′ ∫ 2v 2v

(5.89)

If the medium is charge-free, or ρvm = 0 , then ∇ ⋅ B = 0 , which is always true for the magnetic fields, the second term in (5.89) vanishes. Therefore, the magnetostatic energy stored in a static magnetic field H within a volume v is obtained from (5.89) as

WT =

1 μ 1 2 B ⋅ H dv′ = 0 ∫ H dv′ = B 2dv′ ∫ ∫ 2v 2 v 2 μ0 v

(5.90)

Consequently, the magnetostatic energy per unit volume can be written from (5.90) as

1 μ 1 2 Wv = B ⋅ H = 0 H = B 2 2 2μ0

2

(5.91)

Example 5.17 Find the magnetostatic energy stored in the region between the inner and outer conductors of the coaxial cable in Example 5.2. Assuming the region between the conductors is free space. Solution From Example 5.2, the radii of the inner and outer conductors are a and b respectively. The magnetic field in the region between the conductors is

H = aφ

I 2πρ

, a≤ ρ ≤b

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Substituting H into (5.90), with ρ = ρ ′ , the magnetostatic energy is obtained as

μ0 μ0 I 2 1 2 ′ Wm = ∫ H dv = dv′ 2 v 8π 2 ∫v ρ ′2 In cylindrical coordinate system, dv′ = ρ ′ dρ ′dφ ′ dz′ , then

Wm = 5.8.

1 2π b

μ0 I 2 1 μ0 I 2 ′ ′ ′ ρ φ d d d z = ln(b a ) J ′ ρ π 8π 2 ∫0 ∫∫ 4 0 a

MULTI-POLE EXPANSION FOR THE VECTOR POTENTIAL

Following the same analysis carried in Chapter 4 for the scalar potential, the vector potential can be expanded in infinite series. Assuming that the current source I is at (r′,θ ′,φ′), the field point at (r,θ ,φ ), and the distance between them is R as illustrated in Fig. (5.31). We know from (4.38), 1 R can be expanded in terms of the coordinates of the source point (r ′,θ ′, φ ′) and field point (r,θ , φ ) as +∞ 1 − jn (φ −φ ′ ) = ∑ Ψn e R n = −∞

(5.92)

where

⎧ ′ k ′ ⎪ r (r r ) , r > r (k − n )! n n Pk (cosθ )Pk (cosθ ′) ⎨ Ψn = ∑ 1 k = n (k + n )! ⎪ (r r ′)k , r < r ′ +∞

1

(5.93)

⎩ r′

Pkn (⋅) are associated Legendre polynomials of order k and degree n . Substituting 1 R from (5.92) into (5.57) and using (5.93), the vector potential can be written as

⎧∫ I (r′ r )k Pkn (cosθ ′)e− jn(φ −φ ′ )dl ′, r > r′ (k − n )! n μ0 1 ⎪l (5.94) A= Pk (cosθ ) ⎨ ∑∑ k +1 n − jn (φ −φ ′ ) 4π r n = −∞ k = n (k + n )! dl ′, r < r ′ ⎪ ∫ I (r r ′) Pk (cosθ ′)e ⎩l +∞

+∞

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z

Sameir M. Ali Hamed

Field point (r, θ , φ)

R

r

Current source r′

x

(r′, θ ′, φ ′) Source point

y

Fig. (5.31). Spherical coordinates of the source and field points for a current source.

For a surface current density K and volume current density J , (5.94) can be written respectively, as ⎧∫ K (r′ r )k Pkn (cosθ ′)e− jn(φ −φ ′ )dS ′, r > r′ (k − n)! n μ 1 ⎪ A= 0 ∑∑ Pk (cosθ ) ⎨ S k +1 n − jn (φ −φ ′ ) 4π r n = −∞ k = n (k + n )! dS ′, r < r′ ⎪∫ K (r r′) Pk (cosθ ′)e ⎩S (5.95) +∞

+∞

and ⎧∫ J (r′ r )k Pkn (cosθ ′)e− jn(φ −φ ′ )dv′, r > r′ μ0 1 (k − n)! n ⎪v A= Pk (cosθ )⎨ ∑∑ k +1 n − jn(φ −φ ′ ) 4π r n = −∞ k = n (k + n )! dv′, r < r′ ⎪ ∫ J (r r′) Pk (cosθ ′)e ⎩v (5.96) +∞

+∞

Equations (5.94) – (5.96) are general and give the vector potential at any point (r,θ ,φ ) outside and inside the sphere r = r′ separately due to linear, surface, and volume current densities respectively. Simple vector potential expressions for the cases of the thin-wire straight conductors and thin-wire circular loops can be obtained as special cases from the general expressions in (5.94) – (5.96). 5.8.1. Thin-wire Current-Carrying Conductor The problem is to find general expressions for a thin-wire conductor of length L carrying a constant current I in free space. We assume that the wire is along the

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z-axis and symmetrical about the origin as shown in Fig. (5.32). From the figure, the source point is (r′ = z′,θ ′ = 0,φ′), I = a z I , and dl ′ = dz′ . The field point is any point (r,θ ,φ ) in the space. Substituting r ′ = z′ , θ ′ = 0 , I = a z I and dl ′ = dz′ into (5.94), the vector potential can be written as ⎧+ L 2 k − jn (φ −φ ′ ) dz′, ⎪ ∫ I ( z′ r ) e +∞ +∞ μ0 (k − n )! n ⎪ Pk (cosθ )Pkn (1) ⎨−+LL 22 A = az ∑∑ 4π r n = −∞ k = n (k + n )! ⎪ I (r z′)k +1 e − jn(φ −φ ′ )dz′, ⎪ − L∫ 2 ⎩

r > z′

(5.97) r < z′

(r , θ , φ )

z R r

+L2

(r′, θ ′, φ ′) r′

φ

x

y

−L2

Fig. (5.32). Vector potential for a finite length thin-wire current-carrying conductor.

We have ⎧ n ⎪1, n = 0 Pk (1) = ⎨ ⎪ ⎩0, n ≠ 0

(5.98)

⎧+ L 2 k z′ r ) dz′, r > z′ ( ⎪ ∫ μ0 I +∞ ⎪− L 2 A = az ∑ Pk (cosθ ) ⎨+ L 2 k +1 4π r k = 0 ⎪ (r z′) dz′, r < z′ ⎪ − L∫ 2 ⎩

(5.99)

Consequently, (5.97) reduces to

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Note that +L 2

z ′k ∫ r k dz ′ = −L 2

+L 2

+L 2

+L 2

z ′k z ′k k ′ ( ) d z 1 + − ∫0 r k ∫0 r k dz′ k is odd ⎧0, ⎪ = ⎨ (L 2r )k +1 2r , k is even ⎪ k +1 ⎩

r k +1 ∫ z′k +1 dz′ = −L 2

+L 2

∫ 0

r k +1 k +1 dz′ + (− 1) k +1 z′

+L 2

∫ 0

(5.100a)

r k +1 dz′ z′k +1

⎧ (2r L )k ⎪ , k is odd = ⎨2 r k ⎪ k is even ⎩0,

(5.100b)

Then ⎧ + ∞ (L 2r )2 k P2 k (cosθ ), r>L 2 μ0 I L ⎪ ⎪∑ k = 0 2k + 1 A = az ⎨ 4π r ⎪ + ∞ (2r L )2 k P2 k −1 (cosθ ), r < L 2 −∑ ⎪ ⎩ k =1 2k − 1

(5.101)

Example 5.18 A current element of length L carrying a current I placed along the z-axis and symmetrical about the origin. Using the multi-pole expansion, find the vector potential A at a point (r > L,θ ,φ ) produced by the current element, then find the (a) Vector potential at any point z > L along the axis of the current element. (b) Vector potential at a point on the line perpendicular to the current element at a distance d > L from its center. (c) Magnetic field H at a distance far away from the current element. Solution For multi-pole expansion method, we need only the source point and field point in spherical coordinates. The current element is along z-axis as shown in Fig. (5.33). From the figure the source point is (r′ = z′,θ ′ = 0,φ′), I = a z I , and dl ′ = dz′ .

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(r ,θ ,φ )

z R +L2

r

(r ′, θ ′, φ ′)

r′

y

φ

x

−L2

Fig. (5.33). Geometry of the current element of Example 5.18.

The vector potential can be written directly from (5.101) as

A = az

μ 0 LI +∞ 1 (L 2r )2k P2k (cosθ ), ∑ 4π r k = 0 (2k + 1)

r>L 2

(a) Along the axis of the current element θ = 0 , cosθ = 1 , Pn (1) = 1 for all n , then the vector potential becomes

A = az

μI L⎡ 1 1 2 4 ⎤ 1 + (L 2r ) + (L 2r ) + ⎥ , ⎢ 4π r ⎣ 3 5 ⎦

r > L2

(b) At a distance d > L 2 from the center of the current element, θ = π 2 , cosθ = 0 . Since d > L , the vector potential is in the region r > L 2 , then

A = az

μI L⎡ 1 3 2 4 ⎤ 1 − (L 2d ) + (L 2d ) + ⎥ , ⎢ 2π d ⎣ 6 40 ⎦

d>L 2

Note that for k = 0,1, 2,, P2k (0) = (− 1)k (2k )! (22k k ! k !). (c) At a distance far away from the current element, r >> L 2 , the second term and above in the region r > L 2 can be ignored, and the vector potential reduces to

A = az

μI L 4π r

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Transforming a z to the spherical coordinates, then

A=

μ IL0 ⎛ cosθ sin θ ⎞ − aθ ⎜ ar ⎟ r r ⎠ 4π ⎝

Using (5.45), yields

H =

1 ∇×A μ0

a r raθ r sin θ aφ 1 1 ∂ ∂ ∂ = 2 μ 0 r sin θ ∂r ∂θ ∂φ Ar rAθ 0 1 ⎡ ∂ (rAθ ) ∂ Ar ⎤ − = aφ ∂θ ⎥⎦ μ r ⎢⎣ ∂r Substituting the vector potential into the above equation, the magnetic field produced by the straight wire of length L at a far distant r can be obtained as H = aφ

I L sin θ 4π r 2

5.8.2. Thin-Wire Circular Loop with a Uniform Current Consider a circular thin-wire loop of the radius a and carrying a current I in free space as shown in Fig. (5.34). Let the center of the loop coincides with the origin of the coordinate system, and the loop lies in the xy plane as shown in Fig. (5.34). The source and field points are (r′ = a,θ ′ = π 2 ,φ′) and (r,θ ,φ ) respectively. The current is along φ direction, and the length of the current element is dl = adφ ′ , then

r = a x r sin θ cosφ + a y r sin θ sin φ + a z r cosθ

(5.102a)

r′ = a x r ′ sin θ ′ cosφ ′ + a y r ′ sin θ ′ sin φ ′ + a z r ′ cosθ ′

(5.102b)

R = r − r′

= r 2 + r′2 − 2rr′[sin θ sin θ ′ cos(φ − φ ′) + cosθ cosθ ′]

(5.103)

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Field Point

z

359

(r,θ ,φ )

r Loop

R

a

y

r′

I

(r′,θ ′,φ ′)

x

Source Point

Fig. (5.34). Vector potential for a thin-wire circular loop with a uniform current.

The current in the φ direction, then the current in terms of the source point is

Iaφ = −a x I sin φ ′ + a y I cosφ ′

= −(a ρ cosφ − aφ sin φ )I sin φ ′ + (a ρ sin φ + aφ cosφ )I cosφ ′

(5.104)

= a ρ I sin (φ − φ ′) + aφ I cos(φ − φ ′)

Substituting I = Iaφ , r ′ = a , dl = adφ ′ and θ ′ = π 2 in (5.94), yields A =

μ0 I 4π

+∞

+∞ 2π

∑ ∑ ∫ [a ρ sin (φ − φ ′)+ aφ cos(φ − φ ′)]e

− jn (φ −φ ′ )

n = −∞ k = n 0

(k − n )! P n (cosθ )P n (0) ⎧(a r )k +1, × ⎨ k k (k + n )! k ⎩(r a ) ,

r>a

dφ ′

(5.105)

ra ra

(5.106b)

r a μ I +∞ 1 A = aφ 0 ∑ Pk1 (cosθ )Pk1 (0)⎨ k 2 k =1 k (k + 1) ⎩(r a ) , r < a k +1

(5.108)

Since Pk1 (0) = 0 when k + 1 is odd, then replacing k by 2k − 1 in (5.108), the vector potential for the current loop can be written in the form

⎧(a r ) , r > a μ I +∞ 1 A = aφ 0 ∑ P21k −1 (cosθ )P21k −1 (0)⎨ 2 k −1 4 k =1 k (2k − 1) ⎩(r a ) , r < a 2k

(5.109)

Equation (5.109) gives the multi-pole expansion of the vector potential for any circular thin loop of an arbitrary radius a and a uniform current I . It is clear that from this equation the multi-pole expansion of the magnetic vector potential for the loop does not contain the monopole term (the term with the factor 1 r ).

Magnetostatic Fields

5.9.

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THE MAGNETIC DIPOLE

A small loop of current I and area S is known as a Magnetic Dipole. Likewise the electric dipole, the magnetic dipole has a magnetic dipole moment m , which is a vector quantity. The magnitude of the magnetic dipole moment is SI and its direction is determined by the unit vector a n along the axis of the magnetic dipole. The direction of a n is the same as the direction of the magnetic field at the center of the dipole. Hence, once the direction of the current is known, the direction of a n can be determine by applying the righ-hand rule. The direction of the magnetic dipole moment is defined in Fig. (5.35). Mathematically, the magnetic dipole moment is expressed as

m = an IS = anm an

(5.110)

Current loop

I

S Fig. (5.35). The magnetic dipole.

Consider a magnetic dipole centered at the origin with its axis along the z-axis as shown in Fig. (5.36). The magnetic field of a small magnetic dipole at a point far away from it can be obtained using the vector potential obtained in the previous section for the circular loops. Field Point

z m = azSI = azIπ a2

r

Magnetic dipole

R a

I

x

(r θ φ )

y

r′

(r′ θ ′ φ ′)

Source Point

Fig. (5.36). Magnetic field of a magnetic dipole.



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When the loop is small and the field point is far away from the loop, then a > a ), is obtained as

Ia2 [ar 2 cosθ + aθ sin θ ] = IS 3 [ar 2 cosθ + aθ sin θ ] 3 4r 4π r m [ar 2 cosθ + aθ sin θ ] = 4π r 3

H =

(5.115)

where S is the area enclosed by the loop, and m = SI is the magnitude of the magnetic dipole moment. Equation (5.115) shows that the magnetic field of the magnetic dipole is similar in form to the electric field of the electric dipole that is given by (2.102). The magnetic field lines of the small magnetic dipole and that of the small bar magnet are similar as shown in Fig. (5.37). Hence, a small bar magnet may be considered as a magnetic dipole. Magnetic dipole m

Small bar magnet m

N S

I

Fig. (5.37). Magnetic field lines of a magnetic dipole and a small bar magentic.



Consider the small magnet shown in Fig. (5.38), then its dipole moment is

m = a z M = a z qml

(5.116)

where qm is the pole strength which is equivalent to a magnetic charge, and l is the distance between the poles. The unit vector a z is along the axis of the magnet in the direction from the south (S) pole to the North (N) pole. M = qml is the magnitude of the magnetic dipole moment of the small magnet. The magnetic

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field of a small bar magnet is similar to that of a small loop and it may be written according to (5.115), as

H=

M (a r 2 cosθ + aθ sin θ ) 4π r 3

Field point (r, θ, φ)

(5.117)

y

+

r +qm +

z

-qm

θ

-

Equivalent magnetic dipole

l Small bar magnet

S

N l



Fig. (5.38). A small bar magnet and its equivalent magnetic dipole.

If the length l of the small bar magnet and its pole strength qm are chosen such that it gives the same magnetic field strength of the small loop of Fig. (5.36), then (5.115) and (5.117) are equivalent. It follows that M = m , and hence

qml = SI

(5.118)

Example 5.19 The rectangular loop in Fig. (5.39) lies in the plane y + z = 2 , − ∞ ≤ x ≤ +∞ in free space. Find its magnetic dipole moment, assuming the units are in cm. z

I

(-2, 0, 2)

(0, 0, 2) an (-1, 0, 0)

x

(-2, 2, 0) (0, 2, 0)

Fig. (5.39). Geometry of the problem in Example 5.19.

y

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365

Solution Let the center of the loop be at ( x′, y′, z′) and a n is the unit vector along the axis of the loop. Since ( x′, y′, z′) is the mid-point between (0, 0, 2) cm and (−2, 2, 0) cm , then ( x′, y′, z′) ≡ (−1,1,1) cm . Referring to Fig. (5.39) a n is the unit vector along the line between (−1, 0, 0) cm and (−1,1,1) cm ; hence

an =

(1 − 0)a y + (1 − 0)a z a y + a z = 2 (1 − 0)2 + (1 − 0)2

Alternative Method: Since the equation of the plane is y + z = 2 , the surface of the plane can be described by f (x, y, z ) = y + z − 2 , hence the normal is

an =

∇f ( x , y , z ) a y + a z = ∇f ( x , y , z ) 2

Based on the direction of the current and using the right-hand rule, a n is directed as shown in Fig. (5.39). The area of the loop is S = 2 × 2 2 = 4 2 × 10 −2 m2 . Hence, the dipole moment of the loop is

m = an IS

= (a y + a z ) × 10 × 4 × 10 −2 = 0.4 (a y + a z ) A⋅ m2

5.10. MAGNETIC TORQUE The mechanical torque T produced by a force F on an object about an axis is given by T = r×F

(5.119)

where r is the distance from the axis to the object. Consider a rectangular loop of length l and width w having a magnetic dipole moment m = a z wlI . If the loop is placed in a uniform magnetic flux density B = a y B0 as shown in Fig. (5.40), it will be under the influence of a magnetic force F , which can be determined from (5.75).

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Let FAB , FBC , FCD , and FDA are the forces on the sides AB , BC , CD , and DA respectively. Hence, the total force acting on the loop using (5.75) is

F = ∫ Idl × B = FAB + FBC + FCD + FDA l

=

∫ Idxa x × a y B0 −

AB

∫ Idya z × a y B0 −

BC

∫ Idxa x × a y B0 +

CD

∫ Idya y × a y B0

(5.120)

DA

Evaluating the above integrals, yields

FAB = −FCD = a z IlB0 = Fm FBC = FDA = 0 z

(5.121b)

w

B

C

Fm

l

(5.121a)

B = ay B0

x

y

Fm

A I

D

Axis Fig. (5.40). Magnetic torque acting on a rectangular loop in a uniform magnetic field.

Note that the forces on the side BC and DA are zero. The forces on the sides AB and CD are equal in magnitude and opposite in direction, and therefore, a couple acting on the loop is created. This couple tends to rotate the loop about the axis shown in Fig. (5.40). The total torque acting on the loop is due to the forces on sides AB and CD . Hence,

w ⎛ w ⎞ T = ⎜ − a y ⎟ × FAB + a y × FCD = − ( wa y ) × (lIB0a z ) = wlIa z × a y B0 (5.122) 2 ⎝ 2 ⎠ Since m = wlIa z and B = a y B0 , the magnetic torque acting on the loop becomes

T = m × B N⋅ m

(5.123)

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Although the derivation of (5.123) is for the rectangular loop shown in Fig. (5.40), it is valid for any loop of an arbitrary shape and magnetic dipole moment m placed in an arbitrary uniform magnetic flux density B . Example 5.20 A loop of a magnetic dipole moment m = 2a x + 5a y A⋅ m2 is placed in a uniform magnetic flux density B = (2a x + 3a y + 4a z ) mWeb m2 . Find the torque acting on the loop. Solution Using (5.123), the torque is

T = m × B = (2a x + 5a y ) × (2a x + 3a y + 4a z ) × 10 −3 = 20a x − 8a y − 4a z mN⋅ m Example 5.21 Two small loops of magnetic dipole moments

m1 = 4a x + a z A⋅ m 2 and

m2 = 2a y A⋅ m2 are centered at the origin and (10,10, 0) m , respectively. Find the torque acting on the loop of magnetic moment m 2 . Solution The position vectors of the source and field points are r ′ = 0 , r = xa x + ya y + za z

The vector potential A produced by m1 at the location of m2 can be obtained using (5.113) as

A1 =

( xa x + ya y + za z ) μ0 μ r − r′ = 0 (4a x + a z ) × m1 × 3 3 4π 4π r − r′ ( x2 + y2 + z 2 ) 2

⇒ A1 =

μ 0 ( x − y ) a x − (4 z − x) a y + 4 y a z 3 4π ( x2 + y 2 + z 2 ) 2

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The magnetic flux density due to m1 is

⎛ ∂A ∂A ⎞ ⎛ ∂A ∂A ⎞ ⎛ ∂A ∂A ⎞ B1 = ∇ × A1 = a x ⎜⎜ z − y ⎟⎟ − a y ⎜ z − x ⎟ + a z ⎜⎜ y − x ⎟⎟ ∂z ⎠ ∂y ⎠ ∂z ⎠ ⎝ ∂x ⎝ ∂y ⎝ ∂x Substituting the components of A1 in the above expression, yields

B1

[

]

2 2 2 μ 0 a x 8 x − 4 y + 20 z − 3xz − a y [ 3 ( x − y ) ( z − x)] = 5 4π ( x2 + y 2 + z 2 ) 2 μ a − x 2 − y 2 + 2 z 2 + 3xy + 12 xz + 0 z 5 4π ( x2 + y 2 + z 2 ) 2

[

]

At the point (10,10, 0) m

B1 =

μ 0 4a x + a z mWeb m 2 16π 2

Hence, the torque on the loop of magnetic moment m2 becomes

T = m 2 × B1 = 2a y ×

μ 0 4a x + a z μ × 10 − 3 = − 0 (a x + 4a z ) N⋅ m 16π 16π 2

SOLVED PROBLEMS Solved Problem 5.1 The toroid shown in Fig. (5.41) has a rectangular cross section with the following data: N = 2500 turns; I = 0.5 A , a = 5 cm , b = 8 cm , and t = 4 cm . Find the magnetic flux and the energy stored in the toroid. Solution Using (5.20) the magnetic field and magnetic flux density of the toroid are

H = −aφ

μ NI NI ⇒ B = −aφ 0 2 πρ 2πρ

Magnetostatic Fields

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y

369

y

2πρ dρ

N H I

a

I

x

b

x

ρ

ρ

dρ

(a) t a

b

(b) Fig. (5.41). Geometry of the solenoid of Solved Problem 5.1.

Referring to Fig. (5.41b), the cross section of the toroid is a rectangle with width ( b − a ) and height t and the differential cross-sectional area is dS = − a φ t dρ . Thus, the flux through the toroid is b

μ 0 NI (− a φ )⋅ (− a φ )t dρ = μ 0 NIt ln (b a ) 2πρ 2π a

ψ m = ∫∫ B ⋅ dS = ∫ S

Substituting N = 2500 turns I = 0.5 A , a = 5 cm , b = 8 cm , and t = 4 cm , then

ψm

4π × 10 −7 × 2500 × 0.5 × 4 × 10 −2 = ln (8 5 ) = 4.7 μ Web 2π

Using (5.90), with ρ ′ = ρ and dv ′ = 2π tρ ′dρ ′ , the energy stored in the toroid is

Wm =

μ0 2

∫ v

H dv ′ = 2

μ 0 N 2 I 2t b 1 μ 0 N 2 I 2t ′ ρ d = ln ( b a ) ∫a ρ ′ 4π 4π

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Substituting N = 2500 turns, I = 0.5 A , a = 5 cm , b = 8 cm , and t = 4 cm , Wm becomes

4 π × 10 −7 × ( 2500 ) × ( 0.5 ) × 4 × 10 −2 ln ( 8 5 ) = 2.94 mJ 4π 2

Wm =

2

Solved Problem 5.2 Find the magnetic field at the origin in Fig. (5.42). A

I

B

Origin

a

y D

C

I

x

Fig. (5.42). Geometry of Solved Problem 5.2.



Solution Referring to Fig. (5.42), the magnetic fields due to the straight wires AB and CD at the origin are equal and directed towards − a z . Using the result of Example 5.7 part (b),

H AB = −a z

1 4π a

,

H CD = −a z

1 4π a

The magnetic field H BC due to the semicircle wire BC can be obtained using Biot – savart’s law. Following the same procedure as in Example 5.8, with r = 0 , and r′ = a x a cosφ ′ + a y a sin φ ′ , then

r − r′ − a x cos φ ′ − a y sin φ ′ = 3 a2 r − r′ Since the current in − aφ direction around the semicircle, then

Id l′ = − aφ Iadφ ′ = − Iadφ ′(− a x sin φ ′ + a y cosφ ′)

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371

Using (5.67), yields

dH BC

1 r − r′ Id l′ × 3 4π r − r′ − a x cosφ ′ − a y sin φ ′ 1 Iadφ ′(− a x sin φ ′ + a y cosφ ′)× =− a2 4π =

Simplifying and integrating both sides, the magnetic field at the origin is

H BC = − a z

I

π

4π a ∫

dφ ′ = − a z

0

I 4a

The total magnetic field at the origin in Fig. (5.42) becomes

H = H AB + H CD + H BC = −a z

1 4π a

(2 + π )

Solved Problem 5.3 A constant current I flows through a thin wire loop in the shape of a regular polygon of N sides. If the radius of the circle circumscribing the polygon is a , show that the magnetic field on its axis at a distance d from its center is given by

H = az

I a2 1 sin (2π N ) 2 2 2 2 a 2 + d 2 a cos (π N ) + d 2π N

and the magnetic field at the center of the loop is

H = az

I sin (π N ) 2a π N

Solution Let the polygon plane lies in the xy plane with its center at the origin as shown in Fig. (5.43). Then the angle φn for any side n is given by

φn =

2π (n − 1), N

n = 1, 2,  , N

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Sameir M. Ali Hamed

From Example 5.7, it has been shown that the magnetic field at a distance ρ from the center of a thin wire of length 2 L carrying a current I is given by

H = an

I 2πρ

L L + ρ2 2

z (0, 0, d) Hn ar Side #n

ρ I

b

ψ

φn

x

y

a

b

aw l (bcosφn, bsinφn, 0)

Side #n l

y x

π/N

Fig. (5.43). Geometry of Solved Problem 5.3.

where a n is a unit vector normal to the line joining the center of the wire and the field point. Let l and b as defined in Fig. (5.43), then ρ = b 2 + d 2 and L = l 2 . Let the unit vector along the wire is a w and the unit vector along the line joining the center of the wire and the field point is a r , then

an = aw × ar It can be shown from Fig. (5.43) that

a w = −a x sin φn + a y cosφn a r = −a x cosψ cosφn − a y cosψ sin φn + a z sinψ Consequently,

an = (− a x sin φn + a y cosφn )× (− a x cosψ cosφn − a y cosψ sin φn + a z sinψ ) = a x sinψ cosφn + a y sinψ sin φn + a z cosψ

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Substituting ρ = b 2 + d 2 , L = l 2 and a n into the expression of H n , we get

Hn =

I

l

4π b 2 + d 2

(l 2) + b 2 + d 2 2

(a

x

sinψ cosφn + a y sinψ sin φn + a z cosψ )

Note that a = (l 2) + b 2 . H n is the magnetic field due to any side n = 1, 2,, N . The total magnetic field H is the superposition of the field due to all sides of the polygon, or 2

N

H = ∑ Hn = n =1

I

N ⎤ ⎡ a N cosψ + sinψ ∑ (a x cosφn + a y sin φn )⎥ 2 2 ⎢ z a +d ⎣ n =1 ⎦

l

4π b 2 + d 2

Since φn = 2π (n − 1) N , it can be shown that N

N

n =1

n =1

∑ cosφn = ∑ sin φn = 0 Thus, H reduces to

H = az

I

Nl

4π b + d 2

2

a + d2 2

cosψ

From Fig. (5.43), we have

cosψ =

b b +d 2

2

,

sin (π N ) =

l , 2a

cos(π N ) =

b a

Then H on the axis of the polygon loop at a distance d from its center becomes

H = az

I a2 1 sin (2π N ) 2 2 2 2 2 2 a + d a cos (π N ) + d 2π N

At the center of the loop d = 0 , and H reduces to

H = az

I sin (π N ) 2b π N

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Solved Problem 5.4 A solenoid of length 2 L and radius a , has n turns per unit length. The solenoid carries a current I and centered along the z-axis in the range − L ≤ z ≤ + L . Find the magnetic field at any point z on the axis of the solenoid. Solution The geometry of the solenoid is shown in Fig. (5.44). Since the current density in the solenoid is nI A m , then the loop of width dz′ illustrated in Fig. (5.44) can be considered as a loop of a constant current nIdz′ with its center displaced by a distance z′ from the origin. Thus, using (5.18), the magnetic field dH at any point z on the axis of the loop (z-axis) due to the loop current nIdz′ can be obtained as

dH =a z

nIdz′a 2 2

1

[(z − z′)

2

+ a2

]

3 2

Integrating both sides along the solenoid, yields +L

nIa 2 dz′ H = az ∫ 2 − L (z − z′)2 + a 2

[

]

3 2

Using the substitution z − z′ = a tan θ , then the above integral reduces to θ

nI 2 nI H =az cosθ dθ = − (sin θ 2 − sin θ1 ) ∫ 2 θ1 2 2 where sin θ1 = (L + z ) (L + z )2 + a 2 and sin θ2 = (L − z ) (L − z ) + a 2 . Substituting these values in the above equation, then the exact magnetic field on the axis of the solenoid becomes

H = az

nI ⎡ ⎢ 2 ⎣

L−z

(L − z )2 + a 2

+

⎤ ⎥ (L + z )2 + a 2 ⎦ L+z

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x

x z′

(a, 0, -L)

dz′

z′

(a, 0, +L)

dz′ a

dH

z I

375

z

-L

I

nIdz′

L

z

nIdz′

Fig. (5.44). Geometry of Solved Problem 5.4.

Solved Problem 5.5 A thin toroid with 500 turns is wound around a magnetic material of permeability μ = 4000 μ 0 . The main radius of the toroid is 10 cm and its cross section area is

1 cm 2 . If the current in the toroid is 0.1A , find the energy stored in the toroid. Solution For a thin toroid, the magnetic field is H ≈ NI 2πρ0 . The cross section area of the toroid is S = 1 cm 2 and its volume is v = S × 2πρ0 . The permeability

μ = 4000 μ 0 . The permeability is μ = 4000 μ 0 ; hence, the magnetic energy in the toroid is

Wm =

μ 0μ r 4000 μ 0 N 2 I 2 2 ′ H d v = dv′ 2 ∫v 2 ∫v (2πρ0 )2 2 4000 μ 0 (NI ) = × × S × 2πρ0 2 (2πρ0 )2

Substituting N = 500 , I = 0.1A , and ρ0 = 10 cm , yields

4000 × 4π × 10 −7 (500 × 0.1) × × 1 × 10 − 4 = 1 mJ −2 2 2π × 10 × 10 2

Wm =

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Alternative Solution The magnetic field at ρ ′ , a ≤ ρ ′ ≤ b in the toroid is H = NI 2πρ′ . Using (5.90), with dv′ = 2π tρ ′dρ ′ , the energy stored in the toroid is b

μ μ N 2 I 2t 1 μ 0 N 2 I 2t 2 ′ ρ Wm = 0 ∫ H dv′ = 0 d = ln(b a ) 2 v 4π ∫a ρ ′ 4π Since the toroid is thin, (b − a ) > d .

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Solution Since, l >> d , the magnetic field inside the solenoids can be considered uniform. The magnetic field due to coil 2 is

H2 = a z N 2 I 2 l I1

I

N1 N2

2d

z

l Fig. (6.18). Geometry of the solenoids in Example 6.8.

Then the flux inside coil 1 due to H 2 is

ψ 12 = ∫∫ B2 ⋅ dS1 = μ 0 ∫∫ H 2 ⋅ dS1 = μ 0 H 2 S1 = μ 0 S

S

N2 I 2 × πd 2 l

Then, the mutual inductance between the solenoids is M=

N1ψ 12 NN = μ 0 1 2 × πd 2 I2 l 

Example 6.9 Determine the exact mutual inductance between two coaxial circular thin-loops of radii a and b . The distance between the loops is d and they are in free space. Derive an approximate mutual inductance from the exact expression assuming d >> a, b . Solution Using the Neumann formula, we have M=

μ0 4π

dl1 ⋅ dl 2 R l 2 l1

∫∫

Magnetic Materials and Metamaterials















Electromagnetics for Engineering Students

Referring to Fig. (6.19), we have

dl1 = a (a x sin φ1 − a y cosφ1 )dφ1

dl 2 = b (a x sin φ2 − a y cosφ2 )dφ2

dl1 ⋅ dl 2 = cos(φ2 − φ1 )ab dφ1 dφ2 +∞ 1 = ∑ Ψn e − jn(φ 2 −φ1 ) R n = −∞

where

⎧ (a r )k , r > a ⎪ ( k − n )! n , Ψn = ∑ Pk ( cos θ 2 ) Pkn ( 0 ) ⎨ r k = n ( k + n )! ⎪ 1 (r a )k , r < a 1

+∞

r = b2 + d 2

⎩a

Substituting the above quantities into the formula of M , then μ ab M= 0 4π =

μ 0 ab 4π

2π 2π +∞

∫ ∫ ∑Ψ

n

0 0 n = −∞

cos(φ2 − φ1 ) e− jn(φ2 −φ1 )dφ1dφ2

⎡ e+ jn(φ2 −φ1 ) + e− jn(φ2 −φ1 ) ⎤ − jn(φ2 −φ1 ) Ψ dφ1dφ2 n⎢ ⎥e ∫0 n∑ 2 = −∞ ⎣ ⎦

2π 2π + ∞

∫ 0

This can be written as ⎤ ⎡ 2π 2π + ∞ 2π 2π + ∞ ⎥ ⎢ μ ab M= 0 ⎢ ∫ ∫ ∑ Ψne+ j (n −1)(φ2 −φ1 ) dφ1dφ2 + ∫ ∫ ∑ Ψne− j (n +1)(φ2 −φ1 ) dφ1dφ2 ⎥ 8π ⎢ 0 0 n = −∞ 0 0 n = −∞

   

   ⎥ ⎥⎦ ⎢⎣
1
2


1 and
2 can be simplified as 2π 2π +∞

⎧4π 2Ψ+1 , + j (n −1)(φ 2 −φ1 ) e d φ d φ Ψ = ⎨ 1 2 n ∫0 ∫0 n∑ = −∞ ⎩0, 2π 2π +∞ ⎧4π 2 Ψ−1 ,
2 = ∫ ∫ ∑ Ψne− j (n +1)(φ2 −φ1 ) dφ1dφ2 = ⎨ ⎩0, 0 0 n = −∞
1 =

n = +1 othwise n = −1 othwise



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I1

l1

Loop 1

I2

l2

S1 d a

Loop 2

r I1d12

I1d11

z

b S2

R

(r1 = a, 02 = p/2, f1)

(r, 02, f 2)

x Fig. (6.19). Geometry of the problem in Example 6.9.

Since Ψ+1 = Ψ−1 , then M=

μ 0 ab 1 (I + I 2 ) = μ 0 ab (4π 2Ψ+1 + 4π 2Ψ−1 ) = π μ 0 abΨ1 8π 8π

Substituting n = 1 in the expression of Ψn , yields k ⎧1 ⎪ r (a r ) , r > a 1 1 1 Pk (cosθ 2 ) Pk (0 )⎨ Ψ1 = ∑ 1 k =1 k (k + 1) ⎪ (r a )k , r < a ⎩a +∞

Substituting Ψ1 in the equation of the mutual inductance, taking into consideration that Pk1 (0) = 0 for odd values of k , it follows that the exact mutual inductance between the loops can be expressed as

M=

π μ0 b 2

( (

) )

⎧⎪ a b 2 + d 2 2 k , 1 1 1 P2 k −1 (cosθ 2 ) P2 k −1 (0) ⎨ ∑ 2 k −1 ⎪⎩ b 2 + d 2 a k =1 k (2k − 1) , +∞

b2 + d 2 > a b2 + d 2 < a

where cosθ 2 = d / b 2 + d 2 . For d >> a, b , then b2 + d 2 ≈ d and the terms in the summation for k > 1 can be ignored. Consequently, the approximate mutual inductance is obtained by taking the first term only ( k = 1 ), as

1 2 M ≈ π μ 0 b P11 (cosθ 2 )P11 (0)(a d ) 2


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We have, P11 (0) = −1, P11 (cosθ 2 ) = − sin θ 2 ≈ − b d , then the approximate mutual inductance between the loops becomes

M ≈ π μ0

a 2b 2 2d 3

6.5.4. Energy Stored in the Inductance We know that from (5.90) in Chapter 5 the total magnetostatic energy stored in a static magnetic field H within a volume v is given by

WT =

1 B ⋅ H dv 2 ∫v

(6.71)

Substituting B = ∇ × A into (6.71), yields

WT = Using the vector identity becomes

WT =

1 (∇ × A ) ⋅ H dv 2 ∫v

(∇ × A) ⋅ H − (∇ × H) ⋅ A = ∇ ⋅ (A × H),

1 1 (∇ × H ) ⋅ A dv + ∫ ∇ ⋅ (A × H ) dv ∫ 2v 2v

(6.72) then (6.72)

(6.73)

The volume integral in the second term on the right-hand side can be replaced by a surface integral using the divergence theorem; hence (6.73) becomes

WT =

1 1 (∇ × H ) ⋅ A dv + ∫∫ (A × H) ⋅ dS ∫ 2v 2 S

(6.74)

where S is the surface enclosing the volume v . Let the magnetic fields inside and outside S be Hi and H o respectively, then the surface current will be K = (Ho − Hi ) × an where a n is a unit vector normal to

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S . If S is discrete and composed of the surfaces Si ( i = 1, 2, , N ), the second term on the right-hand side of (6.74) can be written as

WS =

1 1 N × ⋅ dS = A H a A × (H o − Hi ) ⋅ a n dS ∑ n ∫∫ 2 ∫∫ 2 i = 1 S S

(6.75)

Using the vector identity B ⋅ C × A = C ⋅ A × B , (6.75) becomes

WS =

1 N 1 N ⋅ ( − ) × dS = A H H a A ⋅ K dS ∑ ∑ n o i ∫∫ 2 i =1 ∫∫ 2 i = 1 S S

(6.76)

Since K = (Ho − Hi ) × an . On the other hand, if the surface S is continuous and there is no surface current or H o = H i , then WS = 0 and the total energy in (6.74) reduces to

WT =

1 A ⋅ J dv 2 ∫v

(6.77)

where ∇ × H = J has been used in (6.74) to obtain (6.77). Using (5.54) the vector potential at a point described by a position vector r due to a current J′ at a point described by a position vector r′ is

A=

μ 4π

J′(r′)

∫ r − r′ dv′

(6.78)

v

Substituting A from (6.78) into (6.77), yields

WT =

μ 8π

J(r ) ⋅ J′(r′) dvdv′ r − r′ v v

∫∫

(6.79)

For uniform linear currents I and I ′ , we have dv′J′ = I ′dl′ and dvJ = Idl ; hence (6.79) can be written as

WT =

μ I I ′ dl ⋅ dl′ 8π ∫∫ R l l′

(6.80)


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where R = r − r′  r′ and r are the position vectors of the element dl′ and dl respectively. For a system of N coils with currents I i or I j where i = 1, 2,, N , and j = 1, 2,, N we have N

I ′ = ∑ Ii , i =1

N

I = ∑I j ,

dl′ = dl i ,

and dl = dl j

(6.81)

j =1

Substituting I , I ′ , dl′ , and dl from (6.81) into (6.80), we get

WT =

μ 8π

N

N

∑∑ I I ∫∫

dl i ⋅ dl j

i j

i =1 j =1

l l′

R

(6.82)

Using (6.69) with (6.82), yields

WT =

1 N N ∑∑ Ii I j Lij 2 i =1 j =1

(6.83)

where Lij is the mutual inductance between coil i and j . When there is only one coil with current I , then L11 = L (the self-inductance of the coil). In this case, the magnetic energy stored in the inductance of the coil is obtained by substituting i = j , I = I1 and L11 = L in (6.83), as

1 WT = I 2 L 2

(6.84)

Example 6.10 Find the external inductance per unit length of a coaxial cable of Example 5.2 using the stored energy approach. Solution For the result of the coaxial cable of Example 5.2, H in the space between the conductors is obtained as H = aφ

I 2 πρ

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From (6.84), we have L ′ = 2WT I 2 . Using dv = ρ dρ dφ dz , then

μ WT = 0 2

∫H

2

dv =

v

μ0 I 2 8π 2

l 2π b

∫ 0

μ0 I 2 dρ dφ dz = ln ( b a ) ∫∫ ρ 4π 0 a 1

Substituting WT , in the expression L′ = 2WT I 2 , yields L′ = 2

2 μ0 I 2 μ l WT ln(b a ) = 0 ln(b a ) = 2 2 2π I I 4π

Thus, the inductance per unit length of the cable is

L=

L′ μ 0 = ln(b a ) l π 

Example 6.11 Show that for any two magnetically coupled circuits, their self-inductances L11 and L22 are related to their mutual inductance M by

M ≤ L11L22 Solution Let the currents in the coils are I1 and I 2 , then the energy stored in the coils is

WT =

1 2 2 1 I i I j Lij = (I12 L11 + I 22 L22 + I1I 2 L12 + I1I 2 L12 ) ∑∑ 2 i =1 j =1 2

Since L12 = L21 = M , and the energy is

1 1 WT = I12 L11 + I 22 L22 + I1I 2 M ≥ 0 2 2 The minimum energy occurs when dWT dI1 = 0 or dWT dI2 = 0 ; hence

dWT = I1L11 + I 2 M = 0 dI1

⇒

I1 = −I 2 M L11


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Substituting I1 = −I 2 M L11 in the expression of WT , the minimum energy is

WT min =

(

I 22 L22 − M 2 L11 2

)

Since the energy is always positive quantity, WT min ≥ 0 , then L22 − M 2 L11 ≥ 0. Thus

M ≤ L11L22 6.6.

MAGNETIC CIRCUIT

The problems of magnetic flux in the magnetic devices can be analyzed approximately by a method analog to that used to analyze the flow of electric current in the electric circuits problems. To illustrate the concept of the magnetic circuit, consider a magnetic material with a coil of N turns and carrying a current I is wounded as shown in Fig. (6.20). Magnetic material A

ψ 1

B

ψ 2 mmf

N

ψ 3

C

Air gap

I F I1 A emf ≡ mmf

E ≡ REFAB ≡ ℜEFAB

D

(a) B RBCDE ≡ ℜBCDE I3 ≡ ψ3

I2 ≡ ψ 2

C

RBE ≡ ℜBE Rgap ≡ ℜgap

F

E (b)

Fig. (6.20). (a) Magnetic circuit. (b) Analogous electric circuit.

D

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The magnetomotive force mmf in the magnetic circuit which corresponds to the electromotive force emf in the electric circuit is defined as

mmf = F = ∫ H ⋅ dl = NI

At

(6.85)

l

The reluctance is opposition presented by the magnetic circuit to the flux. The reluctance R of the magnetic circuit corresponds to the resistance in the electric circuit and it is directly proportional to the length and inversely proportional to the cross section of the branch of the circuit. For a branch of a magnetic circuit of length l and cross section area S , the reluctance is given by R =

l

μS

(6.86)

At Wb

The reciprocal of the reluctance is the permeance P . A relation similar to Ohm’s law is used to relate the mmf across R with flux ψ passing through it, as F = Rψ

(6.87)

At

If the magnetic circuit has an air gap, fringing occurs at air gaps as illustrated in Fig. (6.21). This fringing causes a decrease in the flux density of air gap. For air gaps of small lengths, the effect of fringing is usually neglected.

Fringing

Core Air gap length

Magnetic flux Fig. (6.21). The fringing of the magnetic flux at the edges.

The Kirchhoff’s current and voltage laws can be applied to analyze magnetic circuit in a similar manner as in the electric circuit. Thus, at any node, the magnetic flux ψ entering the node is equal to the flux out of the node. The magnetic flux in the magnetic circuit corresponds to the current in the electric circuit, then the Kirchhoff’s current law for magnetic circuit at any node is

∑ψ = 0

(6.88)


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For any loop in the magnetic circuit, the Kirchhoff’s voltage law is

∑ mmf = ∑ψ R

(6.89)

which is equivalent to Ampere’s Law

∑ NI = ∑ Hl

(6.90)

Example 6.12 Assuming that the cross section area of all branches of the magnetic circuit shown 2

in Fig. (6.22) is 1 cm and the core is made of iron with μr = 1500. Calculate the total reluctance of the circuit and the flux in the core and in the air gap. ψ1

Magnetic Material

N =200 turn

B

A

10 cm

A

ℜ1

B C

0.2 cm

ψg mmf = NI

8 cm

C

ℜg

D

I = 10 A

10 cm

F

E

D F

E

Fig. (6.22). The magnetic circuit and corresponding electric circuit of Example 6.12.

Solution Referring to Fig. (6.22), since the cross section area of all branches of the circuit, is the same, the path DEFABC can be considered as a single path of reluctance R1

in series with the reluctance of air gap R g .

(a) The length of the path DEFABC is l1 = 2 × (10 + 8)− 0.2 = 35.8 cm and its cross 2

section area is S = 1 cm , then

R1 =

l1 μ 0μ r S

35.8 × 10 −2 4π × 10 − 7 × 1500 × 10 − 4 = 1.898 × 10 6 A − Turns Wb =

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The length and cross-section area of the air gap are l g = 0.2 cm and S = 1 cm

⇒ Rg =

l1 μ0 S

=

2

0.2 ×10 −2 = 1.59 ×10 7 A − Turns Wb −7 −4 4π ×10 ×10

Since the reluctances are in series, the total reluctance is

R = R1 + R g = 1.898 ×106 + 15.91 ×106 = 17.808 ×106 A− Turns Wb  (b) Since R 1 and R g are in series, ψ 1 = ψ g , then

ψ1 = ψ g =

F NI 200 ×10 −4 = = = 1.123 ×10 Wb 6 R R 17.808 ×10

Example 6.13 Assuming that the cross section area of all branches of the magnetic circuit shown in Fig. (6.20) is 200 cm 2 and the core is made of cast steel. The B-H curve of cast steel is shown in Fig. (6.23). Calculate the current required to produce a flux of 6 mWb in the air gap if the number of turns is N = 400 . The lengths of the branches are as follows: Length of the path EFAB = Length of the path BCDE = 100 cm. Length of air gap = 0.25 cm. The length of the path BE = 20 cm. Neglect fringing. Solution Let the reluctances of the branches EFAB, BCDE be R1 and R 3 respectively. The Reluctances of the branch BE are R 2 due to the core and R g due to the air gap. The electric circuit in Fig. (6.24) is analogs to the magnetic circuit in Fig. (6.20). Neglecting the fringing at the air gap, the flux through air gap ψ g is equal to that through the branch BE, or ψ g = ψ 2 = 6 mWb . The cross section area of all 2

branches is S = 200 cm , the flux density in the air gap is Bg =

ψg S

=

6 ×10 −3 = 0.3 Wb m 2 200 ×10 −4


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ψ1

ℜ1

ψ3

ℜ3

B

A

437

C

ψ2 ℜ2

mmf = NI

1

2 ℜg

F

D

E

Fig. (6.23). The electric circuit corresponding to the magnetic circuit of Example 6.13.

The flux density in the branch BE is B2 = Bg = 0.3 Wb m2 , then

Hg =

Bg μ0

=

0.3 4π × 10

5

−7

= 2.386 × 10 A m

From the B-H curve of cast steel, for B2 = 0.3 Wb m2 ⇒ H2 = 150 A m . Applying Ampere’s law on the loop BCDEFB in Fig. (6.23). In this loop NI = 0 . Thus,

∑ NI = 0 = ∑ Hl = H l

g g

+ H 2l2 − H3l3

l g = 0.25 cm , l2 = 20 − 0.25 = 19.75 cmand l3 = 100 cm , then

H3 =

H g l g + H 2l2 l3

−2

5

=

2.386 × 10 × 0.25 × 10 + 150 × 19.75 × 10 100 × 10

−2

−2

= 626 A m

From the B-H curve of cast steel shown in Fig. (6.24), for H 3 = 675 A m

⇒ B3 = 0.92 Wb m 2 , hence the flux in the branch BCDE is −4

−2

ψ 3 = B3S = 0.92 × 200 × 10 = 1.84 × 10 Wb Apply the Kirchhoff’s current law at node B, the flux ψ 1 in the branch EFAB can be obtained. Since ψ 1 entering the node and ψ 2 , ψ 3 are out of the node , then

∑ψ = 0 = ψ

1

−ψ 2 −ψ 3

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⇒ ψ 1 = ψ 2 + ψ 3 = 6 × 10 −3 + 1.84 × 10 −2 = 0.0244 Wb Consequently, the flux density in the branch EFAB is B1 =

0.0244 ψ1 = = 1.22 Wb m 2 S 200 ×10 −4

From the B-H curve of cast steel shown in Fig. (6.24), for B1 = 1.22 Wb m2 ⇒ H1 = 1200 A m Applying Ampere’s law on the loop ABEFA in Fig. (6.23), yields

∑ NI = NI = H l + H l 11

2 2

+ H g lg

lg = 0.25 cm , l2 = 19.75 cm and l3 = 100 cm , then the required current is I=

H 1 l1 + H 2 l 2 + H g l g

N 1200 ×100 ×10 −2 + 150 ×19.75 × 10 −2 + 2.386 ×10 5 × 0.25 ×10 −2 = 400 = 4.57 A 1.6

Cast Steel

B (Wb/m2)

1.2

0.8

Cast Iron 0.4

0

0

1000

2000

H (A/m)

Fig. (6.24). The B-H curve for cast steel and cast iron.

3000

4000

5000

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6.7.


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FREQUENCY-DEPENDENT PERMITTIVITY AND PERMEABILITY

6.7.1. Dispersion The material polarization P and its relation to the relative permittivity ε r are discussed in Chapter 3 under a static electric field condition. Under timeharmonic fields the induced polarization may result in a permittivity that is dependent on the operating frequency. Materials exhibit this property are said to be dispersive. Similarly, if the induced magnetization in a material under timeharmonic fields results in a frequency-dependent permeability, the material is also dispersive. Consider the polarization process discussed in Section 3.4. When the material is placed in a time-harmonic field E = a xE x = a x Ex cos ω t ( ω is the angular frequency) instead of a static field, the electrons move a distance x at the same frequency of the applied field and a time-varying polarization P = a xPx = a x Re Px e j ω t will be induced in the material. In the presence of the external field, the electrons in the atomic structure of the material will be under the following forces:

(

)

1. The force Fr = − m ωo2 x due to the positive charges in the nucleus of the atom (restoring force), which oscillates with a natural frequency ωo = κ m . Where κ is a constant and m is the electrons mass. 2. The force Fd = −mν Le x due to the losses or damping of the system with the electric damping factor ν Le . 3. The force Fe = neE x due to the applied field. Thus, the equation of motion of the electrons can be written as m

d 2x dx + mν Le + −m ωo2 x = neE x 2 dt dt

(6.91)

where n is the number of electrons per atom. The first term on the left-hand side accounts for the acceleration of electrons. Dividing both sides of (6.91) by electron mass m , yields d 2x dx ne + ν Le + ωo2 x = Ex cos ω t 2 dt dt m

(6.92)

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According to (3.47) the polarization of the material is P = a x P x = a x Nnex , where N is the number of induced electric dipoles per unit volume of the material. Multiplying both sides of (6.92) by Nne , and letting ς Le = Nn 2e 2 m ε 0 , (6.92) can be written in terms of P x as d 2Px dP + ν Le x + ωo2 Px = ε 0ς Le Ex cos ω t 2 dt dt

(6.93)

j ωt ς Le is known as the electric coupling factor. Since Px = Re (Px e ), (6.93) can be written in a complex form as

d 2 Px e j ω t dPx e j ω t + ν + ωo2 Px e j ω t = ε 0ς Le Ex e j ω t Le 2 dt dt

(6.94)

Substituting d dt = j ω and d 2 dt 2 = − ω2 into (6.94), the solution can be obtained as

Px =

ς Le ε 0 Ex ω − ω2 + j ων Le 2 o

(6.95)

From (3.48) we have Px = χ eε 0 Ex , where χ e is electric susceptibility. Substituting Px = χ eε 0 Ex into (6.95) and solving for χ e , yields

χ e (ω) =

ς Le ω − ω2 + j ων Le 2 o

(6.96)

2 Using ε = ε 0 (1 + χ e ) with (6.96), and replacing ς Le by ωep , the permittivity of the material at angular frequency ω can be written as

⎛ ⎞ ω2 ⎟ ε(ω) = ε 0 ⎜⎜1 + 2 2 ep ⎟ ω ω j ω ν − + Le ⎠ o ⎝ where ωep =

(6.97)

Nn 2e2 m ε 0 is known as the electric plasma angular frequency.


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The relative permittivity ε r = ε ε 0 can be written as

ε r = ε′− j ε′′ = 1 +

ωep2

(6.98)

ωo2 − ω2 + j ων Le

where

ε′ = 1 + ε′′ =

2 (ωo2 − ω2 ) ωep

(ω − ω ) 2 o

2 2

(6.99a)

+ ω2 ν Le2

2 ωep ων Le

(ω − ω ) 2 o

2 2

(6.99b)

+ ω2 ν Le2

Fig. (6.25) shows the variation of ε′ and ε′′ with ω graphically. Note that ωo is the resonance frequency of the atoms, it follows that the maximum absorption (maximum ε′′ ) takes place at resonance, since the displacement x is a maximum, and hence the collision is a maximum at resonance. This phenomenon is called resonance absorption. Maximum absorption 3

e

,

2 1

e ,,

0 w1

w

0

w

2

w

Fig. (6.25). Variation of ε′ and ε ′′ with the angular frequency ω .

Referring to Fig. (6.25), it is clear that for ω < ω1 and ω > ω2 , d ε′ d ω > 0 , while for ω2 > ω > ω1 , d ε′ d ω < 0 . Dispersion in the region when d ε′ d ω > 0 is called normal dispersion, while for d ε′ d ω < 0 dispersion is called anomalous. Note that the anomalous dispersion is always associated with the resonance absorption.

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6.7.2. Models of Materials Equation (6.93) can be generalized for any material to describe the temporal response of a component of the polarization field of the material to the same component of the applied electric field. Let the applied electric field be E i V/m at angular frequency ω rad/s and the corresponding polarization component be Pi C m 2 , then (6.93) can be re-written as d 2Pi dP 2 + ν Le i + ωo2 Pi = ε 0ωep Ei 2 dt dt

(6.100)

This is known as the Lorentz model. The permittivity of the material based on the Lorentz model is given by (6.97). When the restoring force m ωo2 x is negligible, Lorentz model in (6.100) reduces to the Drude model given by the equation

d 2Pi dP 2 + ν De i = ε 0ωep Ei 2 dt dt

(6.101)

where ν De is the electric damping factor of the Drude model. The permittivity of the material based on the Drude model can be obtained as 2 ⎛ ⎞ ωep ⎟ ε(ω) = ε 0 ⎜⎜1 − 2 ⎟ ⎝ ω − j ων De ⎠

(6.102)

When the acceleration term is small in comparison to other terms, the Lorentz model reduces to the Debye model. The differential equation of the Debye model is given by

ν de

dPi 2 + ωo2 Pi = ε 0ωep Ei dt

(6.103)

where ν de is the electric damping factor of the Debye model. The permittivity of the material based on the Debye model can be obtained as

⎛ ⎞ ωep2 ⎟ ε(ω) = ε 0 ⎜⎜1 + 2 ⎟ + ω j ω ν de o ⎝ ⎠

(6.104)


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The most general model is the two-time-derivative Lorentz metamaterial model. The permittivity of this model is given by

⎛ αeωep2 + j βeωep ω+ γ eω2 ⎞ ⎟ ε(ω) = ε 0 ⎜⎜1 + 2 2 ⎟ ω ω j ω ν − + Le o ⎝ ⎠

(6.105)

where α e , β e , and γ e are constants. Corresponding models of the magnetic susceptibility and permeability of a material can be obtained in the same fashion of the electric field models using the duality between the electric field and magnetic field. 6.8.

METAMATERIALS CONDUCTORS

AND

ARTIFICIAL

MAGNETIC

The term metamaterials was introduced in 1999 by Dr. Rodger Walser and he defined metamaterials as Macroscopic composites having man-made, threedimensional, periodic cellular architecture designed to produce an optimized combination, not available in nature, of two or more responses to specific excitation [92]. Typical metamaterials can be synthesized by embedding various inclusions with novel geometric shapes and forms in some host media as shown in Fig. (6.26) [76] and [92]. The properties of the synthesized metamaterial can be controlled by many parameters such as [76]: 1. 2. 3. 4.

The properties of the host medium. The geometry and composition of the inclusions. The density of the inclusions. The arrangement and alignment of the inclusions.

Host

Inclusions

Fig. (6.26). Typical metamaterials can be synthesized by embedding various inclusions with novel geometric shapes and forms in some host media.

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6.8.1. Material Classifications Electrical properties of the material are determined mainly according to the behavior of ε and μ with frequency. ε and μ may be both positive, either negative or different in sign as shown in Fig. (6.27), depending on the operating frequency [7692]. μ ENG Materials

DPS Materials

ε < 0, μ > 0

ε > 0, μ > 0

Plasmas

Dielectrics

DNG Materials ε < 0, μ < 0 Not found in nature

MNG Materials

ε

ε > 0, μ < 0 Gyrotropic magnetic materials

Fig. (6.27). Classification of materials.

Based on the behavior of ε and μ with frequency, electrical materials can be classified into double positive (DPS), epsilon-negative (ENG), mu-negative (MNG), and double negative (DNG) materials. 6.8.1.1. Double Positive Materials Materials with both ε and μ positive are designated double positive materials. Most dielectrics are DPS materials. Fig. (6.27) shows the classification of materials based on the values of ε and μ . 6.8.1.2. Epsilon-Negative Materials Materials with ε negative and μ positive are designated epsilon-negative materials. Nobel metals like silver and gold have a negative ε at infrared and visible frequencies. ENG materials can be synthesized by an array of parallel thinconducting wires as shown in Fig. (6.28). The radius of each wire is r and the distance between any adjacent wires is a . This array exhibits a high-pass behavior for a plane wave whose electric field is parallel to the wires. Below a certain frequency, the plane wave will experience a total reflection [77]. This behavior is equivalent to the propagation of electromagnetic wave in plasma. If the spacing between the adjacent wires a is much smaller than the wavelength λ

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( a 1 and z < 0 are free space. Find the magnetic flux density and the energy density in the slab. B1 = μ0 H1

z

H 1 = H 0 az

μ1 = μ0 z=1

Slab

μ 2 = μ 0μ r

B2 H3

y

μ3 = μ0



Fig. (6.33). Geometry of Solved Problem 6.1.

Solution Given that H1 = H 0a z A m and the normal is a z , then H t1 = 0 and H n1 = H 0a z . Let the magnetic field in the slab is H 2 = Ht 2 + H n 2 . Applying the boundary conditions at the interface z = 1 , yields

B n1 = B n 2 ,

⇒ μ 0 H n1 = μ 0μ r H n 2 , ⇒ H n 2 = H t1 − H t 2 = 0 ,

H0 az μr

⇒ H t 2 = H t1 = 0

⇒ H 2 = Ht 2 + H n2 =

H0 az μr

Thus, the magnetic flux density B 2 and the energy density W v in the slab are

B2 = μ 0μ r H 2 = μ 0 H 0a z  Wv =

1 2μ 0 μ r

B2

2

=

μ0 H 02 J m 3 2μ r

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Solved Problem 6.2 The region x + 2 y − 2 > 0 is occupied by medium 1 ( μ1 = 6 μ 0 ), while the region x + 2 y − 2 > 0 is occupied by medium 2 ( μ 2 = 3.5 μ 0 ). If the magnetic field in medium 1 is H1 = 11a x + 12a y − 15a z A m and the boundary between the media is current-free, find the magnetic flux density in medium 2. Solution The boundary between the media is the plane x + 2 y − 2 = 0 ,which can be defined by the function f (x, y, z ) = x + 2 y − 2 ; hence the normal to the plane is

an =

∇f (x, y, z ) a x + 2a y = ∇f (x, y, z ) 5

Medium 1 and 2 are illustrated in Fig. (6.34). Let H1 = Ht1 + H n1 , then

H1 ⋅ a n = (H t1 + H n1 ) ⋅ a n = ⇒ H n1 =

35 = H n1 5

35 a n = 7a x + 14a y 5

Since H1 = Ht1 + H n1 , then

H t1 = H1 − H n1 = 4a x − 2a y − 15a z Let H 2 = Ht 2 + H n 2 . Since there is no surface current on the boundary, then H1  and H 2 satisfy

⇒ H t 2 = H t1 = 4a x − 2a y − 15a z B n1 = B n 2 , ⇒ μ1 H n1 = μ 2 H n 2

H t1 − H t 2 = 0 ,

Substituting μ1, μ 2 and H n1 in to above last relation, yield

Hn2 =

6 (7a x + 14a y ) = 12a x + 24a y A m 3.5

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⇒ H 2 = H t 2 + H n 2 = 16a x + 22a y − 15a z A m Hence, the magnetic flux density in medium 2 is B 2 = μ 2 H 2 = 3.5 × 4π × 10 −7 × (16a x + 22a y − 15a z ) = 70.4a x + 96.8a y − 66a z μ Wb m

2

Medium 2

μ2 = 3.5μ0

1

0

y Plane x + 2y – 2 = 0

H2, B2

H1

0.5 an

μ1 = 6μ0 Medium 1

x Fig. (6.34). Geometry of Solved Problem 6.2.



Solved Problem 6.3 The plane 1.5 x + 2 y = 0 separates medium 1 (μ1 = μ 0 ), and medium 2 (μ 2 = 10 μ 0 ), where medium 1 in the region 1.5 x + 2 y > 0 . If the magnetic field in medium 1 is H1 = 4a x + a y − 2a z A m and the surface current on the plane

1.5 x + 2 y = 0 is K12 = 10a z A m , find the magnetic flux density in medium 2. Solution The boundary between the media is the plane 1.5 x + 2 y = 0 ,which can be defined by the function f (x, y, z ) = 1.5x + 2 y . Hence, the unit vector normal to the plane is

an =

∇f ( x , y , z ) 3 x + 4 y = 5 ∇f ( x , y , z )

With reference to Fig. (6.35), the angle φ between the direction of a n and a x is

an ⋅ a x = cosφ = 3 5 , and sin φ = 4 5

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Let H1 = H x1a x + H y1a y + H z1a z and H 2 = H x 2a x + H y 2a y + H z 2a z , then

H1 = H t1 + H n1 = H x1a x + H y1a y + H z1a z = 4a x + a y − 2a z H 2 = H t 2 + H n 2 = H x 2 a x + H y 2a y + H z 2a z Hx2

Medium 2 μ2 = 10μ0

H2

φ

Hy2

Hy1

φ

Plane 1.5x + 2y = 0 at

y

H1

an Hx1

μ1 = μ0 Medium 1

x Fig. (6.35). Geometry of Solved Problem 6.3.

The tangential and normal components in medium 1 are

H t1 = (H x1 sin φ − H y1 cosφ )at + H z1a z  H n1 = (H x1 cosφ + H y1 sin φ )a n  Similarly, in medium 2

H t 2 = (H x 2 sin φ − H y 2 cosφ )at + H z 2a z  H n 2 = (H x 2 cosφ + H y 2 sin φ )a n  Applying the boundary conditions μ1 H n1 = μ 2 H n 2 , yields

H x 2 cosφ + H y 2 sin φ = 10(3H x1 + 4 H y1 ) ⇒ 3H x 2 + 4 H y 2 = 160   Applying the boundary conditions (Ht 2 − Ht1 ) × a n = K , yields

(4H

x2

− 3H y 2 − 4 H x1 + 3H y1 )at × a n + 5(H z 2 − H z1 )a z × a n = 50a z 

Note that at × a n = a z . Equating components in each direction, we get

4 H x 2 − 3H y 2 = 63  H z 2 = H z1 = −2 



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Solving 4 H x 2 − 3H y 2 = 63 with 3H x 2 + H y 2 = 130 obtained above, yields

H x 2 = 34.85  H y 2 = 25.5  ⇒ H 2 = 34.85a x + 25.5a y − 2a z A m Hence, the magnetic flux density in medium 2 is

B 2 = μ 2 H 2 = 10 × 4π × 10 −7 (34.85a x + 25.5a y − 2a z ) = 438.1a x + 320.6a y − 25.1a z μ Wb m 2 Solved Problem 6.4 For the coaxial cable with cross-section shown in Fig. (6.36), find the external inductance of the coaxial cable per unit length. H2 H1

μ2

c a

μ1

y b

μ2

y

μ1

x

x



Fig. (6.36). The cross section of the coaxial cable of Solved Problem 6.4.

Solution The magnetic field in the region a < ρ < b can be obtained using Ampere’s law as H1 = aφ

I 2πρ

, ⇒ B1 = aφ

μ1 I 2πρ

a a,θ ,φ ) due to K b can be obtained using multi-pole expansion given by (5.95) with the source point (r ′ = a,θ ′,φ ′) is on the surface of the sphere and K = K b . Hence for r > a

A=

μ Mz 4π r

+∞

+∞

(k − n )!

∑ ∑ (k + n)! P (cosθ )∫ aφ sin θ ′(a r ) P (cosθ ′)e k

n k

n = −∞ k = n

n k

− jn (φ −φ ′ )

dS ′

S

On the surface of the sphere dS ′ = a 2 sin θ ′dθ ′dφ ′ . Also, it can be shown that aφ = −a ρ I sin (φ − φ ′) + aφ I cos(φ − φ ′) , then

A= ×

μ M z a 2 +∞ +∞ (k − n )! n Pk (cosθ ) ∑∑ 4π r n = −∞ k = n (k + n )!

2π π

∫∫ [− a ρ I sin (φ − φ ′) + aφ I cos(φ − φ ′)]sin θ ′(a r ) P (cosθ ′)e k

n k

− jn (φ −φ ′ )

sin θ ′dθ ′dφ ′

0 0

The integral in the direction of a ρ is zero for all n , while that in the direction of aφ is zero for all n ≠ 1 . For n = 1 , A becomes

A = aφ

π

μ M z a 2 +∞ 1 (a r )k Pk1 (cosθ )∫ Pk1 (cosθ ′)sin 2 θ ′dθ ′ ∑ 2 r k =1 k (k + 1) 0


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For r >> a , all terms of index k > 1 can be ignored and A reduces to π

μ M za2 A = aφ (a r )P11 (cosθ )∫ P11 (cosθ ′)sin 2 θ ′dθ ′ 4 r 0 Since P11 (cosθ ) = − sin θ and P11 (cosθ ′) = − sin θ ′ , then π

μ M z a3 μ M z a3 3 ′ ′ A = aφ θ θ θ = a sin θ d sin sin φ ∫0 3 r2 4 r2 The magnetic field intensity radiated by the sphere at a distance far away is

H=

1 1 ∂ (Aφ sin θ ) 1 ∂ (rAφ )  ∇ × A = ar − aθ r sin θ ∂θ r ∂r μ

Substituting Aφ and simplifying, yields H=

1 M z a3 (ar 2 cosθ + aθ sin θ ) 3 r3

which is the field of a magnetic dipole of dipole moment m = 4πa3 M z 3 , or the volume of the sphere multiplied by the magnetization.

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PROBLEMS 6.1.

A long solenoid with its axis along the z-axis is wound uniformly on a core of a magnetic material of permeability μ = 2000 μ 0 with n turns/m. If the magnetic flux density in the solenoid is 0.5 a z Wb m 2 , find the (a) (b) (c) (d)

6.2.

Magnetization M . Bound surface current density K b . Magnetic flux density in the solenoid when the core is replaced by air. Number of turn/m n when the current in the solenoid is 10 A .

A long solenoid with its axis along the z-axis is made of a thin wire wound uniformly around a magnetic core of permeability μ = μ r μ 0 . The current through the solenoid produces a magnetic flux density 0.1a z Wb m 2 in the solenoid. When the magnetic core is replaced by free space, the same current produces a magnetic flux density of 0.1a z mWb m 2. Find the (a) Relative permeability μ r . (b) Magnetization M . (c) Bound surface current density K b .

6.3.

A magnetic material in a shape of a cylinder with its axis along the z-axis, has a permeability of μ = μ 0μ r . If the magnetic field in the material is

H = H 0a z A m . Find the (a) Magnetization M . (b) Bound surface current density K b on the cylindrical surface of the material. (c) Bound volume current density J b . 6.4.

A ferromagnetic cylinder of the radius a with its axis along the z-axis is magnetized with magnetization M = a z ρ M z . If the permeability of the cylinder is μ , find the bound surface K b and volume J b current densities.

6.5.

Using multi-pole expansion show that the magnetic vector potential at any point ( r ,θ , φ ) produced by the disk of Example 6.1 can be written as

⎧(a r ) , r > a μ M z t +∞ 1 P21k −1 (cosθ )P21k −1 (0)⎨ ∑ 2 k −1 4 k =1 k (2k − 1) ⎩(r a ) , r < a 2k

A = aφ

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6.6.

Using the result obtained in Problem 6.5, find the magnetic field intensity of the disk and show that it behaves like a magnetic dipole when a > a ) on the (a) Plane z = 0  (b) Axis of the sphere along the z-axis.

6.8.

The magnetic field inside an infinitely long cylinder of radius a at a radial distance ρ from its axis is given by

H = az

C 2 (a − ρ 2 ) A m 2

where C is constant. Find the stored energy per unit length of the cylinder. 6.9.

An air-core solenoid of radius 3 mm and length 12 cm has 400 turns and carries a current of 1.5 A . Assuming uniform field in the solenoid, find the (a) Inductance of the solenoid. (b) Stored magnetic energy in the solenoid.

6.10. For the two solenoids in Fig. (6.18), if 2 N1 = N2 = 400 turns and I1 = 3I 2 = 0.6 A , l = 20 cm and d = 0.5 cm . Find the (a) Mutual inductance of the solenoids. (b) Total stored magnetic energy in the solenoids. 6.11. A thin toroid core of permeability μ = 1000 μ 0 , mean radius ρ0 = 10 cm , and cross section area S = 0.8 cm 2 is wound with two coils of thin wires one over the other, with N1 = 1200 and N2 = 1000 turns. Find the (a) Self-inductance of each coil. (b) Mutual inductance of the coils. (c) Coupling coefficient. (d) Total stored magnetic energy in the toroid.

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6.12. Three magnetically coupled circuits have self-inductances 5 m H , 80 m H and 20 m H . If the coupling coefficient between any two circuits is 0.8, find the (a) The mutual inductance between each circuit. (b) The total magnetic energy stored in the circuits. 6.13. Show that the exact mutual inductance between two concentric thin circular loops of radii a and b ( b > a ) can be written as

⎤ n ⎡ (2n )! (a 4b)n ⎥ M = 2π μ 0 b∑ ⎢ n =1 2n − 1 ⎣ n! n! ⎦ +∞

2

6.14. For the loops in Problem 6.13, if b >> a , find an approximate mutual inductance between the loops. 6.15. Find the approximate mutual inductance of a two concentric thin circular loops of radii 1 cm and 12 cm in air. 6.16. Find the approximate mutual inductance of two coaxial thin circular loops of radii 1.2 cm and 1.6 cm, if the loops are in the air a distance 20 cm apart. 6.17. A thin-wire rectangular loop of sides a and b is placed parallel to infinitely-long thin-wire current carrying conductor as shown in Fig. (6.38). Show that the mutual inductance between the loop and the wire is given by

M=

μb ln(1 + a d ) 2π

6.18. For the coaxial cable with cross-section shown in Fig. (6.39), find the external inductance of the coaxial cable per unit length. 6.19. The cross section of the core of a long solenoid of length l >> b and N turns is shown in Fig. (6.40). Find an expression for the inductance of the solenoid. 6.20. The toroid shown in Fig. (6.41) has a rectangular cross section. If the mean radius of the toroid ρ 0 is much greater than b − a , show that the


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461

inductance L and the stored magnetic energy in the toroid Wm can be approximated by

L≈

μ N 2t ⎛ b − a ⎞ μ N 2 S ⎟= ⎜ π ⎝ b + a ⎠ 2πρ0

and

Wm ≈

μ N 2 I 2t ⎛ b − a ⎞ μ N 2 I 2 S ⎜ ⎟= 2π ⎝ b + a ⎠ 4πρ0

where t and S are the thickness and cross section area of the toroid respectively. Loop

b

b

I

a

d b

a μ1

μ

μ

μ2

a Fig. (6.40). Problem 6.19

Fig. (6.39). Problem 6.18

Fig. (6.38). Problem 6.17

6.21. Let the toroid shown in Fig. (6.42) has a rectangular cross section with the following data: μ = 4000 μ 0, N = 2500 turns; I = 0 . 25 A, a = 11 cm , b = 12 cm , g = 0.2 cm , and thickness t = 4 cm . Find the inductance and the energy stored in the toroid. N

N H

H μ

a

b

μ

ρ

a

b Air gap

g

ρ

Fig. (6.41). Problems 6.20 and 6.29

Fig. (6.42). Problems 6.21, 6.30, and 6.31

6.22. In Fig. (6.43), μ1 = μ 0 and μ 2 = 200 μ 0 . The magnetic field in medium 1 is

H1 = 3a x − a y − 4a z A m

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Find the (a) Magnetic field in medium 2 (b) Angle that the magnetic field in medium 2 makes with normal. (c) Energy density in each region. 6.23. In Fig. (6.43), let μ1 = 2 μ 0 and μ 2 = 3μ 0 . The magnetic field in medium 1 is

H1 = 2a x − 5a z A m If the current density on the plane z = 0 is K = 2a y A m , find the (a) Magnetic field in medium 2 (b) Energy density in each medium. 6.24. A slab of permeability μ 2 = 2000 μ 0 is placed in a magnetic field H1 = 2az A m as shown in Fig. (6.44). The regions z > 1 and z < 0 are free space. Find the magnetic flux density and the energy density in the slab. z H1

zz

z=1

Medium 1 μ1 σ2

Medium 2 μ2

Fig. (6.43). Problem 6.22 and 6.23

μ1 = μ0

Slab

x

H1

μ2 μ3 = μ0

y

Fig. (6.44). Problem 6.24

6.25. For the two-layer magnetic slab shown in Fig. (6.45), the magnetic field in medium 1 is

H1 = 20a x − 4a y + 12a z A m If the surface current densities on the boundaries between the media are K12 = K 23 = K 34 = 0 , find the magnetic flux densities B2 , B 3 and B4 .


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6.26. For the two-layer K = −8a x + 10a y A m

463

magnetic slab shown in Fig. (6.45), if and K 23 = K 34 = 0 , find the magnetic flux

densities B2 , B 3 and B4 . H1 z

Air K12

Infinitely extended slab

B2

ε 0, μ 1 = μ 0 ε0, μ2 = 4μ0

y

K23 B3 B4

K34 Air

ε0, μ3 = 2μ0 ε 0, μ 4 = μ 0

Fig. (6.45). Problems 6.25 and 6.26



6.27. The plane x + y = 0 separates medium 1 (μ1 = 2 μ 0 ), and medium 2 (μ 2 = 5 μ 0 ), where medium 1 occupies the region x + y > 0 . If the magnetic field in medium 1 is H1 = 3a x + a y − a z A m and the boundary between the media is current-free, find the magnetic flux density in medium 2. 6.28. The plane 4 x − 3 z = 0 separates medium 1 (μ1 = μ r1μ 0 ), and medium 2 (μ 2 = μ r 2μ 0 ), where medium 1 in the region 4 x − 3z > 0 . If the magnetic fields in medium 1 and 2 are H1 = 9a x − 14a y + 2a z A m and

H 2 = 20a x + 24a y A m , find (a) The surface current on the boundary between the media. (b) μ r 2 μ r1 . 6.29. If the toroid shown in Fig. (6.41) has a circular cross section with the following data: N = 600 turns; I = 0.5 A , a = 6 cm , and b = 6.6 cm . The B-H curve of the magnetic core of the toroid can be described by the 2 function B = 1.4 H (64 + H ) Wb m . Find the magnetic flux density and flux through the toroid. 6.30. The toroid shown in Fig. (6.42) has a circular cross section with the following data: N = 400 turns; I = 0.25 A , a = 8 cm , b = 9 cm and
= 1 mm . The magnetic core of the toroid is made of the cast steel. Find

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the magnetic flux density and flux through the magnetic core and the air gap. 6.31. The toroid shown in Fig. (6.42) has a circular cross section with the following data: N = 300 turns; a = 10 cm , b = 11 cm and
= 1 mm . The magnetic core of the toroid is made of the cast steel. If the flux density 2 through the air gap is 0.5 Wb m , find the current in the toroid. 6.32. The cross section area of all branches of the magnetic circuit shown in Fig. (6.46) is S = 7 cm 2 . If N1 = 150 turns, I 2 = 0 A , l1 = 24 cm, l2 = 8 cm and
= 1 mm . The magnetic core of the circuit is made of the cast steel. Neglecting fringing, find the current I1 required to establish a flux of

1.4 × 10 −4 Wb in the air gap. 6.33. The cross section area of all branches of the magnetic circuit shown in Fig. (6.46) is S = 1.5cm 2 . If N 2 = 300 turns, I1 = 0 A , l1 = 15cm ,

l2 = 5cm and
= 2 mm . The magnetic core of the circuit is made of the cast iron. Neglecting fringing, find the current I 2 required to establish a flux of 42 μ Wb in the left arm. 6.34. The cross section area of all branches of the magnetic circuit shown in Fig. (6.46) is S = 1cm 2 . If N1 = 200 turns, N 2 = 130 turns, I1 = 2 A ,

l1 = 18cm , l2 = 6cm and
= 1.5 mm . The magnetic core of the circuit is made of the cast iron. If the magnetic flux in the right arm is 3.6 μ Wb , neglecting fringing, find the current I 2 and the flux density in the air gap. Magnetic Material

I1

l1

N1

I2

S

g

N2

Fig. (6.46). Problems 6.32, 6.33, and 6.34

l1

l 2




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465

CHAPTER 7

Time-Varying Electromagnetic Fields Abstract: Maxwell’s equations are essential mathematical tools in the analysis of electromagnetics and antennas problems. In this chapter, Maxwell's equation for general time-varying electromagnetic fields are derived from the basic laws of electromagnetics, and presented in integral and point forms. The special cases for timeharmonic and static fields are obtained from the general equations. The vector potential concept that has been introduced in Chapter 5 for static fields, is generalized in this chapter for time-varying fields. The electric and magnetic vector potentials are important quantities in determining the electromagnetic fields radiated from electric and magnetic current sources. By solving Helmholtz equations, general formulations for the electric and magnetic vector potentials are presented in terms of electric and magnetic current sources respectively. The chapter includes also the application of multi-pole expansion technique to obtain the vector potential for some time-varying fields problems, derivation of boundary conditions for time-varying fields, derivation of Poynting vector, and discussion of electromagnetic power flow. The topics of the chapter are supported by numerous illustrative examples and figures in addition to solved problems and homework problems at the end of the chapter.

Keywords: Ampere’s law, boundary conditions, displacement current, electric field intensity, electric flux density, electric vector potential, Faraday’s law, Gauss’ laws, magnetic field intensity, magnetic flux density, magnetic vector potential, Maxwell’s equations, Poynting vector, time-varying fields. INTRODUCTION Maxwell’s equations are the basis of electromagnetic theory and essential mathematical tools in the analysis of electromagnetic and antennas problems. They relate the fundamental electromagnetic time-varying fields quantities, electric field intensity E V/m, magnetic field intensity H A/m, electric flux density D C/m2, magnetic flux density B Wb/m2, current density, and charge density, in a given medium. Principally, Maxwell’s equations represent the basic laws of electromagnetics, Ampere’s law, Faraday’s law, and Gauss’ laws for electric and magnetic fields combined in four compact equations. In this chapter, Maxwell’s equations are presented in integral and point forms for general time-varying fields, in addition to the special cases for time-harmonic and static fields. Helmholtz equation and its solution based on the concept of vector potential are also presented. The chapter includes the application of multi-pole expansion technique to obtain the vector potential for some time-varying fields problems. Boundary conditions for time-varying fields and Poynting vector are Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

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discussed at the end of the chapter. Extensive discussions on Maxwell's equations are available in many texts e.g. [18, 19, 21, 25, 41, 45, 52]. Advanced treatment of Maxwell's equations and the topics related to the time-varying fields, boundary conditions, and Poynting vector is available in [92-105]. 7.1.

FARADAY’S LAW OF INDUCTION

Faraday had showed experimentally in 1831 that a magnetic field could produce an electric current. An electromotive force e induces between the terminals of a conductor when it moves in a constant magnetic field of flux density B with velocity v or when it placed in a time varying magnetic field B . This induced electromotive force is proportional to the relative change of the magnetic flux ψ with respect to the time t . These observations constitute Faraday’s law of induction which is one of the essential laws in electromagnetic. Mathematically, Faraday’s law is stated as

e=−

dψ dt

(7.1)

The magnetic flux ψ is the flux that passes through a surface S enclosed by a closed path l . The minus sign is due to Lenz’s law, which states that the induced electromotive force e is always opposing the direction of original flux producing it. The electromotive force (emf) is induced between the terminal of the conductor as the result of the relative change of the magnetic flux ψ with respect to the time t . This relative change takes place by the one of the following: 1. A stationary conducting loop is placed in a time varying field, which is known as transformer action 2. A conducting loop is moving in a static field, which is known as motional action. 3. A conducting loop is moving in a time varying field (both transformer and motional action). Transformer action: In which a stationary loop is placed in a time-varying magnetic field as shown in Fig. (7.1) . The induced electromotive force is

Ve = −

dψ ∂B • dS = − ∫∫ dt ∂t S

(7.2)

Time-Varying Electromagnetic Fields

Electromagnetics for Engineering Students

Stationary

conducting

S

l

Ve

467

-

B

+

I B

is out of the

Fig. (7.1). A stationary loop in a time-varying magnetic field.

The induced emf by transformer action in (7.2) is referred to as transformer emf. Motional action: In this case, a moving conductor is cutting a static flux at a velocity v as shown in Fig. (7.2). The force F on a charge Q moving at a velocity v in a static magnetic flux B is given by

F = Qv × B Conducting wire

(7.3)

Moving conductor

I Ve

B F

v l

I

+

I B is out of the page Fig. (7.2). A moving conductor in a static magnetic field.

The associated electric field E = F Q , or

E =v ×B

(7.4)

Therefore, the induced emf between the terminals of the moving conductor is

V e = ∫ E ⋅ dl = ∫ v × B ⋅ dl l

l

(7.5)

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This emf is known as motional emf. Transformer and motional action: In this case, a loop is moving in a time varying field and the total emf V e is due to both transformer and motional action. Assuming that the time-varying magnetic flux density is B , the velocity of the conductor is v and the area enclosed by the loop l is S . Therefore, the total induced electromotive force is the superposition of transformer and motional emfs, or

V e = − ∫∫ S

∂B • dS + v × B ⋅ dl ∫l ∂t

(7.6)

Example 7.1 A rectangular loop is rotating about its axis at an angular frequency ω in a static magnetic flux density B as shown in Fig. (7.3). If the width of the loop is d and its length is l , find the induced electromotive force between the terminals of the loop. ω

b c d

B

d

θ = ωt

a

B

c b ω

l d

Fig. (7.3). A rectangular loop rotating about its axis in a static magnetic field.

Solution The loop is rotating about its axis with angular frequency ω , then at any instant t the angle between the direction of B and the plane of the loop is θ = ω t . Since B is a static and the loop is moving, then referring to Fig. (7.3), the induced electromotive force can be obtained from (7.5), as

Ve = ∫ v × B ⋅ dl = ∫ v × B ⋅ dl + ∫ v × B ⋅ dl + ∫ v × B ⋅ dl + ∫ v × B ⋅ dl l

ab

bc

cd

da

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From Fig. (7.3) we can write

v × B = v B sin θ = v B sin (ω t )au  where a u is the unit vector perpendicular to both v and B . Note that au is parallel to the paths ab and cd and perpendicular to bc and da, thus

∫ v × B ⋅ dl = ∫ v × B ⋅ dl = l v B sin (ωt )

ab

cd

∫ v × B ⋅ dl = ∫ v × B ⋅ dl = 0

bc

da

Then the induced electromotive force becomes

Ve = ∫ v × B ⋅ dl + ∫ v × B ⋅ dl + ∫ v × B ⋅ dl + ∫ v × B ⋅ dl = 2l v B sin (ω t ) ab

bc

cd

da

From Fig. (7.4) the magnitude of the velocity v can be obtained as

v =

(d 2)ω t = ω d t

2

ms

Substituting v = ωd 2 in the above expression of Ve , yields

Ve = 2l v B sin (ω t ) = ld B sin (ω t ) = S B sin (ω t ) V where S = ld is the area of the loop. ω

c ωt

d/2 b Fig. (7.4). The rectangular loop of Example 7.1.

v B

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Example 7.2 A conductor slides freely over conducting rails as shown in Fig. (7.5) in a varying magnetic flux density B = a z B0 cos(ω t − β x ) with velocity v = a xυ , where B0 , ω, and β are real constants. Find the induced electromotive force between the terminals of the conductor assuming the conductor starts from the point x = x 0 . B is out of the page

Moving conductor y

Conducting rails

v

B

l

x x = x0

x

Fig. (7.5). A conductor slides freely over conducting rails in a time-varying magnetic field.

Solution The conductor is moving in a time-varying magnetic flux density, then both transformer and motional emfs will be induced between the terminals of the moving conductor. Replace x by x′ in B where x is the displacement over the rails in time t as illustrated in Fig. (7.5) , then B = a z B0 cos(ω t − β x′) and

v = a xυ , using (7.2) Ve

= − ∫∫ S l x

∂B • dS + v × B ⋅ dl ∫l ∂t l

= ∫ ∫ B 0 ω sin ( ω t − β x ′ )a z ⋅ a z dx ′dy + ∫ a xυ × a z B 0 cos ( ω t − β x ′ ) ⋅ a y dy 0 x0

0

⇒ Ve = −

⇒ Ve =

2 ω lB0

β

B0 ω l

β

[cos(ω t − β x′)] xx

l

+ ∫υB0 cos(ω t − β x )a y ⋅ a y dy 0 0

⎡ ⎛ x + x0 ⎞⎤ ⎡ ⎛ x − x0 ⎞⎤ sin ⎢ω t − β ⎜ ⎟ sin β ⎜ ⎟ − υ lB0 cos(ω t − β x ) ⎝ 2 ⎠⎥⎦ ⎢⎣ ⎝ 2 ⎠⎥⎦ ⎣

Time-Varying Electromagnetic Fields

7.2.

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DISPLACEMENT CURRENT

Consider the circuit shown in Fig. (7.6). If the capacitor C is initially uncharged, then when SW is switched on, a current flows with an initial value I 0 = V0 R . This current decays exponentially with time such that its value at any instant t is

I = I0 e

−

t RC

A

(7.7)

At the point P , electrons flow causing a current. However, between the plates, no electrons can flow in the dielectric. By Kirchhoff’s current law a current must flow between the plates. This current is not in the form of electrons flow; rather it takes the form of a time-varying electric field. This type of current is known as the displacement current. To show that a changing electric field is equivalent to a current, consider the circuit of Fig. (7.6) again. The rate of change of the charge q at P with respect to the time t is

Id =

dq dt

(7.8)

where q is the total charge on the plates of the capacitor. Applying Gauss’s theorem, the total charge can be expressed as

q = ∫∫D ⋅ dS = ε ∫∫E ⋅ dS s

(7.9)

s

SW

I R

+ P S

V0

Fig. (7.6). Displacement current.

ε -

D = εE

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where D and E are the time varying electric flux density and field intensity between the plates of the capacitor. S is the closed surface shown in Fig. (7.6). Substituting q from (7.9) into (7.8), yields

Id =

d d ∂D ∂E ⋅ dS = ε ∫∫ ⋅ dS D ⋅ dS = ε ∫∫E ⋅ dS = ∫∫ ∫∫ dt s dt s ∂t ∂t s s

(7.10)

The current I d can be expressed in terms of the corresponding time-varying displacement current density J d as

I d = ∫∫ J d ⋅ dS

(7.11)

s

Comparing (7.10) and (7.11), the displacement current can be expressed as

Jd =

∂D ∂E =ε ∂t ∂t

(7.12)

Thus, the current between the plates takes the form of change in the electric field with respect to the time. If the electric field is stationary or ∂E ∂t = 0 , then no current will flow. If the dielectric between the plates of the capacitor is lossy with conductivity σ , the time varying conduction current density J c is related to the electric field E between the plates by

J c = σE

(7.13)

Hence, the total current density in the lossy dielectric between the plates becomes

J = Jc + Jd 7.3.

(7.14)

MAXWELL’S EQUATIONS IN INTEGRAL FORM

Faraday’s law states that the induced electromotive force V e between the terminals of a conductor in a time-varying magnetic field H with flux ψ and flux density B , is proportional to the rate of change of ψ with respect to the time as stated mathematically in (7.1). Thus,

Ve = −

dψ dt

(7.15)

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It is known from (5.26) that

ψ = ∫∫ B ⋅ dS

(7.16)

S

Substituting ψ from (7.16) into (7.15), yields

Ve = −

⎞ ⎛ ⎞ d⎛ ⎜ ∫∫B ⋅ dS ⎟ = −⎜ ∫∫ ∂B ⋅ dS ⎟ ⎟ ⎟ ⎜ ∂t dt ⎝⎜ S ⎠ ⎝S ⎠

(7.17)

Ve is related to the electric field intensity E along the conductor l by Ve = ∫ E ⋅ dl

(7.18)

l

Comparing (7.17) and (7.18), leads to

⎞

⎛

∂B ∫ E ⋅ dl = −⎜⎜ ∫∫ ⋅ dS ⎟⎟ ⎝




S

∂t

⎠

(7.19)

This is Maxwell’s equation that is derived from Faraday’s law. Ampere’s law states that the line integral of a magnetic field H along a closed path l enclosing a current is equal to the magnitude of the current. If the magnitude of the current is I , then Ampere’s law can be expressed mathematically by

∫H

•

dl = I

(7.20)

l

If the surface area enclosed by the closed path l is S , then the current I is

I = ∫∫ J • dS

(7.21)

S

where J is the total time varying current density through S . From (7.20) and (7.21), we can write

∫H l

•

dl = ∫∫ J • dS S

(7.22)

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The current density J is given by

J = Jc + Jd

(7.23)

Hence, using (7.22) and (7.23), we can write ⎛

∫ H ⋅ dl = ∫∫ ⎜⎝ J l

c

+

S

∂D ⎞ ⎟ ⋅ dS ∂t ⎠

(7.24)

This is Maxwell’s equation in an integral form that is derived from Ampere’s law. If a charge Q

e

is enclosed within a surface S , then the electric flux density D is

related to the enclosed charge Q e by Gauss’s law for electric flux, which can be expressed as

∫∫D

•

dS = Q e

(7.25)

S

If the charge Q

e

is distributed within a volume V with volumetric charge density


e , then

Q e = ∫
e dV

(7.26)

Eliminating Q e from (7.25) and (7.26), yields

∫∫ D S

•

dS = ∫
e dV

(7.27)

This is Maxwell’s equation in an integral form that is derived from Gauss’s law for electric flux. Similarly, from Gauss’s law for magnetic flux, and taking into consideration that the magnetic charge is zero Q m = 0 , Maxwell’s equation from Gauss’s law for magnetic flux can be obtained as

∫∫B S

•

dS = 0

(7.28)

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Summarizing the above analysis, Maxwell’s equations for time-varying electromagnetic fields in integral form are

∫
E

•

dl = − ∫∫ ∂B ⋅ dS ∂t

(7.29a)

S

⎛

∫ H ⋅ dl = ∫∫ ⎜⎝ J l

c

+

S

∫∫ D S

•

∂D ⎞ ⎟ ⋅ dS ∂t ⎠

dS = ∫
e dv

∫∫B

•

(7.29b) (7.29c)

dS = 0

(7.29d)

S

7.4.

CONSTITUTIVE RELATIONS AND PARAMETERS

In Chapters 3 and 5 we discussed the concepts of linearity, isotropy, and homogeneity for dielectric and magnetic materials and the interaction of static electromagnetic fields with these materials. These concepts are still valid for timevarying fields. The expressions describing the interaction of time-varying electromagnetic fields with materials are referred to as the constitutive relations. For linear, isotropic, and homogeneous material the constitutive relations are

D = εE ,

B =μH ,

E = σJ

(7.30)

The permittivity ε , permeability μ , and conductivity σ of the material are known also as the constitutive parameters of the material. For a linear anisotropic material, the permittivity and permeability depend on the directions of field components and the constitutive relations are given by D = εE ,

B = μH ,

E = σJ

(7.31)

where ε is a matrix known as the permittivity dyadic, μ is the permeability dyadic, and σ is the conductivity dyadic. In this chapter and the remainder of the book, we will restrict our discussion and analysis to only the linear, isotropic, and homogeneous material. Thus, the simple constitutive relations given in (7.30) will be used.

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7.5.

Electromagnetics for Engineering Students

Sameir M. Ali Hamed


MAXWELL’S EQUATIONS IN POINT FORM

7.5.1. General Time-Varying Fields Applying Stokes’ theorem on the left-hand side of (7.29a), the line integral can be replaced by a surface integral and the equation becomes

∫
E

•

dl = ∫∫ ∇ × E • dS = − S

⎞ d⎛ ⎜⎜ ∫∫ B • dS ⎟⎟ dt ⎝ S ⎠

(7.32)

Then

∇×E = −

∂B ∂t

(7.33)

Equation (7.33) is Maxwell’s equation that represents Faraday’s law in point form. Similarly, by changing the line integral on the left-hand side of (7.29b) to surface integral, we get

∫H l

•

∂D ⎞ ⎛ dl = ∫∫ ∇ × H • dS = ∫∫ ⎜ J c + ⎟ ⋅ dS ∂t ⎠ S S ⎝

(7.34)

Consequently, Maxwell’s equation in point form that represents Ampere’s law is obtained as

∇ × H = Jc +

∂D ∂t

(7.35)

Application of divergent theorem on the right-hand side of (7.29c), the surface integral changes to volume integral, and the equation can be written as

∫∫D S

•

dS = ∫ ∇ • D dv = ∫
e dv

(7.36)

From (7.36), we can write Gauss’s law for electric flux as

∇ • D =
e

(7.37)

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477

Similarly, from (7.29d) we can write Gauss’s law for magnetic flux as ∇•B = 0

(7.38)

Summarizing, we can write Maxwell’s electromagnetic fields in point form as

equations

∂B ∂H = −μ ∂t ∂t ∂D ∂E ∇×H = Jc + = σE + ε ∂t ∂t ∇×E = −

∇ • D =
e

for

time-varying

(7.39a) (7.39b) (7.39c) (7.39d)

∇•B = 0

Maxwell’s equations assume that the time varying electric field E and magnetic field H propagate in free space with the speed of light [45]. The equations reveal that existence of a time-varying electric field E induces the magnetic field H , DQGYLFHYHUVD 7.5.2. Time-Harmonic Fields If J c  E and H are time-harmonic fields with angular frequency ω , then we can define the complex spatial quantities J c  E and H that related to instantaneous time varying quantities J c  E and H by


e = Re [ρ e e j ω t ]

J c = Re [J c e

j ωt

]

E = Re [E e ] H = Re [H e j ω t ] j ωt

(7.40a) (7.40b) (7.40c) (7.40d)

J c  E and H are not functions of time. Substituting ε E = D and μ H = B into (7.39a) – (7.39d), and using ∂ ∂t = j ω, Maxwell’s equations for time-harmonic fields can be written in terms of the complex quantities J c  E and H as

∇ × E = − j ωμ H ∇ × H = J c + j ωε E = (σ + j ωε )E

(7.41a) (7.41b)

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∇ ⋅ E = ρe ε

(7.41c)

∇⋅H = 0

(4.41d)

The source for the fields in (7.41a) - (7.42b) is the electric current density J c and electric charge density ρ e . Since there is no magnetic charge and current in nature, the magnetic current density M and charge density ρ m does not appear (set to zero) in the equations. Although the current density M does not exist in the real world, it may appear in some electromagnetic problems as a result of the application of the Equivalence Principle For more details on the Equivalence Principle the student may refer to [106] and [107]. In this case, the original problem is replaced by an equivalent problem that gives the same solution as the original one. The equivalent problem is modeled in terms of equivalent electric and magnetic current densities. Therefore, the concept of magnetic current density has been used extensively in the analysis of electromagnetic and antenna problems, despite that it does not exist in the real world. Let us assume that the magnetic current density M and charge density ρ m exist and J c = 0 and ρe = 0 , then Maxwell’s equation for these magnetic sources can be written using the duality between the electric and magnetic fields. For more information on the duality principle the student may refer to [14]. By duality principle, the solution to a known electric (magnetic) field problem can be used to solve a related magnetic (electric) field problem by just replacing the electric (magnetic) field variables with the corresponding magnetic (electric) field variables. The variables of the original problem and the variables of the corresponding dual problem are listed in Table 7.1. Using this duality principle, Maxwell's equations for magnetic sources M and ρ m can be obtained starting from (7.41a) – (7.41c) by replacing E  H , J c , ρ e , ε and in (7.41a) – (7.41c) with H , − E , M , ρ m , and μ respectively. Doing this, yields

∇ × E = −M − j ωμ H

(7.42a)

∇ × H = j ωε E

(7.42b)

∇⋅E = 0

(7.42c)

∇⋅H = ρm μ

(7.42d)

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479

Table 7.1. Duality principle. &RUUHVSRQGLQJ 9DULDEOHVRIWKH 'XDO3UREOHP

 9DULDEOHVRIWKH  2ULJLQDO3UREOHP 

E H Jc ρe M

H −E M



ρm

− Jc − ρe μ

ρm

ε 7.5.3. Static Fields

For static fields ω = 0 , and (7.41a) – (7.4d) reduce to ∇×E = 0 ∇×H = J ∇ • D = ρe ∇•B =0

(7.43a) (7.43b) (7.43c) (7.43d)

Table 7.2 summarizes the different form of Maxwell’s equations for electric current source. Table 7.2. Maxwell’s equations in different forms. Integral Form

General TimeVarying Fields

d⎛

⎞

∫ E ⋅ dl = − dt ⎜⎜⎝ ∫∫B ⋅ dS ⎟⎟⎠


S

⎛

∫ H ⋅ dl = ∫∫ ⎜⎜⎝ J l

S

c

+

∂D ∂t

∫∫D ⋅ dS = ∫


e

S

∫∫B ⋅ dS = 0 S

⎞ ⎟⎟ ⋅ dS ⎠

dv

∇×E = −

∂B ∂t

∇× H = Jc +

∇ ⋅ D =
e ∇ ⋅B = 0

∂D ∂t

Point form Time-Harmonic Fields

Static Fields

∇ × E = − j ωμ H

∇×E = 0

∇ × H = J + j ωε E

∇×H = J

ρe ε

∇ ⋅ D = ρe

∇⋅H = 0

∇⋅B = 0

∇⋅E =

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Example 7.3 The displacement current density in a material with ε = 5 ε 0 , μ = μ 0 , σ = 0 is

(

)

J d = a x 5 cos 2π × 10 t − β z A m 8

2

Find D , E , B , and H . Solution The displacement current is given by J d = ∂D ∂t , then −8

(

)

5 × 10 8 2 D = ∫ J d dt = a x sin 2π × 10 t − 2π z 3 A m 2π ⇒ E = D ( 5 ε 0 ) = a x 180 sin (2π × 10 8 t − 2π z 3 ) V m Using Faraday’s law, we have

ax ∂B ∂ = −∇ × E = ∂t ∂x

ay ∂ ∂y 0

Ex

az ∂ ∂E ∂E = −a y x + a z x ∂z ∂z ∂y 0

Since ∂E x ∂y = 0 , then

(

)

∂B ∂E 2π 8 = −a y x = a y180 × cos 2π × 10 t − 2π z 3 ∂t ∂z 3 8 ⇒ B = a y120π ∫ cos 2π × 10 t − 2π z 3 dt

(

−8

(

)

8

)

= a y 60 × 10 sin 2π × 10 t − 2π z 3 T The magnetic field intensity is

H =

(

)

120π B 8 = ay sin 2π × 10 t − 2π z 3 A m 8 −7 μ0 2π × 10 × 4π × 10 3 8 = ay sin 2π × 10 t − 2π z 3 A m 2π

(

)

Time-Varying Electromagnetic Fields

Electromagnetics for Engineering Students

Example 7.4 The instantaneous electric and magnetic fields in free space are given by

E = a x120π sin(ω t − 2 z ) V m H = a y cos ( ω t − 2 z − π 4 ) A m (a) Write the spatial complex fields. (b) Find the displacement current density at 1 GHz in free space. Solution jϕ

(a) Let E = E e Using E = Re[E e

[

j ωt

]

] , then

E = Re E e j ω t = Re ⎡ E e j (ω t +ϕ ) ⎤ = E cos(ω t + ϕ ) ⎢⎣ ⎥⎦ Since E = a x 120π sin ( ω t − 2 z ) V m

⇒ E cos ( ω t + ϕ ) = 120π sin ( ω t − 2 z ) = 120π cos ( ω t − 2 z − π 2 ) ⇒ E = 120π V m  ⇒ ϕ = −2 z − π 2  Hence, the spatial complex electric field is E = 120π e

− j (2 z +π 2 )

Vm

Similarly, the spatial magnetic field can be obtained as

H=e

− j ( 2 z −π 4 )

Am

(b) In free space ε = ε 0 , the displacement current density is

Jd =

∂D ∂E =ε = a x120π ωε 0 cos(ω t − 2 z ) ∂t ∂t

At 1 GHz 9

J d = a x120π × 2π × 10 × 8.85 × 10

(

9

)

−12

(

9

cos 2π × 10 t − 2 z

= a x 20.98 cos 2π × 10 t − 2 z A m

2

)

481

482

7.6.

Electromagnetics for Engineering Students

Sameir M. Ali Hamed


TIME-VARYING VECTOR AND SCALAR POTENTIALS

7.6.1. General Time-Varying Fields The concept of the magnetic vector potential has been introduced in Chapter 5 and defined by (5.45) and (5.48). Using the definition of the vector potential, the magnetic field H e resulted from a time varying electric current source J can be expressed as

He =

1 ∇× A μ

(7.44)

where A is the instantaneous time-varying vector potential. For J = 0 and M ≠ 0 , the electric flux density satisfy ∇ ⋅ D = 0 , then as in the case of the electric current source, the electric field E m that is resulted from a magnetic current source M can be expressed in terms of a vector F as

1 E m = − ∇ ×F ε

(7.45)

where F is the time-varying electric vector potential that results from magnetic current sources. Subscripts e and m indicate that the quantity is due to electric or magnetic current source respectively. On the other hand, it is known from the vector properties that, the curl of the gradient of a scalar quantity is zero. Let G and
e be a time varying vector and scalar quantities respectively, such that G = ∇
e , then

∇ × ( ∇
e ) = ∇ × G = 0

(7.46)

This last equation shows that if the curl of a vector G is zero then this vector is equal to a gradient of a scalar
e . Using (7.44) along with B = μ H , (7.39a) can be written as

∂A ⎞ ⎛ ∇ × ⎜E + ⎟=0 ∂t ⎠ ⎝

(7.47)

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483

Comparing (7.46) and (7.47), and replacing G by E + ∂A ∂t , then for electric current source, the relation between the magnetic vector potential A and the scalar quantity
e can be written as

Ee +

∂A = −∇
e ∂t

(7.48)

where
e is the scalar potential due to the electric current source. As in the case of static fields, (7.44) is not enough to define A . Both the curl and the divergence of A are needed to define A uniquely. The divergence of A for time harmonic fields can be chosen as

∇ ⋅ A = − με

∂
e ∂t

(7.49)

This is known as the Lorentz gauge. For static condition ∂
e ∂t = 0 , and (7.49) reduces to Coulomb gauge given by (5.48). Using the duality property, the relation between the electric vector potential F and magnetic scalar potential
m for the magnetic field can be written as

Hm+

∂F = −∇
m ∂t

(7.50)

and the divergent of F is

∇ ⋅ F = − με

∂
m ∂t

(7.51)

7.6.2. Time-Harmonic Fields For time-harmonic fields with a time-dependent factor e j ω , ∂ ∂t = j ω and A and F can be replaced by their spatial vectors A and F respectively. The relation between instantaneous time-varying vector potentials and corresponding spatial potentials A and F are

A = Re [A e j ω t ] F = Re [F e j ω t ]

(7.52a) (7.52b)

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e = Re [Φ e e j ω t ]


m = Re [Φ m e

j ωt

]

(7.52a) (7.52a)

Consequently, for time-harmonic fields (7.44) and (7.45) become

1 He = ∇ × A μ 1 Em = − ∇×F ε

(7.53) (7.54)

and the Lorentz gauge for electric and magnetic current sources become

∇ ⋅ A = − j ωμε Φ e ,

∇ ⋅ F = − j ωμε Φ m

(7.55)

Similarly, (7.48) and (7.50) can be rewritten respectively as

1 ∇ (∇ ⋅ A ) − j ω A j ωμε 1 H m = −∇Φ m − j ω F = ∇ (∇ ⋅ F ) − j ω F j ωμε

E e = −∇Φ e − j ω A =

(7.56) (7.57)

Once A and Φ e are known, the electric field E e due to an electric current source can be obtained from (7.56). Using Maxwell’s equation (7.42a), the corresponding magnetic field H e can be obtained. Similarly, E m and H m can be obtained from (7.57) and (7.42b) respectively. The total electromagnetic fields due the presence of both electric and magnetic current sources become

1 1 ∇ (∇ ⋅ A ) − j ω A − ∇ × F j ωμε ε 1 1 H = He + Hm = ∇ (∇ ⋅ F ) − j ω F + ∇ × A j ωμε μ E = Ee + Em =

Example 7.5 If the instantaneous vector potential and scalar potential in free space are

(

8

)

A = a z 0.001y cos 3 × 10 t cos z Wb m 

(7.58) (7.59)

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e = 3 × 10 5 y sin (3 × 10 8 t ) sin z V Find E and H . Solution From given A and
e , we can get

∇
e

∂A = − a z × 3 × 10 5 y sin (3 × 10 8 t ) cos z V m ∂t ∂
e ∂
e ∂
e = ax +ay +az ∂z ∂x ∂y  5 8 = a y 3 × 10 sin (3 × 10 t )sin z + a z 3 × 10 5 sin (3 × 10 8 t )cos z V m

Substituting A and
e into (7.48), the electric field can be obtained as

E =-

∂A − ∇
e = − a y 3 × 10 5 sin (3 × 10 8 t ) sin z V m ∂t

The magnet field can be obtained from A directly as

ax ay az ∂A z ⎞ 1 1 ⎛ ∂A H = ∇ × A = ∂ ∂x ∂ ∂y ∂ ∂z = ⎜ a x z − a y ⎟ μ μ⎝ ∂y ∂x ⎠ 0 0 Az Since ∂A z ∂x = 0 , then

H = ax

1 ∂A x 0.001 = cos (3 × 10 8 t ) cos z = 795.5 cos (3 × 10 8 t ) cos z A m −7 μ 0 ∂y 4π × 10

Alternatively, H can be obtained from E using Maxwell’s equation (7.39a) as

1 1 ∂E y 3 × 10 5 ∂H sin (3 × 10 8 t ) cos z = − ∇ × E = −a x = ax −7 μ0 μ 0 ∂z ∂t 4π × 10 −3 10 ⇒ H = ax cos (3 × 10 8 t ) cos z = 795.5 cos (3 × 10 8 t ) cos z A m −7 4π × 10

486

7.7.

Electromagnetics for Engineering Students

Sameir M. Ali Hamed


THE ELECTRIC AND MAGNETIC HERTZ VECTORS

The time varying magnetic vector potential A can be expressed in terms of a vector
e as follows

∂
e ∂t

A = με where


e

(7.60)

is known as the electric hertz vector. Using (7.49) with (7.60), we get


e = −∇ ⋅
e

(7.61)

Substituting A and
e from (7.60) and (7.61) respectively into (7.48) and (7.44), the electric field E e and magnetic field H e can be expressed in terms of


E e = − με

∂2
∂t

He =ε

e

2

+ ∇ (∇ ⋅
e )

e

as

(7.62)

∂ (∇ ×
e ) ∂t

(7.63)

Similarly, the time varying electric vector potential F can be expressed in terms of a vector
m as

F = με where


m

∂
m ∂t

(7.64)

is known as the magnetic hertz vector. Using (7.51) and (7.64) we get

Φ m = −∇ ⋅
m

(7.65)

Using (7.45) and (7.50) along with (7.64) and (7.65), the electric field E m and magnetic field H

m

can be expressed in terms of


m

as

∂ (∇ ×
m ) ∂t ∂2
m + ∇ (∇ ⋅
= − με ∂t 2 E m = −μ

Hm

(7.66)

m

)

(7.67)

Time-Varying Electromagnetic Fields

Electromagnetics for Engineering Students

For time-harmonic fields with angular frequency ω ,





m

e

487

= Re [Π e e j ω t ] and

= Re [Π m e j ω t ], where Π e and Π m are the complex spatial vectors

corresponding to instantaneous quantities


and
m respectively. Also, we

e

have ∂ ∂t = j ω and ∂ ∂t = − ω . Thus, (7.62), (7.63), (7.66), and (7.67) for time-harmonic fields become 2

2

2

E e = ω 2 με Π e + ∇ ( ∇ ⋅ Π e )

(7.68)

H e = j ωε ∇ × Π e E m = − j ωμ ∇ × Π m

(7.69) (7.70)

H m = ω 2 με Π m + ∇ ( ∇ ⋅ Π m )

(7.71)

Example 7.6 For a time-harmonic electric current source J , show that

(∇ 2 + ω 2 με )
e = −

1 J j ωε

Solution Substituting H e from (7.69) into Maxwell’s equation (7.41b), yields

∇×∇× Πe = −

1 J + Ee j ωε

Using the vector identity

A × B × C = ( A ⋅ C )B − ( A ⋅ B )C Then

∇ (∇ ⋅ Π e ) − ∇ 2 Π e = −

1 J + Ee j ωε

Substituting E e from (7.68) into the last equation and re-arranging, yields

(∇ 2 + ω 2 με )Π e = j

1 J ωε

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HELMHOLTZ EQUATION AND ITS SOLUTION

Taking the curl for both sides of (7.41a) and eliminating H using (7.41b), yields

∇ × ∇ × E = ω 2 με E − j ωμ J

(7.72)

Let β = ω με , where β is known as the propagation constant or the wave number, then (7.72) becomes

∇ × ∇ × E = β 2 E − j ωμ J

(7.73)

Taking the curl of (7.41a) and using (7.53), then

∇ × ∇ × E = − j ω ∇ × ∇ × A = − j ω[∇ ( ∇ ⋅ A ) − ∇ 2 A ]

(7.74)

Equating the right-hand sides of (7.73) and (7.74) and substituting E from (7.56), we can get

− j ω β 2 A − β 2 ∇Φ e − j ω μJ = − j ω [∇ ( ∇ ⋅ A ) − ∇ 2 A ]

(7.75)

From (7.55) we have β 2∇Φ e =  − j ω∇(∇ ⋅ A). Hence, (7.75) reduces to

∇ 2 A + β 2 A = −μ J

(7.76)

Equation (7.76) is known as the non-homogenous Helmholtz equation. Similar equation can be obtained for the magnetic current source M starting from (7.42a) as

∇ 2F + β 2F = −εM

(7.77)

The following analysis presents the solution of Helmholtz equation for the electric current source J . Assume infinitesimal sinusoidal time varying current source J of length dl =  along the z-axis as shown in Fig. (7.7). The current density can be expressed as J = a z J z and the vector potential of the current source is 

A  = a z Az . Substituting J and A  into (7.76), yields

∇ 2 Az + β 2 Az = − μ J z

(7.78)

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z dS = a r r 2 sin θ dθ dφ  r

r dθ


dθ 

θ

r sin θ dφ 

dφ 



+ 2 

− 2



y

r sin θ

φ

x

Fig. (7.7). Spherical coordinates system for an infinitesimal current element.



This is valid only inside the current source. Outside the current source J z = 0 , and (7.78) reduces to

∇ 2 Az = − β 2 Az

(7.79)

  2  ∂ ⎛⎜ ∂Az ⎞⎟ 1 ∂ ⎜⎛ 2 ∂Az ⎞⎟ 1 1 ∂ Az ∇ A = 2 ⎜r + 2 sin θ + ∂θ ⎠⎟ r 2 sin 2 θ ∂φ 2 r ∂r ⎝ ∂r ⎠⎟ r sin θ ∂θ ⎝⎜

(7.80)

In spherical coordinates system 2

 z



Since the current distribution is uniform, then Az varies with r and constant with 

2



2

θ and φ , so ∂Az ∂θ = ∂ Az ∂φ = 0 , and (7.80) reduces to  1 ∂ ⎛ 2 ∂Az ⎞ ⎜ ⎟⎟ = − β 2 Az r ⎜ 2 ∂r ⎠ r ∂r ⎝

(7.81)

Let 

Az =

z r

(7.82)

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then (7.81) can be written as

∂2 ∂r

2

z

+β 2 z = 0

(7.83)

The general solution of (7.83) can be written in the form

 z = C 1 e − j β r + C 2 e + jβ r

(7.84)

It is assumed that the fields from the current element propagate outward with phase retardation − β r (the minus sign indicates that the propagation is outward). The presence of the factor e + jβr contradict the assumption that the propagation is outward, then C 2 = 0 and the solution reduces to

 z = C 1 e − jβ r

(7.85)

Substituting  z into (7.82), the solution of Helmholtz equation becomes

Az = C1

e − jβr r

(7.86)

The constant C1 can be determined from the boundary conditions of the problem. Since C1 remains the same for static or time-varying fields, let us consider the static field case to find C1 . In this case, ω = 0 → β = 0 , then (7.78) and (7.86) reduce to

∇2 Az = − μ J z 

Az =

(7.87)

C1 r

(7.88)

The boundary condition is that the total current of the current element is I z  . Integrating both sides of (7.87) over the unit volume of the current element, then

∫ ∇ A dv = − μ ∫ J dv = − μ I  2

v

 z

z

v

z

(7.89)

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Using the Divergence Theorem on left-hand side of (7.89), the volume integral can be transformed to surface integral as follows

∫∫∇A

 z

⋅ dS = − μ I z 

(7.90)

S

Differentiating both sides of (7.88) with respect to r , yields 

∂Az C = − 21 ∂r r 

(7.91)



Since ∇Az = ar ∂Az ∂r , then (7.91) can be written as

∇Az = −a z

C1 r

2

(7.92)



Substituting ∇Az and dS = ar sinθ dθ dφ into (7.90), yields 2π π

− C1 ∫∫ sin θ dθ dφ = −4π C1 = − μ I z 

(7.93)

0 0

From this last equation, the constant C1 is obtained as

C1 =

μ Iz  4π

(7.94)

Substituting C1 in (7.86), the vector potential due the infinitesimal element of length dl =  and current I z along z-axis is obtained as

A z = dA z =

μ I z e − jβr dl 4π r

(7.95)

The total vector potential for a current source of length L along the z-axis with current distribution I = az I z can be obtained by integrating the vector potential dAz along the entire length L of the current source, then

μ e − jβr dl Az = Iz 4π ∫L r

(7.96)

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Based on (7.96), the vector potential A = a x Ax + a y Ay + a z Az at (x, y, z ) produced

by an arbitrary linear electric current I = a x I x + a y I y + a z I z at (x′, y′, z ′) shown in Fig. (7.8a), can be expressed as

A ( x,y,z ) =

z

μ 4π

∫

l′

I e ( x ′, y ′, z ′ )

Source Point (x′, y′, z′)

R

(x′, y′, z′) I dl′

l

(7.97a)

z

Field Point (x, y, z)

Source Point

e − jβR dl ′ R

Field Point (x ,y, z)

R

K dS′ r

S

r′

r

r′

y

y

x

x (a)

(b)

z Source Point R (x′,y′,z′) J dv′ v

Field Point (x,y,z)

r r′

y x (c) Fig. (7.8). Rectangular coordinates system for the vector potential. (a) linear current element. (b) Surface current element. (c) Volume current element.

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For the surface electric current density K e shown in Fig. (7.8b), and volume current density J shown in Fig. (7.8c), (796) becomes

μ A ( x,y,z ) = 4π

e − jβR ∫S K e ( x ′ , y ′, z ′ ) R dS ′

(7.97b)

and

μ A ( x,y,z ) = 4π

∫

v

J ( x ′, y ′, z ′ )

e − jβR dv ′ R

(7.97c)

R is the distance between the source point (x′, y′, z ′) and the field point (x, y, z ). Similar expressions for the electric vector potential F can be written as ε e − jβR ′ ′ ′ ( ) I x , y , z dl ′ m 4π ∫l ′ R ε e − jβR ′ ′ ′ ( ) F ( x,y,z ) = K x , y , z dS ′ m 4π ∫S R ε e − jβR ′ ′ ′ F ( x,y,z ) = M( x , y , z ) dv ′ 4π ∫v R F ( x,y,z ) =

(7.98a) (7.98b) (7.98c)

where I m , K m , and M are the magnetic linear, surface, and volume current respectively. Example 7.7 The current distribution along an infinitesimal electric current source  positioned symmetrically along the z-axis is given by I = a z I ο . Find the vector potential in rectangular, cylindrical, and spherical coordinates systems. Solution The geometry of the problem is shown in Fig. (7.9). The current distribution is given by I ( x ′,y ′,z ′ ) = a z I ο and the vector potential is given by

A ( x,y,z ) =

μ 4π

∫

l′

I ( x ′, y ′, z ′ )

e − jβR dl ′ R

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where (x′, y′, z ′) represents the source point in the interval (−  2 ≤ z′ ≤ +  2) and (x, y , z ) the observation point at a distance r from the origin. The source is along the z-axis, then dl ′ = dz ′ . Substituting I ( x ′,y ′,z ′ ) = a z I ο and dl ′ = dz ′ into above equation we get +

A ( x,y,z ) = a z

μ 4π

 2

∫Iο −

 2

e − jβR dz ′ R

(x, y , z )

z R

r



+ 2

( 0 ,0 ,z ′ )

θ

y 

− 2

x

φ

Fig. (7.9). Coordinates system for the problem in Example 7.7.

From the geometry of the problem

R=

(x − x′)2 + (y − y′)2 + (z − z′)2

The current source is infinitesimally small ( z ′ ⎪ −L 2 ×⎨ +L 2 ⎪ j ( β r ) h ( 2 ) ( β z ′ ) dz ′ , r < z ′ ∫ k ⎪ k −L 2 ⎩

(7.102)

We have Pkn (1) = 1 for n = 0 , and Pkn (1 ) = 0 for n ≠ 0 . Consequently,

A = −a z j

β μ I +∞ ∑ ( 2 k + 1) Pk ( cosθ ) 4π k = 0 +L 2 ⎧ (2) ⎪ h k ( β r ) ∫ j k ( β z ′ ) dz ′ , r > z ′ ⎪ −L 2 ×⎨ +L 2 ⎪ j ( β r ) h ( 2 ) ( β z ′ ) dz ′ , r < z ′ ∫ k ⎪ k −L 2 ⎩

(7.103)

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(r ,θ ,φ )

z R + L 2

r

(r ′, θ ′, φ ′)

r′

y

φ

x

− L 2

Fig. (7.11). Vector potential for a finite length thin-wire current-carrying conductor.

7.9.2. Infinitesimal Electric Dipole with a Time-Harmonic Current A conductor of an infinitesimal electrical length ( β L 0 . Note that 2π

⎧0,

∫ cos (φ − φ ′ ) cos[n (φ − φ ′ )]dφ ′ = ⎨⎩π , 0

n ≠1 n =1

(7.113)

Using (7.113), all terms satisfying n ≠ 1 in (7.112) vanish. Thus, the exact vector potential of the loop at any point ( r ,θ ,φ ) reduces to

A = −aφ j

β μ aI 2

+∞

( 2 k + 1)

∑ k ( k + 1) P

1 k

( cosθ ) Pk1 ( 0 )

k =1

⎧ h ( 2 ) ( β r ) j k ( β a ), r > a × ⎨ k( 2 ) ⎩ h k ( β a ) j k ( β r ), r < a

(7.114)

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7.9.4. Magnetic Dipole with a Time-Harmonic Current A small circular loop is called a magnetic dipole. The vector potential in (7.114) can be used to obtain the fields of a magnetic dipole of a uniform time-harmonic current I . Let the electrical size of the loop be βa 1 are very small compared to that corresponds to k = 1 and can be ignored. Thus, the vector potential in the region r > a for magnetic dipole can be approximated by

μ I ( βa ) 1 A = −aφ j P1 ( cosθ ) P11 ( 0 ) h1( 2 ) ( βr ) 4 2

(7.116)

Since P11 ( cosθ ) = − sin θ , P11 ( 0 ) = −1, and using

h1( 2 ) ( βr ) = −

e − jβr βr

1 ⎤ ⎡ ⎢1 + jβr ⎥ ⎣ ⎦

(7.117)

Then (7.116) reduces to

A ≈ aφ

jβ μ Ia 2 1 ⎞ e − jβr ⎛ sin θ ⎟, ⎜1+ 4 r ⎝ jβr ⎠

r>a

(7.118)

Since Ar = Aθ = 0 , then H φ = 0 , and the other magnetic fields components radiated from the small loop are obtained as

Hr

∂ (sin θAφ ) 1 ∂θ μ r sin θ r>a 2 jβ a I e − jβ r ⎛ 1 ⎞ ≈ cosθ 2 ⎜ 1 + ⎟, jβ r ⎠ 2 r ⎝ ≈

(7.119)

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Hθ

≈− ≈−

1 ∂ ( rAφ μ r ∂r

( βa ) 2 I 4

Sameir M. Ali Hamed


)

e − jβr sin θ r

⎛ 1 1 ⎜⎜ 1 + − jβ r ( β r ) 2 ⎝

⎞ ⎟⎟ , ⎠

r>a

(7.120)

The associated electric field components can be obtained by application of Maxwell’s equation (7.41b). 7.10. BOUNDARY CONDITIONS Consider time varying electromagnetic fields at the interface between two media shown in Fig. 7.13. The electrical properties of medium 1 and 2 are (ε1, μ1, σ1 ) and (ε 2 , μ 2 , σ2 ) respectively. The electromagnetic fields propagating in medium 1

are E1 and H 1 , while those in medium 2 are E1 and H 1 . a n is the unit vector in the direction normal to the interface. The normal (n) and tangential (t) electromagnetic field quantities current densities in each medium are shown in Fig. (7.13). The electromagnetic fields in medium 1 and 2, respectively are

E 1 = E t 1 + E n1,

H 1 = H t 1 + H n1

(7.121)

E 2 = E t 2 + E n 2,

H 2 = H t2 + H n2

(7.122)

The boundary conditions governing the behavior of these time-varying fields at the interface is determined by the properties of the two media as follows: 1. When medium 1 and 2 are dielectrics with finite conductivities σ1 and σ 2 : in this case, σ1 and σ 2 are finite and both or either of them may be zero. The boundary conditions at the interface are the same as for static electromagnetic fields discussed in chapters 3 and 6 provided that no electric current (K = 0)  magnetic current (M S = 0) , electric charge (
es = 0 ) and magnetic
ms = 0 exist along the boundary between the media. Hence, the boundary conditions for this case are

an × (E 2 − E1 ) = 0 , ⇒ E t 2 = E t1 , (M S = 0) an • (ε 2 E 2 − ε1 E1 ) = 0 , ⇒ D n 2 = D n1 , (
es = 0 ) an × (H 2 − H 1 ) = 0 , ⇒ H t 2 = H t1 , (K = 0)
ms = 0 an • (μ 2 H 2 − μ1 H 1 ) = 0 , ⇒ B n 2 = B n1 ,

(7.123) (7.124) (7.125) (7.126)

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E 1 , H 1, D 1, B 1, J 1 E n1, H n1, D n1, B n1, J n1

503

Medium 1 ε1,, μ1, σ1

E t1, H t1, D t1, B t1, J t1 E t2, H t2, D t2, B t2, J t2 an E n2, H n2, D n2, B n2, J n2 E 2 , H 2, D 2, B 2, J 2

Fig. (7.13). Boundary conditions for time-varying fields.

Medium 2 ε 2 , μ 2, σ 2



For M S = 0 along an interface between two dielectrics the boundary condition (7.123) states that the tangential components of the time-varying electric field across the interface between the two dielectrics are continuous. Equation (7.124) states that when there is no surface charge density (
es = 0 ) along an interface between two media of finite conductivities, the normal components of the electric field across the interface are discontinuous, while the normal components of the electric flux density are continuous. Equation (7.125) states that for K = 0 along an interface between two media of finite conductivities, the tangential components of the time-varying magnetic fields across the interface are continuous. Equation (7.126) states that when there is no magnetic charge density
ms = 0 along the interface between two media of finite conductivities, the normal components of the magnetic field across the interface are discontinuous, while the normal components of the magnetic flux density are continuous. 2. When medium 1 and 2 are dielectrics with finite conductivities such that K ≠ 0 M S ≠ 0 ,
es ≠ 0 and
ms ≠ 0 , the boundary conditions are

an × (E 2 − E1 ) = −M S a n • ( ε 2 E 2 − ε 1 E 1 ) =
es an × (H 2 − H 1 ) = K a n • ( μ 2 H 2 − μ 1 H 1 ) = ρ ms

(7.127) (7.128) (7.129) (7.130)

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3. When medium 1 is a perfect electric conductor (PEC), or σ1 ≈ ∞ and medium 2 is a dielectric: Assuming that the current density J1 in medium 1 is finite, then E1 = J1 σ1 = 0 . Consequently, from (7.37a) ∂H 1 ∂t = 0 , and hence H 1 does not depend on time. This implies that no time-varying magnetic field can exist within a PEC (medium 1). Thus, H 1 = 0 and the boundary conditions for this case, provided that no magnetic charge and magnetic current densities exist along the interface, become

an × E 2 = 0 , ⇒ Et 2 = 0 , (M S = 0) a n • E 2 =
es ε 2 , ⇒ E n 2 =
es ε 2 an × H 2 = K , ⇒ H t 2 = K (
ms = 0 ) an • H 2 = 0 , ⇒ H n 2 = 0,

(7.131) (7.132) (7.133) (7.134)

It was mentioned in Chapter 6 that perfect magnetic conductors cannot exist in nature, and they can be synthesized artificially. As in the case of PECs, no timevarying electromagnetic fields can exist within PMCs. Therefore, if medium 1 is a PMC, the boundary conditions provided that no electric charge and electric current densities exist along the interface, become

an × E 2 = −M S , ⇒ Et 2 = −M S a n • E 2 = 0 , ⇒ E n 2 = 0 , (
es = 0 ) a n × H 2 = 0, ⇒ H t2 = 0 , (K = 0) a n • H 2 =
ms μ 2 , ⇒ H n 2 =
ms μ 2

(7.135) (7.136) (7.137) (7.138)

Example 7.8 The plane z = 0 separates medium 1 and 2, where medium 1 occupies the region z > 0 . The constitutive parameters of medium 1 and 2 are ( ε1 = 1.5 ε 0 , μ1 = μ 0 ,

σ = 10 −3 S m ) and ( ε 2 = 4 ε 0 , μ 2 = μ 0 , σ = 2 × 10−3 S m ) respectively. If the electric field in medium 1 is

E1 = (8a x + 4a y + 16a z ) cos (2π ×108 t ) V m find E 2 

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Solution Since the interface is z = 0 and medium 1 occupies the region z > 0 , then the normal is an = −a z . Hence,

E n 1 = 16 a z cos (2π × 10 8 t ) V m E t 1 = (8a x + 4 a y ) cos (2π ×10 8 t ) V m Since there are no charges along the interface, then from (7.123)

E t 2 = E t 1 = (8a x + 4 a y ) cos (2π ×10 8 t ) V m From the boundary condition (7.123), D n 2 = D n1, we get

E n 2 = ( ε 1 ε 2 )E n 1 = (1.5 4 ) × 16 a z cos (2π × 10 8 t ) = 6 a z cos (2π × 10 8 t ) V m Then,

E 2 = E t 2 + E n 2 = (8a x + 4 a y + 6 a z ) cos (2π ×10 8 t ) V m 7.11. ELECTROMAGNETIC POWER 7.11.1. Poynting Theorem and Poynting Vector Let the time varying electromagnetic fields E and H propagate through a medium of volume v and enclosed by a surface S . If the electrical properties of the medium are (ε, μ, σ) , then from (7.37a) and (7.37b), we have

∂H ∂t ∂E ∇×H = J + ε ∂t ∇ × E = −μ

(7.139a) (7.139b)

Finding the dot product of both sides of (7.139b) with E , then

ε ∂E 2 ∂E E ⋅ ∇ × H = E ⋅ J + εE ⋅ = σE + ∂t 2 ∂t

2

(7.140)

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The dot product of both sides of (7.139a) with H , leads to

∂H μ ∂H H ⋅ ∇ × E = −μ H ⋅ =− ∂t 2 ∂t

2

(7.141)

Subtracting (7.140) from (7.141), then

ε ∂E μ ∂H 2 H ⋅∇×E − E ⋅∇× H = − − − σE 2 ∂t 2 ∂t 2

2

(7.142)

Applying the vector identity B ⋅ ∇ × A − A ⋅ ∇ × B = ∇ ⋅ (A × B) on the left-hand side of (7.142), yields

∇ ⋅E × H = −

∂ ⎛ε 2 μ 2⎞ 2 ⎜ E − H ⎟ −σ E ∂t ⎝ 2 2 ⎠

(7.143)

Integrating both sides of (7.143) over the volume v

ε ∂

∫ ∇ ⋅ E × H dv = − 2 ∂t ∫ E v

2

dv −

v

μ ∂ 2 2 H dv − σ ∫ E dv ∫ 2 ∂t v v

(7.144)

The volume integral in the left-hand side of (7.144) can be changed to surface integral over the surface S using the divergence theorem, then

ε ∂

∫ E × H dS = − 2 ∂t ∫ E S

v

2

dv −

μ ∂ 2 2 H dv − σ ∫ E dv ∫ 2 ∂t v v

(7.145)

The quantity on the left-hand side is the total instantaneous power of the electromagnetic fields crossing the surface S . The first term on the right-hand side of (7.145) is the instantaneous power stored in the electric field and the second term is the instantaneous power stored in the magnetic field. The third term on the right-hand side is the power dissipated in the medium. This equation is known as the Poynting theorem. The integrand of the left-hand side is a vector quantity represents the total instantaneous power density and known as the Poynting vector. Thus, the instantaneous Poynting vector is

T =E ×H

(7.146)

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The total instantaneous power P crossing the surface S can be obtained as

P = ∫∫T ⋅ dS = ∫∫ E × H ⋅ dS S

(7.147)

S

7.11.2. Average Power for Time-Harmonic Fields For sinusoidal varying fields, the electric field and magnetic field are expressed by the complex spatial fields E and H , where

E = Re [E e j ω t ] = (Ee j ω t + E ∗ e − j ω t ) 2 H = Re [H e j ω t ] = (He j ω t + H ∗ e − j ω t ) 2

(7.148) (7.149)

Substituting E and H from (7.148) and (7.149) into (7.146), then

T =

E × He

j2ωt

∗

∗ − j2ωt

+E ×H e 4

∗

+

∗

E× H + E × H 4

(7.150)

This equation can be written as

1 1 T = Re [E × He j 2 ω t ] + Re [E × H ∗ ] 2 2

(7.151)

The average Poynting vector Ta can be obtained by integrating the instantaneous Poynting vector over one period divided by the period. Using (7.151) T

Ta

ω = T dt 2 π ∫0 ⎡2π ω ⎤ ω ⎡2π ω ⎤ ω Re ⎢ ∫ E × He j 2 ω t dt ⎥ + Re ⎢ ∫ E × H ∗ dt ⎥ 4π ⎢⎣ 0 ⎥⎦ 4π ⎢⎣ 0 ⎥⎦ 2 j 2ω 2π ω ω = Re [E × He j 2 ω t ]0 + Re [E × H ∗ t ] 02 π ω 4π 4π =

(7.152)

The first term in (7.152) vanishes and the average Poynting vector becomes

[

1 ∗ Ta = Re E × H 2

]

(7.153)

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where E and H are the maximum values of the sinusoidal time varying fields. The average power Pa  crossing the surface S can be obtained by integrating the components of Ta along the direction of S over S , then

Pa = ∫∫ Ta ⋅ dS = S

[

]

1 ∗ Re E × H ⋅ dS ∫∫ 2 S

(7.154)

Example 7.9 If the spherical electromagnetic field components from an electric current source of length L in free space are given by

I ο Lβ e − jβ r , sin θ E θ = jη ο 4π r

I ο Lβ e − jβ r sin θ Hφ = j 4π r

where η ο is the free space characteristic impedance and I ο the current in the source. Find the Poynting vector and the total power radiated from the current source. Solution The time average Poynting vector is given by (7.153). Substituting E and H in (7.153), we get the average Poynting vector as

I ο L2 β 2 1 ∗ Ta = Re [aθ Eθ × a φ H φ ] = a rη ο sin 2 θ 2 2 2 32 π r 2

The total power radiated from the current source can be found by integrating the Poynting vector through a sphere surrounds the current source

Prad = ∫∫ Ta • dS S

Substituting dS = a r r 2 sin θ dθ dφ and Ta into the last equation, yields 2

Prad = ηο

2

Iο L β 32π

2

2 2π π

∫ ∫ sin 0 0

3

θ dθ dφ

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Evaluating the double integral, the total radiated power from the current source can be found as

Prad

π Iο L = ηο 3 λ

2

SOLVED PROBLEMS Solved Problem 7.1 The instantaneous magnetic flux density produced on the plane of a currentcarrying circular loop of radius a , placed on the xy plane with its center at the origin of the coordinate, is given by −12

10 B = az cos(1500πt ) T 1 + 25 ρ where ρ is the radial distance in cylindrical coordinates. Find the (a) Total flux in z direction passing through the loop. (b) Electric field at any point ρ within the loop. Solution (a) From Fig. (7.14) dS = a z ρ dρdφ , the flux density passing through the loop is 2π a

−12

10 ψ = ∫∫B ⋅ dS = ∫∫ cos(1500πt ) a z ⋅ a z ρ dρdφ 1 + 25 ρ S 0 0 ⇒ψ =

2π × 10 25

−12

a

1 ⎞ ⎛ cos(1500πt )∫ ⎜1 − ⎟ dρ + 1 25 ρ ⎝ ⎠ 0

2π × 10 −12 1 ⎡ ⎤ ⇒ψ = cos(1500πt ) ⎢ ρ − ln(1 + 25 ρ )⎥  25 25 ⎣ ⎦0 a

This reduces to

ψ=

2π × 10 −12 1 ⎡ ⎤ cos(1500πt ) ⎢a − ln(1 + 25a )⎥ 25 25 ⎣ ⎦

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z dS = az ρdρ dφ

B

S

a

ρ

y dφ

φ

x

Fig. 7.14. Geometry of the circular loop of Solved Problem 7.1.

(b) Let the radial distance for the given B is represented by ρ ′ , the electric field at any point ρ within the loop of radius ρ in the direction of φ can be obtained using 2π ρ

−12

10 ψ = ∫∫B ⋅ dS = ∫∫ cos(1500πt ) a z ⋅ a z ρ ′ dρ ′dφ 1 + 25 ρ ′ 0 0 S

Using Faraday’s law ρ

d 1 ∫
E • dl = − dt ∫∫S B ⋅ dS = ∫0 1 + 25 ρ ′ a z ⋅ a z ρ ′ dρ ′ 2π × 1500π × 10 sin (1500πt ) ⎡ 1 ρ − ln(1 + 25 ρ )⎤⎥ = ⎢ 25 25 ⎣ ⎦ −12

⇒ ∫ E • dl = 2πρEφ =


2π × 1500π × 10 sin (1500πt ) ⎡ 1 ⎤ − ρ ln(1 + 25 ρ )⎥ ⎢ 25 25 ⎣ ⎦ −12

Then

1 −12 ⎡ ⎤ E = aφπ × 60 × 10 ⎢1 − ln(1 + 25 ρ )⎥ sin (1500πt ) V m ⎣ 25 ρ ⎦

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Solved Problem 7.2 The instantaneous electric flux density in free space produced by oscillating charge placed at the origin is given by

D = ar

9 ε0 r

2

cos(ω t − β r ) C m

2

where β = ω μ 0ε 0 . Find the time-average charge density that produces this electric flux density. Solution Using Maxwell’s equation (7.39c), the charge density ρ v is given by

ρv = ∇ ⋅ D = ∇ ⋅ ar

9 ε0 cos(ω t − β r ) r2

In spherical coordinate

∇ ⋅ D = 9 ε0 ∇ ⋅ ar

cos(ω t − β r ) 1 ∂ ⎡ cos(ω t − β r ) 2 ⎤ = 9 ε0 2 ⎢ ⋅r ⎥ 2 r r ∂r ⎣ r2 ⎦

Then

ρv =

9β ε 0 sin (ω t − β r ) r

Solved Problem 7.3 A time-varying voltage source of V = 10 cos ( ωt ) V is connected across parallel −4 plate capacitor with polystyrene (ε =2. 56 ε 0, σ = 3. 7 ×10 S m) between the plates. Assume a small plate separation of 2 cm and no field fringing, determine the maximum values of the displacement and conduction current densities within the polystyrene at (a) 1 MHz. (b) 100 MHz.

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Solution The electric field across the capacitor is obtained as E=

V d

10 cos ( ω t ) d = 500 cos ( ω t ) V m =

(a) The maximum electric field across the capacitor is Emax = 500 V m , then the maximum conduction current density is −4

2

J c = σ Emax = 3.7 × 10 × 500 = 0.185 A m  The displacement current in the polystyrene is

Jd =

∂D = −500 ωε sin ( ω t ) = −2.56 × 500 ωε 0 sin ( ω t ) ∂t

The maximum displacement current density at 1 MHZ is

J d = 2.56 × 500 ωε 0

= 2.56 × 500 × 10 6 × 8.85 × 10 −12 = 1.1328 × 10 − 2 A m 2

(b) The maximum conduction current density at 100 MHz is the same as 2 previous, or J c = 0.185 A m  The maximum displacement current density at 100 MHZ is

J d = 2.56 × 500 × 100 × 10 6 × 8.85 × 10 −12 = 1.1328 A m 2 Note that at 1 MHz J c > J d , while at 100 MHz J c < J d . Solved Problem 7.4 A magnetic field in free space is given by

H = a z H 0 cos(βx )sinω t A m where β is a constant. Find the associated electric field and the charge density.

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Solution In free space σ = 0 , then Maxwell’s equations (7.39b) becomes ∇ × H = ∂D ∂t

ax ay az ∂H z ⎞ 1 ⎛ ∂H z ∂E 1 ⇒ = ∇ × H = ∂ ∂x ∂ ∂y ∂ ∂z = ⎜ a x −ay ⎟ ε⎝ ∂t ε ∂y ∂x ⎠ 0 0 Hz Since ∂H z ∂y = 0

⇒

∂E 1 ∂H z βH 0 = −a y = ay sin ( βx )sinω t ∂t ε 0 ∂x ε0

In free space η 0 = μ 0 ε 0 , then the electric field is

E = ay

βH 0 ε0

sin (βx )∫ sinω t dt = −a yη0 H 0 sin ( βx )cosω t V m

The charge density
e can be obtained using ∇ • D =
e , thus


e = ∇ ⋅ ε 0 E = 0 Solved Problem 7.5 The fields radiated from an antenna at the origin of the coordinates are given by

30 sin θ sin ( ω t − β r ) V m r 1 H = aφ sin θ sin (ω t − β r ) A m 4πr

E = aθ

Find the instantaneous and the time average power densities. Then find the total average power radiated by the antenna. Solution The instantaneous power density is the Poynting vector, then T = E ×H 

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Substituting E and H , yields

30 1 sin θ sin ( ω t − β r )a θ × a φ sin θ sin ( ω t − β r ) r 4πr 30 = a r 2 sin 2 θ sin 2 ( ω t − β r ) VA m 2 4r

T

=

The time average Poynting vector is given by (7.153) as follows

1 Ta = Re [E × H ∗ ] 2 The spatial components E and H can be obtained from the given fields as

E = aθ

j 30 j − jβ r − jβ r sin θ e , H = aφ sin θ e r 4πr

Substituting E and H into the above average Poynting vector expression as

[

]

1 30 2 ∗ Ta = Re aθ Eθ × aφ H φ = a r sin θ 2 2 4πr The total power radiated from the current source can be found by integrating the Poynting vector through a sphere surrounds the current source

P = ∫∫ Ta • dS S

2

where dS in the spherical coordinates systems is given by dS = ar r sin θ dθ dφ . Substituting Ta and dS into the expression of Pr , yields

30 Pr = 4π

2π π

π

π

1 ⎡ ⎤ sin θ dθ dφ = 15∫ sin θ dθ = −15⎢cosθ − cos3 θ ⎥ = 20 W . ∫∫ 3 ⎣ ⎦0 0 0 0 3

3

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PROBLEMS 7.1.

The rectangular loop shown in Fig. (7.3) has 6000 turns and rotating about its axis at an angular frequency 100π in a static magnetic flux density B . The width of the loop is 10 cm and its length is 20 cm . If the maximum induced electromotive force between the terminals of the loop is 240 V, find B .

7.2.

A conductor slides freely over conducting rails as shown in Fig. (7.2) in a varying magnetic flux density B = a z 0 . 25 cos ( 100 π t− π x ) Wb m 2 with velocity v = 4 a x m s . If the rails separation is l = 10 cm , find the induced electromotive force between the terminals of the conducting rails assuming the conductor starts from the point x = 0 .

7.3.

If the free conductor in Problem 7.1 is fixed at

x = 1 m and

B = a z 0.25 cos (100π t − π x ) Wb m 2 , find the induced electromotive force between the terminals of the conducting rails at t = 1 ms . 7.4.

In

Problem

7.1

if the magnetic flux density is constant at B = a z 0.25 Wb m and the conductor slides freely over conducting rails 2

with velocity v = 4 a x m s , find the induced electromotive force between the terminals of the conducting rails. 7.5.

Consider the circuit shown in Fig. (7.6). The region between the parallel plates is filled with free space. Find the following immediately after S is closed between the parallel plates: (a) Instantaneous time-varying electric field E . (b) Instantaneous time-varying magnetic field H . (c) The power delivered into the capacitor.

7.6.

The displacement current in a medium with permittivity 4 ε 0 , permeability

μ 0 , and conductivity σ = 0 S m, is a z 8 cos (2π ×10 8 t − β x ) C m 2 . Find β , D , E , H , and B .

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A dielectric material has permeability μ 0 , permittivity 10 ε 0 , and conductivity σ = 20 nS m. Find the frequency at which the magnitudes of the conduction and displacement current will equal. At this frequency determine the instantaneous displacement current in the material.

7.8.

For a time-harmonic magnetic current source M , show that

(∇ 2 + ω 2 με )
m = j where
7.9.

m

1 M ωμ

is the magnetic hertz vector?

Using the results of Example 7.6 find the electric hertz vector in terms of the current density, and then find the electric and magnetic field components radiated from a small electric current element Idl positioned along the z-axis and symmetrically about the origin. (Hint: find first the vector potential A ).

7.10. If the scalar Φ and instantaneous vector potentials A are given by

(

Φ = −y x + t

)

(

μ 0ε 0 V , A = y x μ 0ε 0 + t

)

Wb m

(a) Show that

∇ ⋅ A = − μ 0ε 0

∂Φ ∂t

(b) Find the B , H , E , and D . (c) Find the instantaneous Poynting vector. 7.11. The instantaneous electric field inside a source-free, homogenous, isotropic and linear medium is given by

E = [a x 10 ( x + y ) + a y b ( x − y ) ] cos ( ω t ) V m where b is constant. Find b . 7.12. The instantaneous magnetic flux density in free space is given by

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B = a x Bx cos(2 y )sin (ω t − π z ) + a y By cos(2 x )cos(ω t − π z ) T where Bx and B y are constants. Assuming there are no sources at the field points x , y determine the displacement current. 7.13. The displacement current density within a source-free cube centered about the origin is given by 2

J d = a x yz + a y y + a z xyz A m

2

Each side of the cube is 1 m long and the medium within it is free space. Find the displacement current leaving, in the outward direction, through the surface of the cube. 7.14. The instantaneous electric flux density in free space produced by oscillating charge placed at the origin is given by

⎛

1

+ aθ D = ⎜⎜ a r 2 ⎝ βr

1⎞ ⎟⎟ sin θ cos(ω t − β r ) C m2 r⎠

where β = ω μ 0ε 0 . Find the time-average charge density that produces this electric flux density. 7.15. A time varying electric field is given by the following spatial complex expression

E = − j 30a x + j10a y V m If the frequency is 1 GHz, find the (a) Associated instantaneous time-varying magnetic field H . (b) Instantaneous Poynting vector T . (c) Time average Poynting vector Ta . 7.16. A time-varying magnetic field is given by the spatial complex expression

H = a x j 2e

−j

4π + 60 z 3

Am

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If the frequency is 100 MHz, find the (a) Associated instantaneous time-varying electric field E . (b) Instantaneous Poynting vector T . (c) Time average Poynting vector Ta . 7.17. The electromagnetic fields of a wave in free space are given by

E = a z 100 cos⎛⎜ ω t − ⎝

4π 3

5 4π ⎞ ⎛ x⎟ V m, H = ay cos⎜ ω t − 6π 3 ⎠ ⎝

⎞ x⎟ A m ⎠

If the frequency is 200 MHz , find the time average power crossing a 2

surface of 4 m on the yz plane. 7.18. The electromagnetic fields of a wave in free space in spherical coordinate are given by

10 4π ⎞ ⎛ r⎟ V m sin θ cos⎜ ω t − r 3 ⎠ ⎝ 4π ⎞ 1 ⎛ H = aφ sin θ cos⎜ ω t − r⎟ A m 12π r 3 ⎠ ⎝

E = aθ

Find the (a) Instantaneous Poynting vector. (b) Spatial complex fields. (c) Time average power leaving the sphere of radius r0 = 100 m and centered at the origin. 7.19. The electromagnetic fields of a wave in free space in spherical coordinate are given by

4π ⎞ 5 4π x ⎟ V m, H = a y E = a z 100 cos ⎛⎜ ω t + cos ⎛⎜ ω t + 3 ⎠ 6π 3 ⎝ ⎝ Find the (a) Instantaneous Poynting vector.

x ⎞⎟ A m ⎠

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(b) Corresponding spatial complex fields. (c) Time average power crossing the surface bounded by 0 ≤ y ≤ 2 m , z = 2 m. 7.20. The time-harmonic instantaneous electric field in free space is given by

E = a x E0 sin ⎜⎛ ω t − β r + ⎝

π⎞

⎟ V m 2⎠

where E0 is a real constant and β = ω μ 0 ε 0 . (a) Write an expression for the spatial electric field. (b) Find the corresponding spatial magnetic field. (c) Determine the time-average Poynting vector. 7.21. The time-harmonic instantaneous electric field inside a source-free coaxial line with inner and outer radii a and b , respectively, that is filled with a homogeneous dielectric ε = 2.25 ε 0 , μ = μ 0 and σ = 0 , is given by

E = aρ

100

ρ

cos(108 πt − β z ) V m

where β = ω με . Determine the (a) Magnetic field. (b) Displacement current density. (c) Time-average Poynting vector. 7.22. Show that the time-harmonic spatial electric field radiated in free space by an infinitesimal linear current element is given by

⎡ cosθ ⎛ jβ sin θ ⎛ 1 ⎞ 1 1 ⎜⎜ 1 + E = ⎢a r − 2 2 ⎜1+ ⎟ + aθ 2 ⎝ r ⎝ jβ r ⎠ jβ r β r ⎣

⎞ ⎤ e − jβ r ⎟⎟ ⎥ E 0 V m r ⎠⎦

where E 0 = I 0 Lη 0 2π , I 0 is the current through the element, L its length, η 0 = μ 0 ε 0 , and β = ω μ 0ε 0 . Find the average Poynting vector.

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7.23. The current distribution along an infinitesimal electric current source  positioned symmetrically along the y-axis is given by I = a z I ο . Find the vector potential in rectangular, cylindrical, and spherical coordinates systems. 7.24. Show that the time-harmonic spatial electric field radiated in free space by a current-carrying small circular loop of radius a placed on the xy plane with its center at the origin, is given by

η 0 β 2 a 2 I 0 sin θ ⎛ 1 ⎞ e − jβ r V m E = aφ ⎜1+ ⎟ 4 jβ r ⎠ r ⎝ where

I 0 is the current through the loop, η 0 = μ 0 ε 0

and

β = ω μ 0 ε 0 . Then find the (a) Instantaneous electric field. (b) Instantaneous Poynting vector. (c) Average Poynting vector. 7.25. An electric line source of infinite length and constant current, placed along z-axis, radiates in free space at large distance from the source a timeharmonic spatial magnetic field

H = aφ

H0

ρ

e

− jβ ρ

Am

where H 0 is a constant and β = ω μ 0ε 0 . Find the (a) Spatial and instantaneous electric field components. (b) Average Poynting vector. 7.26. Repeat Problem 7.23 if the current source  is positioned symmetrically along the x-axis. 7.27. Find the electromagnetic field components in cylindrical coordinate system for the current source in Problem 7.23. 7.28. Find the electromagnetic field components in spherical coordinate system for the current source in Problem 7.23.




Electromagnetics for Engineering Students, 2017, 521-582

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CHAPTER 8

Propagation of Uniform Plane Waves Abstract: Maxwell’s equations are applied in this chapter to investigate the behavior of electromagnetic plane waves propagation in unbounded homogeneous media. The behavior of electromagnetic waves propagating in a homogeneous medium is determined by its electrical properties (permittivity, permeability, and conductivity) and the operating frequency. General formulas for electromagnetic field components and basic characteristics such as loss tangent, attenuation constant, phase constant, propagation constant, intrinsic impedance, phase velocity, wavelength, and skin depth are derived for an arbitrary medium assuming plane wave propagation. These general formulas are used to study the behavior of uniform plane waves propagating in lossless dielectrics, lossy dielectrics, and good conducting media. Moreover, the concepts of wave polarization and group velocity are also discussed in this chapter. The topics of the chapter are supported by numerous illustrative examples and figures in addition to solved problems and homework problems at the end of the chapter.

Keywords: Attenuation constant, circular polarization, elliptical polarization, group velocity, intrinsic impedance, linear polarization, loss tangent, phase constant, phase velocity, propagation constant, skin depth, wave polarization. INTRODUCTION In communication systems, information is transmitted from point to point in a certain medium in a form of time varying electromagnetic energy. This energy propagates through the medium as a wave at high frequencies. The characteristics of the transmitted electromagnetic waves and the performance of the communication system are determined by the properties of the medium through which the information has been transmitted and the frequency of operation. The medium may be a dielectric material, magnetic material, conducting material, metamaterial or combination of these materials. However, the actual behavior of the medium depends on its constitutive parameters (ε, μ, σ) and the frequency of operation. The transmission medium may be bounded or unbounded. In the bounded media, electromagnetic waves are forced to propagate through a specific bounded path, such as ferrite cores, transmission lines, and waveguides. In unbounded media, electromagnetic waves propagate freely through the media without any obstacles. The unbounded media may be a homogeneous with constant constitutive parameters as free space, or an anisotropic as in the case of the ionosphere, where the relative permittivity is a function of position and frequency.

Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

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In this chapter, Maxwell’s equations are used to analyze the behavior of timeharmonic plane waves propagating in unbounded charge-free homogeneous media. The analysis is carried out for lossless dielectrics, lossy dielectrics, and good conducting media. The concepts of wave polarization and group velocity are also discussed. An extensive discussion on the topic of plane waves is available in many references, e.g. [18, 19, 21, 45, 93]. The student may refer, for example, to [41, 92, 98, 99] for more advanced treatment of this topic. 8.1.

THE WAVE EQUATION

Assuming that time-varying fields E and H propagate in a charge-free medium (
e = 0 ), characterized by the properties (ε, μ, σ). Maxwell’s equations governing the behavior of these fields in the medium, from (7.39a) – (7.39b), are

∇ × H = J + ε ∂E ∂t ∇ × E = − μ ∂H ∂t ∇ ⋅E = 0 ∇⋅H = 0

(8.1a) (8.1b) (8.1c) (8.1d)

Taking the curl for both sides of (8.1b) and using (8.1a), we obtain ∇×∇×E = −μ

∂⎛ ∂E ⎞ ⎜J +ε ⎟ ∂t ⎝ ∂t ⎠

(8.2)

Using A × B × C = ( A ⋅C ) B − ( A ⋅B ) C on the left-hand side of (8.2), yields ∇(∇ ⋅ E ) − ∇ 2E = − μ

∂⎛ ∂E ⎞ ⎜J +ε ⎟ ∂t ⎝ ∂t ⎠

(8.3)

Since ∇ ⋅ E = 0 from (8.1c), and J = σ E , then (8.3) reduces to

∇2E − με

∂ 2E ∂E − μσ =0 2 ∂t ∂t

(8.4)

Equation (8.4) is known as the wave equation for the electric field. Taking the curl of (8.1a) and following a similar analysis as above, we can obtain the wave equation for the magnetic field, as

∇ 2 H − με

∂ 2H ∂H − μσ =0 2 ∂t ∂t

(8.5)

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SOLUTION OF THE WAVE EQUATION IN UNBOUNDED MEDIA

The instantaneous time-harmonic fields E and H in (8.4) and (8.5) can be replaced by the corresponding complex spatial fields E and H , using (7.40c) and (7.40d) respectively. Assuming the time dependence factor is e j ωt , it follows that 2 2 2 ∂ ∂t = j ω and ∂ ∂ t = − ω . Hence, (8.4) and (8.5) in terms of E and H , are

∇2E − j ωμ(σ+ j ωε )E = 0 ∇2H − j ωμ(σ+ j ωε )H = 0

(8.6b)

γ 2 = j ωμ(σ+ j ωε )

(8.6c)

(8.6a)

Letting

Equations (8.6a) and (8.6b) become

∇ 2E − γ 2 E = 0 ∇2H − γ 2 H = 0

(8.7a) (8.7b)

To determine the electromagnetic fields E and H in the medium, we need to determine one of them by solving (8.7a) or (8.7b), then using Maxwell’s equations, the other field can be determined. In the following analysis, we will determine E first by solving (8.7a). In rectangular coordinates system E is written as

E = a x Ex + a y E y + a z Ez

(8.8)

where E x , E y , and Ez are spatial field components in the x , y , and z directions, respectively. The Laplacian operator ∇ 2 , is given by

∂2 ∂2 ∂2 ∇ = 2+ 2+ 2 ∂x ∂y ∂z 2

(8.9)

Substituting E from (8.8) into (8.7a), we get the following equations

∂ 2 Ex ∂ 2 E x ∂ 2 E x + 2 + 2 = γ 2 Ex 2 ∂y ∂z ∂x

(8.10a)

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∂2 Ey

+ 2 + 2 = γ2 Ey ∂x 2 ∂y ∂z ∂ 2 Ez ∂ 2 Ez ∂ 2 Ez + 2 + 2 = γ 2 Ez 2 ∂x ∂y ∂z

(8.10b) (8.10c)

Assuming that the wave propagates in the z direction, the electric field has only one component in the x direction i.e. ∂ 2 E x ∂x 2 = ∂ 2 E x ∂y 2 = 0 , and E y = E z = 0 , then (8.10a) – (8.10c) reduce to

∂ 2 Ex − γ 2 Ex = 0 2 ∂z

(8.11)

The solution of (8.11) can be written in the form

Ex = Ae γ z + Be − γ z

(8.12)

where A and B are constants. Equation (8.12) shows that the electric field consists of two waves, one traveling in the negative z-direction with the factor e γ z and the other in the positive z direction with the factor e − γ z . Replacing the constants A and B by their values at z = 0 , which are assumed to be E − and E + respectively, then

Ex = E + e− γ z + E −e γ z

(8.13)

E − and E + are complex and may be written in polar form as −

E − = E − e jϕ ,

E + = E + e jϕ

+

(8.14)

Consequently, (8.13) becomes +

Ex = E + e− γ z + jϕ + E − eγ z + jϕ

−

(8.15)

The magnetic field can be obtained from Maxwell’s equation (7.41a) as ax ay az j j j ⎡ ∂Ex ∂E ⎤ ∇× E = ∂ ∂x ∂ ∂y ∂ ∂z = − az x ⎥ H= ⎢a y ∂y ⎦ ωμ ωμ ωμ ⎣ ∂z Ex 0 0

(8.16a)

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Since ∂Ex ∂y = 0 , then

H = ay

j ∂Ex ωμ ∂z

(8.16b)

Substituting E x from (8.15) into (8.16b), the magnetic field component can be obtained as

Hy =

(

)

+ − γz γ E + e− γ z + jϕ − E − e γ z + jϕ = H −e + H +e− γ z j ωμ

(8.17a)

where

H− = −

γ E− , j ωμ

H+ =

γ E+ j ωμ

(8.17b)

γ is given by (8.6c). It is a complex quantity known as the propagation constant and can be written in the form

γ=

j ωμ(σ+ j ωε ) = α+ jβ = γ e jϑ

(8.18a)

where α is known as the attenuation constant, and β as the phase constant. γ and ϑ are given respectively by

γ = α2 + β 2

(8.18b)

ϑ = tan −1 (β α )

(8.18c)

and

The attenuation constant α is expressed in Neper per meter (Np/m) or decibel per meter (dB/m), while the phase constant β is expressed in radian per meter (rad/m). To find the relation between Np/m and dB/m, consider the attenuation in one meter in the direction of propagation ( z = 1 m ), then

[

]

dB m = 20 log(e− α z ) = 20 log e− α×(1 m ) = 20 × (− α )× log(e) = −8.68 α (8.19)

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Consequently,

α (dB m) = 8.68α (Np m)

(8.20)

Note that for a wave to propagate in a medium, the phase constant must be greater than zero, or β > 0 . Using (8.15) and (8.17a), the complex spatial fields E and H can be written as

E = a x (E + e − α z e − jβz + E −e + α z e + jβz )

H = ay

(8.21)

γ (E +e−α ze− jβz − E −e+α ze+ jβz ) j ωμ

(8.22)

The instantaneous time-harmonic electric field can be obtained by substituting (8.21) into (7.40c) as

E = a x Re ⎡⎢ E + e − α z e j (ω t − βz +ϕ ) + E − e + α z e j (ω t + βz +ϕ )⎤⎥ ⎣ ⎦ +

−

(8.23a)

or

E = a x E + e − α z cos(ω t − βz + ϕ + )

(

+ a x E − e + α z cos ω t + βz + ϕ −

)

(8.23b)

Similarly, the instantaneous time-harmonic magnetic field can be obtained using (8.22) and (7.40d) as

H = ay

Re ⎡ E + e − α z e j (ω t − βz +ϕ ωμ ⎢⎣ γ

−

+ϑ − π 2

) − E − e+ α z e+ j (ω t + βz +ϕ

−

+ϑ − π 2

)⎤ ⎥⎦

(8.24a)

or

H

(

)

(

)

γ + −α z E e sin ω t − βz + ϕ + + ϑ ωμ γ − +α z − ay E e sin ω t + βz + ϕ − + ϑ ωμ = ay

(8.24b)

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CHARACTERISTICS OF HOMOMOGENOUS MEDIA

The behavior of electromagnetic waves propagating in a homogeneous medium is determined by its the electrical properties and the operating frequency. General formulas for basic characteristics of such media will be presented in this section. 8.3.1. Loss Tangent Consider a homogeneous medium characterized by ε = ε 0 ε r , μ = μ 0μ r and conductivity σ , where ε r and μ r are the relative permittivity and permeability of the medium respectively. Both the conduction current J c = σ E and displacement current J d = j ωε E exist in the medium. In this case, we can rewrite (7.41b) as

∇ × H = J c + J d = (σ+ j ωε 0ε r )E = J

(8.25)

Imaginary Loss tangent = tanδ′

Jd



J δ′ Jc

Real

Fig. (8.1). The loss tangent of a lossy medium.

Equation (8.25) can be written in the form σ ⎞ ⎛ ∇ × H = j ωε 0 ⎜ ε r − j ⎟E ωε 0 ⎠ ⎝

(8.26)

The factor between the brackets in (8.26) represents the complex dielectric constant of the medium. We may define the complex dielectric constant as

⎛ σ ⎞ ⎟⎟ = ε r − j ε′′ ε′r = ε r ⎜⎜1 − j ωε ε 0 r ⎠ ⎝

(8.27)

where ε′′ = σ ωε 0 . The imaginary part in the brackets, i.e. the factor σ ωε 0 ε r is referred to as the loss tangent of the medium and equal to the ratio of the

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magnitude of the conduction current J c to the magnitude of the displacement current J d as illustrated in Fig. (8.1). Mathematically, the loss tangent is expressed in terms of the loss angle δ′ , as

tanδ′ =

ε′′ σ = ε r ωε 0ε r

(8.28)

8.3.2. Propagation, Phase, and Attenuation Constants The relation between propagation γ , phase β , and attenuation α constants is given by (8.18). From that relation, we have

α+ jβ =

j ωμ(σ+ j ωε )

(8.29)

Squaring both sides of (8.29), yields

α 2 − β 2 + j 2β α = j ωμ(σ+ j ωε )

(8.30)

Equating real parts on both sides of (8.30), then

β 2 − α 2 = ω2με

(8.31)

The amplitudes in both sides of (8.31) are equal, then

(α − β ) 2

2 2

(

+ 4 β 2 α 2 = ω 2μ 2 σ 2 + ω 2 ε 2

)

(8.32)

This simplifies to 2

⎛ σ ⎞ β + α = ω με ⎜ ⎟ + 1 ⎝ ωε ⎠ 2

2

2

(8.33)

Adding (8.31) and (8.33), the phase constant can be obtained as

⎡ ⎛

β = ω με 2 ⎢ ⎜⎜1 + ⎢⎣ ⎝

σ ⎞ ⎤ ⎟ + 1⎥ ω2ε 2 ⎟⎠ ⎥ ⎦ 2

1 2

(8.34)

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Subtracting (8.31) from (8.33), the attenuation constant is obtained as

⎡ α = ω με 2 ⎢ ⎢⎣

1

2 ⎛ σ ⎞ ⎤ ⎜⎜1 + 2 2 ⎟⎟ − 1⎥ ⎝ ω ε ⎠ ⎥⎦ 2

(8.35)

Substituting the properties of the medium ( ε = ε 0ε r ,μ = μ 0μ r , σ ) in (8.34) and (8.35), and using β 0 = ω μ 0ε 0 , the phase constant β and the attenuation constant α for an arbitrary medium become 1

β = β0

2 ⎡ ⎤2 ⎛ ⎞ σ ⎟⎟ + 1⎥ μ r ε r 2 ⎢ 1 + ⎜⎜ ⎢ ⎥ ωε ε ⎝ 0 r⎠ ⎣ ⎦

⎡ ⎤ ⎛ σ ⎞ ⎟⎟ − 1⎥ α = β 0 μ r ε r 2 ⎢ 1 + ⎜⎜ ⎢ ⎥ ⎝ ωε 0ε r ⎠ ⎣ ⎦ 2

(8.36)

1 2

(8.37)

8.3.3. Intrinsic Impedance The intrinsic impedance of the medium is the ratio of the transverse electric field to the transverse magnetic field, for a wave propagating in the positive z direction, or minus the ratio of the transverse electric field to the transverse magnetic field, for a wave propagating in the negative z direction. Letting the intrinsic impedance denoted by ηc , then using (8.17b) and (8.18a), we have

ηc =

E+ E− j ωμ ωμ j (π = − = = e + − H H γ γ

2 −ϑ )

(8.38a)

where ϑ is given by (8.18c). Equation (8.38a) can be written as

η c = η c e jφ

(8.38b)

c

The magnitude and phase φc of η c are

ηc =

μr εr ωμ ωμ 0μ r = = 120π 1 γ β 2 + α2 (σ ωε 0ε r )2 + 1 4

[

]

(8.38c)

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φc = π 2 − ϑ

(8.38d)

Using (8.18c), (8.28) , (8.36) and (8.37) along with (8.38d), it can be shown that

φc = π 2 − ϑ = δ′ 2

(8.38e)

8.3.4. The Phase Velocity and Wave Length The phase velocity of a wave is the speed at which the points of constant phase of the wavefront travel in the direction of propagation. From (8.23b), the phase of the wave traveling in the positive z direction is given by Θ = ω t − βz + ϕ +

(8.39)

Differentiating both sides of (8.39) with respect to the time, yields

dΘ dz = ω− β = ω − βυ p dt dt

(8.40)

where υ p = dz dt is the phase velocity. Since the phase is constant, then d Θ dt = 0 , and the phase velocity is obtained as

υp =

ω

(8.41)

β

Substituting β from (8.36) into (8.41), the phase velocity can be expressed as 2 ⎡ ⎤ ⎛ σ ⎞ υ0 ⎢ ω ⎟⎟ + 1⎥ υp = = 1 + ⎜⎜ ⎥ β ωε ε μrεr 2 ⎢ ⎝ 0 r⎠ ⎣ ⎦

− 12

(8.42)

where υ0 is the velocity of plane waves in free space. The wavelength can be obtained using the relation λ = 2π β with (8.36) as

λ=

2π

β

=

β0

2 ⎡ ⎤ 2π ⎢ 1 + ⎛⎜ σ ⎞⎟ + 1⎥ ⎜ ωε ε ⎟ ⎥ μrεr 2 ⎢ ⎝ 0 r⎠ ⎣ ⎦

−

1 2

(8.43)

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8.3.5. Skin Depth To explain the concept of skin depth, let us consider the electric field traveling in the positive z direction through a medium as shown in Fig. (8.2). The positive z-directed spatial electric field from (8.21) can be written as

E = a x E x = a x E + e − α z e − jβ z

(8.44)

The amplitude Ex = E + e− α z is decaying as z increases by the factor e − α z as shown in Fig. (8.2). The depth in the medium along the z direction at which the amplitude reaches 1 e of its maximum value is known as the skin depth. Assuming that the amplitude reaches 1 e of its maximum value at z = δ , then

E + e− αδ = E + e−1

(8.45)

Consequently, the skin depth is obtained from (8.45) as

δ=

1 α

(8.46)

Substituting α from (8.37) into (8.46), a general expression for the skin depth of the medium can be written as

δ=

2 ⎡ ⎤ ⎛ ⎞ 1 σ ⎢ 1+ ⎜ ⎥ ⎜ ωε ε ⎟⎟ − 1⎥ μrεr 2 ⎢ ⎝ 0 r⎠ ⎣ ⎦

β0

−

x |Ex| = |E+| e-αz

|E+|

εr,μσ

|E+| e-1

z = 1/α

y Fig. (8.2). Illustration of the skin depth.

δ

z

1 2

(8.47)

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Example 8.1 A non-ferromagnetic material has an intrinsic impedance of 188e− jπ 15 Ω . Find its (a) Loss tangent. (b) Dielectric constant. (c) Complex permittivity. Solution Given that ηc = ηc e jφc = 188 e j π 15 Ω , then η c = 188 and φc = π 15 . From (8.382), we have

φc = δ′ 2 ⇒ δ′ =

2π 15

(a) The loss tangent is

tanδ′ = tan(2π 15) = 0.4454 σ ε′′ ⇒ = = 0.4454 ωε r ε 0 ε r (b) Substituting σ (ωε r ε 0 ) = 0.4454 with μ r = 1 in (8.38c), we can write

ηc = 120π

μr εr

[(σ ωε ε ) + 1] 2

1 4

= 188

0 r

⇒

⎛ 188 ⎞ =⎜ ⎟ 2 (0.4454) + 1 ⎝ 120π ⎠ 1

εr

2

Thus, ⎛ 120π ⎞ εr = ⎜ ⎟ ⎝ 188 ⎠

2

1

(0.4454)2 + 1

= 3.68

(c) From (a) above, we have ε′′ ε r = 0.4454

⇒ ε′′ = 0.4454εr = 0.4454× 3.68 = 1.638

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Thus, the complex permittivity is

ε ′ = ε 0 (ε r − j ε ′′) = ε 0 (3.68 − j1.638 ) = (3.25 − j1.45)×10 −11 F m 8.4.

UNIFORM PLANE WAVES IN UNBOUNDED MEDIA

The behavior of uniform plane waves propagating in different media will be discussed in the following sections. The media are classified based on their loss tangent ( σ ωε 0 ε r ) as illustrated in Table 8.1. Table 8.1. Classification of media based on the loss tangent ( σ ωε 0 ε r ). Loss Tangent

Medium

σ ωε 0 ε r = 0 , (σ = 0)

The medium is lossless (perfect dielectric) and only the displacement current exists in the medium.

σ ωε 0 ε r ≠ 0 , (σ ≠ 0)

The medium is lossy and both the displacement and conduction currents exist in the medium.

σ ωε0 ε r > 1

The medium is a good conductor and the conduction current is dominant in the medium.

8.4.1. Lossless Media For a lossless medium σ = 0 , ε = ε 0ε r and μ = μ r μ 0 , where ε r and μ r are the relative permittivity and permeability of the medium respectively. Substituting σ = 0 in (8.36) and (8.37), results in α = 0 and

β = ω μ 0μ r ε0ε r = β0 μ r ε r

(8.48)

From (8.18a), the propagation constant for lossless medium is obtained as

γ = jβ = j ω μ 0μ r ε 0ε r = jβ0 μ r ε r

(8.49)

where β0 = ω μ 0ε 0 is the phase constant in the free space. With α = 0 , it follows from (8.18b) and (8.18c) that γ = β and ϑ = π 2 . Substituting these

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quantities into (8.23b) and (8.24b), the instantaneous electromagnetic fields propagating in a lossless medium can be expressed as

E = a x ⎡⎢ E + cos(ω t − βz + ϕ + ) + E − cos(ω t + βz + ϕ − )⎤⎥ ⎣ ⎦ H = ay

1 ⎡ + E cos(ω t − βz + ϕ + ) − E − cos(ω t + βz + ϕ − )⎤⎥ ηc ⎢⎣ ⎦

(8.50) (8.51)

The electric and magnetic fields traveling in the positive z direction at instants ωt = 2nπ ( n is an integer), assuming that ϕ + = 0 are shown in Fig. (8.3). Note that for a wave propagating in a lossless media: 1. The electric field and magnetic field traveling in the positive z direction are in phase as clear from (8.50) and (8.51). 2. The electric field and magnetic field traveling in the negative z direction are out of phase by 180 as clear from (8.50) and (8.51). 3. The electric field, magnetic field, and the direction of propagation are orthogonal as shown in Fig. (8.3). This mode of propagation is known as the transverse electric and magnetic fields (TEM), since both the electric and magnetic fields are always transverse to the direction of propagation. 4. The wave propagates in lossless media without attenuation and the amplitudes of the fields remain constant since the attenuation constant is zero. 5. The characteristic impedance η c of the medium is real. y

Hy

Ex

x

z

Fig. (8.3). Electromagnetic fields traveling in the positive z -direction in a lossless medium, assuming + ϕ = 0 and ωt = 2nπ (n = 0, 1, 2, …).

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Substituting σ = 0 in (8.38c), the intrinsic impedance of a lossless medium is obtained as

ηc = μ 0μ r ε 0ε r = η0 μ r ε r

(

(8.52)

)

where η0 = μ 0 ε 0 = 4π ×10 −7 10−9 36π = 120π Ω is the intrinsic impedance of free space. Similarly, the velocity of electromagnetic waves in the lossless medium is obtained from (8.42) as

υp =

υ 1 = 0 μ 0μ r ε 0 ε r μrεr

(

(8.53)

)

where υ0 = 1 4π ×10−7 × 10−9 36π = 3 ×108 m s is the wave velocity in free space. The wavelength in lossless medium can be written as

λ=

2π

β

=

υ0 f μrεr

(8.54)

Since σ = 0 in the lossless dielectric, the skin depth is theoretically infinity and the wave penetrates into the medium without any attenuation. Example 8.2 The instantaneous electric field of a uniform plane wave traveling in an unbounded non-ferromagnetic lossless dielectric is given by

E = a y10−3 cos(2π ×108 t − 2πz ) V m Find the (a) (b) (c) (d) (e)

Phase velocity of the wave. Dielectric constant of the medium. Wavelength. Corresponding instantaneous magnetic field. Spatial electric field and corresponding magnetic field.

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Solution (a) Since f = 108 Hz and β = 2π , the phase velocity of the wave in the lossless dielectric is

υ p = ω β = 108 m s (b) Let the dielectric constant is ε r , then from (8.53) with μ r = 1, we have

υ p = υ0

μ r ε r = 3 × 108

ε r = 108 ⇒

εr = 3 ⇒ εr = 9

(c) The wavelength can be computed as

λ=

2π

β

=1m

(d) The instantaneous magnetic field can be obtained using (8.1b), then ax ay az ∂E ∂E y ⎤ ∂H 1 1 1⎡ = − ∇ × E = − ∂ ∂x ∂ ∂y ∂ ∂z = − ⎢− a x y − a z ∂t ∂z ∂x ⎥⎦ μ μ μ⎣ Ey 0 0

Since ∂E y ∂x = 0 ⇒

∂H 1 ∂E y 2π −3 = ax = ax 10 sin 2π ×108 t − 2πz ∂t μ ∂z μ

(

⇒ H = ax

)

2π 10 −3 ∫ sin 2π ×108 t − 2πz dt μ 0μ r

= −a x

(

)

2π 10 −3 cos 2π ×108 t − 2πz 2π ×10 × 4π ×10 −7

(

8

⇒ H = −a x 7.95 ×10 −6 cos(2π ×108 t − 2πz )

)

Am

(e) The relation between the instantaneous and spatial electric fields is given by

E = Re[Ee j ω t ]

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Let E = a y E0e− j (βz+ϕ o ) , then

E = Re⎡Ee j ω t ⎤ = a y Re ⎡ E0e j (ω t − βz −ϕ ) ⎤ = a y E0 cos(ω t − βz − ϕ0 ) 0

⎢⎣ ⎥⎦ ⎥⎦ ⇒ a y E0 cos(ω t − βz − ϕ0 ) = a y10−3 cos(2π × 108 t − 2πz) ⎢⎣

⇒

E0 = 10 −3 and ϕ 0 = 0

Consequently, the spatial electric field is

E = a y 10 −3 e − j 2π z Similarly, the spatial magnetic field can be obtained from H in (d) above, or alternatively using (7.41a), as follows

ax ay az j j j ∂E y H= ∇× E = ∂ ∂x ∂ ∂y ∂ ∂z = −a x ωμ ωμ ωμ ∂z 0 Ey 0

⇒ H = −a x

2π j ∂E y = −a x 10 −3 e− j 2πz = −a x 7.95 × 10 − 6 e− j 2πz A m ωμ ∂z ωμ 0μ r

Example 8.3 The instantaneous electromagnetic fields of a uniform plane wave traveling in a lossless dielectric are given by

E = a x 80π cos(2π × 109 t − 40πz ) V m H = a y cos(2π ×109 t − 40πz ) A m Find the (a) Phase velocity of the wave. (b) Constants ε r and μ r of the medium. Solution (a) Since ω = 2π × 109 and β = 40π , the phase velocity of the wave is

υ p = ω β = 5 ×107 m s

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(b) From (8.53), we have

υ p = υ0

μ r ε r = 3 × 108

μ r ε r = 5 × 107 ⇒

μ r ε r = 6 ⇒ μ r ε r = 36

The intrinsic impedance of the medium is η = 80π . Hence,

η = E x H y = 80π = 120π μ r ε r ⇒ ε r = 81 2

⇒ εr = 9

⇒ μr = 4 εr 9 ⇒ μr = 4 εr 9 = 4

8.4.2. Lossy Media The lossy medium is characterized by (ε = ε 0ε r ,μ = μ 0μ r , σ ≠ 0), which can be considered as an arbitrary medium with σ ≠ 0 . Therefore, the phase constant, the attenuation constant, the phase velocity, the wavelength, the intrinsic impedance, and the skin depth of a lossy medium can be determined from (8.35) - (8.38), (8.42), and (8.43) respectively. The same can be said for electromagnetic fields, then the instantaneous time-harmonic electric and magnetic fields in a lossy medium, can be expressed using (8.23b), (8.24b) and (8.38d), as

E = a x E + e − α z cos(ω t − βz + ϕ + )

(

+ a x E − e + α z cos ω t + βz + ϕ −

H

= ay − ay

E+

ηc E−

ηc

)

(

e − α z cos ω t − βz + ϕ + − φc

(

e + α z cos ω t + βz + ϕ − − φc

(8.55)

) (8.56)

)

Substituting φc = δ′ 2 from (8.38e) into (8.56) , where δ′ is the loss tangent angle of the medium, the magnetic field in (8.56) becomes

H

= ay − ay

E+

ηc E−

ηc

(

e − α z cos ω t − β z + ϕ + − δ′ 2

(

e + α z cos ω t + β z + ϕ − − δ′ 2

)

)

(8.57)

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The electromagnetic fields traveling in the positive z direction through a lossy medium at ωt = 2nπ ( n is an integer) assuming that ϕ + = 0 , are shown in Fig. (8.4). Note that for a wave propagating in a lossy medium: 1. The electric field and magnetic field traveling in the positive z direction are out of phase by the angle δ′ 2, where δ′ is the loss tangent angle of the medium. 2. The electric field and magnetic field traveling in the negative z direction are out of phase by the angle π − δ′ 2. 3. The electric field, magnetic field, and the direction of propagation are orthogonal as shown in Fig. (8.4), or the fields propagate as TEM waves. 4. The amplitudes attenuate as z increases by a factor e − α z for the wave traveling in the positive z direction, and by a factor e + α z for the wave traveling in the negative z direction. 5. The intrinsic impedance of the lossy medium is complex.

y Hy

E

x

x

z

Fig. (8.4). Electromagnetic fields traveling in the positive z direction in a lossy medium, assuming ϕ + = 0 and ωt = 2nπ (n = 0, 1, 2, …).

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Example 8.4 A mixture of graphite powder in wax is found to have a dielectric constant of 10 − j12 at 9 GHz . Determine the conductivity of the medium and the phase constant, attenuation constant and intrinsic impedance at 9 GHz . Solution Since ε′r = 10 − j12 at 9 GHz and μ r = 1, then from (8.27), we have

ε′r = ε r − j

σ = 10 − j12 ωε 0

Equating real and imaginary parts on both sides, yields

ε r = 10 ,

σ = 12 ωε 0

⇒ σ = 12 ωε 0

⇒ σ = 12 × 2 × π × 9 × 109 × 8.85 × 10−12 = 6.007 S m The phase constant can be computed from (8.36) as 1

β = β0

2 ⎡ ⎤2 ⎛ ⎞ σ ⎟⎟ + 1⎥ μ r ε r 2 ⎢ 1 + ⎜⎜ ⎢ ⎥ ωε ε ⎝ 0 r⎠ ⎣ ⎦ 1

4π × 10 − 7 × 8.85 × 10 −12 × 10 ⎡ 2 2 = 2π × 9 × 10 × 1 + (12 10 ) + 1⎤ ⎢ ⎥ ⎣ ⎦ 2 = 675.38 rad m 9

Similarly, the attenuation constant from (8.36) is 1

4π × 10 − 7 × 8.85 × 10 −12 × 10 ⎡ 2 2 α = 2π × 9 × 10 × 1 + (1.2) − 1⎤ ⎢ ⎥ ⎣ ⎦ 2 = 316.33 Np m 9

The intrinsic impedance can be determined as follows

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2π × 9 × 109 × 4π × 10 −7 j ωμ 0μ r ωμ 0μ r (β + j α ) = (675.38 + j316.33) = 2 ηc = γ α + β2 (675.38)2 + (316.33)2 = 88.36 + j 40.45 Ω 8.4.3. Good Dielectrics In the case σ ωε 0 ε r 1, and the conduction current is dominant while the displacement current can be ignored. Under this condition, (8.6a) reduces to

∇2E − j ωμσ E = 0

(8.78)

Since J = σ E , then (8.78) can be written as

∇2J − j ωμσ J = 0 Equation (8.79) is known as the diffusion equation.

(8.79)

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8.5.1. Current Flow in a Semi-Infinite Conducting Slab Consider the solution of the diffusion equation for a semi-infinite conducting slab that fills the space z > 0 as shown in Fig. (8.6). In rectangular coordinate system, the diffusion equation is written as

∂ 2J ∂ 2J ∂ 2J + + = j ωμσ J ∂x 2 ∂y 2 ∂z 2

(8.80)

If the electric field causing J is in the space z < 0 , and given by E = a x Ex , then J has only one component in the x direction, or J = a x J x . Since the slab is infinite in both x and y directions, then ∂ 2 J x ∂y 2 = ∂ 2 J x ∂x 2 = 0 . Consequently, (8.80) reduces to ∂2 J x (8.81) = γ2 J x ∂z 2 where γ 2 = j ωμσ . The solution of (8.81) can be written in the form

J x = Ae γ z + Be − γ z

(8.82)

where A and B are constants to be determined from the boundary conditions of the problem. γ can be written as

γ=

jπ

j ωμσ = ωμσ e 4 =

ωμσ (1 + j ) = (1 + j ) 2 δ

(8.83)

where δ = 2 ωμσ is the skin depth of the conducting slab. Substituting (8.83) into (8.82), yields

J x = Ae (1+ j )z δ + Be − (1+ j )z δ

(8.84)

At z → ∞ , the current approaches zero, this imply that A in (8.84) must be zero. If we assume that the current at the surface z = 0 is J s , then B = J s , and J x in (8.84) becomes

J x = J s e − (1+ j )z δ

(8.85)

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z

Semi-infinite conducting slab

σ

σ

δ

δ

y

y w (b)

E

(a) Fig. (8.6). Diffusion current through a semi-infinite conducting slab.

The total current in the width w shown in Fig. (8.6b), is obtained by integrating the current density over the cross-section of the width as ∞w

∞w

0 0

0 0

I = ∫∫ J x dydz = ∫∫ J s e

− (1+ j )z δ

wJ s e− (1+ j )z δ dydz = − (1 + j ) δ

z =∞

= z =0

wδ Js (1 − j ) (8.86) 2

8.5.2. Surface and A.C. Resistances of a Conductor The resistance of a width w of the conducting slab shown in Fig. (8.7), along l in the x direction can be determined by computing the dissipated power using 2

P= I R=

1 2 J dv ∫ σv

(8.87a)

The power dissipated in the width w per unit length in the x-direction is 2

2

w2 δ 2 J s wδ J s (1 − j ) R = P= I R= R 2 2 2

(8.87b)

P can be obtained also as

P=

∞w l

∞

2 1 1 wl δ 2 wl 2 2 J s (8.87c) J dv = ∫∫∫ J s e−(1+ j )z δ dxdydz = ∫ J s e−2 z δ dz = ∫ 2σ σ 0 σv σ000

Equating (8.87b) and (8.87c), the resistance is obtained as

R=

l wδ σ

(8.88a)

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x

x l

σ

l

δ

y

y

te


w

w

z

z

(a)

(b)

Fig. (8.7). (a) Resistance of a semi-infinite conducting slab. (b) Resistance of a unit cross section w×te and length l.

Consequently, the resistance of the width w per length l in the x direction of the conducting slab is

R=

l l ωμ = wδ σ w 2σ

(8.88b)

The expression (8.88b) reveals that the effective thickness of the slab is equal to the skin depth δ . To explain this, consider the resistance of a unit of the cross section w× te shown in Fig. (8.7b), where te is the effective thickness of the slab, then the resistance per length l in the x direction is R=

l wt e σ

(8.89)

It is clear from (8.88a) and (8.89), that the effective thickness of the slab is te = δ . If w = 1 and l = 1 in (8.89), then R in (8.89) becomes

Rs =

1 ωμ = δσ 2σ

(8.90)

Rs which is the resistance of a unit width and unit length of the semi-conducting slab is identical to the resistance of a conducting material obtained in (8.77). The result obtained for the conducting slab in (8.90), can be generalized for any conductor of conductivity σ , and the resistance Rs is defined as the surface resistance of the conductor, and its unit is ( Ω m 2 ). Therefore, the effective

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resistance of good conducting wire of length l and circumference Cs is known as the A.C. resistance of the wire. The resistance per unit length of the wire can be obtained from (8.90) as

Rac =

π fμ σ

1 1 = Cs δ σ Cs

(8.91)

The corresponding high-frequency internal inductance of the wire using (8.77) is

Lin =

1 Rac = ω 2Cs

μ π fσ

(8.92)

Fig. (8.8) shows the effective cross section area for good conducting wires of circular and rectangular cross sections. a

δ

Cs = 2πa

δ

b

a Cs = 2(a+b) (a)

(b) Cs δ (c)

Fig. (8.8). Effective cross section area for good conducting wire at high frequencies. (a) circular wire. E rectangular wire. (c) Effective cross section.

Example 8.7 A copper rod of a rectangular cross section as shown in Fig. (8.8b), has a = 5 cm , b = 1 cm . If the length of the rod is L = 10 cm , find the resistance at (a) f = 0 Hz (D.C.). (b) f = 1 kHz . (c) f = 10 GHz .

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Solution (a) For copper σ = 5.8 × 10 7 S m . At frequency f = 0 Hz , the resistance is the D.C. resistance Rdc and is obtained as

Rdc =

L 10 = −2 = 3.448 ×10 −4 Ω −2 7 ab σ 10 × 5 ×10 × 5.8 ×10

(b) At f = 1 kHz , the resistance can be obtained using (8.91) as

L π fμ L π fμ π ×103 × 4π ×10 −7 10 Rac = = = Cs 2σ 2(a + b ) 2σ 2(1 + 5) ×10 −2 5.8 ×107 = 6.878 ×10 −4 Ω (c) Similarly, the resistance at f = 10 GHz is

Rac =

L π fμ L π fμ π ×1010 × 4π ×10 −7 10 = = Cs 2σ 2(a + b ) 2σ 2(1 + 5) ×10 −2 5.8 ×10 7

= 0.02175 Ω Note that the A.C. resistance at f = 10 GHz is much greater than that of the D.C. resistance and the ratio at f = 10 GHz is Rac Rdc = 63 . 8.6.

POWER FLOW

Poynting vector and the power of a time-harmonic uniform plane wave propagating in a medium characterized by ( ε = ε 0ε r , μ = μ 0μ r , σ ), can be obtained using the average Poynting vector for time-harmonic fields obtained in (7.153). The spatial field component E and H for a general medium can be written using (8.21), (8.22) and (8.38a), as + − E = a x ⎡⎢ E + e − α z e − j (βz −ϕ ) + E − e + α z e + j (βz +ϕ )⎤⎥ ⎦ ⎣ + 1 ⎡ + − α z − j (βz −ϕ −ϑ +π 2 ) − + α z + j (βz +ϕ − +ϑ −π 2 )⎤ − H = ay E e e E e e ⎥⎦ ηc ⎢⎣

(8.93) (8.94)

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Substituting (8.93) and (8.94) into (7.153), the average Poynting vector is obtained as

Ta = a z − az + az

E+

2

2η c E−

e − 2 α z Re e j (π 2 −ϑ )

[

]

[

]

2

2η c

e + 2 α z Re e j (π 2 −ϑ )

E+ E− 2η c

[

Re e + j (2 βz +ϕ

−

− ϕ + −ϑ + π 2

(8.95)

) − e − j (2 βz +ϕ

−

−ϕ + +ϑ − π 2

)

]

Finding the real part in each term of (8.95), it simplifies to

⎧+ E + 2 e − 2 α zsin ϑ , (+ z ) Direction ⎪ 2 1 ⎪ − +2 α z (− z ) Direction Ta = a z ⎨− E e sin ϑ , 2η c ⎪ + − + − ⎪⎩−2 E E sin 2 βz − ϕ + ϕ cos ϑ , Cross Term

(

)

(8.96)

ϑ is the phase of the propagation constant. The first term in (9.96) represents the average Poynting vector propagating in the positive z direction, the second term is the average Poynting vector propagating in the negative z direction, and the third term is the cross-coupling between them. From (8.18c) and (8.38c), we have

ϑ = tan −1 (β α ) and ηc = ωμ

α 2 + β 2 . Substituting these quantities into (9.96),

it can be written as

⎧+ E + 2 e − 2 α z , (+ z ) Direction ⎪ β ⎪ − 2 +2α z (− z ) Direction (8.97) , Ta = a z ⎨− E e 2 ωμ ⎪ + − + − ⎪⎩− 2 (α β ) E E sin 2 βz − ϕ + ϕ , Cross Term

(

)

The general expressions (8.96) or (8.97) can be used to obtain the average Poynting vector for the individual cases of lossless, lossy, and good dielectric media.

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8.6.1. Lossless Media In this case, α = 0 and ϑ = π 2 → δ′ = 0 , then the cross-coupling term in (8.96) vanishes, and the average Poynting vector becomes + 2 1 ⎧⎪+ E , Ta = a z ⎨ 2ηc ⎪− E − 2 , ⎩

(+ z ) Direction (− z ) Direction

(8.98)

8.6.2. Lossy Media For a lossy medium of loss tangent tanδ′ = σ ωε 0ε r , ϑ = (π − δ′) 2 , the average Poynting vector in (8.96) becomes

⎧+ E + 2 e − 2 α z cos (δ′ 2), (+ z ) Direction 2 1 ⎪⎪ − +2α z (− z ) Direction (8.99) cos (δ′ 2), Ta = a z ⎨− E e 2η c ⎪ − 2 E + E − sin (2 βz − ϕ + + ϕ − ) sin (δ′ 2) , Cross Term ⎪⎩ Note that the phase of the intrinsic impedance η c is φc = π 2 − ϑ = δ′ 2 . 8.6.3. Good Dielectrics In a good dielectric, we have σ ωε 0ε r ωm , so that the information can be transmitted at the carrier frequency, which is designed to meet the communication systems requirements. The resulted signal is then transmitted through the communication channel as a group of signals. The group velocity of these signals can be defined as the velocity of the envelope of modulated group of signals. In other words, it is the speed of energy in the direction of propagation. To derive an expression for the group velocity, consider the time-harmonic waves E c and E m traveling along z axis with phase constants β m and β c , respectively, and given by

Ec = Eo cos(ωc t − βc z ) Em = Eo cos(ωm t − β m z )

(8.108) (8.109)

where ωc and ω m are the angular frequencies, of the waves E c and E m respectively, and ωc >> ωm . Superimposing waves E c and E m , yields

E = Ec + Em

⎡ (ω + ωm ) (β c + β m ) ⎤ ⎡ (ωc − ωm ) (β c − β m ) ⎤ (8.110) = 2 Eo cos⎢ c t− z ⎥ cos⎢ t− z⎥ 2 2 2 2 ⎣ ⎦ ⎣ ⎦

Let

β=

βc + βm

,

2 Δβ β c − β m , = 2 2

ωc + ω m 2 Δ ω ωc − ω m = 2 2 ω=

(8.111a) (8.111b)

where β and ω are the average phase constant and angular frequency for the two waves, and Δβ and Δ ω are the phase constant and angular frequency differences between the two waves. Substituting β , ω , Δβ , and Δ ω from (8.111a) and (8.111b) into (8.110), then E can be expressed as

⎛ Δ ω Δβ ⎞ ⎛ ω β t− z ⎟ cos⎜ t − 2 2 ⎠ ⎝⎜ 2 ⎝ 2

E = 2 Eo cos⎜

⎞ z ⎟⎟ ⎠

(8.112)

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Note that E in (8.112) is an amplitude modulated wave which propagates with average phase constant β and average angular frequency ω . The amplitude of the signal in (8.112) can be written as

Δω ⎞ ⎛ Δβ A = 2 Eo cos ⎜ z− t⎟ 2 ⎠ ⎝ 2

(8.113)

The speed at which the amplitude A travels is the velocity of the two waves as a group and referred to as the group velocity υ
. The group velocity from (8.113) can be obtained as

υ
=

Δ ω Δβ Δ ω = 2 2 Δβ

(8.114)

Since ωc >> ωm , then Δ ω → 0 and the group velocity satisfies

⎛ Δβ ⎞ dβ = lim ⎜ ⎟= υg Δ ω → 0 ⎝ Δ ω ⎠ d ω 1

(8.115)

Consequently, the group velocity is obtained as

υg =

dω dβ

(8.116)

SOLVED PROBLEMS Solved Problem 8.1 A non-ferromagnetic material has an intrinsic impedance of 120e − jπ 6 Ω . If the phase constant of the material is β = 0.5 rad m , find  the (a) (b) (c) (d) (e)

Attenuation constant. Loss tangent. Skin depth for the material. Dielectric constant. Complex permittivity

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Solution Given that ηc = ηc e jφc = 120 e j π 6 Ω , then ηc = 120 and φc = π 6 . From (8.38d), we have

φc = π 2 − ϑ ⇒ ϑ = π 3 Using (8.18c), then

ϑ = tan −1 ( β α ) ⇒ β α = tan(ϑ ) = tan(π 3) = 1.734 (a) Since β = 0.5 , the attenuation constant is

⇒ α=

β 0.5 = = 0.288 Np m 1.734 1.734

(b) From (8.57), the relation between loss tangent angle δ′ and ϑ is ϑ = (π − δ′) 2 . Hence,

δ′ = π − 2ϑ = π 3 ⇒ tanδ′ = tan(π 3) = 1.733 (c) The skin depth can be obtained as

δ=

1 1 = = 3.476 m α 0.288

(d) Given that ηc = 120 , then using (8.38c) , we can write

ηc = 120π

μr εr

[(σ ωε ε ) + 1] 2

1 4

= 120 Ω

r 0

The loss tangent from (8.28) is σ ωε r ε 0 = 1.733, then the dielectric constant ε r can be obtained as

1 εr

[(1.733) + 1] 2

1 4

=

120 120π

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⎛ 120π ⎞ ⇒ εr = ⎜ ⎟ ⎝ 120 ⎠

2

1

(1.733)2 + 1

569

= 4.937

(e) From (b), we have tan(δ′) = ε′′ εr = 1.733

⇒ ε′′ = 1.733ε r = 1.733× 4.937 = 8.556 Then, the complex permittivity is

ε′ = ε 0 (ε r − j ε′′) = ε 0 (4.937 − j8.556 ) = (4.369 − j 7.572 ) × 10 −11 F m Solved Problem 8.2 The dust has a complex dielectric constant of 3 − j 0.01 at 10 GHz . Determine the attenuation, phase constant, phase velocity, and intrinsic impedance of dust at 10 GHz .

Solution Given that ε′ = 3 − j 0.01 at 10 GHz , then σ 0.01 = = 0.0033 ωε 0ε r 3

⇒

σ ⎛ 0.01 ⎞ n2 , or n2 n1 < 1 , it follows from (9.78) that sin θ i n2 , then θi < θt for all θi . When θ i   increased gradually from zero, the rate of the corresponding increment in θt is larger, and at a certain value for θi , θt reached π 2 , which is the maximum possible value for the transmission angle in medium 2 as shown Fig. (9.7a). Let the incidence angle that corresponds to θt = π 2 , is θ i = θ c , then using Snell’s law, θ c is obtained as

sin θc =

n2 ε = 2, n1 ε1

(

θc = sin −1 (n2 n1 ) = sin −1 ε 2 ε1

)

(9.80)

The angle θ c is referred to as the Critical Angle. If the incidence angle is increased beyond the critical angle, the wave is reflected back from the boundary to medium 1. This phenomenon is known as the Total Internal Reflection.

Reflection, Transmission, and Negative Refraction

Incident wave

Reflected wave

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611

Incident wave

θi = θc

θi < θc (ε1, μ1, σ1 = 0)

θi

(ε1, μ1, σ1 = 0)

θc

Medium 1

Medium 1 x

Wave parallel to the interface

θt

(ε2, μ2, σ2 = 0)

(ε2, μ2, σ2 = 0)

Medium 2

z

x

Medium 2

z

(a)

(b) Incident wave

Total internal reflection

θi > θc θi

(ε1, μ1, σ1 = 0)

θi

Medium 1 x

(ε2, μ2, σ2 = 0) Medium 2

z (c) Fig. (9.7). Total internal reflection and critical angle.

Consider the incident, reflected and transmitted waves at the interface between two media such that the refractive index of medium 1 n1 is greater than that of medium 2 n2 . For perpendicular and parallel polarizations cases, the reflected and transmitted time-average Poynting vectors can be written using (9.54b), (9.54c),

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(9.61b) and (9.61c) along with (9.78) and (9.80) in terms of the incident and critical angles as ⎫ ⎪ ⎪ ⎪ 2 ⎤⎬ 2 2 ⎡ τ E0 ⎢ sin θ i ⎛ sin θ i ⎞ ⎥ ⎪ ⎟ + a z 1 − ⎜⎜ Tt⊥ = ⊥ ax 2η 2 ⎢ sin θ c sin θ c ⎟⎠ ⎥ ⎪ ⎝ ⎣ ⎦⎪ ⎭ 2 2 ⎫ ρ|| E0 ⎪ [a x sin θi − a z cosθi ] Tr|| = 2η1 ⎪ ⎪ 2 2 ⎡ 2 ⎤⎬ τ || E0 ⎢a x sin θ i + a z 1 − ⎛⎜ sin θ i ⎞⎟ ⎥ ⎪ Tt|| = ⎜ sin θ ⎟ ⎥ ⎪ 2η 2 ⎢ sin θ c c ⎠ ⎝ ⎪ ⎣ ⎦⎭ 2

Tr⊥ =

ρ ⊥ E0 2η1

2

[a x sin θi − a z cosθi ]

(9.81)

(9.82)

Equations (9.81) and (9.82) reveal that when a plane wave from a lossless medium of refractive index n1 incident to a lossless medium of refractive index n2 < n1 , the behavior of the transmitted and reflected waves depends on the angle of incidence θi and the critical angle θ c . To illustrate this, let us examine the following three cases: 9.4.2.1. Case 1: θi < θc In this case, part of the incident power is reflected back to medium 1 and the other part is transmitted to medium 2 as illustrated in Fig. (9.7a). The Poynting vectors of reflected and transmitted waves for this case are given by (9.81) and (9.82). 9.4.2.2. Case 2: θ i = θ c When θi = θ c , the reflection coefficient for perpendicular and parallel polarization are ρ⊥ = 1 and ρ|| = −1 respectively, while the corresponding transmission coefficients are τ ⊥ = 2 and τ || = 2η 2 η1 . Substituting these quantities into (9.81) and (9.82), the transmitted average Poynting vectors, in this case, becomes ⎧ τ ⊥ 2 E0 2 , ⊥ Polarized Wave ⎪a x 2η 2 ⎪ Tt = ⎨ 2 2 ⎪ τ || E0 , || Polarized Wave ⎪a x 2η 2 ⎩

(9.83)

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613

which indicates that the transmitted wave when θi = θ c , propagates in the x  direction as illustrated in Fig. (9.7b), i.e. parallel to the interface between the two media. The wave propagating parallel to the interface between the two media, is known as the surface wave. Moreover, it is clear from (9.81) and (9.82) that the reflected power densities for perpendicular and parallel polarizations are equal to the respective incident power densities under the condition θi = θ c , or

ρ ⊥ E0⊥

⊥ r

T =

2

a x sin θ c − a z cosθ c =

2η1

Tr|| =

ρ || E0|| 2η1

2

a x sin θ c − a z cosθ c =

E0⊥

2

= Ti⊥

2η1

E0||

(9.84)

2

2η1

= Ti||

(9.85)

9.4.2.3. Case 3: θ i > θ c When θi > θ c , then sinθi > sinθc . Letting sinθi sinθc = ζ , the reflected and transmitted average power densities become ⎫ ρ E0 [a x sin θi − a z cosθi ]⎪ T = ⊥ 2η1 ⎪ ⎬ 2 2 τ ⊥ E0 ⊥ 2 Tt = a x ζ+ a z j ζ −1 ⎪ ⎪ 2η 2 ⎭ 2 2 ⎫ ρ|| E0 [a x sin θi − a z cosθi ]⎪ Tr|| = 2η1 ⎪ ⎬ 2 2 τ || E0 ⎪ || 2 Tt = a x ζ+ a z j ζ −1 ⎪ 2η 2 ⎭ 2

2

⊥ r

[

]

[

]

(9.86)

(9.87)

It is clear that the reflected power is real for both cases of perpendicular and parallel polarizations, and propagates with components in the x direction and the negative z direction LH in medium 1 only. On the other hand, the transmitted power is real in the x direction and imaginary in the z direction, which means that there is no real power crossing the interface to medium 2. Since ζ = sinθi sinθc , it can be shown that the transmitted fields for the case θi > θ c are attenuated in the direction of z , by a factor α = jβ 2 1 − (sin θi sin θ c ) = jβ 2 1 − ζ 2 = β 2 ζ 2 − 1 2

(9.88)

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Therefore, the transmitted wave decays rapidly and vanishes after a short distance in z direction, while propagates in the x direction parallel to the interface. Based on this analysis, it is clear that when θi > θ c , the incident power is totally reflected back to medium 1 and no real power transmitted to medium 2 as shown in Fig. (9.7c). The total internal reflection concept has been used in many applications such as in the radio wave transmission by reflection from ionosphere and transmission through dielectric waveguides and optical fiber cables. 9.5.

TOTAL TRANSMISSION AND BREWSTER ANGLE

When a uniform plane wave is incident on the boundary between two lossless media as illustrated in Fig. (9.8), a fraction of the wave may reflect back to the first media and the other part is transmitted into the second media. For some particular angles of incidence, the incidence wave can be totally transmitted into the second medium depending on the wave polarization. Hi

Incident wave

θi

Ei

(ε1, μ1, σ1 = 0) Medium 1

Plane z = 0

(ε2, μ2, σ2 = 0)

x

Medium 2

θt z

Transmitted wave

Fig. (9.8). Reflection and transmission at the interface between two lossless media.

9.5.1. Perpendicular Polarization Case With reference to Fig. (9.8), η1 = μ1 ε1 and η2 = μ 2 ε 2 . Therefore, reflection coefficients for perpendicular polarization from (9.50), can be written as

ρ⊥ =

cosθi − μ1ε 2 μ 2ε1 cosθt cosθi + μ1ε 2 μ 2ε1 cosθt

(9.89)

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For total transmission, the reflection coefficient must be zero. Substituting ρ⊥ = 0 in (9.89), we get cosθ i = μ1ε 2 μ 2ε1 cosθ t

(9.90)

Solving (9.74) and (9.90) simultaneously for θ i , yields sin 2 θ i =

μ1ε 2 μ 2ε1 − 1 (μ1 μ 2 )2 − 1

(9.91)

Let the incidence angle for the total transmission in the case of perpendicular polarization be θ i = θ b⊥ , then from (9.91) θ b⊥ can be obtained as

⎡ μ1ε 2 μ 2ε1 − 1 ⎤ ⎡ ⎤ μ1ε 2 μ 2ε1 − 1 −1 ⎥ = sin ⎢ ⎥ 2 ⎢⎣ (μ1 μ 2 ) − 1 ⎥⎦ ⎣ (μ1 μ 2 − 1)(μ1 μ 2 + 1) ⎦

θb⊥ = sin −1 ⎢

(9.92)

The angle of incidence θi = θb⊥ is known as the Brewster angle for the perpendicular polarization wave. If we assume that ε1 = ε 2 , (9.92) reduces to

⎡

μ2 ⎤ ⎥ ⎣ μ1 + 1 ⎦

θb⊥ = sin −1 ⎢

(9.93)

For materials with ε1 ≠ ε 2 and μ1 = μ2 , sin θ b⊥ = ∞ . In this case, θ b⊥ cannot exit, since sin θ b⊥ must be less than one. Hence, for a perpendicularly polarized plane wave with θ i = θ b⊥ , there is no reflected wave and all the power is transmitted into medium 2, provided that ε1 = ε 2 or μ1 ≠ μ 2 . 9.5.2. Parallel Polarization Case In this case, the reflection coefficient from (9.51) is given by

ρ|| =

cosθt − μ1ε 2 μ 2ε1 cosθi cosθt + μ1ε 2 μ 2ε1 cosθi

(9.94)

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In the case of total transmission ρ|| = 0 . Hence, from (9.94) for the parallel polarization case we have

cosθ t = μ1ε 2 μ 2ε1 cosθ i

(9.95)

Following the same analysis as in the perpendicular case, (9.74) and (9.95) are solved for θ i to obtain sin 2 θ i =

1 − μ 2ε1 μ1ε 2 2 1 − (ε1 ε 2 )

(9.96)

Let the incidence angle for the total transmission in the case of parallel polarization be θ i = θ b|| , then from (9.96) θb|| can be obtained as

⎡ μ1ε 2 μ 2ε1 − 1 ⎤ ⎥ 2 ⎢⎣ (ε 2 ε1 ) − 1 ⎥⎦

θb|| = sin −1 ⎢

(9.97)

The angle of incidence θ i = θ b|| for total transmission is the Brewster angle for the parallel polarization wave. For non-ferromagnetic materials μ1 ≈ μ 2 ≈ μ 0 and (9.97) reduces to

⎡

ε2 ⎤ ⎥ ⎣ ε 2 + ε1 ⎦

θb|| = sin −1 ⎢

(9.98)

When a parallel polarized uniform plane wave is incident on the boundary between medium 1 and medium 2 with an incidence angle θ i = θ b|| , there will be no reflected wave and all the power will be transmitted into medium 2. Example 9.4 A parallel-polarized uniform plane wave traveling in a medium having ε1 = 4 ε 0 , μ1 = μ 0 , σ1 = 0 impinges on an interface formed by the medium and air. Find the angle of incidence that will allow complete (a) Reflection of the wave to the medium. (b) Transmission of the wave to air.

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617

Solution Medium 1 is lossless with ε1 = 4 ε 0 , μ1 = μ 0 , and medium 2 is air . (a) The angle of incidence that will allow complete reflection is the critical angle. From (9.75) the critical angle is obtained as

(

θc = sin −1 ε 2 ε1

)

(

)

= sin −1 ε 0 4 ε 0 = sin −1 (0.25) = 14.47 

(b) The angle of incidence that will allow complete transmission is the Brewster angle. From (9.98) the Brewster angle for parallel polarization case θ b|| is obtained as ⎞ 1 ⎟ = sin −1 ⎛⎜ 1 ⎞⎟ ⎟ ⎝ 5⎠ ⎝ (ε1 ε 2 ) + 1 ⎠  = 26.6 ⎛

θb|| = sin −1 ⎜⎜

9.6.

NORMAL INCIDENCE

In the previous sections, the reflection and transmission coefficients are derived for oblique incidence of electromagnetic waves on an interface between two media. When the angle of incidence θi = 0 , the direction of propagation is normal to the boundary between the media and the incidence is said to be normal incidence. The reflection and transmission coefficients for the normal incidence case can be obtained from the perpendicular polarization coefficients given in (9.50) and (9.51), or from the parallel polarization given in (9.57) and (9.58) by letting the angle of incidence equal to zero. 9.6.1. Reflection and Transmission Coefficients For perpendicular polarization case, from (9.73), θi = 0 implies that θt = 0 . Substituting θt = θi = 0 , in (9.50) and (9.51), the reflection and transmission coefficients ρ⊥ and τ ⊥ for the normal incidence from medium 1 to medium 2 becomes

ρ⊥ θ

i

=θ t = 0

=

η 2 − η1 η 2 + η1

(9.99)

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τ⊥ θ

i =θ t = 0

Sameir M. Ali Hamed

=

2η2 η 2 + η1

(9.100)

For Parallel polarization case, substituting θt = θi = 0 , in (9.57) and (9.58), ρ|| and τ || for the normal incidence from medium 1 to medium 2 are

ρ|| θ

i =θ t = 0

τ || θ

=θ t = 0

i

η2 − η1 η 2 + η1 2η 2 = η 2 + η1 =

(9.101) (9.102)

It is clear from (9.99) and (9.101), ρ|| = ρ ⊥ , and from (9.100) and (9.101) τ || = τ ⊥ . Therefore, the reflection and transmission coefficients for normal incidence are independent of the polarization, and can be written as

η 2 − η1 η 2 + η1 2η 2 τ= η 2 + η1 ρ=

(9.103) (9.104)

The coefficient τ and ρ are dimensionless and may be complex as in the oblique incident cases. The reflection coefficient can be written as

ρ = ρ e jϕ

(9.105)

where ρ and ϕ are the magnitude and phase of the reflection coefficient respectively. The expression of the incident, reflected and transmitted electric fields for the normal incidence case in the general coordinate system uvw with plane of incidence uw and interface at w = 0 , can be obtained from (9.23a) – (9.23c) by substituting θt = θ r = θi = 0 , and removing the superscript ⊥ , as

Ei = a v E0 e − γ1 w

(9.106a)

E r = a v ρ E0 e + γ1 w

(9.106b)

Et = a vτ E0 e − γ 2 w

(9.106c)

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Assuming the plane of incidence is the xz plane, the interface between the media is the xy plane ( z = 0) and the incident electric field is perpendicular to the plane of incidence as shown in Fig. (9.9). For normal incidence, the incident, reflected and transmitted electric fields can be obtained from (9.106a) – (9.106c) in the xyz coordinate system by letting (u , v, w) → ( x, y, z ) , then

Ei = a y E0e− γ1 z

(9.107a)

Er = a y ρ E0e+ γ1 z

(9.107b)

Et = a yτ E0e− γ 2 z

(9.107c)

The corresponding magnetic fields are Hi = a x

E0

e − γ1 z

(9.108a)

η1 ρ E0 + γ z H r = −a x e η1 τ E0 − γ z Ht = a x e η2

(9.108b)

1

(9.108c)

2

Reflected wave

Incident wave Hr Ei

(ε1, μ σ 

γi

Medium 1

Er

(ε1, μ σ 

Hi

γr

Medium 1

z=0

z=0 x

x (ε2, μσ  Medium 2

(ε2, μσ 

γt Ht

Et⊥

Medium 2

Transmitted wave z Fig. (9.9). Normal incidence of plane waves on the interface between two media.

z

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The time-average Poynting vectors for the incident, reflected, and transmitted waves can be obtained by letting θt = θr = θi = 0 in (9.54a) – (9.54c). Doing this and removing the superscript ⊥ , yields 2

E Ti = a z 0 2η1 2

(9.109a)

2

ρ E0 2 Tr = −a z = −a z ρ Ti 2η1 2

(9.109b)

2

τ E0 η 2 = a z 1 τ Ti Tt = a z η2 2η 2

(9.109c)

The electric and magnetic fields in medium 1 are obtained by superimposing the incident and reflected fields. Thus, the electric field in medium 1 is

(

)

(

E1 = a y E0 e− γ1 z + ρ e+ γ1 z = a y E0e− α1 z e− jβ1 z + ρ e2 α1 z e+ jβ1 z

)

(9.110)

The corresponding magnetic field is H1 = a x

E0

η1

(e

− γ1 z

)

− ρ e + γ1 z = a x

E0

η1

(

e − α1 z e − jβ1 z − ρ e 2 α1 z e + jβ1 z

)

(9.111)

The transmitted electric and magnetic fields into medium 2 are given by (9.107c) and (9.108c) respectively. When medium 1 is a perfect dielectric, α1 = 0 and γ1 = jβ1 . Hence, (9.110) and (9.111) reduce to

(

E1 = a y E0 e− jβ1 z + ρ e+ jβ1 z H1 = a x

E0

η1

(e

− jβ 1 z

)

− ρ e + jβ 1 z

)

(9.112) (9.113)

9.6.2. Standing Wave Ratio The standing wave ratio is defined as the ratio of the magnitude of the maximum electric field E1 max to the magnitude of the minimum electric field E1 min in medium 1. Mathematically, the standing wave ratio is given by

SWR =

E1 max E1 min

=

H1 max H1 min

(9.114)

Reflection, Transmission, and Negative Refraction

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Substituting ρ from (9.105) into (9.112) and (9.113), the electric and magnetic fields in medium 1 become

(

)

(9.115)

e − jβ1 z 1 − ρ e + j (2 β1 z +ϕ )

(9.116)

E1 = a y E0 e− jβ1 z 1 + ρ e+ j (2 β1 z +ϕ ) H1 = a x

E0

η1

(

)

The maximum value of E1 occurs when H1 is minimum at z = zmax where e + j (2 β1 z +ϕ ) = 1 as clear from (9.115) and (9.116). Let the maximum magnitude of E1 be E1 max and the minimum magnitude of H1 be H1 min , then

E1 max = E0 (1 + ρ H1 min = z max =

E0

η1

)

(9.117a)

(1 − ρ )

(9.117b)

2 nπ − ϕ , 2 β1

n = 0,1,

(9.117c)

Similarly, The minimum value E1 min of E1 and the maximum value H1 max of H1

occur at the same point z = zmin where e + j (2 β1 z +ϕ ) = −1. Thus, from (9.115) and (9.116), we obtain

E1 min = E0 (1 − ρ

E0 (1 + ρ ) Z1 (2n − 1)π − ϕ , = n = 1, 2, 2 β1 H1 max =

zmin

)

(9.118a) (9.118b) (9.118c)

Note that both E1 max and H1 min occur at the points z max , while E1 min and H1 max occur at the points zmin . Using the definition in (9.114) with (9.117a) and (9.118a), the standing wave ratio in medium 1 can be expressed as

SWR =

1+ ρ 1− ρ

(9.119)

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Since 0 ≤ ρ ≤ 1, then 1 ≤ SWR ≤ ∞ . When SWR = 1 , this means that the incident wave is completely transmitted to medium 2. On the other hand, SWR = ∞ corresponds to the complete reflection of the incident wave to medium 1. In the above analysis, we assumed that medium 1 is a perfect dielectric, while medium 2 is arbitrary medium with propagation constant γ2 = α2 + jβ2 . In the following, special cases of medium 2 as a perfect conductor and a perfect dielectric will be considered. 9.6.2.1. Case 1: Medium 2 is a Perfect Conductor In this case, η2 → ∞ and ρ = −1, which means that the fields are completely reflected back to medium 1 and no fields transmitted to medium 2. Consequently, from (9.105) and (9.119), ρ = 1, ϕ = 0 , and SWR = 1 . Putting ρ = 1 and ϕ = 0 in (9.115) and (9.116), the electric and magnetic fields in medium 1 reduce to E1 = a y 2 E0 cos(β1 z ) 2E H1 = a x j 0 sin (β1 z ) η1

(9.120) (9.121)

The points of the maximum electric field (minimum magnetic field) occur at zmax =

nπ

β1

=

nλ1 , 2

n = 0,1,

(9.122)

and the points of the minimum electric field (maximum magnetic field) at

zmin =

(2n − 1)λ1 , 4

n = 1, 2,

(9.123)

Equation (9.120) and (9.121) show that the electric and magnetic field resembles standing waves in medium 1. The corresponding instantaneous fields are

E1 = Re[E1 e j ω t ] = a y E0 cos(β1 z )cos(ω t ) 1 2

H 1 = Re[H1 e j ω t ] = −a x 1 2

2 E0 sin (β1 z )sin (ω t ) Z1

(9.124) (9.125)

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When medium 1 is a lossless dielectric and medium 2 is a perfect electric conductor, there are no fields transmitted to medium 2. 9.6.2.2. Case 2: Medium 2 is a Perfect Dielectric In this case, η2 = μ 2 ε 2 and η1 = μ1 ε1 are real. Consequently, ρ is real according to (9.103). We can differentiate between two cases based on the values of η1 and η2 as follows: 1. The case η2 > η1 . In this case, using (9.103) and (9.105), it is clear that ρ > 0 and ϕ = 0 . Thus, based on (9.115) and (9.116) the fields in medium 1 are

(

E1 = a y E0 e− jβ1 z 1 + ρ e+ j 2 β1 z H1 = a x

E0

η1

(

e − jβ1 z 1 − ρ e + j 2 β1 z

)

(9.126)

)

(9.127)

The maximum and minimum electric fields occur at points given by (9.122) and (9.123) respectively. To illustrate the behavior of the electric and magnetic fields in medium 1, the magnitude of spatial electric E1 and magnetic H1 fields are plotted in Fig. (9.10) for the case of medium 1 is polystyrene (η1 = 235.6 Ω) , while medium 2 is free space (η 2 = 377 Ω) , taking E0 = 1 V m . Since η2 > η1, the maxima of the electric field and minima of the magnetic field occur at nλ1 2 , while the minima of the electric field and maxima of the magnetic field at (2n − 1)λ1 4 , where n = 1, 2,, as clear from Fig. (9.10). h > h

2

1

5.4

1.2

4.5

1

1

|E | (V/m)

1.5

0.9 0.6 -1.5

|H | (mA/m)

6.3

1.8

3.6 -1.25

-1

-0.75

-0.5

-0.25

0

2.7

z/l1

Fig. (9.10). The magnitude of spatial electric E1 and magnetic H1 fields in medium 1, when medium 1 is polystyrene (η1 = 235.6 Ω) , while medium 2 is free space (η2 = 377 Ω) , with E0 = 1 V m .

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2. The case η2 < η1. In a similar manner as in case 1, it can be shown that ρ < 0 and ϕ = π . Substituting these quantities in (9.115) and (9.116) , yields

(

E1 = a y E0 e− jβ1 z 1 − ρ e+ j 2 β1 z H1 = a x

E0

η1

(

)

e − jβ 1 z 1 + ρ e + j 2 β 1 z

(9.128)

)

(9.129)

In this case, the points of the maximum electric fields occur at zmax =

(2n − 1)π 2β1

=

(2n − 1)λ1 ,

n = 1, 2,

4

(9.130)

and the points of the minimum electric fields at zmin =

nπ

β1

=

n λ1 , 2

n = 0,1,

(9.131)

The magnitude of spatial electric E1 and magnetic H1 fields for the case η2 < η1 are plotted in Fig. (9.11). Medium 1 is free space (η1 = 377 Ω ), while medium 2 is polystyrene ( η2 = 235.6 Ω), and E0 = 1 V m . 2

3.4

1.1

2.8

1

1

|E | (V/m)

1

2.2

0.8 0.5 -1.5

|H | (mA/m)

h < h

1.4

-1.25

-1

-0.75 z/l1

-0.5

-0.25

0

1.6

Fig. (9.11). The magnitude of spatial electric E1 and magnetic H1 fields in medium 1, when medium 1 is free space (η1 = 377 Ω) , while medium 2 is polystyrene (η2 = 235.6 Ω) , with E0 = 1 V m .

From (9.107c) and (9.108c) with γ2 = jβ2 , the fields in medium 2 are

E2 = Et = a yτ E0e− jβ 2 z

(9.132)

Reflection, Transmission, and Negative Refraction

H 2 = Ht = a x

Electromagnetics for Engineering Students

τ E0 − jβ e η2

625

(9.133)

2z

Note that the fields in medium 1 are standing wave fields with SWR > 1 , while those in medium 2 are traveling wave fields. Example 9.5 A uniform plane wave propagating in a medium having a relative permittivity of 4 is incident normally on a dielectric medium with dielectric constant of 9. Assuming both media are non-ferromagnetic and lossless, determine the (a) Reflection and transmission coefficients. (b) Percentage of power densities that is reflected and transmitted. (c) Positions in the medium of relative permittivity 4 where the electric field maxima and minima occur at 3 GHz. Solution Let the medium of relative permittivity 4 be medium 1, then ε1 = 4 ε 0 , and ε 2 = 9 ε 0 . The intrinsic impedances of medium 1 and 2 are

η1 = 120π η2 = 120π

ε r = 120π

4 = 60π Ω

ε r = 120π

9 = 40π Ω

(a) ρ and τ from (9.103) and (9.104) respectively are

η 2 − η1 40π − 60π = = −0.2 η 2 + η1 40π + 60π 2η 2 2 × 40π = = 0.8 τ= η2 + η1 40π + 60π

ρ=

(b) The ratio of the reflected to incident power densities, from (9.109b), is 2

Tr = ρ Ti ⇒ Tr Ti = ρ

2

Then the Percentage of the reflected power density is 2

ρ ×100 = 0.04 ×100 = 4%

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The ratio of the transmitted to incident power density, from (9.109c), is

Tt Ti =

η1 2 60π 2 3 2 τ = τ = τ 40π 2 η2

Then the Percentage of the transmitted power density is

3 2 3 τ × 100 = × 0.82 × 100 = 96% 2 2 (c) Since η2 < η1, then the maxima positions of electric field in the medium of relative permittivity 4 can be obtained from (9.130) as zmax =

(2n − 1)λ1 , 4

n = 1, 2,

where λ1 at f = 3 GHz is

λ1 = υ f =

3 × 108 = 0.5 m 4 × 3 × 109

Then z max =

(2n − 1) 8

m,

n = 1, 2, 

Similarly, from (9.131) the positions of minima are zmin =

n λ1 n = m, 2 4

n = 0,1,

Example 9.6 A uniform plane wave propagating in a free space is incident normally upon a lossless dielectric medium with dielectric constant of 4. In the free-space medium, a standing wave is formed. If the normalized magnetic of the incident electric field is E0 = 1 V m , find the (a) Voltage standing wave ratio. (b) Shortest distance in wavelength from the interface where the first maximum in the electric field standing wave pattern will occur.

Reflection, Transmission, and Negative Refraction

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(c) Maximum value of the electric field standing wave pattern in free space. (d) Shortest distance in wavelength from the interface where the first minimum in the electric field standing wave pattern will occur. (e) Minimum value of the electric field standing wave pattern in free space. Solution Let free space be medium 1, then ε1 = ε 0 , and ε 2 = 4 ε 0 . The intrinsic impedances of medium 1 and 2 are

η1 = η0 = 120π Ω

η2 = 120π

ε r = 120π

4 = 60π Ω

(a) The reflection coefficient is given by ρ = (60π − 120π ) (60π + 120π ) = −1 3, then the voltage standing wave ration in free space is found as

SWR =

1+ ρ 1+ −1 3 = =2 1− ρ 1− −1 3

(b) Since η2 < η1, the maxima of the electric field from the interface occur at zmax = (2n − 1)λ1 4 , the shortest distance is the first point ( n = 1 ), then the shortest distance where the maximum electric field occurs is zmax = λ1 4 . (c) Substituting zmax = λ1 4 and E0 = 1 V m in (9.126), the maximum value of the electric field in medium 1 is found as

(

)

E1 max = a y E0 e− jβ1 z mas 1 − ρ e+ j 2 β1 z mas = 1 − ρ e+ jπ = 1 + − 1 3 = 1.33 V m (d) Similarly, the minima of the electric field from the interface occur at zmax = n λ1 2 , and the shortest distance is the first point ( n = 1 ), then the shortest distance where the minimum electric field occurs is zmin = λ1 2 . (e) Substituting zmin = λ1 2 and E0 = 1 V m in (9.128), the minimum value of the electric field in medium 1 is found as

(

) (

)

E1 min = a y E0 e− jβ1 z min 1 − ρ e+ j 2 β1 z min = 1 − ρ e+ j 2π = 1 − − 1 3 = 0.67 V m

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Example 9.7 A uniform plane wave traveling in a free space is incident normally upon a perfect conducting planar surface z = 0 . If the spatial electric field of the wave is given by Ei = a x E0 e − jβ 0 z

Find the (a) Reflected electric field. (b) Incident and reflected magnetic fields. Solution In the general coordinate system uvw , the incident electric field is given in (9.106a) by

Ei = a v E0 e − γ1 w Since the interface is z = 0 , then w = z , u = x and v = y . Comparing with the given problem, yields av = a x and z = w . Substituting av = a x and z = w in (9.106b), the reflected electric field is obtained as

E r = a x ρE0 e + jβ1 z The wave is incident from free space to a perfect conductor. Hence, ρ = −1 and the wave is completely reflected back to free space. in this case, E r is Er = a x ρE0 e + jβ 0 z = −a x E0 e + jβ 0 z

The magnetic field can be obtained using H=

1

η

ar × E

For the incident wave ar = a z , and η = η1 = η0 , then H i is Hi =

1

η0

a z × Ei = a y

E0

η0

e − jβ 0 z 

Reflection, Transmission, and Negative Refraction

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For the reflected wave ar = −a z , then H r is Hr = −

9.7.

1

η0

a z × Er = a y

E0

η0

e + jβ 0 z 

NEGATIVE REFRACTIVE INDEX

Metamaterials and the general classification of materials have been introduced in Chapter 6. The DNG materials are the class of materials that characterized by a negative permittivity and permeability, while in DPS materials both permittivity and permeability are positive. The negative refraction phenomenon occurs when an oblique plane wave incidents on the interface between a DPS and DNG materials. Consider a uniform plane wave traveling in a DPS medium characterized by ε1, μ1, σ1 = 0 impinges on an interface formed by this medium and a DNG medium characterized by − ε 2 , − μ 2 , σ2 = 0. Enforcing the boundary conditions at the interface as has been done in Section 9.3.1, it can be shown that the Snell's laws of reflection and refraction, in this case, are respectively sin θi = sin θ r → θi = θ r

β1 sin θi = β 2 sin θt

→

μ1

(9.134a)

ε1 sin θi = − μ 2 − ε 2 sin θt (9.134b)

This reveals that the Snell's law of reflection for the DNS-DNG interface is the same as that for the DPS-DPS interface. On the other hand, the Snell's law of refraction for DPS-DNG interface from (9.134b) is

μ ε n sin θi =− 2 2 =− 2 sin θt μ1 ε1 n1

(9.135)

where n1 and n2 are the magnitudes of the refractive indices of the DPS and DNG media respectively. Since the refractive index of the DPS medium is positive, it is clear from (9.135) that the refractive index of the DNG medium is negative. Accordingly, the transmission angle θt is also negative, which means that θt is on the same side of the interface normal as the incidence angle is, as shown in Fig. (9.12).

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Reflected wave

Incident wave

θi

θr

Sameir M. Ali Hamed

Incident wave

Reflected wave

θi

DPS

θr

(ε1, μ1, σ1 = 0)

(ε1, μ1, σ1 = 0) x

(ε2, μ2, σ2 = 0)

θt

θt

DNG

Transmitted z Wave

z

(a)

DPS

(ε2, μ2, σ2 = 0)

x

DPS

Transmitted Wave

(b)

Figure (9.12). Reflection and transmission. (a) DPS-DNG interface. (b) DPS-DPS interface.

For an oblique incidence on a DPS-DNG interface with perpendicular and parallel polarizations cases, the incident, reflected and transmitted time-average Poynting vectors are obtained by substituting θt = − θt in (9.81) and (9.82), as ⊥ i

T

Tr⊥ Tt⊥

Ti|| Tr|| Tt||

⎫ ⎪ ⎪ 2η1 ⎪ 2 ⊥ 2 ρ ⊥ E0 [a x sin θi − a z cosθi ] ⎪⎬ = 2η1 ⎪ 2 || 2 ⎪ τ || E0 [a x sin θt − a z cosθt ]⎪ =− 2η 2 ⎪ ⎭ 2 ⎫ E0|| ⎪ [a x sin θi + a z cosθi ] = ⎪ 2η1 2 ⎪ || 2 ρ|| E0 [a x sin θi − a z cosθi ] ⎪⎬ = 2η1 ⎪ 2 2 ⎪ τ || E0|| [a x sin θt − a z cosθt ]⎪ =− 2η 2 ⎪ ⎭ =

E0⊥

2

[a x sin θi + a z cosθi ]

(9.136)

(9.137)

Reflection, Transmission, and Negative Refraction

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631

Similarly, from (9.49c) the propagation vector for the transmitted wave into the DNG medium is

βt =

ω

υ0

n2 (a x sin θt − a z cos θt )

(9.138)

where υ 0 is the wave velocity in free space. It is clear from (9.136), (9.137), and (9.138) that when a plane wave is incident on the DPS-DNG interface the propagation vector β t and Poynting vectors ( Tt⊥ and Tt|| ) of the transmitted wave into the DNG medium point in the opposite directions. The negative refractive index (NRI) property of the DNG material leads to several applications in communications systems [92] and [76-79], as in the 1. 2. 3. 4. 5. 6. 7.

Phase compensation of beams. Dispersion compensation of transmission lines. Sub-wavelength focusing. Microwave circuits. Phase shifters. Broadband power divider. Antenna systems.

SOLVED PROBLEMS Solved Problem 9.1 The instantaneous electric field of a uniform plane wave travels in free half-space y ≤ 0 , is given by

E i = (6a y − 8a z ) sin (ω t − 4 y − 3z ) V m The wave impinges on a dielectric half-space occupying the region y ≥ 0 , at y = 0 . If the constitutive parameters of the dielectric are ε 2 = 4 ε 0 , μ 2 = μ 0 , σ 2 = 0 . Find the (a) Polarization of the incident wave. (b) Angle of incidence. (c) Reflected and transmitted electric fields.

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Solution (a) Let free space be medium 1 (ε1 = ε 0 , μ1 = μ 0 , σ 2 = 0) and the dielectric be medium 2 (ε 2 = 4 ε 0 ,μ 2 = μ 0 , σ 2 = 0) . From the expression of the incident electric field in medium 1, the corresponding spatial electric field is

Ei = − j (6a y − 8a z ) e− j (4 y +3 z ) V m The interface between the two media is the plane y = 0 , then the normal to the interface is the vector a y . The propagation constant vector is γ i = j (4a y + 3a z ). Therefore, the plane yz is the plane of incidence, since it

contains both γ i and a y . Ei is in the yz plane, which is the plane of incidence, then the polarization of the incident wave is parallel. The interface is the plane y = 0 , ⇒ w = y , which implies (uvw) → (zxy). (b) Since the wave polarization is parallel, then the incident wave is equivalent to E||i given by (9.15b). Comparing E||i and Ei , yields ⇒ cosφ = 1,

sin φ = 0,

⇒ φ = 0

− j (6a y − 8a z ) e− j (4 y +3 z ) = E0|| (− a z cosθi + a y sin θi ) e− γ1 (z sinθ i + y cos θ i ) ⇒ E0|| = − j10  γ1 (z sin θi + y cosθi ) = jβ1 (z sin θi + y cosθi ) = j (4 y + 3z ) ⇒ β1 sin θi = 3,

β1 cosθi = 4

⇒ β1 = 42 + 32 = 5

Then the angle of incidence is

θ i = sin −1 (3 5) = 36.87  (c) Since φ = 0 and the polarization is parallel, then from Table 9.1 with E0|| = j10 , θ r = θi , γ1 = jβ1 , γ 2 = jβ2 , w = y , u = z and v = x , the reflected and transmitted electric fields from the first row under parallel polarization, are

E||r = − j (6a z + 8a y ) e− j (4 z −3 y )

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E||t = j10τ || (− a z cosθt + a y sin θt )e− jβ 2 (z sinθ t + y cos θ t ) The transmission angle θt is obtained using Snell’s law as

μ 0ε1 sin θi = μ 0ε 2 sin θt ⇒ θt = sin −1 (3 10 ) = 17.5  The reflection and transmission coefficients from (9.57) and (9.58) are cos17.5 − 4 cos 36.87  = −0.253 cos17.5 + 4 cos 36.87  2 cos 36.87 τ || = = 0.627 cos17.5 + 4 cos 36.87

ρ|| =

The phase constant β 2 from Snell’s law is

β1 sin θi = β 2 sin θt ⇒ β 2 = β1

4 ε0 με sin θi =5 0 2 =5 = 10 ε0 μ 0ε1 sin θt

Substituting β 2 , θt , ρ|| and τ || in the above expressions for reflected and transmitted electric fields, yield

E||r = j (1.52a y + 2.02a z ) e− j (4 y −3 z ) V m

E||t = − j (5.97a y − 1.88a z ) e− j (4 y +9.17 z ) V m The corresponding instantaneous fields are

E r|| = Re[E||r e j ω t ] = (1.52a y + 2.02a z ) sin(ω t − 4 y + 3z ) V m Et|| = Re[E||t e j ω t ] = (5.97a y − 1.88a z ) sin (ω t − 4 y − 9.17 z ) V m Solved Problem 9.2 In Fig. (9.13), show that the apparent depth of the object is given by

d′ =

t cosθi ε1 − sin θ i 2

+

(d − t )cosθi ε 2 − sin 2 θ i

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Electromagnetics for Engineering Students Air ε0, μ 

Sameir M. Ali Hamed

θi

Glass ε0ε1, μ 

d′

θ1

t d

Water

θ2

ε0ε2, μ 

Object



Fig. (9.13). Geometry of Problem 9.2.

Solution Referring to Fig. (9.14), we have d ′ = (z + y )cot θi

z = t tanθ1  y = (d − t )tanθ2 

Substituting z and y in the expression of d ′ , yields

⎡ sin θ1 cosθi sin θ 2 cosθi ⎤ d ′ = [t tan θ1 + (d − t )tan θ 2 ]cot θi = ⎢t + (d − t ) ⎥ cosθ 2 sin θi ⎦ ⎣ cosθ1 sin θi From Snell’s law

sin θ1 =

sin θ 2 =

1 sin θi ε1

1 ε1 sin θ1 = sin θi ε2 ε2

cosθ1 and cosθ2 are obtained as cosθ1 = 1 − sin 2 θ1 = 1 − (1 ε1 )sin 2 θi cosθ 2 = 1 − sin 2 θ 2 = 1 − (1 ε 2 )sin 2 θi

Reflection, Transmission, and Negative Refraction

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635

Substituting sinθ1 , sin θ2 , cosθ1 and cosθ2 in the above expression of d ′ , then the apparent depth is obtained as

d′ = t

cosθ i ε1 − sin θ i 2

Air ε0, μ  Glass ε0ε1, μ 

cosθ i ε 2 − sin 2 θ i

θi

d′

t

θ1 z

Water

+ (d − t )

y

d

θ2

ε0ε2, μ  Fig. (9.14). Illustration of the solution for Solved Problem 9.2.



Solved Problem 9.3 A uniform plane wave, is incident from free space normal to the surface of a material having ε r = 4 , μ r = 1, and σ = 103 S m . If the amplitude of the wave is 10 V m and its frequency is 3 MHz , determine the power dissipated per unit of surface area in 1 mm penetration of the material. Solution Medium 1 is free space, and medium 2 have εr = 4 , μ r = 1, and σ = 103 S m . Let us first examine medium 2 if it can be considered as a good conductor at given frequency,

σ 103 = = 1.5 × 10 6 >> 1 6 −12 ωε r ε 0 2π × 3 × 10 × 4 × 8.85 × 10 Then medium 2 can be considered as a good conductor at 3 MHz, with attenuation constant α 2 where

α2 ≈ π f μσ = π × 3 ×106 × 4π ×10−7 ×103 = 108.9 Np m



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The intrinsic impedance of medium 2 can be approximated as

ωμ 2π × 3 × 106 × 4π × 10 −7 (1 + j ) = η2 ≈ = 0.1089 (1 + j ) Ω 2σ 2 × 103 The transmitted electric field is

Et = a yτ E0e− γ 2 z = a yτ E0e− α 2 z e− jβ 2 z Ht = a x

τ E0 − α z − jβ e e η2 2

2z

The average power density in medium 2 is 2

2

2

2

τ E0 − 2 α 2 z ⎡ 1 ⎤ τ E0 1 Tt = Re Et × H∗t = a z e Re ⎢ ∗ ⎥ = e− 2 α 2 z W m2 η × 2 2 2 0 . 1089 ⎣ 2⎦

[

]

where

τ =

2η2 2 × 0.1089 (1 + j ) = 8.17 × 10 − 4 = η2 + η1 0.1089 (1 + j ) + 377

⎡1⎤ ⎡ ⎤ 1 1 ⎡ 1− j ⎤ = Re ⎢ = Re ⎢ ∗ ⎥ = Re ⎢ ⎥ ⎥ ⎣ 0.1089 × 2 ⎦ 0.1089 × 2 ⎣ 0.1089 (1 − j )⎦ ⎣η2 ⎦ Substituting E0 = 10 V m , Re [1 η2∗ ] and τ in the expression of Tt above, the average power density becomes 2

Tt =

8.17 × 10 − 4 × 10 2 × 0.1089 × 2

2

e − 2 α 2 z = 153e − 2 α 2 z μW m 2

The power dissipated per unit of the surface area PDd along 1 mm penetration of the material is

⎞⎟ PDd = ⎛⎜ e− 2 α 2 z − e− 2 α 2 z = 0 = 1 mm z z ⎝ ⎠ −6 − 2×108.9×0.001 = 153 × 10 1 − e = 30 μW m2

(

)

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Solved Problem 9.4 A right hand circularly polarized plane wave of frequency 100 MHz , have

Ei = (a y − ja z ) e− j 6π z V m is traveling in a lossless non-ferromagnetic medium. The plane wave is incident normally on a planar interface formed by the lossless medium and free space. If the interface is on the plane z = 0 , determine the (a) (b) (c) (d)

Dielectric constant of the medium. Reflection and transmission coefficients for x and y components. Polarization of the reflected wave. Polarization of the Transmitted wave.

Solution (a) The phase constant in medium 1 is β1 = 6π . Since medium is lossless and non-ferromagnetic, let the dielectric constant of medium 1 is ε1 then

β1 = 6π = 2πf μ 0ε1ε 0 =

2π × 100 × 106 ε1 3 × 108 2

⎛ 3 × 3 × 108 ⎞ ⎟ = 81 ⇒ ε1 = ⎜⎜ 6 ⎟ × 100 10 ⎝ ⎠ (b) the given electric field can be considered as a superposition of the following fields

Eix = a x e − j 6π z V m



Eiy = − ja y e− j6π z V m



Since the incidence is normal, the reflection ρ ant transmission τ coefficients are independent of the polarization, then ρ and τ for Eix and Eiy are the same and given by

ρ=

η2 − η1 η0 − η0 = η2 + η1 η0 + η0

81 = 0.8 81

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τ=

Sameir M. Ali Hamed

2η2 2η0 = = 1.8 η2 + η1 η0 + η0 81

Since the interface on the plane z = 0 , then w = z and (uvw) → (xyz). Assume that the plane of incidence is the xz plane, then in this case, Eix is in the plane of incidence and Eiy is perpendicular to the plane of incidence. Accordingly, the polarization of Eix is parallel while that of Eiy is perpendicular. Since Eix is parallel polarized, the reflected and electric field can be obtained as follows: Referring to Table 9.1 for φ = 0 under parallel polarization, the incident electric field is E||i = E0|| (− au cosθ i + a w sin θ i ) e − γ1 (u sin θ i + w cos θ i )

Substituting θi = 0 and (uvw) → (xyz), then comparing with Eix , yields

− a x E0|| e − γ1 z = a x e − j 6π z ⇒ E0|| = −1, γ1 = j 6π

(

)

γ 2 = j (β 2 β1 )β1 = j μ 0ε 0 μ 0ε1ε 0 6π = j 6π 9 = 2π / 3 Then the corresponding reflected and transmitted electric fields are obtained by substituting θi = 0 , E0|| = −1, γ1 = j 6π , γ 2 = 2π / 3 , ρ|| = 0.8 , and τ || = 1.8 in the first row under parallel polarization in Table 9.1, as

Erx = − ρ|| E0|| (au cosθr + aw sin θr )e− γ1 (u sinθ r − w cos θ r ) = 0.8 a xe+ j 6π z

Etx = τ || E0|| (− au cosθt + aw sin θt )e− γ 2 (u sinθ t + w cos θ t ) = 1.8a x e− j 2πz 3 Similarly, Eiy is perpendicularly polarized. Hence, from Table 9.1 for φ = 0 under perpendicular polarization, the reflected and transmitted fields are

Ery = − j 0.8a y e+ j 6πz

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Ety = − j1.8a y e− j 2πz / 3 (c) The total reflected electric field is

Er = Erx + Ery = 0.8a xe+ j 6π z − j 0.8a y e+ j 6πz = 0.8 (a x − ja y ) e+ j 6πz  The polarization of the reflected wave is the same as incident wave, or right hand circularly polarized. (d) The total transmitted electric field is

Et = Etx + Ety = 1.8a x e− j 2πz 3 − j1.8a y e− j 2πz 3 = 1.8 (a x − ja y ) e− j 2πz 3  The transmitted wave is also right-hand circularly polarized.

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PROBLEMS 9.1.

A plane wave in a lossless medium characterized by ε = 1.5 ε 0 , μ = μ 0 and fills the space z ≥ 0 , having

Ei = (20a y − 50a z ) e− j (5 y − 2 z ) V m is incident on a lossless medium that fills the region z ≤ 0 , and characterized by ε = 4 ε 0 , μ = μ 0 , Find the frequency of the wave, the reflected and transmitted magnetic fields. 9.2.

A uniform plane wave traveling in a lossless medium having ε = 4 ε 0 and μ = 2 μ 0 , impinges on an interface formed by the medium and air. Find the angle of incidence for perpendicular and parallel polarization cases, that will allow complete (a) Reflection of the wave to the medium. (b) Transmission of the wave to air.

9.3.

A parallel-polarized plane wave is incident on an interface formed by nonferromagnetic lossless media and free space. Show that the relation between the critical angle θ c and the Brewster angle θ b is given by cosec 2 θb = 1 + cosec 2 θ c 

Then show that the critical and Brewster angles always satisfy θ b < θ c for interfaces formed by a non-ferromagnetic lossless media and free space. 9.4.

In Fig. (9.15), show that the apparent depth of the object is given by

d′ = 9.5.

d cosθ i ε1 − sin 2 θ i

A uniform plane wave traveling in a non-magnetic lossless media, with relative permittivity ε r is obliquely incident on a free-space medium. If the angle of incidence is 29.9 and its polarization is parallel, find ε r so that the wave will be completely (a) Reflected.

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(b) Transmitted. Air ε0, μ 

θi d′

θ1

Image

d

Water ε1, μ 

Object Fig. (9.15). Problem 9.4.

9.6.



The spatial electric field of a uniform plane wave in free space is given by

Ei = (12a x + 9a y + 5a z ) e− j (3 x − 4 y ) V m If the wave is incident on a planar interface z = 0 formed by the dielectric medium with (ε 2 = 2.56 ε 0 ,μ 2 = μ 0 , σ 2 = 0) . Determine the (a) (b) (c) (d) 9.7.

Polarization of the wave with respect to the plane of incidence. Angle of incidence. Reflected and transmitted magnetic fields. Time-average Poynting vector of the transmitted wave.

The instantaneous electric field of a uniform plane wave travels in free half- space z < 0 , is given by

Ei = 3a x cos(ω t + y − 3z ) V m The wave incidents on a dielectric half-space z ≥ 0 , at z = 0 . If the dielectric constitutive parameters are (ε = 2.25 ε 0 , μ = μ 0 , σ = 0) . Find the (a) (b) (c) (d)

Incidence angle. Polarization of the incident wave. The reflected and transmitted magnetic fields. The average Poynting vector of the reflected and transmitted waves.

642

9.8.

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A uniform plane wave is incident on a non-magnetic lossless dielectric slab of thickness t and permittivity ε 2 as shown in Fig. (9.16). The slab is surrounded by a nonmagnetic lossless medium with permittivity ε1 . Show that the wave displacement d due to the presence of the slab is given by

d=

t

(sin θc

sin θi ) − 1 2

where θ c is the critical angle for a wave from the slab to the medium, and θi is the angle of incidence. 9.9.

Referring to Fig. (9.16), show that in Problem 9.8, the phase difference between the incidence point A and the exit point B, is given by

⎛ t + d tan θi ⎞ ⎟⎟ Δφ = β1d ⎜⎜ ⎝ t tan θi − d ⎠  where β1 is phase constant of the medium. Incident wave ε 1, μ 

t

ε 2, μ  ε 1, μ 

θi A

θ1 B

θ2

d

Fig. (9.16). Problem 9.8 and 9.9.

9.10. In Problem 8, let ε 2 = 4 ε 0 , and the medium surrounding the slab is free space, find θ1 and θ 2 when the angle of incidence is θ i = 30   9.11. A uniform plane wave is incident at an angle θ i = 30  at point A on a slab of two lossless dielectric layers as shown in Fig, (9.17). The slab is

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surrounded by a nonmagnetic lossless medium of permittivity ε1 . If ε1 = ε 0 , ε 2 = 4 ε 0 and ε 3 = 2.56 ε 0 , find θ1 and θ 2 . 9.12. For the two-layer dielectric slab in Fig. (9.17), assume that ε1 > ε 3 , determine the condition on ε1 and ε 3 for a total internal reflection at point B. 9.13. For the two-layer dielectric slab in Fig. (9.17), if ε1 = 4 ε 0 , ε 2 = 9 ε 0 and ε3 = 2.56 ε 0 , find the angle of incidence θ i at A that permits total internal reflection at point B. 9.14. The 30 to 60 dielectric prism shown in Fig. (9.18) is surrounded by free space. (a) Find the minimum value of the prism’s dielectric constant so that there is no time-average power density transmitted across the hypotenuse when a plane wave is incident on the prism at point A as shown in the figure. (b) For the dielectric constant found in (a), find θ t . Incident wave

ε 1, μ 

θi A

ε2, μ 

θ1 B

θ2

ε3, μ  ε 1, μ 

Fig. (9.17). Problems 9.11, 9.12, 1nd 9.13.



9.15. A uniform plane wave is incident at point A on a dielectric prism shown in Fig. (9.19). If the prism is surrounded by free space, show that the incident angle can be written in terms of the dielectric constant of the prism ε r  as

⎛

⎞ 1 ⎟ ⎟ − 1 2 ε r ⎠ ⎝

θi = tan −1 ⎜⎜

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Sameir M. Ali Hamed

45°

30°

θi

A

Incident wave

A

θt 45°

εr

60°

θt Fig. (9.18). Problem 14.

Fig. (9.19). Problem 15.

9.16. In an optical fiber cable, the light is launched at an angle θi < θ A , where θ A is known as the acceptance angle, so that the wave propagates in the cable by total internal reflection as shown in Fig. (9.20). If the cable is surrounded by free space, show that

(

θ A = sin −1 ε1 − ε 2 ε0

)

ε0ε2

Cladding Core

θi

ε0ε1

θA Fig. (9.20). Problem 9.16.



9.17. A plane wave whose electric field is given by Ei = a x 200 e− j 6πx passes normally from a material having, ε r = 4 , μ r ≈ 1, and σ = 0 to a material having ε r = 9 , μ r = 4 , and σ = 0 . Write complete expressions for the instantaneous incident, reflected, and transmitted electric and magnetic fields. 9.18. A plane wave of 200 MHz traveling in free space, impinges normally on a block of material characterized by ε r = 9 , μ r = 4 , and σ = 0 . Determine: (a) The intrinsic impedances in the material and free space. (b) The propagation constant of the material and free space. (c) The reflection and transmission coefficient at the interface between free space and the material.

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CHAPTER 10

Rectangular Waveguides Abstract: The transmission media may be unbounded as free space or bounded. The behavior of electromagnetic waves in unbounded media has been dealt with in Chapter 8. The bounded media are structures that are constructed from conducting or/and dielectric materials such as transmission lines, waveguides, and optical fiber cables. Waveguides are constructed from metallic tubes filled with a dielectric material or dielectric slabs with a rectangular, circular or elliptical cross section. This chapter deals with a metallic waveguide of rectangular cross section. The general equations that govern the electromagnetic waves propagation in the waveguide are derived starting from Maxwell's equations. These general equations are used to analyze the behavior of electromagnetic waves and power flow in the metallic waveguide of a rectangular cross section. The topics of the chapter are analyzed in details and supported by numerous illustrative examples and figures in addition to solved problems. Homework problems are included at the end of the chapter.

Keywords: Attenuation constant, cut-off frequency, cut-off wavelength, group velocity, phase constant, phase velocity, propagation constant, transverse electric field, transverse electric and magnetic field, transverse magnetic field, wave impedance. INTRODUCTION It has been mentioned in Chapter 8 that the information in communication systems is transmitted through different media in a form of electromagnetic energy at high frequencies. These media which are referred to as transmission media may be unbounded or bounded. The behavior of electromagnetic waves in unbounded media has been dealt with in Chapter 8. In the bounded media the electromagnetic energy is guided through structures that are constructed from conducting or/and dielectric materials. These guiding structures must be designed properly to transfer the electromagnetic energy efficiently without degradation in the performance of the communications system. The common bounded media are the transmission lines, waveguides, and optical fiber cables. The optical fiber cables are constructed from two coaxial dielectrics of cylindrical geometry and different permittivity. The inner dielectric is called the core and the outer one surrounds the core and called the cladding. The energy is transmitted through the optical fiber at optical frequencies by the total internal reflection at Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

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the interface between the core and cladding [109, 110]. The transmission lines, which will be discussed in details in Chapter 13, are made from two or more conductors separated by a dielectric material. The waveguides are constructed from metallic tubes filled with a dielectric material or dielectric slabs with a rectangular, circular or elliptical cross section. Although the transmission lines are sometimes classified as special waveguides, they are different from them in many aspects. The main differences between them can be summarized as follows: 1. Geometrical Structure: The waveguide may be a metallic tube filled with a dielectric material or a dielectric slab/rod with a rectangular, circular, or elliptical cross section. The most common waveguides are those of rectangular and circular cross sections and they are referred to as rectangular and circular waveguides respectively. On the other hand, the transmission line is constructed from two metallic lines or more, separated by a dielectric material. Fig. (10.1) shows some types of waveguides. 2. Propagation Mode: The electromagnetic waves in the waveguides, always has at least one of the electric or magnetic field components in the direction of propagation, while the transmission lines can support all modes of propagation.

(a)

(b)

(c)

Fig. (10.1). (a) Rectangular waveguide. (b) Coaxial cable. (c) Circular waveguide.

3. Operating Frequency Range: The waveguide permits the propagation of electromagnetic waves only if their frequencies are higher than a certain frequency that known as the cut-off frequency. Electromagnetic waves with frequencies lower than the cut-off frequency cannot propagate in the waveguide. Since the cut-off frequency is inversely proportional to the dimensions of the cross section of the waveguide, the cut-off frequency must be high enough, so that practical waveguides can be realized. Therefore, the waveguides cannot transmit DC signals or signals with low frequencies. On the other hand, there are no cut-off frequencies for transmission lines and they can transmit signals in a wide range of frequencies including DC signals.

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4. Attenuation: The attenuation in the waveguides is much less than in the transmission lines at the high frequencies. The attenuation in the transmission lines increases with the frequency due to the high ohmic and dielectric losses. This makes the transmission lines are not useful for power transmission beyond certain frequencies and the waveguides are more efficient for the electromagnetic power transmission at these high frequencies. 5. Wave Velocity: The wave velocity in the waveguide depends on the wave frequency, while the wave velocity in a lossless transmission line is constant at all frequencies and depends only on its geometry and material. 6. Wave Impedance: the wave Impedance of the waveguide depends on the operating frequency, while that of a lossless transmission line depends only on the geometry and the material of the line. In this chapter, the general formulations that govern the electromagnetic waves propagation in the waveguide are derived starting from Maxwell's equations. These general equations are used to analyze the behavior of electromagnetic waves and power flow in the metallic waveguide of a rectangular cross section. The topic of rectangular waveguides is covered also in references [18, 19] and [93]. For more advanced treatment on the topic of rectangular waveguides, their method of excitation and applications, the student may refer to [92, 98, 111-122]. 10.1. PROPAGATION MODES Assuming that the longitudinal axis of the waveguides is parallel to z-axis, then the waves will travel in the direction of z-axis and the xy plane is the transverse plane. The instantaneous electromagnetic fields E
and H
in the waveguide are considered to be time-harmonic fields with the time dependence factor e j ω t and propagation constant
= jβ
. The electromagnetic fields E
and H
are defined in terms of the spatial components E s and H s as follows

Eg ( x, y, z, t ) = Etg + a zE zg = Re[Ese j ωt ]

(10.1)

H g ( x, y, z, t ) = H tg + a z H zg = Re[Hse j ωt ]

(10.2)

where E t
and H t
are the transverse instantaneous electric and magnetic fields.

E z
and H z
are the instantaneous electric and magnetic fields in the direction of propagation respectively.

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The spatial fields E s and H s can be expressed as

Es (x, y, z ) = E(x, y ) e

− jβ g z

H s ( x, y , z ) = H ( x , y ) e

(10.3)

− jβ
z

(10.4)

where E and H are complex vectors that are independent of z. These vectors can be decomposed into transverse components Et and Ht , and z-directed components Ez and H z as follows

E = Et + az Ez H = Ht + a z H z

(10.5) (10.6)

The electromagnetic fields propagate through the waveguide in different modes based on the geometry and the material of the waveguide. 10.1.1. Transverse Electric and Magnetic Fields (TEM) In this case, both the electric field and magnetic field are perpendicular to the direction of propagation and there are no any field components in the direction of propagation. Fig. (10.2) shows the rectangular and cylindrical coordinates for the TEM electromagnetic fields. The TEM fields satisfy

Ez = H z = 0 E = Et H = Ht y Direction of propagation

Hy

(10.7a) (10.7b) (10.7c) y

Ez = Hz = 0 Direction of propagation

Ey

z

Eρ Hφ

z

Ex

Hρ

Eφ

Hx Transverse Plane

(a)

x

Transverse Plane

(b)

Fig. (10.2). TEM mode: (a) rectangular coordinates. (b) Cylindrical coordinates.

x

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10.1.2. Transverse Electric Field (TE) When only the electric field is always perpendicular to the direction of propagation and has no component in the direction of propagation as shown in Fig. (10.3), the mode of propagation is known as transverse electric field (TE) mode. TE mode fields satisfy

Ez = 0 , H z ≠ 0 E = Et H = Ht + a z H z y Direction of propagation

z

Hy

(10.8a) (10.8b) (10.8c) y

Ez = 0 Direction of propagation

Eρ

Ey

Hz

Hφ

z

Ex

Hz

Hρ

Eφ

Hx Transverse Plane

x

Transverse Plane

(a)

x

(b)

Fig. (10.3). TE mode: (a) Rectangular coordinates. (b) Cylindrical coordinates.

10.1.3. Transverse Magnetic Field (TM) In this case, the magnetic field is always perpendicular to the direction of propagation and has no component in the direction of propagation, while the electric field has a component in the direction of propagation as illustrated in Fig. (10.4), the mode of propagation is known as transverse magnetic field (TM) mode. TM mode fields satisfy

Ez ≠ 0 , H z = 0 E = Et + az Ez H = Ht

(10.9a) (10.9b) (10.9c)

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Hz = 0

y Direction of propagation

z

Hy

Sameir M. Ali Hamed

y Direction of propagation

Ey

Eρ

Ez

Ez

Hφ

z

Ex

Hρ

Eφ Hx

x

Transverse Plane

x

Transverse Plane

(a)

(b)

Fig. (10.4). TM mode: (a) Rectangular coordinates. (b) Cylindrical coordinates.

10.1.4. Hybrid Mode In this case, both the electric field and magnetic field has components in the direction of propagation in addition to transverse components. This mode is known as the Hybrid Mode or HE mode and can exist only in the dielectric slabs waveguides. Figure (10.5) shows rectangular and cylindrical coordinates for the HE electromagnetic fields. The HE fields satisfy

Ez ≠ 0 , H z ≠ 0 E = Et + az Ez H = Ht + a z H z

(10.10a) (10.10b) (10.10c)

y Direction of propagation

z

Hy

y Direction of propagation Ey

Ez

Ez

Hφ

z

Ex

Hz

Eρ

Hz

Hρ

Eφ

Hx Transverse Plane

(a)

x

Transverse Plane

(b)

Fig. (10.5). HE mode: (a) Rectangular coordinates. (b) Cylindrical coordinates.

x

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10.2. GENERAL ELECTROMAGNETIC FIELDS EQUATIONS It is required to find the electromagnetic fields propagating in a waveguide which is made of a good electric conducting material (σ ≈ ∞) and filled with a lossless dielectric of permittivity ε = εrε0 and permeability μ. The electromagnetic fields propagating in the waveguide can be obtained by solving Maxwell\s equation and applying the boundary conditions at the boundaries of the waveguide. Since the fields inside the waveguide propagate in a source-free and charge-free medium, then the fields E
and H
inside the waveguide satisfy Maxwell’s equations for source-free and charge-free medium which can be written as

∇ × E
= − j ωμ H
∇ × H
= j ωε E
∇ ⋅ E
= 0 ∇⋅H
= 0

(10.11a) (10.11b) (10.11c) (10.11d)

The operator ∇ can be written in the form two components, one is the transverse component ∇t = a x ∂ ∂x + a y ∂ ∂y and the other in the direction of propagation

a z ∂ ∂z , as ∇ = ∇t + a z

∂ ∂z

(10.12)

Substituting E
, H
, and ∇  from (10.1), (10.2), and (10.12) respectively, into (10.11a) - (10.11d) , yields

∂⎞ ⎛ − jβ z − jβ z ⎜ ∇t + a z ⎟ × (Et + a z Ez )e g = − j ωμ (H t + a z H z )e g ∂z ⎠ ⎝ ∂⎞ ⎛ − jβ z − jβ z ⎜ ∇t + a z ⎟ × (H t + a z H z )e g = j ωε(Et + a z Ez )e g ∂z ⎠ ⎝ ∂⎞ ⎛ − jβ z ⎜ ∇t + a z ⎟ ⋅ (Et + a z Ez )e g = 0 ∂z ⎠ ⎝ ∂⎞ ⎛ − jβ z ⎜ ∇t + a z ⎟ ⋅ (H t + a z H z )e g = 0 ∂z ⎠ ⎝

(10.13a) (10.13b) (10.13c) (10.13d)

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Equating the transverse components and z components in both sides of equations (10.13a) – (10.13d), we can obtain the general equations that govern the electromagnetic waves propagation in the waveguide as

∇ t × Et e

− jβ
z

= −a z j ωμ H z e

− jβ
z

∂ − jβ z − jβ z − jβ z × Et e
+ ∇t × a z Ez e
= − j ωμ Ht e
∂z − jβ z − jβ z ∇t × H t e
= a z j ωε Ez e
∂ − jβ z − jβ z − jβ z a z × H t e
+ ∇t × a z H z e
= j ωε Et e
∂z ∂ − jβ z − jβ z ∇ t ⋅ Et e
= − E z e
∂z ∂ − jβ z − jβ z ∇t ⋅ H t e
= − H z e
∂z

az

(10.14a) (10.14b) (10.14c) (10.14d) (10.14e) (10.14f)

Since Et , Ht , Ez , and H z are independent of z , we can put ∂ ∂z = − jβ g in (10.14a) – (10.14f) and suppress the factor e Doing this, we obtain

− jβ
z

in both sides of each equation.

∇t × Et = −az j ωμ H z − jβ
a z × Et + ∇t × a z Ez = − j ωμ Ht ∇t × Ht = a z j ωε Ez − jβ
a z × Ht + ∇t × a z H z = j ωε Et ∇ t ⋅ E t = jβ g E z ∇ t ⋅ H t = jβ g H z

(10.15a) (10.15b) (10.15c) (10.15d) (10.15e) (10.15f)

Taking the curl in the transverse direction of both sides of (10.15d), then

− jβ
∇t × a z × Ht + ∇t × ∇t × a z H z = j ωε ∇t × Et

(10.16)

Using the vector identity A × B × C = (A ⋅ C)B − (A ⋅ B)C, (10.16) becomes − jβ
[(∇t ⋅ H t )a z − (∇t ⋅ a z )H t ] + [(∇t ⋅ a z )a z − (∇t ⋅ ∇t )a z ]H z

= j ωε (− a z j ωμ H z )

(10.17)

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Note that in (10.17), ∇t ⋅ a z = 0 and ∇ t ⋅ ∇ t = ∇ t2 . Substituting ∇ t ⋅ H t from (10.15f) into (10.17) and letting β = ω με with simplification, the following wave equation can be obtained

(

)

∇t2 H z + β 2 − βg2 H z = 0

(10.18)

Similarly, taking the curl of both sides of (10.15b) and using (10.15e), a similar wave equation for the electric field Ez can be obtained as

(

)

∇t2 Ez + β 2 − βg2 Ez = 0

(10.19)

where β is the phase constant of plane waves propagating in the medium that fills the waveguide, and β
is the phase constant of the waves E
, H
in the waveguide. Taking the curl of (10.15c) and using (10.15f), we get

jβg ∇t H z − ∇t2Ht = j ωε ∇t × a z Ez

(10.20)

Vector multiplication of both sides of (10.15d) by a z with simplification, yields

jβ g H t + ∇ t H z = j ωε a z × Et

(

(10.21)

)

Multiplying (10.21) by − jβ g and add the result to (10.21), we get

βg2Ht − ∇t2Ht = − jβg × j ωε a z × Et + j ωε ∇t × a z Ez

(10.22)

Multiplying (10.15d) by ( j ωε ), yields

− jβg × j ωε a z × Et + j ωε ∇t × a z Ez = β 2Ht

(10.23)

From (10.22) and (10.23), the wave equation for Ht can be obtained as

(

)

∇t2Ht + β 2 − βg2 Ht = 0

(10.24)

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Following the same analysis using (10.15a), (10.15b) (10.15d) and (10.15e), the wave equation for Et can be obtained as

(

)

∇t2Et + β 2 − βg2 Et = 0

(10.25)

Equations (10.15a) – (10.15f), (10.18), (10.19), (10.24), and (10.25) give the general relations between the transverse electromagnetic fields Et and Ht and the field components in the direction of propagation Ez and H z for an arbitrary waveguide. The TEM mode case equations listed in the left column of Table 10.1 are obtained by letting Ez = H z = 0 in (10.15a) – (10.15f). The equations governing the TE and TM modes cases listed in Table 10.1 are obtained by letting Ez = 0 for the TE case and H z = 0 for the TM case in (10.15a) – (10.15f). The behavior of the electromagnetic fields in the waveguide is determined by (10.18) and (10.19) for the TE and TM modes respectively. Therefore, the starting point in the analysis of the characteristics of the electromagnetic fields in the waveguide is to solve these equations. The steps showing how to find the electromagnetic fields in the waveguides for different modes of propagation are detailed in the next sections. Table 10.1. General equations for the TEM, TE, and TM Modes. 

I II III

IV

TEM Ez = H z = 0

TE Ez = 0 , H z ≠ 0

TM H z = 0 , Ez ≠ 0

∇ t × Et = 0

∇t × Et = −a z j ωμ H z

∇ t × Et = 0

a z × Et =

ωμ

βg

Ht

a z × Et =

∇t × H t = 0

a z × Ht = −

ωε

βg

ωμ

βg

a z × Et −

Ht

∇t × H t = 0

Et

a z × Ht +

j

βg

∇t × a z H z = −

j

βg

∇t × a z Ez =

ωμ Ht jβ g

∇t × H t = a z j ωε Ez ωε

βg

Et

a z × Ht = −

ωε

βg

Et

V

∇ t ⋅ Et = 0

∇ t ⋅ Et = 0

∇t ⋅ Et = jβg Ez

VI

∇t ⋅ Ht = 0

∇t ⋅ Ht = jβ g H z

∇t ⋅ Ht = 0

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10.2.1. TEM Mode Fields In the waveguides supporting TEM mode waves, the potential gradient between the conductors of the waveguides can be used to find the transverse electric field, then the equations in the left column in Table 10.1 can be used to find other electromagnetic fields components. If the electric potential at any point is V , then

Et = −∇tV

(10.26)

Using equation TEM-V in Table 10.1, Laplace equation can be obtained as

∇t V = 0 2

(10.27)

Laplace equation in (10.26) along with the equations TEM-I to TEM-VI in Table 10.1 can be used to analyze the behavior of the TEM mode waves propagating in a waveguide as in the following steps: 1. Solving Laplace equation in (10.27) to determine the distribution of the electric potential V in the waveguide. 2. Substituting V into (10.26) to determine E t , then the time-harmonic electric field in the waveguide can be obtained using

Eg = Re ⎡Et e ⎢⎣

− j ( β g z −ω t )

⎤ ⎥⎦

(10.28)

3. Substituting E t in equation TEM-II in Table 10.1, then Ht can be determined from E t as

⎛β ⎞ Ht = ⎜⎜
⎟⎟ a z × Et ⎝ ωμ ⎠

(10.29)

and the time-harmonic magnetic field can be obtained from

H g = Re ⎡Ht e ⎢⎣

− j ( β g z −ω t )

⎤ ⎥⎦

(10.30)

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4. The instantaneous Poynting vector is given by

1 T = E
× H
∗ 2

(10.31)

Since the fields in the waveguide are time-harmonic, then the average Poynting vector Ta and the average power Pa crossing the area S in the waveguide can be obtained using (7.153) and (7.154) respectively as

[

]

[

]

1 1 Ta = Re Es × H∗s = Re E × H∗ 2 2 1 Pa = ∫∫ Ta ⋅ dS = ∫∫ Re E × H∗ ⋅ dS 2 S S

[

]

(10.32) (10.33)

5. The wave impedance of the TEM waveguide can be obtained using

Vo2 Zo = 2Pa

(10.34)

where Vo is the electric potential difference between the conductors. 6. Applying Ampere’s law, the current in the conductors of the waveguide can be obtained. Assume that l is a closed path enclosing the conductor of the current I , then the current can be obtained using

I = ∫ H
⋅ dl

(10.35)

l

Example 10.1 Using the TEM waveguide equations find the electromagnetic fields in the a long parallel-plates line. Determine the intrinsic impedance of the line and the current in each plate. The separation between the plates is d and the width of each plate is w such that w >> d . One plate is grounded and the other is maintained at a potential Vo . The line is filled with a dielectric with a permittivity ε . Solution The geometry of the problem is shown in Fig. (10.6). It is appropriate to use the rectangular coordinates. Assuming the axis of the line is parallel to z-axis, then the wave travels in the z direction.

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The operators ∇ and ∇ t2 can be written as

∂2 ∂2 ∂ ∂ ∂ ∂ 2 , ∇t = 2 + 2 ∇ = ax + ay + az = ∇t + a z ∂x ∂y ∂z ∂z ∂x ∂y The electric potential V between the plates satisfies Laplace equation in (10.7). Assuming the planes of the plates are parallel to xy plane, then the Laplace equation can be written as

∇t2V =

∂ 2V ∂ 2V + =0 ∂x 2 ∂y 2

y

V = V0

-w/2

z

ε, μ0

d

w/2

x

V=0

Fig. (10.6). The geometry of the problem of Example 10.1.

Since w >> d the edge fringing can be ignored, and V can be considered as constant with x and varies with y only, then ∂V ∂x = ∂ 2V ∂x 2 = 0 and Laplace equation reduces to

∂ 2V =0 ∂y 2 Integrating twice with respect to y , the solution can be obtained as

V ( y ) = K1 y + K2 where K1 and K2 are constants to be determined from the boundary conditions.

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The boundary conditions are: at y = 0 , V = 0 and at y = d , V = V0 . Applying these boundary conditions, then K1 = Vo d and K2 = 0 . Substituting K1 and K2 in the above solution, the potential between the plates can be obtained as

V (y) =

Vo y d

Consequently, Et can be determined as follows

E t = −∇ tV = −a x

∂V ∂V V + ay = −a y o ∂x ∂y d

From (10.28), the instantaneous electric field in the line can be obtained as ⎡

E g = Re ⎢− a y

Vo − j (ω t − β g z ) ⎤ e ⎥ d ⎦⎥

⎣⎢ V = −a y o cos(ω t − β g z ) d

From (10.29), we have

Ht = −

Vo d

y

β
Vo ⎛ β
⎞ ⎜⎜ ⎟⎟ a z × a y = a x ωμ 0 d ⎝ ωμ 0 ⎠ y

Ey Hx

dS = az dxdy

Hx = 0

D

Hx

A

d

C B

x -w/2

0 (a)

w/2

d

x -w/2

0

w/2

(b)

Fig. (10.7). The parallel-plates transmission line. (a  ) The electric  and magnetic fields  in  the xy plane. (b) Application of Ampere’s law.

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The instantaneous magnetic field is by substituting Ht in (10.30), as

⎡ β V − j (ω t − β H g = Re ⎢a x g o e ⎢⎣ ωμ 0 d

g z)

⎤ β V ⎥ = a x g o cos(ω t − β g z ) ωμ 0 d ⎥⎦

Note that the electric and magnetic fields are always orthogonal and uniformly distributed in the xy plane for all z as shown in Fig. (10.7a). The average power transmitted through the line can be obtained using (10.33) as

Pa =

⎤ β
Vo jβ
z 1 1 ⎡ Vo − jβ
z ∗ d e e dS × ⋅ = − × ⋅ Re E H S Re a a y x ⎢ ⎥ d 2 ∫∫ 2 ⎣∫∫ ωμ 0 d S S ⎦

[

]

where dS = a z dxdy , then 2 ⎡d +w 2 ⎤ 1 β
w 1 ⎛V ⎞ β
Pa = ⎜ o ⎟ Re ⎢ ∫ ∫ a z ⋅ a z dxdy ⎥ = Vo2 2 ⎝ d ⎠ ωμ 0 ⎣⎢ 0 − w 2 ωμ 0 d ⎦⎥ 2

Using (10.34), the intrinsic impedance of the line is

ωμ 0 d Vo2 d d = = μ0 ε = η Zo = 2 Pav β
w w w where η = μ 0 ε is the plane wave impedance for the medium between the plates. The current in the plates can be determined using Ampere’s law by choosing the path ABCD shown in Fig. (10.7b) as B

I = ∫ H g ⋅ d l = ∫ ax l

β g Vo

C

cos(ω t − β g z ) ⋅ d l + ∫ a x

β g Vo

cos(ω t − β g z ) ⋅ d l ωμ 0 d ωμ 0 d B A β g Vo β V + ∫ ax cos(ω t − β g z ) ⋅ d l + ∫ a x g o cos(ω t − β g z ) ⋅ d l ωμ 0 d ωμ 0 d C D A D

Since the magnetic field outside the plates is zero, the last term will be zero. The integral along the path BC is equal and opposite in sign to the integral along the path CD, and they cancel each other. The remaining integral is that in the first

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term only in which d l = a x dx . Thus, the current in the upper and lower plates are I and − I respectively, where

I=

+w 2 β g Vo β Vw cos(ω t − β g z ) ∫ a x ⋅ a x dx = g o cos(ω t − β g z ) ωμ 0 d ωμ 0 d −w 2 V = o cos(ω t − β g z ) Zo

Example 10.2 Using the TEM waveguide equations find the electromagnetic fields in the coaxial cable if its inner and outer radii are a and b respectively. Assume that the cable is long, and the outer conductor is grounded while the inner conductor is at a potential Vo . The space between the conductors is filled with a dielectric with a permittivity ε and permeability μ 0 . Determine the intrinsic impedance of the cable and the current in the inner and outer conductors. Solution The geometry of the problem is shown in Fig. (10.8). It is suitable to use the cylindrical coordinate systems. Assuming the axis of the line is parallel to the zaxis, and then the wave travels in the z-direction. The electric potential V between the plates satisfies Laplace equation in (10.27). Writing the operator ∇t in cylindrical coordinates, the Laplace’s equation becomes

1 ∂ ⎛ ∂V ⎞ 1 ∂ 2V =0 ⎜ρ ⎟+ ρ ∂ρ ⎝⎜ ∂ρ ⎟⎠ ρ 2 ∂φ 2 y b

B, Y0 a

z V = V0 V=0 Fig. (10.8). The Geometry of the problem in Example 10.2.

x

2

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Since the cable is long, and V can be considered constant with φ and varies with

ρ only, then ∂V ∂φ = ∂ 2V ∂φ 2 = 0 and Laplace equation reduces to 1 ∂ ⎛ ∂V ⎞ ⎟=0 ⎜ρ ρ ∂ρ ⎝⎜ ∂ρ ⎠⎟ Integrating twice with respect to ρ , yields

V (ρ ) = K1 ln ρ + K2 The boundary conditions are: at ρ = a , V = V0 and at ρ = b , V = 0 . Applying these boundary conditions, then K1 = Vo ln(a b) and K 2 = −V0 ln b ln (b a) . Substituting K1 and K2 in the above solution, the potential between the inner and outer conductor can be obtained as

()

V ρ = V0 ln (b a) ln (b a) Consequently, Et can be determined as follows

⎛ ∂V 1 ∂V Et = −∇tV = −⎜⎜ a ρ + aφ ρ ∂φ ⎝ ∂ρ

⎞ Vo ⎟⎟ = a ρ ρ ln (b a) ⎠

From (10.28), the instantaneous electric field in the cable is

⎡

E g = Re ⎢a ρ ⎣

Vo Vo − j (ω t − β g z ) ⎤ = aρ cos(ω t − βg z ) e ⎥ ρ ln (b a) ρ ln (b a) ⎦

Substituting Et in (10.29), yields Ht = −

β Vo βg Vo 1 Vo a z × a ρ = aφ g = aφ d ωμ 0 ωμ 0 ρ ln (b a) μ 0 ε ρ ln (b a)

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The instantaneous magnetic field in the line can be obtained by substituting Ht in (10.30), then ⎡

H g = Re ⎢aφ ⎢⎣

⎤ 1 1 Vo Vo − j (ω t − β g z ) e cos(ω t − β g z ) ⎥ = aφ μ 0 ε ρ ln (b a) μ 0 ε ρ ln (b a) ⎥⎦

The electric and magnetic fields are always orthogonal and uniformly distributed in the xy plane for all z as shown in Fig. (10.9a). The average power transmitted through the line can be obtained using (10.33) as

⎤ β Vo Vo 1 ⎡ − jβ z jβ z Pa = − Re ⎢∫∫ e g a y × ax g e g ⋅ dS ⎥ 2 ⎣ S ρ ln (b a) ωμ 0 ρ ln (b a) ⎦ where dS = a z ρdφdρ as clear from Fig. (10.9b), then 2

2 ⎡ b 2π 1 ⎤ π Vo 2 Vo πβg 1 1 ⎡ Vo ⎤ βg ⋅ φ ρ a a = = Re d d Pa = ⎢ ⎢ ⎥ z z μ 0 ε ln(b a) 2 ⎣ ln(b a) ⎥⎦ ωμ 0 ⎣∫a ∫0 ρ ⎦ ln(b a) ωμ 0

Using (10.34), the intrinsic impedance of the cable is

Zo =

y

Vo2 ln(b a) = μ0 ε 2 Pav 2π dS = az ρdφ dρ

y

Eρ Hφ

dφ

x

ρ

a

φ

x

b

(a)

(b)

Fig. (10.9). The coaxial transmission line. (a) The electric and magnetic fields in the xy plane. (b) The elementary surface area dS.

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The currents in the inner and outer conductors are I and − I respectively. I can be determined using Ampere’s law by choosing a closed circle of radius ρ around the inner conductor, then d l = aφ ρ dφ . Consequently, 2π

βg

Vo cos(ω t − β g z ) ⋅ aφ ρ dφ ωμ 0 ρ ln(b a) 0 1 2π Vo = cos(ω t − β g z ) μ 0 ε ln(b a)

I = ∫ H g ⋅ d l = ∫ aφ l

Example 10.3 For the coaxial cable in Example 10.2, find the capacitance and inductance per unit length of the cable. Solution Let the cable length be l , then the capacitance per unit length is C l = Qa Vl , where Qa is the total charge crossing the surface S = 2πρ l . Applying Gauss law, with E = a ρ Vo ρ ln(b a ) as obtained in Example 10.2, and from Fig. (10.10a), dS = a ρ ρ dφdz , then l 2π

Q V = (ε V )∫∫ E ⋅ dS = ε ∫ ∫ S

a 0

⇒

1 2π ε l a ρ ⋅ a ρ ρ dφdz = ln(b a )  ρ ln(b a )

C l=

2π ε  ln(b a )

The inductance is L = Φ I , H = aφ Vo [ μ ε ρ ln (b a)] , and from Fig. (10.10b),

dS = aφ dρ dz . Applying Gauss law for magnetic flux, we get μ Φ I = (μ 0 I )∫∫ H ⋅ dS = 0 2π S

l b

1

μ0 l

∫∫ ρ aφ ⋅ aφ dρ dz = 2π ln(b a ) a a

μ ⇒ L l = 0 ln(b a ) 2π



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y

Sameir M. Ali Hamed

y

dS = aZ Z d dz

dS = a dZ dz

l aZ

d

Z d dz

Z

z

a

dS a

z

S

(a)

b

x

(b)

x

dZ

dz

Fig. (10.10). The unit surface area for the problem in Example 10.3.

10.2.2. TE Mode Fields In the waveguides that support TE mode fields, H z ≠ 0 and Ez = 0 . The component H z can be obtained by solving (10.18), and then the TE fields equations in Table 10.1 can be used to analyze the behavior of the electromagnetic waves in the waveguide. The steps are as follows: 1. Equation (10.18) can be written as

∇t2 H z + β c2 H z = 0

(10.36a)

βc2 = β 2 − βg2

(10.36b)

where

Once the geometry and the boundary conditions of the waveguide are known, (10.36a) can be solved for H z . 2. The next step is to find Et from H z as illustrated in the following analysis. Taking the curl of both sides of equation TE-I in Table 10.1, yields

∇t × ∇t × Et = (∇t ⋅ Et )∇t − ∇t2Et = −∇ t × a z j ωμ H z

(10.37)

Using (10.36b), (10.25) can be written as

∇ t2Et = − β c2Et

(10.38)

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From equation TE-V in Table 10.1, we have ∇t ⋅ Et = 0 . Hence, using (10.38) and substituting ∇t ⋅ Et = 0 in (10.37), Et can be expressed in term of H z as

Et = − j

ωμ

β c2

∇t × a z H z

(10.39)

The TE mode instantaneous electric field in the waveguide is obtained from

E g = Re ⎡Et e ⎢⎣

− j ( β g z −ω t )

⎤ ⎥⎦

(10.40)

3. Using equation TE-II in Table 10.1, H t can be determined from Et as

Ht =

βg a z × Et ωμ

(10.41)

and the instantaneous magnetic field in the waveguide can be obtained from

H g = Re ⎡(Ht + a z H z ) e ⎢⎣

− j ( β g z −ω t )

⎤ ⎥⎦

(10.42)

4. Instantaneous Poynting vector, average Poynting vector, and the average power can be obtained using (10.31), (10.32) and (10.33) respectively. 5. The wave impedance Z TE is given by the ratio of the one components of the transverse electric field to the associated perpendicular transverse magnetic field. From (10.41), the wave impedance for the TE mode can be written as

ZTE =

ωμ

β


(10.43)

6. The phase constant for the TE mode electromagnetic fields in the waveguide can be obtained from (10.36b) as

βg2 = β 2 − βc2

(10.44)

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10.2.3. TM Mode Fields In this case, Ez ≠ 0 and H z = 0 . The component Ez is obtained by solving (10.19). Following similar procedure as in TE case, the TM fields equations in Table 10.1 are used to analyze the behavior of the electromagnetic waves propagating in the waveguide, as follows: 1. Equation (10.19) can be written in the form

∇t2 Ez + β c2 Ez = 0

(10.45)

Depending on the geometry of the waveguide and the boundary conditions, (10.45) is solved for E z , and hence the other TE mode fields components can be obtained as outlined in the following points. 2. Taking the curl of both sides of equation TM-III in Table 10.1, we get

∇t × ∇t × Ht = (∇t ⋅ Ht )∇t − ∇t × Ht = ∇t × a z j ωε Ez 2

(10.46)

Using (10.24) and (10.36b) we can write

∇t × Ht = −βc2Ht 2

(10.47)

Substituting ∇t ⋅ Ht = 0 from TM-VI into (10.46) and using (10.47), Ht is obtained from E z as

Ht = j

ωε

β c2

∇t × a z Ez

(10.48)

and the instantaneous magnetic field in a TM waveguide becomes

H g = Re ⎡Ht e ⎢⎣

− j ( β g z −ω t )

⎤ ⎥⎦

(10.49)

3. Using equation TM-IV in Table 10.1, E t is obtained as

Et = −

βg ωε

a z × Ht

(10.50)

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and the corresponding instantaneous electric field in the TM waveguide is

E g = Re ⎡(Et + a z Ez ) e ⎢⎣

− j ( β g z −ω t )

⎤ ⎥⎦

(10.51)

4. The expressions of the instantaneous Poynting vector, average Poynting vector, and the average power for the TM mode waves are the same as those for TE and TM waves given by (10.31), (10.32) and (10.33) respectively. 5. The wave impedance Z TM is the ratio of the one components of the transverse electric field to the associated perpendicular component of the transverse magnetic field. Accordingly, the characteristics impedance for the TM mode fields in a waveguide can be obtained from (10.50) as

Z TM =

βg ωε

(10.52)

6. The phase constant for the electromagnetic fields in the waveguide that is supporting TM mode fields has the same expression as in the case of TE mode given by (10.44). 10.3. ELECTROMAGNETIC WAVEGUIDES

FIELDS

IN

THE

RECTANGULAR

The rectangular cross section waveguides can support either TE or TM fields and cannot support TEM mode fields. The coordinates of the rectangular waveguide are shown in Fig. (10.11). The wave propagates in the direction of the z-axis, which is parallel to the longitudinal axis of the waveguide. The transverse components are along x- and y-axes. Therefore, the quantities Et and H t can be written as

Et = a x E x + a y E y

(10.53)

Ht = a x H x + a y H y

(10.54)

The operators ∇ , ∇ t and ∇ t2 in the rectangular coordinates are given by

∇ = ax

∂ ∂ ∂ ∂ + ay + az = ∇t + a z ∂x ∂y ∂z ∂z

(10.55)

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∂ ∂ + ay ∂x ∂y 2 2 ∂ ∂ ∇t2 = 2 + 2 ∂x ∂y

∇t = a x

(10.56) (10.57)

10.3.1. TE Mode Fields in the Rectangular Waveguide The geometry of the waveguide is shown in Fig. (10.11). The dimensions of the cross section of the waveguide are a and b , where a > b . The walls of the waveguide are made from a good electric conducting material (σc ≈ ∞), and it is filled with a lossless dielectric material of permittivity ε and permeability μ. Since we are interested in the TE fields, we will use (10.36a) to find H z , then using (10.39) – (10.41) along with (10.53) and (10.54) the other TE electromagnetic field components in the rectangular waveguide can be obtained. Equation (10.36a) in the rectangular coordinates can be written as

∂2 H z ∂2 H z + + βc2 H z = 0 2 2 ∂x ∂y

(10.58)

y σc ≈ ∞

(0, b, 0) ε, μ, σd ≈ 0

(a, 0, 0)

x

z Fig. (10.11). The geometry of the rectangular waveguide.

Equation (10.58) can be solved using the method of separation of variables. Using this method, H z is expressed in terms of independent functions U and W , where U and W are functions of x and y respectively. Hence,

H z (x, y ) = U (x)W ( y )

(10.59)

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Substituting H z from (10.59) into (10.58), and dividing both sides of the result by UW , yields

1 ∂ 2U 1 ∂ 2W + = −βc2 2 2 U ∂x W ∂y

(10.60)

The constant β c2 can be written in terms of other constants k x and k y as

βc2 = kx2 + k y2

(10.61)

Since U and W are independent, then (10.60) can be separated into the following equations ∂ 2U + k x2U = 0 2 ∂x ∂ 2W + k y2W = 0 2 ∂y

(10.62a) (10.62b)

The solution of (10.62a) and (10.62b) can be written respectively as

U ( x) = A cos(k x x) + B sin( k x x)

(10.63a)

W ( y ) = C cos(k y y ) + D sin( k y y )

(10.63b)

Substituting (10.63a) and (10.63b) in (10.59), H z becomes

H z ( x, y ) = [ A cos( k x x) + B sin( k x x)] [C cos( k y y ) + D sin( k y y )] where A , B , C , D , k x , and

(10.64)

k y are constants. Using (10.39) with (10.53),

components of E t can be obtained from H z as

a x Ex + a y E y

ωμ ⎛ ∂ ∂ ⎞ ⎜a + a y ⎟⎟ × a z H 2 ⎜ x β c ⎝ ∂x ∂y ⎠ ωμ ⎛ ∂H z ∂H z ⎞ ⎟ = − j 2 ⎜⎜ − a y + ax βc ⎝ ∂x ∂y ⎟⎠ =−j

(10.65)

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Equating the quantities in x and y directions in both sides of (10.65), yields Ex

ωμ ∂H z β c2 ∂y (10.66a) ωμ = jk y 2 [A cos( k x x) + B sin(k x x)] C sin(k y y ) − D cos( k y y )

=−j

[

βc

Ey = j

]

ωμ ∂H z β c2 ∂x

⎛ ωμ ⎞ = − jk x ⎜⎜ 2 ⎟⎟[A sin (k x x ) − B cos(k x x )] C cos(k y y ) + D sin (k y y ) ⎝ βc ⎠

[

]

(10.66b)

Using (10.41) with (10.54), we can write

axH x + a y H y =

βg β a z × (a x Ex + a y E y ) = g (a y Ex − a x E y ) ωμ ωμ

(10.67)

Equating the quantities in x and y directions in both sides of (10.67), yields

Hx

=−

βg ωμ

Ey

β = jkx g2 [A sin(k x x) − B cos(k x x)][C cos(k y y ) + D sin(k y y )] βc Hy

=

βg ωμ

(10.68a)

Ex

β = jk y g2 [A cos(k x x) + B sin(k x x)][C sin(k y y ) − D cos(k y y )] βc

(10.68b)

The constants in (10.64) – (10.68) are determined from the boundary conditions. The boundary conditions are that the electric fields on the walls of the waveguide are continuous and equal to zero. Thus,

E x ( y = 0 ) = Ex ( y = b ) = 0 E y (x = 0 ) = E y (x = a ) = 0

(10.69a) (10.69b)

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Imposing the boundary conditions Ex ( y = 0) = 0 and E y (x = 0) = 0 on (10.66a) and (10.66b), yields B=D=0

(10.70)

While the boundary conditions Ex ( y = b) = 0 and E y (x = a ) = 0 on (10.66a) and (10.66b), yields

sin (k x a )= 0

(10.71)

sin (k yb )= 0

(10.72)

It follows that from (10.71) and (10.72)

k x = mπ a , k y = nπ b ,

m = 0,1, 2,

(10.73)

n = 0,1, 2,

(10.74)

Equations (10.73) and (10.74) show that there are an infinite number of modes of TE fields that can exist in the waveguide depending on m and n . To distinguish between modes, the symbol TE mn will be used instead of TE . Substituting k x and k y from (10.73) and (10.74) in (10.61), yields

βc2 = (mπ a)2 + (nπ b)2

(10.75)

Substituting β c from (10.75) into (10.44), the propagation constant in the waveguide can be obtained as


= jβg = (mπ a) + (nπ b) − β 2 2

2

(10.76)

Substituting the constants B , D , k x and k y in (10.64), (10.66a), (10.66b), (10.68a) and (0.68b1), and letting AC = H mn , the components of Et , H t and H z can be determined.

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Substituting the resulted components into (10.5) and (10.6), and using (10.3) and (10.4) the spatial TE mn electromagnetic fields components in a rectangular waveguide can be obtained as

Exs = j E ys = − j

ωμ nπ − jβ z H mn cos(mπ x a )sin (nπ y b )e g 2 βc b

(10.77a)

ωμ mπ − jβ z H mn sin (mπ x a )cos(nπ y b )e g 2 βc a

(10.77b)

Ezs = 0

(10,77c)

H xs = j

βg mπ − jβ H mn sin (mπ x a )cos(nπ y b)e 2 βc a

H ys = j

βg nπ − jβ H mn cos(mπ x a )sin (nπ y b)e 2 βc b

H zs = H mn cos(mπ x a )cos(nπ y b)e

gz

(10.77d)

gz

(10.77e)

− jβ
z

(10.77f)

Note that H mn is in ( A m ). Assuming that H mn is real, the instantaneous TE mn electromagnetic fields components propagating in a rectangular waveguide at an angular frequency ω can be written from (10.77a) – (10.77f) using (10.1) and (10.2), as E xg = −

E yg =

(

) (

) (

ωμ nπ H cos mπ x a sin nπ y b sin ω t − β g z β c2 b mn

)

ωμ mπ H sin (mπ x a )cos(nπ y b )sin (ω t − β g z ) β c2 a mn

E z
= 0

(10.78a) (10.78b) (10.78c)

H xg = −

βg mπ F sin (mπ x a )cos(nπ y b )sin (ω t − βg z ) βc2 a mn

(10.78d)

H yg = −

βg nπ F cos(mπ x a )sin (nπ y b)sin (ω t − βg z ) βc2 b mn

(10.78e)

H z
= Fmn cos(mπ x a )cos(nπ y b )cos(ω t − β
z )

(10.78f)

Rectangular Waveguides

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10.3.2. Characteristics of TE Mode Fields in the Waveguide 10.3.2.1. Cut-off Frequency The propagation constant in (10.76) can be written as


=

(m a )2 + (n b)2 − β 2

= βc2 − β 2

(10.79)

Based on the value of β (10.79) we can distinguish between the following cases for
 1. Cut-off : in this case β c = β and
= 0 . The fields do not propagate. 2. Evanescent: when β c > β ,
= β
is real, and hence the fields  attenuate  and vanish rapidly. 3. Propagation: when β > β c ,
= jβ g  is imaginary,  and the fields propa  gate through the waveguide. Therefore, the fields will propagate in the waveguide only when
is non-zero imaginary quantity, which corresponds to β > β c or f > ( β c 2π με ) . It follows from (10.79) that for propagation, the frequency of waves must satisfy f > f c where

fc =

1 2 με

(m a )2 + (n b)2 = υ (m a )2 + (n b)2 2

(10.80)

υ = 1 με is the phase velocity of uniform plane waves in an unbounded medium of permittivity ε and permeability μ . The electromagnetic waves with frequencies greater than f c  will propagate through the waveguide, while those of frequencies less than f c will evanescence. The frequency f c is referred to as the cut-off frequency, at which the propagation constant is zero. For the TE mn rectangular waveguide, the cut-off frequency from (10.80) can be expressed as f cTE mn =

1 2 ( m a ) 2 + ( n b) 2 με

(10.81)

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The corresponding cut-off wavelength can be obtained using λc = υ f cTE mn , as

λc =

2

(10.82)

(m a ) + (n b) 2

2

10.3.2.2. Phase and Propagation Constants The phase constant for waves in a rectangular waveguide with TE mn mode fields can be obtained using (10.79) and the relation
= jβ
as

βg = β 2 − (mπ a )2 − (nπ b )2

(10.83)

Using (10.81), (10.83) can be written in terms of the cut-off frequency as

β g = β 1 − ( f cTE

f

mn

)

2

(10.84)

It is clear from (10.84) that, when the operating frequency is less than the cut-off frequency, the phase constant becomes imaginary and the waves in the waveguide attenuate rapidly and evanescence. The propagation constant in (10.79) can be expressed also in terms of the cut-off frequency in the form
= 2π (με )2

2

1

f cTE mn − f 2

(10.85)

10.3.2.3. Wavelength Using (10.83), the wavelength for TE mn mode fields is λg =

2π

βg

= =

2π ω με − (mπ a ) − (nπ b ) 2

2

λ

2

(10.86)

1 − ( f cTE mn f )

2

where λ = 2π β is the wavelength of uniform plane waves in an unbounded medium of permittivity ε and permeability μ .

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10.3.2.4. Wave Impedance Using (10.43) along with (10.83) and (10.84) the wave impedance of TE mn mode fields in a rectangular waveguide can be expressed in the form

ωμ

Z TE mn =

ω2με− (mπ a ) − (nπ b ) 2

2

=

η

(

1 − f


TE mn

f

(10.87)

)

2

where η = μ ε is the intrinsic impedance of the medium of the waveguide 10.3.2.5. Phase Velocity Because of reflections of the electromagnetic waves from the walls of the waveguide, they can be considered as a group of uniform plane waves in the waveguide propagate with phase constant β g . The velocity of any of these uniform plane waves in the medium of the waveguide is υ = 1 με . The phase velocity is the velocity of the constant phase fronts of the waves in the guide. The phase velocity of the TE mode fields in a rectangular waveguide can be obtained as

υp =

ω

βg

=

ω

β 2 − (mπ a )2 − (nπ b )2

=

(

1− f

υ TE nm c

f

)

2

(10.88)

It is clear from (10.88) that the phase velocity of waves propagating in the waveguide always satisfies υ p > υ . If the medium in the waveguide is free space, then υ = υ0 , where υ0 is the speed of light in free space. Note that υ p > υ0 does not mean that the speed of the electromagnetic energy in the waveguide is greater than the speed of light, because υ p is the velocity of the constant phase fronts of the wave, and the electromagnetic energy in the waveguide does not propagate at this speed. 10.3.2.6. Group Velocity The reflection of the electromagnetic waves from the walls of the waveguide makes the transmitted energy in the waveguide travels as a group of waves. It has

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been shown that in Chapter 8, the group velocity is given by d ω dβ . Therefore, differentiating both sides of (10.83) with respect to ω , yields

dβg ωμε = 2 2 dω ω2με − (mπ a ) − (nπ b )

(10.89)

It follows from (10.81) and (10.89) that the group velocity of TE mode fields in a rectangular waveguide can be expressed as

υ
=

(

1 2 2 ω2με− (mπ a ) − (nπ b ) = υ 1 − f cTE nm f ωμε

)

2

(10.90)

Equation (10.90) shows that the group velocity in the waveguide is less than the uniform plane wave velocity υ . It can be shown that from (10.88) and (10.90) the phase and group velocities are always satisfying

υ = υυ


(10.91)

10.3.2.7. Poynting Vector and Power Flow Instantaneous Poynting of TE mn mode electromagnetic fields is given by (10.31), and can be written for a rectangular waveguide as

1 1 T = E g × H g∗ = (a xE xg + a yE yg )× (a x H x∗g + a y H y∗g + a z H z∗g ) 2 2 1 = (a zE xg H y∗g − a yE xg H z∗g − a zE yg H x∗g + a xE yg H z∗g ) (10.92) 2 1 1 = a z (E xg H y∗g − E yg H x∗g ) + H z∗g (a xE yg − a yE xg ) 2 2 Using E x
= ZTE mn H y
and E y
= − Z TE mn H x
, (10.92) can be written in the form

⎛ E x
2 + E y
⎜ T = az ⎜ 2Z TE mn ⎝

2

⎞ 1 ⎟ + H ∗ (a E − a E ) y x
⎟ 2 z
x y
⎠

(10.93)

Note that, the component of Poynting vector in the direction of propagation is always real. Using the time average Poynting vector for time-harmonic fields

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677

given by (10.32), it can be shown that the time average Poynting vector for TE mn mode fields in a rectangular waveguide, is Exs 2 + E ys 1 ∗ Ta = Re E s × H s = 2 2Z TE mn

[

]

2

(10.94)

Consequently, the average power through the waveguide becomes

PaTE mn = ∫∫ Ta ⋅ dS = S

1 2Z TE mn

∫∫ (E

2 xs

+ E ys

2

)a ⋅ dS

(10.95)

z

S

The unit vector of the cross section area is always in the direction of propagation, or dS = a z dxdy. Substituting dS , Exs and E ys from (10.77a) and (10.77b) respectively into (10.95), yields TE mn a

P

⎛ ωμ nπ ⎜ = H mn 2Z TE mn ⎜⎝ β c2 b 1

⎛ ωμ mπ ⎜ + H mn 2Z TE mn ⎝⎜ β c2 a 1

⎞ ⎟⎟ ⎠

2b a

⎞ ⎟⎟ ⎠

∫∫ cos (mπ x a )sin (nπ y b) dxdy 2

2

0 0

(10.96)

2b a

∫∫ sin (mπ x a )cos (nπ y b) dxdy 2

2

0 0

Note that a

a

0

0

2 2 ∫ cos (mπ x a ) dx = ∫ sin (mπ x a ) dx = a 2 , m ≠ 0 b

(10.97a)

b

∫ cos (nπ y b) dy = ∫ sin (nπ y b) dy = b 2 , 2

2

0

n≠0

(10.97b)

0 b

a

∫ cos (mπ x a ) dx = a , ∫ cos (nπ y b) dx = b , 2

2

0

m = 0 , n = 0 (10.97c)

0

a

b

∫ sin (mπ x a ) dx =∫ sin (nπ y b) dy = 0, 2

0

2

0

m =0, n = 0

(10.97d)

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Using (10.97a) – (10.97d), the integrals in (10.96) for all combination of m and n can be evaluated. Replacing β c and Z TE mn from (10.84) and (10.87) respectively, with some algebraic manipulations, the average power through the waveguide that supports TE mn mode fields can be expressed in terms of the waveguide dimensions, the cut-off frequency, and the operating frequency, as

⎧0, ⎪ H 2 2 0n ⎪ 1 − f cTE 0 n f , 2 TE 0 n ⎪ fc f 2 abη ⎪ 2 = ⎨ H m0 1 − f cTE m 0 f , 4 ⎪ TE m 0 2 f f ⎪ c 2 ⎪ 1 H mn 2 1 − f cTE mn f , ⎪ 2 TE mn 2 f ⎩ fc

(

PaTE mn

)

(

)

(

)

m = 0, n = 0

(

)

m = 0, n ≠ 0

(

)

m ≠ 0, n = 0

(

)

(10.98)

m ≠ 0, n ≠ 0

Example 10.4 Write expressions for the instantaneous and the spatial electromagnetic fields in an air-filled rectangular waveguide supporting the TE10 mode. Hence, find the average power flowing in the waveguide if the operating frequency is higher than the cut-off frequency by 25%. Solution For the TE10 , m = 1 and n = 0 , which gives βc = π a from (10.75). Substitute these values into (10.78a) – (10.78b), we obtain the instantaneous fields components as

E x
= E z
= H y
= 0 , E yg = (a ωμ π ) H10 sin (π x a )sin (ω t − β g z ) E xg = −(aβ g π )H10 sin (π x a )sin (ω t − β g z ) H z
= H10 cos(π x a )cos(ω t − β
z ) The spatial field components from (10.77a) – (10.77b) are

E xs = E zs = H ys = 0

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E ys = − j (a ωμ π )H10 sin (π x a ) e

− jβ g z

H xs = + j (aβg π )H10 sin (π x a ) e

H zs = H10 cos (π x a )e

679

− jβ g z

− jβ
z

Using (10.95), the average power can be obtained as 2

TE 10 c

P

b a ⎛ ωμ π ⎞ a 3b 2 1 2 2 ⎜ ⎟ ( ) = = H x a dxdy π β g H10 Z TE10 sin 10 ⎟ 2 2 ∫∫ ⎜ 2Z TE10 0 0 ⎝ βc a 4π ⎠ 3 2 ab 2 2 2 = μ f H10 1 − f cTE10 f

=

η0 abη0 4

(

(f

H10 TE 10 c

2

f

)

2

)

(

1 − f cTE10 f

)

2

Since f = 1.25 f cTE10 2

⇒ P

TE 10 c

ab × 120π H10 2 = 1 − (1 1.25) 2 4 (1 1.25) 2 = 88.4 ab H10 W

Alternatively, substituting m = 1 and n = 0  in (10.98), the same result for PcTE10 above, can be obtained. Example 10.5 In an air-filled rectangular waveguide with a = 2.286 cm and b = 1.016 cm, the ycomponent of the TE mode is given by

(

Eyg = 5sin (2π x a ) sin 2π × 2 × 1010 t − βg z Determine (a) (b) (c) (d)

The mode of operation The cut-off frequency The phase constant The wave impedance

)

Vm

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Solution (a) The operating frequency is f = 20 GHz , a = 2.286 cm and b = 1.016 cm. From the given electric field component E yg we have

2π x a = mπ x a ⇒ m = 2 0 = nπ y b ⇒ n = 0 Hence, the mode of operation is TE 20 mode. (b) The cut-off frequency can be computed using (10.81), as

f

TE 20 c

1 = 2 μ 0ε 0

(m a ) + (n b) 2

2

3 × 108 2 = 13.12 GHz = 2 2.286 × 10 − 2


(c) The phase constant can be computed using (10.84) as

βg

(

)

2

= 2π f μ 0ε 0 1 − f cTE 20 f 2π × 2 × 1010 2 = 1 − (13.12 20 ) = 316.22 rad/m 8 3 × 10

(d) The wave impedance from (10.87) is Z TE mn =

(

μ0 ε0

1 − f cTE nm f

)

2

=

120π 1 − (13.12 20 )

2

= 499.8 Ω

Example 10.6 If the waveguide in Example 10.5 is filled with polystyrene ( ε r = 2.56 ) instead of air, find the cut-off frequency in this case, then find the average power transmitted through the waveguide relative to that of an air-filled case. Solution From the results of Example 10.5, we have f = 20 GHz , a = 2.286 cm, b = 1.016 cm, m = 2 and n = 0 , then the cut-off frequency when ε r = 2.56 becomes

f cTE 20 =

1 2 μ 0ε 0ε r

(m a )2 + (n b)2

=

3 × 108 2 = 8.2 GHz −2 2 2.56 2.286 × 10


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681

TE

20 20 Let the average power for polystyrene case is Pa ( poly )  Pa ( poly )  can be obtained by

substituting f cTE 20 = 8.2 GHz  f = 20 GHz , η = η0 (10.98), as 2

TE 20 a ( poly )

P

abη H 20 = 4 f cTE 20 f

(

)

2

(

1− f

TE 20 c

f

)

2

ε r , m = 2 and n = 0 , in 2

abη0 H 20 2 = 1 − (8.2 20 ) 2 4 ε r (8.2 20 ) TE

Let the average power for the air-filled waveguide be Pa (air20)  Substituting

f cTE 20 = 13.12 GHz  f = 20 GHz , η = η0 , m = 2 and n = 0 , in (10.98), then 2

TE 20 a ( air )

P

2

H 20 H 20 abη0 abη0 2 2 = 1 − (13.12 20 ) = 1 − (13.12 20 ) 2 2 4 (13.12 20 ) 4 (13.12 20 )

The average power of the polystyrene case to that of air case is 20 PaTE ( poly ) 20 PaTE ( air )

1 ⎛ 13.12 ⎞ = ⎜ ⎟ 2.56 ⎝ 8.2 ⎠

2

1 − (8.2 20 ) = 1.93 2 1 − (13.12 20 ) 2

Example 10.7 A tunnel is modeled as an air-filled metallic rectangular waveguide with dimensions a = 8 m and b = 16 m. For the mode TE01, find the cut-off frequency, and the cut-off wavelength, when the operating frequency is 18.75 MHz. Determine whether the tunnel will pass: (a) A 1.5 MHz AM broadcast signal. (b) A 90 MHz FM broadcast signal. Solution Given that a = 8 m and b = 6 m, m = 0 and n = 1 , then the cut-off frequency can be computed using (10.81) as

f cTE 01 =

1 2 μ 0ε 0

(m a )2 + (n b)2

=

1 3 × 108 = 9.375 MHz = 2×8 2b μ 0ε 0

682

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Since m = 0 and n = 1 , the cut-off wavelength is λc =

2

(m a ) + (n b)2 2

= 2b = 6 m

(a) Since the AM signal frequency is lower than the cut-off frequency, the AM signal will not pass through the tunnel. (b) The FM signal will pass through the tunnel because its frequency is 90 MHz, which is higher than the cut-off frequency of the tunnel. 10.3.3. TM Mode Fields in the Rectangular Waveguide In this case, the rectangular waveguide shown in Fig. (10.11) supports TM mode electromagnetic fields. The starting point is to obtain Ez by solving (10.45). Once Ez is known, the other electromagnetic fields components can be determined using (10.48) – (10.50), (10.53) and (10.54). Equation (10.45) in rectangular coordinate system can be written as

∂ 2 Ez ∂ 2 Ez + 2 + βc2 Ez = 0 2 ∂x ∂y

(10.99)

Following the same analysis as in the TE mode case, E z can be obtained by solving (10.99) as

Ez (x, y ) = [ A′ cos( k x x) + B′ sin( k x x)][C ′ cos(k y y ) + D′ sin( k y y )] (10.100) where A′ , B′ , C ′ , D′ , k x , and k y are constants. The components of Ht can be obtained using (10.48) and (10.54) as

axH x + a y H y

ωε ⎛ ∂ ∂ ⎞ ⎜a + a y ⎟⎟ × a z Ez 2 ⎜ x β c ⎝ ∂x ∂y ⎠ ωε ⎛ ∂E ∂E ⎞ = j 2 ⎜⎜ − a y z + a x z ⎟⎟ βc ⎝ ∂x ∂y ⎠ = j

(10.101)

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Equating the quantities in the x and y directions in both sides of (10.101) and substituting E z from (10.100), H t components are obtained as Hx

ωε ∂Ez β c2 ∂y ωε = jk y 2 [A′ cos(k x x ) + B′ sin (k x x )] − C ′ sin (k y y ) + D′ cos(k y y )

(10.102a)

ωε ∂Ez β c2 ∂x ωε = − jk x 2 [− A′ sin (k x x ) + B′ cos(k x x )] C ′ cos(k y y ) + D′ sin (k y y )

(10.102b)

= j

[

βc

Hy

]

=−j

[

βc

]

Equation (10.50) can be written as

a x Ex + a y E y = −

βg ωε

(a H y

x

− axH y )

(10.103)

Using (10.102a) and (10.102b) along with (10.103), the components of E t can be obtained as Ex

Ey

=

βg ωε

Hy

(10.104a)

β = jkx g2 [A′ sin (k x x ) − B′ cos(k x x )][C ′ cos(k y y ) + D′ sin (k y y )] βc ⎛ βg ⎞

⎟⎟ H x = −⎜⎜ ⎝ ωε ⎠ ⎛β ⎞ = jk y ⎜⎜ g2 ⎟⎟ [A′ cos(k x x ) + B′ sin (k x x )] C ′ sin (k y y ) − D′ cos(k y y ) ⎝ βc ⎠

[

]

(10.104b)

The boundary conditions are given in (10.69a) and (10.69b). Applying the boundary conditions Ex ( y = 0) = 0 and E y (x = 0) = 0 on (10.104a) and (10.104b), yields A′ = C ′ = 0

(10.105)

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Applying the boundary conditions Ex ( y = b ) = 0 and E y (x = a ) = 0 on the equations (10.104a) and (10.104b), the constants k x and k y can be determined as

k x = mπ a ,

m = 0,1, 2,

(10.106)

k y = nπ b ,

n = 0,1, 2,

(10.107)

which are the same results as for the TE mode case. This indicates that there are an infinite number of modes depending on the values of m and n . The mode mn will be denoted by TM mn . Substituting k x and k y into (10.61), yields

βc2 = (mπ a )2 + (nπ b)2

(10.108)

Substituting A′ = C ′ = 0 , k x and k y in (10.100), (10.102a), (10.102b), (10.104a), and (10.104b), and letting B ′D′ = E mn , then the components of E t and H t , in addition to Ez for the TM fields can be determined. Substituting E t , H t and Ez (10.5) and (10.6), and using (10.3) and (10.4), the spatial TM mn mode electromagnetic fields components in the rectangular waveguide can be obtained as

βg mπ − jβ z Emn cos(mπ x a )sin (nπ y b) e 2 βc a β nπ − jβ z E ys = − j g2 Emn sin (mπ x a )cos(nπ y b) e βc b

Exs = − j

g

g

E zs = Emn sin (mπ x a )sin (nπ y b) e H xs = j H ys = − j

− jβ
z

(10.109a) (10.109b) (10.109c)

ωε nπ − jβ z Emn sin (mπ x a )cos(nπ y b ) e g 2 βc b

(10.109d)

ωε mπ − jβ z Emn cos(mπ x a )sin (nπ y b ) e g 2 βc a

(10.109e)

H zs = 0

(10.109f)

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Phase velocity Emn is th peak value of E zs and its dimension is in (V m). Assume that E mn is real, the instantaneous TM fields corresponding to (10.109a) – (10.109f) can be obtained by substituting these fields components into (10.1) and (10.2), as

βg mπ E cos(mπ x a )sin (nπ y b) sin (ω t − βg z ) βc2 a mn β nπ E yg = g2 E sin (mπ x a )cos(nπ y b) sin (ω t − βg z ) βc b mn E z
= E mn sin (mπ x a )sin (nπ y b ) cos(ω t − β
z ) ωε nπ H xg = − 2 E sin (mπ x a )cos(nπ y b ) sin (ω t − β g z ) β c b mn ωε mπ H yg = 2 E cos(mπ x a )sin (nπ y b ) sin (ω t − β g z ) β c a mn E xg =

H z
= 0

(10.110a) (10.110b) (10.110c) (10.110d) (10110e) (10.110f)

10.3.4. Characteristics of the TM Mode Fields in the Waveguide 10.3.4.1.

Cut-off Frequency

Following the same analysis as for the TE case, the cut-off frequency of a rectangular waveguide that supports TM mn fields can be obtained as

f cTM mn =

1 2 με

(m a )2 + (n b)2

=

υ 2

(m a )2 + (n b)2

(10.111)

and the cut-off wavelength for the waveguide will be

λc = 2

(m a )2 + (n b)2

(10.112)

10.3.4.2. Phase and Propagation Constants From (10.44) and (10.111), the phase constant can be written as

β g = β 2 − (mπ a )2 − (nπ b )2 = β 1 − ( f cTM

mn

f)

2

(10.113)

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The propagation constant is obtained from the relation
= jβ
as

(mπ a )2 + (nπ b)2 − β 2


=

= 2π (με )2 f cTM mn − f 2 2

1

(10.114)

10.3.4.3. Wavelength The wavelength for TM mn fields can be obtained as

λg =

2π TM mn g

β

=

2π

ω με − (mπ a ) − (nπ b ) 2

2

2

=

(

1− f

λ TM mn c

f

)

2

(10.115)

10.3.4.4. Wave Impedance The wave impedance of the rectangular waveguide that supports TM mn fields can be determined from (10.52) and (10.113), as 2 ω2με − (mπ a ) − (nπ b ) = = η 1 − ( fcTM mn f ) ωε 2

Z TM mn

2

(10.116)

Note that the wave impedance of the TM fields is different from that of the TE fields given by (10.87). By multiplying both sides of (10.87) by corresponding sides of (10.116), it can be shown that for the same waveguide, the wave impedances Z TE mn and Z TM mn satisfy the relation

η = Z TE Z TM mn

(10.117)

mn

where η is the intrinsic impedance of a uniform plane wave when propagates in the medium of the waveguide. 10.3.4.5. Phase Velocity and Group Velocity The phase velocity υ p and group velocity υ
are similar in form to that for the TE mode case and can be written respectively as

υp =

ω

β 2 − (mπ a )2 − (nπ b )2

=

(

1− f

υ TM mn c

f

)

2

(10.118)

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υg =

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2 1 β 2 − (mπ a )2 − (nπ b)2 = υ 1 − ( f cTM mn f ) ωμε

687

(10.119)

10.3.4.6. Poynting Vector and Power Flow Following the same analysis as in the TE mn fields case, it can be shown from (10.31) that the instantaneous Poynting vector for TM mn mode fields in a rectangular waveguide is given by

⎛ E 2+E 2 ⎞ 1 1 xg yg ⎟ ∗ − E zg (a x H y∗g − a y H x∗g ) T = Eg × H g = a z ⎜ ⎟ ⎜ 2 2Z TM mn 2 ⎝ ⎠

(10.120)

As in the case of TE wave, Poynting vector in the direction of propagation is always real. The average power transmitted through the rectangular waveguide that supports TM mn waves can be obtained using (10.33) as follows

PaTM mn =

1 2Z TM mn

∫ ∫ (E b a

2 xs

+ E ys

0 0

2

) dxdy Z 2

TM mn

∫ ∫ (H b a

2 xs

0 0

+ H ys

2

) dxdy

(10.121)

Substituting Exs and E ys from (10.109a) and (10.109b) respectively into (10.121), yields 2

E ⎛ β mπ ⎞ b a ⎟⎟ ∫∫ cos2 (mπ x a )sin 2 (nπ y b ) dxdy = mn ⎜⎜ g2 2Z TM mn ⎝ β c a ⎠ 0 0 (10.122) 2 2 Emn ⎛ β g nπ ⎞ b a 2 ⎟⎟ ∫∫ sin (mπ x a )cos2 (nπ y b ) dxdy ⎜⎜ 2 + 2Z TM mn ⎝ β c b ⎠ 0 0 2

TM mn a

P

Following the same analysis as in the TE mode case, the average power through the waveguide that supports TM mn can be obtained as TM mn a

P

abπ 2 ⎛ β g ⎜ = 8Z TM mn ⎜⎝ β c

2

2

Emn ⎞ 2 ab 2 TM 21 (10.123) ⎟⎟ Emn = ( ) 1 − f f c 2 8η ( f cTM 21 f ) ⎠

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Example 10.8 Determine the cut-off frequencies for the modes TM11 , TM12 , TM21 , and TM21 in a rectangular waveguide. Assume that the cross section dimensions of the waveguide are a and b , such that a = b . Find the normalized cut-off frequencies of the modes with respect to the lowest cut-off frequency. Solution Let υ = 1 με , since a = b , then from (10.111), the cut-off frequency for any mode TM mn is

f cTM mn =

υ m2 + n2 2a

The cut-off frequency for the mode TM11 is

f cTM11 =

υ υ 1+1 = 2 2a

Similarly, the respective cut-off frequencies for the TM12 , TM21 , and TM21 modes are

f cTM12 =

υ υ υ 5 , f cTM = 5 , f cTM = 2 2a 2a a 21

22

It is clear that the lowest cut-off frequency is f cTM11 . Let the normalized cut-off frequency denoted by f cNTM mn  Normalizing with respect to the lowest cut-off frequency gives

f cNTM11 = 1, fcNTM12 = 5 2 , fcNTM 21 = 5 2 , f cNTM 22 = 2 Example 10.9 An air-filled rectangular waveguide of cross section dimensions a = 2. 286 cm, and b = 1. 016 cm , operates at 25 GHz. Find all TM modes that may exist in the waveguide.

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Solution Since a = 2.286 cm and b = 1.016 cm , and the medium in the waveguide is air, then from (10.111), the cut-off frequency is =

f cTM mn

1 2 μ 0ε 0

(m a )2 + (n b )2 2

= 100 ×

3 × 108 ⎛ m ⎞ ⎛ n ⎞ ⎟ +⎜ ⎜ ⎟ 2 ⎝ 2.286 ⎠ ⎝ 1.016 ⎠

2

The modes in the waveguide must satisfy 25 GHz > f cTM mn . Therefore, 2

2 × 25 × 109 × 10− 2 ⎛ m ⎞ ⎛ n ⎞ > ⎜ ⎟ +⎜ ⎟ 8 3 × 10 ⎝ 2.286 ⎠ ⎝ 1.016 ⎠ 2

5 ⎛ m ⎞ ⎛ n ⎞ ⇒ > ⎜ ⎟ +⎜ ⎟ 3 ⎝ 2.286 ⎠ ⎝ 1.016 ⎠

2

2

The only sets (n, m) that satisfy the above inequality are (m = 1, n = 1) , (m = 1, n = 2) , and (m = 1, n = 3). Consequently, the TM modes that may exist in the waveguide are TM11 , TM12 , TM13 . Example 10.10 An air-filled rectangular waveguide operates at 30 GHz. If the cut-off frequency of the mode TM21 is 18 GHz, for this mode find (a) The wavelength. (b) The phase constant. (c) The phase velocity. (d) The wave impedance. Solution Given that f = 30 GHz , f cTM 21 = 18 GHz , and the waveguide is filled with air.

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(a) The wavelength can be obtained from (10.115) as

λg

=

(

λ

)

2

=

1 1 f μ 0ε 0 1 − f TM 21 f c

1 − f cTM 21 f 3 × 108 1 = = 1.25 cm 10 3 × 10 1 − (18 30 )2

(

)

2

(b) The phase constant is

βg

= β 1 − ( f cTM 21 f ) = β 1 − (18 30 ) = 2π f μ 0ε 0 × 0.8 2π × 3 × 1010 = × 0.8 = 160π rad m 3 × 108 2

2

(c) The phase velocity is

υ

υ 2 2 1 − (18 30 ) 1 − ( f cTM f ) 1 1 3 × 108 = × = = 3.75 × 108 m s μ 0ε 0 0.8 0.8

υp =

=

21

(d) The wave impedance is

Z TM 21

= η 1 − ( f cTM 21 f ) = η 1 − (18 30 ) = 0.8 × μ 0 ε 0 = 0.8 × 277 = 301.6 Ω 2

2

10.4. DOMINANT MODE The dominant mode is the mode of the lowest cut-off frequency. The cross section dimensions of the waveguide are a and b as shown in Fig. (10.11), such that a > b . Lowest cut-off frequencies occur at the lowest modes TE 01 , TE10 , TM 01 , and TM10 . The modes TM10 and TM 01 does not exist because when n = 0 or m = 0 , the TM mode electromagnetic fields vanish as evident from the set of equations (10.109a) – (10.109f). The cut-off frequency for the TE10 mode from (10.81) is f
TE10 = υ 2a , while the cut-off frequency for the mode TE 01 is

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f cTE 01 = υ 2b . Since a > b , then f
TE 01 > f
TE10 . Thus, the lowest possible cut-off frequency of the rectangular waveguides is f
TE10 = υ 2a . It follows that the dominant mode that can propagate through the metallic rectangular waveguides is the TE10 mode. The instantaneous electromagnetic fields components of the dominant mode in a rectangular waveguide of Fig. (10.11) can be obtained by putting m = 1 and n = 0 in (10.78a) – (10.78f) as

E xg = E zg = H yg = 0 E yg = (a ωμ π ) H10 sin (π x a) sin(ω t − β g z ) H xg = − (aβ g π ) H10 sin (π x a) sin(ω t − β g z ) H zg = H10 cos (π x a) cos(ω t − β g z )

(10.124a) (10.124b) (10.124c) (10.124d)

Note that the dominant mode fields characterized by only one electric field component in the y direction, and one transverse magnetic component in the x direction, in addition to the magnetic field component in the direction of propagation. Consider the fields in (10.78a) – (10.78f) at the instant t = 0 sec, on the orthogonal planes A, B, and C shown in Fig. (10.12). Where Plane A: z = λ
4 , 0 ≤ x ≤ a ,

0 ≤ y ≤ b , Plane B: x = a 2 , 0 ≤ y ≤ b , − λ
4 ≤ z ≤ λ
4 , and Plane C: y = b , 0 ≤ x ≤ a , − λ
4 ≤ z ≤ λ
4 . The electric and magnetic fields at the instant t = 0 sec are: Plane A:

E yg = − (a π ) ωμ H10 sin (π x a)

(10.125a)

H xg = (a π ) β g H10 sin (π x a)

(10.125b)

H zg = 0

(10.125c)

E yg = −(a π ) ωμ H10 sin (2π z λg )

(10.126a)

H xg = (a π ) β g H10 sin (2π z λg )

(10.126b)

H zg = 0

(10.126c)

Plane B:

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Plane C:

E yg = − (a π ) ωμ H10 sin (π x a) sin (2π z λg )

(10.127a)

H xg = (a π ) β g H10 sin (π x a) sin (2π z λg )

(10.127b)

H zg = H10 cos (π x a) cos (2π z λg )

(10.127c)

Using the above equations, the field lines on the planes A, B, and C can be constructed as shown in Fig. (10.12). The characteristics of the dominant mode are obtained by letting m = 1 and n = 0 in the all relevant TE mn mode expressions and the results are summarized in Table 10.1. TE10 Fields E y

y

H

Plane A (z = λ10/4)

b

z

x 0

a

a/2

Plane B (x = a/2)

b

λg /4

-λg /4 - λg /4

λg /4 z y

+

+

C

+ +

+ + +

b

+

A

B x a/2

z

a Plane C (y = b)

a x

Fig. (10.12). TE 10 mode electric and magnetic fields in a rectangular waveguide.

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Table 10.2. The characteristics of the dominant mode in a rectangular waveguide. Parameter

Mathematical Expression

Cut-off frequency

1

f

TE10 c

Parameter

2a με

Cut-off Wavelength

2a

λ


λ

Wavelength λ


Average Power

abη H 10

P

Phase Velocity υ


jβ 1 − f cTE10 f

)

Group Velocity υ


(

f

1 − f cTE

10

2

2

)

(

υ

)

2

1 − f cTE10 f 2

(

1 − f cTE

(f

4

)

10

(




η

TE10 a

β 1 − f cTE

Propagation Constants

Wave Impedance Z TE10

f

(

Phase Constants β


Mathematical Expression

(

TE10 c

f

TE10

1− f c

(

10

υ 1 − f cTE

10

f

f

)

)

f

)

2

2

2

)

2

Note: β = 2π f με , λ = 2π β , η = μ ε , and υ = 1 με .

2

Example 10.11 An AM signal at 12 GHz propagates in a rectangular waveguide of square cross section with side 2 cm. If the mode of operation is the TM11 mode, find the group velocity of the AM signal. Solution The signal frequency is f = 12 GHz , and given that a = b = 2 cm . Then the group velocity from (10.119) can be obtained as

υ
=

(

1 1 − f cTM11 f μ 0ε 0

)

2

(

= 3 × 108 1 − f cTM11 12 × 109

)

2

The cut-off frequency f cTM11 is

f cTM11 =

1 1 2 μ 0ε 0

(m a )2 + (n a )2

=

1 3 × 108 = 10.6 GHz = 2a 2 × 10 − 2 2

1 μ 0ε 0

⇒ υ
= 3 × 108 1 − (10.6 12) = 1.406 × 108 m s 2

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Example 10.12 Write expressions for the electromagnetic fields in a rectangular waveguide at the instant t = 0 sec on the plane z = λ
4 , 0 ≤ x ≤ a , 0 ≤ y ≤ b for the cases: (a) TE 20 mode. (b) TM31 mode. Hence, Sketch the electric and magnetic field lines for both modes. Solution For the TE 20 mode, m = 2 and n = 0 . From (10.108), β c = 2π a . Hence, at t = 0 sec, using (10.77a) – (10.77f), the fields on the plane z = λ
4 , are

E x
= E z
= H y
= H z
= 0

E y
= −(2a ωμ π ) H 20 sin (2π x a ) H x
= (2aβ
π ) H 20 sin (2π x a ) Similarly, for the TM31 mode, m = 3 and n = 1 . Letting E0 = 3πβ
E31 aβc2  the field expressions on the planes z = λ
4 , 0 ≤ x ≤ a , 0 ≤ y ≤ b can be written as

E z
= H z
= 0 E x
= E0 cos(3π x a )sin (π y b) E y
= (3aE0 b)sin (3π x a )cos(π y b) H x
= −(aE0 3bZ TM 31 )sin (3π x a )cos(π y b) H y
= (E0 ZTM 31 )cos(3π x a )sin (π y b) The sketches for the modes TE20 and TM31 are shown in Fig. (10.13). y

E H

TE20

y

b

b

b/2

b/2

TM31

x 0

a/4

a/2

3a/4

a

x 0

a/3

2a/3

a

Fig. (10.13). Electric and magnetic fields lines on the plane z = λg 4 , 0 ≤ x ≤ a , 0 ≤ y ≤ b for the TE20 and TM31 modes in the rectangular waveguide of Example 10.12.

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10.5. POWER LOSS AND ATTENUATION IN WAVEGUIDES The derivation of electromagnetic field expressions in the waveguides, assumes ideal waveguides of their walls are made of conducting materials with infinite conductivity (σ c = ∞) and filled with lossless media (σ d = 0). The practical waveguides are made of conducting materials with finite conductivities (σ c ≠ ∞) and field with media of small loss (σ d ≠ 0). Although the waveguides transmit the electromagnetic power efficiently, some of the transmitted power may be lost in the waveguide due to the finite conductivity of the walls of the waveguide, and the dielectric loss in the medium of the waveguide. The power loss in the waveguide is small and has no effect on the electromagnetic fields expressions presented in this previous sections. The power loss in the waveguide increases with waveguide length and can be expressed in terms of the overall all attenuation constant α
, where

α  = α
+ α

(10.128)

α
is the attenuation constant due to the finite conductivity of the walls of the waveguide and α
is the attenuation constant due to the loss of the dielectric in the waveguide. 10.5.1. Attenuation Constant α
Assuming that the power loss is small and its effect is restricted only to the propagation constant
= jβ g  Then the propagation constant in the expressions (10.77a) – (10.77f), (10.78a) – (10.78f), (10.109a) – (10.109f), and (10.110a) – (10.110f), is to be modified to include the overall attenuation constant α
as follows


′ = α g + jβ g

(10.129)

The attenuation due to the dielectric loss α
satisfies α d + jβ g′ =

(mπ a )2 + (nπ b )2 − ω2με d

(10.130)

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where ε d is the complex permittivity of the medium in the waveguide. The complex permittivity can be written as

εd = ε− j

⎛ σ ⎞ = ε⎜⎜1 − j d ⎟⎟ ω ωε ⎠ ⎝

σd

(10.131)

where σ
ωε is the loss tangent of the medium. Substituting ε d from (10.131) into (10.130), and squaring both sides of (10.131), then equating the real and imaginary parts in both sides of the result, yields

αd2 − βg′2 = (mπ a ) + (nπ b ) − ω2με = −βg2 2 α d β g′ = ωμ σ d 2

2

(10.132) (10.133)

Since the attenuation is small in waveguides, αd2 > α ). Therefore, (10.136) reduces to

Pa′ = Pa e −2 α
z

(10.137)

The power loss per unit length due to the finite conductivity of the waveguide walls Pc can be expressed as the rate of the power decrease with the distance. For waves traveling along the z-axis as shown in Fig. (10.14), Pc can be written as Pc = −

ΔPa′ dP′ = − a = α c Pa e − 2 α c z = 2 α c Pa′ Δz Δz → 0 dz

(10.138)

Consequently, the attenuation constant α
is found as

α
=

Pc 2 Pa′

(10.139)

Assuming α
does not depend on z, then at z = 0, Pa′ = Pa , where Pa is the total input power to be transmitted through the waveguide. Thus, (10.139) becomes

α
=

Pc 2 Pa

(10.140)

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Referring to Fig. (10.14), the power dissipated Pd along the waveguide at any point z can be determined by subtracting the power Pa′ at a distance z from the power Pa at the input of the waveguide. Therefore, using (10.137), the dissipated power along the waveguide can be expressed as

(

)

(

)

Pd = Pa − Pa′ = Pa 1 − e −2(α + α
)z = Pa′ e 2(α + α
)z − 1

(10.141)

The average power loss per unit length Pc can be computed from the induced currents due to the tangential magnetic fields on the internal surface of the waveguide walls. Assuming the tangential magnetic field is H tan as shown in Fig. (10.14), the associated electric field Etan is also tangential to the internal walls of the waveguide and perpendicular to the magnetic field. The mathematical relation between H tan and Etan can be written as

Etan = Zc H tan

(10.142)

where Z c is the intrinsic impedance of the waveguide wall and can be written from (8.76) as

Zc =

1

σc δ

(1 + j )

(10.143)

P′a Pc = -Δ P′a /Δz Δz → 0

Pa P′a

ΔP′a an

z

Walls of the Guide

Δz Fig. (10.14). Power loss in the walls of the waveguide

z = 0

Etan

Htan

z

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where δ = 1 π f μ σ c is the skin depth. The average power of the tangential electromagnetic fields can be obtained as

⎤ 1 ⎡ Pc = Re ⎢∫∫ Etan × H∗tan ⋅ dS⎥ 2 ⎣S ⎦

(10.144)

Note that the average power of the tangential fields and S are both perpendicular to the walls of the waveguide. Considering that the unit vector in the direction of normal to the walls is a n , then dS = andl , where dl is the unit length along the circumference of the waveguide. Therefore, using (10.142) – (10.144), the average power loss can be written as

Pc =

1 2 Re[Z c ]∫ H tan dl 2 l

1 = 2 δσ c

∫H

2 tan

(10.145)

dl

l

The current density on the internal surface of the waveguide walls is given by

K = an × H = H tan

(10.146)

From (10.145) and (10.146), the average power loss expression becomes

Pc =

1 2 δσ c

∫K

2

dl

(10.147)

l

Substituting P
from (10.147) into (10.140), the attenuation constant α
for a general metallic waveguide can be expressed as

α
=

RS 2 Pa

∫K

2

dl

(10.148)

l

where RS = 1 (δσ c ) is the surface impedance per unit length of the waveguide walls. The power Pa is given by (10.98) or (10.123) for TE and TM mode fields respectively.

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10.6. POWER LOSS IN THE RECTANGULAR WAVEGUIDES For the rectangular waveguide the unit vector normal a n is as follows: Table 10.3. Unit vectors normal to the waveguide walls. Wall x = 0, 0 ≤ y ≤ b x = a, 0 ≤ y ≤ b y = 0, 0 ≤ x ≤ a y = b, 0 ≤ x ≤ a

an

an = +a x an = −a x a n = +a y an = −a y

Using (10.146) and Table 10.3, the spatial surface current density on the walls of the waveguide can be obtained as follows: Current density on the wall y = 0 , 0 ≤ x ≤ a

K x =0 = +a x × (a x H xs + a y H ys + a z H zs ) = +a z H y s − a y H zs

(10.149a)

Current density on the wall y = b , 0 ≤ x ≤ a

K x =a = −a x × (a x H xs + a y H ys + a z H zs ) = −a z H ys + a y H zs

(10.149b)

Current density on the wall x = 0 , 0 ≤ y ≤ b

K y =0 = +a y × (a x H xs + a y H ys + a z H zs ) = −a z H xs + a x H zs

(10.149c)

Current density on the wall x = a , 0 ≤ y ≤ b

K y =b = −a y × (a x H xs + a y H ys + a z H zs ) = +a z H x s − a x H zs

(10.149d)

Note that K y =0 = −K y =b and K x = 0 = −K x = a . Since only the magnitude of the current density is needed in (10.148), it is necessary to compute the current density in all four walls, instead, the current density on walls x = 0 and y = 0 can be computed and multiplied by two. Substituting the current densities from (10.149a) – (10.149d) into (10.147), putting in mind that dl = dx for walls y = 0

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and y = b , and dl = dy for walls x = 0 and x = a , the average power loss in the rectangular waveguide can be obtained as a b ⎤ 2 2 1 ⎡ dx + ∫ 2 K dy ⎥ Pc = ⎢∫ 2 K 2δσ c ⎢⎣ 0 y=0 x=0 ⎥⎦ 0 a b 2 2 1 ⎡ 2 dx + ∫ H ys 2 + H zs = ⎢ ∫ H xs + H zs δσ c ⎢⎣ 0 y=0 0

(

)

(

)

⎤ dy ⎥ x=0 ⎥⎦

(10.150)

Equation (10.150) is general for all metallic rectangular waveguides.

10.6.1. Power Loss in the Waveguides Supporting TE Mode Fields Substituting H xs , H ys , and H zs from (10.109d) – (10.109f) into (10.150), the power loss per unit length of a waveguide that supporting TE modes becomes TE mn c

P

⎡⎛ β ⎢⎜⎜
2 = ∫ δ σ c 0 ⎢⎝ βc ⎣ b 1 ⎡⎛ β
⎢⎜ + δ σ c ∫0 ⎢⎜⎝ βc2 ⎣ 1

a

2 ⎤ ⎞ ⎟⎟ sin 2 (mπ x a ) + (Fmn )2 cos2 (mπ x a )⎥ dx ⎥⎦ ⎠ 2 ⎤ ⎞ ⎟⎟ sin 2 (nπ y b ) + (Fmn )2 cos2 (nπ y b )⎥ dy ⎥⎦ ⎠

mπ H mn a nπ H mn b

(10.151)

Using the formulas (10.97a) – (10.97d), integrals in (10.151) can be evaluated for all possible combinations of m and n . Simplifying the results, the power loss per unit length in a rectangular waveguide supporting TE mode can be expressed as

PcTE mn

⎧0, ⎪H 2 ⎪ 0 n b ( f f cTE 0 n ) 2 + 2a , ⎪ 2 δσ c 2 ⎪ = ⎨ H m0 a ( f f cTE m 0 ) 2 + 2b , ⎪ 2 δσ c ⎪ 2 2 2 ⎪ H mn ⎧ ( f f TE mn ) 2 − 1 ⎡ a (m a ) + b (n b ) ⎤ + a + b ⎫, ⎬ c ⎢ (m a ) 2 + (n b ) 2 ⎥ ⎪ 2 δ σ c ⎨⎩ ⎦ ⎣ ⎭ ⎩

m = 0, n = 0

[

]

m = 0, n ≠ 0

[

]

m ≠ 0, n = 0

[

]

m ≠ 0, n ≠ 0 (10.152)

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Substituting the average power of the TE mode fields from (10.98) and PcTE mn from (10.152) into (10.140), the attenuation factor for the rectangular waveguide supporting TE mode, after some algebraic manipulations, can be written as

α cTE mn

⎧0, ⎪ 1 ⎪ ⎪η δ σ c ⎪⎪ =⎨ 1 ⎪η δ σ c ⎪ ⎪ 2 ⎪η δ σ c ⎪⎩

( ab 1 − ( f a + 2b ( f ab 1 − ( f

) f) f) f)

b + 2a f cTE 0 n f TE 0 n c TE m 0 c TE 0 n c

m = 0, n = 0

2 2

,

m = 0, n ≠ 0

,

m ≠ 0, n = 0

2 2

⎡ bm 2 + an 2 ⎢ 2 2 1 − f cTE mn f 2 2 ⎢⎣ b m + a n

(

) + (a + b)( f 1− (f 2

TE mn c

TE mn c

f f

)

) 2

2

⎤ ⎥, ⎥⎦

(10.153)

m ≠ 0, n ≠ 0

Example 10.13 A waveguide of a square cross section transmits electromagnetic power at TE 22 mode. The waveguide is made of copper ( σ c = 5.7 × 107 S m ) and filled with polystyrene ( ε r = 2.56, σ d = 10 −17 ). The operating frequency is double the cut-off frequency of the waveguide. If the amplitude of the electric field components are 100 V/m and their frequency is 16 GHz., find (a) (b) (c) (d) (e)

The amplitude of H zs . The power loss per unit length in the walls of the waveguide. The average transmitted power through the waveguide. The attenuation constant of the waveguide. The power dissipated in 150 cm along the waveguide.

Solution Given that a = b , f cTE 22 f = 0.5 , ε r = 2.56 and Ex = Ex = 100 V m . (a) The amplitude of H zs for the TE 22 mode is H 22 , then from (10.77a)

Exs max = E ys

max

=

ωμ 0 nπ ωμ mπ H mn = 20 H mn = 100 2 βc b βc a

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Since a = b and m = n = 2 , then

βc =

(mπ a ) + (nπ b) 2

⇒

⇒

H 22

f

2

TE 22 c

=2 2

π a

⇒

aβ c

π

=2 2

βc

2 = = 2 μ 0ε 0ε r π a μ 0ε 0ε r



2 f cTE 22 βc a βc2 aβ c = 100 × = 100 × × = 100 × × 2π μ 0 2π f η mπ ωμ 0 f = 100 × 2 ×

2.56 1 × = 0.3 A m 120π 2



(b) The power loss per unit length can be obtained by substituting a = b and m = n = 2 in (10.152) , as 2

TE 22 c

P

H = mn 2 δσ c

[(

⎧⎪ TE ⎨ f f c mn ⎪⎩ 2

) − 1] ⎡⎢ a (m a )

5 a H mn 5 a H 22 = = 2 δσ c 2

2

2

2 ⎫⎪ + b (n b ) ⎤ ⎥ + a + b⎬ 2 2 ⎪⎭ ⎣ (m a ) + (n b ) ⎦ 2

π f μ0 σc

Since the operating frequency is f = 16 GHz = 2 f cTE 22 ,

⇒ f cTE 22 = f 2 = 8 × 109 =

⇒ a=

2 a μ 0ε 0ε r

2 × 3 ×108 = 0.033 m = 3.3 cm 8 ×109 2.56

Substituting f = 16 GHz , a = 0.033 m , σ c and μ 0 in the above equation,

5 × 0.033 × (0.3) = 2

2

⇒P

TE 22 c

π × 16 × 109 × 4π × 10 − 7 = 24.73 × 10 −5 W 7 5.7 × 10 

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(c) The average transmitted power through the waveguide at 16 GHz from (10.98), is 2

PaTE 22

(

)

H 22 2 abη = 1 − f cTE mn f 2 8 f cTE mn f  2 2 0.033 120π 0.3 2 = × × 1− 1 2 2 8 2.56 1 2

(

(

)

)

( ) ( )

( )

⇒ PaTE 22 = 10 mW 

(d) The overall attenuation is α  = α
+ α . The attenuation constant due to conduction loss α
at 16 GHz for the TE 22 mode can be computed using (10.140) as α cTE 22 =

Pc P TE 22 24.73 × 10 −5 = 1.236 × 10 − 2 Np m = −0.1073 dB m  = c TE 22 = −2 2 Pa 2 Pa 2 × 1 × 10

22 Alternatively, α TE  can be computed by substituting a = b , m = n = 2 , and c

= 1.236 × 10 −2 Np m . f cTE 22 = f 2 in (10.153), as α TE c 22

The dielectric attenuation for the TE 22 mode is 22 α TE ≈ d

ησ d

(

2 1 − f cTE 22 f

)

2

= =

μ 0 ε 0ε r 2

(

σd

1 − f cTE 22 f

120π × 10 −17

( )

2.56 1 − 1 2

2

)

2

= 1.36 × 10 −15 Np m

The overall attenuation constant is −2 22 22 αgTE 22 = αcTE 22 + αTE ≈ αTE ≈ 1.236 ×10 Np m d c 22 22 Since α TE In dB per meter, α
b . Find the range of frequencies that permits propagation of the dominant mode only. If the waveguide is operated at a frequency higher than the cut-off frequency of the dominant mode by 25%, and lower than the cut-off frequency of the next higher-order mode to dominant mode by 25%, find the dimensions of the waveguide. Then, compute the (a) Group velocity in the waveguide. (b) Wave impedance. (c) The wavelength of waves at 15 GHz in the waveguide.

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Solution The dominant mode in rectangular waveguides is the TE10 mode and its cut-off frequency is

f cTE10 =

υ0 2a

The next higher mode is TE 01 and its cut-off frequency is

f cTE 01 =

υ0 2b

Hence, the range of frequencies that permits propagation of the dominant mode only is

υ0 υ < f < 0 2a 2b The operating frequency f satisfies f = 1.25 f cTE10 = 0.75 f cTE 01 . Thus,

f = 1.25 f cTE10 =

1.25υ0 0.75υ0 = 0.75 f cTE 01 = 2a 2b

⇒

3 b= a 5

(a) The group velocity is

υg = ⇒ υg =

(

1 × 1 − f cTE10 f μ 0ε 0

)

2

1 1 − (1 1.25 ) 2 = 3 × 10 8 × 0.6 = 1.8 × 10 8 m s μ 0ε 0

(b) The wave impedance is

Z TE10 =

(

η

1 − f cTE10 f

)

2

=

η

1 − (1 1.25 ) 2

=

120π = 200π Ω 0.6

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(c) The wavelength for waves propagate at 15 GHz in the waveguide is

λ
=

λ 1 − (1 1.25)

2

=

1 1 3 × 108 1 × = × = 3.33 cm 9 f μ 0ε 0 0.6 15 × 10 0.6

Solved Problem 10.3 An air-filled rectangular waveguide, operating at the dominant mode TE10 , with cross section dimensions a = 1.58 cm and b = 0.8 cm , is used as a phase shifter at 15 GHz. 'etermine the length of the waveguide that introduces a total phase shift of 150 degrees. Solution Given that a = 1.58 cm , b = 0.8 cm , f = 15 GHz ,and the operation mode is TE10 . Hence, the cut-off frequency is

f cTE10 =

υ0 3 × 108 = = 9.49 GHz 2a 2 × 1.58 × 10− 2

The phase constant is

β g = 2π f μ 0 ε 0 1 − ( f cTM

f

10

)

2

2π × 9.49 × 10 9 1 − (9.49 15 ) 2 8 3 × 10 = 154 rad m =

Converting 150 to radians, then

150 =

150

π

= 47.73 rad

Assume the length that introduces 150 = 47.73 rad is z , then

β g z = 47.73



47.73 ⇒ z= = 0.31 m = 31 cm 154  The required length is 31 cm.

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Solved Problem 10.4 An X-band rectangular waveguide, with dimensions a = 2.286 cm and b = 1.016 cm and filled with a dielectric material having a relative permittivity of 2.25. If the waveguide is to be used as a delay line, find the length of the waveguide so that the total delay it presents at 10 GHz is 2 μs . Solution Since ε r = 2.25 , a = 2.286 cm and b = 1.016 cm , then the cut-off frequency is

1 1 fc = 2 μ 0ε 0ε r

(m a )2 + (n a )2

=

1.5 × 10 8 2.25 × 10

−2

(m 2.286)2 + (n 1.016)2

The X-band frequency range extends from 8.2 GHz to 12.4 GHz , the cut-off frequency must be lower than 8.2 GHz , then f c < 8.2 GHz . Thus,

8.2 × 109 > 1010

(m 2.286)2 + (n 1.016)2 

⇒ 3.514 > m2 + 5.0625 n2 The only values of m and n that satisfy the above inequality are ( m = 0 , n = 0 ) and ( m = 1 , n = 0 ). Since there is no modes correspond to ( m = 0 , n = 0 ), the waveguide must operate at the mode TE 10 . Hence, the cut-off frequency is

f cTE10 =

1010 = 4.38 GHz 2.286

Let the length of the waveguide that presents the time delay t = 2 μs is z , then

z = υg t

=

(

t × 1 − f cTE10 f μ 0ε 0

)

2

= 2 × 10 − 6 × 3 × 10 8 1 − (4.38 10 ) 2 = 539.55 m

.

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PROBLEMS 10.1. In an air-filled rectangular waveguide with a = 2.286 cm and b = 1.016 cm, the y-component is given by

E yg = 5sin(2πx a) sin(3π ×1010 t − βg z ) V m Find the (a) (b) (c) (d)

Mode of operation. Propagation constant. Wave impedance. Average power transmitted through the waveguide.

10.2. An air-filled waveguide, operating at 6 GHz has the spatial electric field

Ez = 10sin(100π x) sin(50π y ) V m If the cross-section of the waveguide is square, determine the (a) (b) (c) (d)

Mode of operation Cut-off frequency Phase velocity Average power transmitted through the waveguide.

10.3. Write expressions for the TM11 electromagnetic fields in a rectangular waveguide at the instant t = 0 second. Draw the field pattern of this mode on the plane z = λ
4 , 0 ≤ x ≤ a , 0 ≤ y ≤ b at the instant t = 0 second. 10.4. An air-filled rectangular waveguide is to be used to transmit signals of a carrier frequency of 6 GHz. Find the cross section dimensions of the waveguide, if the cut-off frequency of the dominant mode is lower by 25% than the carrier frequency, and that of the next higher-order mode is at least 25% higher than the carrier frequency. 10.5. An air-filled rectangular waveguide has dimensions a = 2.286 cm and b = 1.016 cm . If the waveguide operates at 10 GHz, find for the dominant mode, the (a) Cut-off frequency.

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(b) Phase and group velocities. (c) Wave impedance. 10.6. The cross section dimensions of a rectangular waveguide are a and b . Compute the cut-off frequencies of the modes: TE 01 , TE10 , TE11 , TE 02 , TE 20 , TM11, TM12 , and TM 22 normalized to the cut-off frequency of the dominant mode, if (a) a = 3b . (b) a = b . 10.7. The cross section dimensions of an air-filled rectangular waveguide are 5 cm and 2 cm . If the following electric field components propagate in the waveguide

E zg = 20 sin (40πx ) cos(50πy ) cos(30π ×109 t − βg z ) V m Find the (a) Propagation mode. (b) Phase constant in the waveguide. 10.8. An air-filled rectangular waveguide has dimensions 4 cm × 2 cm , transmits a power of 2 mW at the dominant mode. If the operating frequency is 10 GHz, find the amplitude of the electric field in the waveguide. 10.9. If the cut-off frequency of an air-filled rectangular waveguide is 5 GHz at TE10 mode and 12 GHz at TE 01 mode. Find the (a) Dimensions of the waveguide (b) Next three TE mode cut-off frequencies. 10.10. An air-filled rectangular waveguide has dimensions a = 1cm and b = 3 cm operates at TE 21 mode. If the operating frequency is higher than the cutoff frequency by 20%, find the (a) Operating frequency. (b) Phase and group velocities.

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10.11. Two signals of carrier frequencies 10 GHz and 12 GHz propagate through an air-filled rectangular waveguide at the TE10 mode. If the cross section of the waveguide has dimensions a = 2 cm and b = 1cm , find the time required for each signal to travel a distance of 10 m down the waveguide. 10.12. An air-filled rectangular waveguide, operating at the dominant mode TE10 , with cross section dimensions a = 2.286 cm and b = 1.016 cm , is used as a phase shifter at 12 GHz. 'etermine the length of the waveguide that introduces a total phase shift of 100 degrees. 10.13. An air-filled rectangular waveguide with cross section dimensions a and b. if the waveguide operates at 4 GHz (a) Determine the allowed modes in the waveguide when a = 8.0 cm and b = 6.0 cm . (b) For b = a , find the range of values of a which will allow propagation through the waveguide in the TE10 mode but not in the TE11, TM11, or any higher modes. (c) For the waveguide of part b, determine the surface current densities on the four walls of the waveguide. 10.14. An air-filled rectangular waveguide is operated at 10 GHz. If the dimension of the waveguide are a = 2.286 cm and b = 1.016 cm , calculate the maximum power that can be transmitted without causing dielectric breakdown of air. The air will break-down when the maximum electric field is 3 MV m . 10.15. An air-filled rectangular waveguide, operating at the dominant mode TE10 , with cross section dimensions a = 2.25 cm and b = 1.125 cm (a) Find the cut-off frequency. (b) If the cut-off frequency of the dominant mode to be reduced by a factor of 2 by replacing the air by a dielectric medium while the physical dimensions of the waveguide stay the same, find relative permittivity of the dielectric. 10.16. The operation frequency of a rectangular waveguide of dimensions 2.286 cm ×1.016 cm , is 9 GHz. If the waveguide is made of copper

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(σ c = 5.8 × 10 7 S m) and filled with polystyrene, find the attenuation constant and the total power loss in 50 m along the waveguide for the dominant mode. For polystyrene (ε r = 2.56, σ d ωε r ε 0 = 3 × 10 −4 ) . 10.17. A rectangular waveguide is made of copper (σ c = 5.8 × 10 7 S m) , and filled with polystyrene (ε r = 2.56, σ d = 10 −17 ) , has dimensions a = 4.2 cm and b = 2.6 cm . If the waveguide is used to transmit a power of 1.2 kW at 30 GHz, compute the dissipated power in a 60 m of the guide length at the mode TE10 . 10.18. A rectangular waveguide is made of aluminum (σ c = 3.5 × 10 7 S m) and filled with Teflon (ε r = 2.6, σ d = 10 −15 ) , operates at the TE10 mode. If the dimensions of the waveguide are a = 4.2 cm and b = 1.5 cm , calculate attenuation constant and the total power loss in dB over 1.5 m of the waveguide at 4 GHz. 10.19. Show that the attenuation constant due to the finite conductivity σ c of a rectangular waveguide of a square cross section with side a , operating in a TM mn mode at a frequency higher than its cut-off frequency by 25%, is given by

α

TM mn c

(m = 10

2

+ n2

(12a )

3 2

)

1 4

σc

10.20. A rectangular waveguide is made of copper (σ c = 5.8 × 10 7 S m) and filled with Teflon (ε r = 2.6, σ d = 10 −15 ) . A wave of 9 GHz is propagating in the waveguide at the TE10 mode. If the dimensions of the waveguide are a = 4.2 cm and b = 1.5 cm , calculate the total power loss in a length of 40 m of the waveguide. 10.21. A waveguide of a square cross section with side a = 4 cm , is filled with a dielectric of relative permittivity ε′r = 16 (1 − j10 −4 ) . If the waveguide is operated in the TM21 mode at a frequency higher than the cut-off

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frequency by 10%. Find the attenuation constant and the distance that the wave travel down the waveguide before its amplitude is reduced by 5%.

Electromagnetics for Engineering Students, 2017, 715-750

715

CHAPTER 11

Circular Waveguides Abstract: This chapter is a continuation of Chapter 10 and devoted to discuss the waveguides of a circular cross-section. The general formulations for the metallic waveguides that are presented in Chapter 10, will be applied in this chapter to analyze the behavior of electromagnetic waves and power flow in the metallic waveguide of circular cross-section. The topics of the chapter are analyzed in details and supported by numerous illustrative examples and figures in addition to solved problems. Homework problems are included at the end of the chapter.

Keywords: Attenuation constant, cut-off frequency, cut-off wavelength, dominant mode, group velocity, phase constant, phase velocity, power loss, Poynting vector, propagation constant, wave impedance. INTRODUCTION The waveguides are one of the transmission media that used to transmit electromagnetic energy in communication systems at high frequencies. The waveguide may be a metallic tube filled with a dielectric material or a dielectric rod with a rectangular, circular, or elliptical cross section. The most common waveguides are those of rectangular and circular cross sections. The general formulations for the metallic waveguides in addition to the analysis of the rectangular waveguides are presented in Chapter 10. This chapter is a continuation of Chapter 10 and devoted to discuss the waveguides of a circular cross-section. The general formulations introduced in Chapter 10 will be used in this chapter to analyze the behavior of electromagnetic fields in the waveguides of the cylindrical structure, which are referred to as circular waveguides. More detials on cylindrical waveguides are available in [50, 92, 98, 111-117]. 11.1. ELECTROMAGNETIC FIELDS It is appropriate to use the cylindrical coordinate system in the analysis of the circular waveguides. Equations (10.1) – (10.6) give the general forms of the electromagnetic fields in a waveguide. For circular waveguides, Et and Ht are

Et = a ρ Eρ + aφ Eφ Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

(11.1a)

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H t = a ρ H ρ + aφ H φ

(11.1b)

Hence, the spatial field components becomes

Es = a ρ Eρs + aφ Eφs + a z Ezs = (Et + a z Ez ) e

− jβ g z

H s = a ρ H ρs + aφ Hφs + a z H zs = (Ht + a z H z )e

− jβ g z

(11.2a) (11.2b)

The operators ∇ t and ∇ t2 in cylindrical coordinate are given by

∂ 1 ∂ + aφ ∂ρ ρ ∂φ 2 ∂ 1 ∂ 1 ∂2 ∇ t2 = + + ∂ρ 2 ρ ∂ρ ρ 2 ∂φ 2 ∇t = a ρ

(11.3) (11.4)

11.1.1. TE Waves in a Circular Waveguide The geometry of the waveguide is shown in Fig. (11.1). The radius of the waveguide cross section is a , and filled with a lossless dielectric material of permittivity ε and permeability μ. The walls of the waveguide are made of a perfect electric conducting material (σ ≈ ∞). For TE mode waves in the waveguide shown in Fig. (11.1), (10.18) is solved first to obtain H z . The other field components can be determined using the general relations (10.39), and (10.41). In cylindrical coordinate, (10.18) can be written as

∂ 2 H z 1 ∂H z 1 ∂2H z + + 2 + βc2 H z = 0 2 2 ∂ρ ρ ∂ρ ρ ∂φ

(11.5)

Using the method of separation of variables, H z can be expressed in terms of independent functions R and T such that

H z (ρ ,φ ) = R(ρ )T (φ )

(11.6)

Substituting H z from (11.6) into (11.5) and dividing both sides by RT , yields

1 ⎛ 2 ∂2R 1 ∂ 2T ∂R ⎞ 2 2 ⎜⎜ ρ ⎟ ρ ρ β + = − + c ∂ρ ⎠⎟ R ⎝ ∂ρ 2 T ∂φ 2

(11.7)

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y σc ≈ ∞ a

ρ

φ

x

ε, μ, σd ≈ 0

z

Fig. (11.1). The geometry of the rectangular waveguide.

Since both sides in (11.7) are independent of each other, then each side must equal to a constant quantity. If we let this constant to be ν 2 , (11.7) can be separated into the following equations

ρ2

∂2R ∂R +ρ + ρ 2 βc2 −ν 2 R = 0 2 ∂ρ ∂ρ 2 ∂ T + ν 2T = 0 2 ∂φ

(

)

(11.8a) (11.8b)

Equation (11.8a) is known as Bessel Differential Equation and its solution is

R(ρ ) = A′′Jν (βc ρ ) + B′′Yν (βc ρ )

(11.9a)

Where A′′ and B′′ are constants. Jν (βc ρ ) and Yν (β c ρ ) respectively, are the first and second type Bessel functions of order ν . The solution of (11.8b) can be written in the form T (φ ) = C ′′ cos[ν (φ − φo )]

(11.9b)

Where C ′′ and φo are constants. Substituting R from (11.9a) and T from (11.9b) into (11.6), the solution of (11.5) becomes

H z (ρ ,φ ) = [ A′′Jν (βc ρ ) + B′′Yν (βc ρ )] C′′ cos[ν (φ − φo )]

(11.10)

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Using (10.39) with (11.1a) and (11.3), then

⎛ ωμ ⎞ ⎛ 1 ∂ ⎞ ∂ a ρ Eρ + aφ Eφ = − j ⎜⎜ 2 ⎟⎟ ⎜⎜ a ρ + aφ ⎟×a H ρ ∂φ ⎟⎠ z z ⎝ β c ⎠ ⎝ ∂ρ ⎛ ωμ ⎞ ⎛ ∂H z 1 ∂H z ⎞ = − j ⎜⎜ 2 ⎟⎟ ⎜⎜ − aφ + aρ ⎟ ρ ∂φ ⎟⎠ ∂ρ ⎝ βc ⎠ ⎝

(11.11)

Equating the quantities in the ρ and φ directions on both sides of (11.11), the transverse components E ρ and Eφ can be written as

Eρ = j

ν ⎛ ωμ ⎞ ⎜ ⎟[ A′′Jν (βc ρ ) + B′′Yν (βc ρ )] C′′ sin[ν (φ − φo )] ρ ⎝⎜ βc2 ⎠⎟

⎛ ωμ ⎞ ∂J (β ρ ) ∂Y (β ρ ) Eφ = j⎜⎜ 2 ⎟⎟[ A′′ ν c + B′′ ν c ] C′′ cos[ν (φ − φo )] ∂ρ ∂ρ ⎝ βc ⎠

(11.12a) (11.12b)

From (10.41), we have

⎛ β
⎞ ⎟⎟ a z × (a ρ E ρ + a φ Eφ ) a ρ H ρ + a φ H φ = ⎜⎜ ωμ ⎠ ⎝ ⎛ β
⎞ ⎟⎟ (a φ E ρ − a ρ Eφ ) = ⎜⎜ ωμ ⎝ ⎠

(11.13)

Using (11.12a) – (11.13), the components H ρ and H φ can be obtained as follows

⎛β ⎞ = −⎜⎜ g ⎟⎟ Eφ ⎝ ωμ ⎠ (11.14a) ⎛ βg ⎞ ∂Jν (β c ρ ) ∂Yν (β c ρ ) + B′′ ]C′′ cos[ν (φ − φo )] = − j ⎜⎜ 2 ⎟⎟[ A′′ ∂ρ ∂ρ ⎝ βc ⎠ ⎛β ⎞ Hφ = ⎜⎜ g ⎟⎟ Eρ ⎝ ωμ ⎠ (11.14b) ν ⎛ βg ⎞ = j ⎜⎜ 2 ⎟⎟[ A′′Jν (β c ρ ) + B′′Yν (β c ρ )]C ′′ sin[ν (φ − φo )] ρ ⎝ βc ⎠

Hρ

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The constants A′′ , B′′ , C ′′ , D′′ , and ν can be determined by applying the boundary conditions of the problem. Since the variation of all quantities in (11.10), (11.12a), (11.12b), (11.14a) and (11.14b) are periodic in φ with period 2π , ν must be an integer and we can let ν = m , where m = 0,1, 2, . Note That: 1. when ρ = 0 the fields are finite. 2. The tangential electric fields on the walls of the waveguide (ρ = a) vanishes or Eφ (a, φ , z ) = 0 . In (11.10), since the fields at ρ = 0 are finite, and Ym (βc ρ ) → −∞ as ρ → 0 , this imply that the term B′′Ym (βc ρ ) must vanish, then B′′ = 0 . Applying the boundary condition Eφ (a, φ , z ) = 0 on (11.10) and substituting B′′ = 0 , yields

∂J m (β c ρ ) = J m′ (β c ρ ) ρ = a = 0 ∂ρ ρ =a

(11.15)

Where J m′ (β c ρ ) is the first derivative of Bessel function of the first type with

respect to ρ . J m′ (β c ρ ) has an infinite number of roots that satisfy (11.15).

Assuming that, the root number n of Bessel function of order m is χ ′mn , then

βc ρ = χ′mn . Consequently, βc =

χ ′mn a

(11.16)

As for the rectangular waveguides, there are an infinite number of modes that can propagate in the circular waveguides depending on m and n . Substituting χ′mn from (11.16) into (10.44), the phase constant of TE waves in the circular waveguide can be obtained as

β g = β 2 − (χ ′mn a )2

(11.17)

Substituting B ′′ = 0 and ν = m in (11.10), (11.12a), (11.12b), (11.14a) and (11.14b), and letting A′′C ′′ = H mn , then Et , Ht and H z can be determined.

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Substituting Et , Ht and H z into (11.2a) and (11.2b), the spatial components of the TE waves in the circular waveguide of Fig. (11.1) are obtained as

Eρs = j

ωμ m

− jβ z H mn J m (χ ′mn ρ a ) sin [m(φ − φo )] e g

β ρ ωμ χ ′ − jβ z Eφs = j 2 mn H mn J m′ (χ ′mn ρ a ) cos[m(φ − φo )] e g βc a Ezs = 0 β χ′ − jβ z H ρs = − j g2 mn H mn J m′ (χ ′mn ρ a ) cos[m(φ − φo )] e βc a β m − jβ z Hφs = j g2 H mn J m (χ′mn ρ a ) sin[m(φ − φo )] e βc ρ − jβ z H zs = H mn J m (χ ′mn ρ a ) cos[m(φ − φo )] e
2 c

g

g

(11.18a) (11.18b) (11.18c) (11.18d) (11.18e) (11.18f)

where H mn has a dimension of ( A m ). The corresponding instantaneous components can be obtained by substituting the above components of E s , H s into (10.1) and (10.2), as H J (χ ′ ρ a ) sin [m(φ − φo )] sin (ω t − β g z ) β c2 ρ mn m mn ωμ χ ′ Eφg = − 2 mn H mn J m′ (χ ′mn ρ a ) cos[m(φ − φo )] sin (ω t − β g z ) βc a E z
= 0

E ρg = −

ωμ m

βg χ′mn H J ′ (χ′ ρ a ) cos[m(φ − φo )] sin (ω t − βg z ) βc2 a mn m mn β m H φg = − g2 H mn J m (χ′mn ρ a ) sin[m(φ − φo )] sin (ω t − βg z ) βc ρ H z
= H mn J m (χ ′mn ρ a ) cos[m(φ − φo )] cos(ω t − β
z )

H ρg =

(11.19a) (11.19b) (11.19c) (11.19d) (11.19e) (11.19f)

Note that J m′ (u ) = ∂J m (u ) ∂u , where u = χ mn ρ a  The roots χ′mn of J m′ (χ′mn ) = 0 are computed and listed in Table 11.1 for some values of m and n .

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Table 11.1. The roots χ′mn of J m′ (x ) = 0 for P andQ n=1

m=0

m=1

m=2

m=3

m=4

m=5

m=6

m=7

m=8

3.8318

1.8412

3.0542

4.2012

5.3175

6.4155

7.5013

8.5777

9.6474

n=2

7.0156

5.3315

6.7062

8.0153

9.2824

10.5199

11.7349

12.9324

14.1155

n=3

10.1735

8.5363

9.9695

11.3459

12.6819

13.9872

15.2682

16.5294

17.7740

n=4

13.3237

11.7060

13.1704

14.5859

15.9641

17.3129

18.6375

19.9419

21.2291

n=5

16.4706

14.8636

16.3475

17.7888

19.1960

20.5755

21.9317

23.2681

24.5872

11.1.2. Characteristics of the TE Mode Fields in the Waveguide 11.1.2.1. Cut-off Frequency Following the same analysis as in the case of the rectangular waveguides, the cutoff frequency of a circular waveguide of radius a and supports TE mn waves is obtained by letting β g = 0 in (11.17), then solving the resulting equation for frequency, yields

f cTE mn =

χ ′mn 2π με a 1

(11.20)

The corresponding cut-off wavelength is

λc =

2π a χ ′nm

(11.21)

11.1.2.2. Phase and Propagation Constants Using (11.17) and (11.21) the phase constant of the circular waveguide can be written in terms of cut-off frequency, as

βg = β 1 − ( fcTE Where υ = 1

mn

f

)

2

(11.22)

με and β = ω με .From the relation
= jβ g , the propagation

constant for TE mn mode waves in a circular waveguide can be expressed as


=

(χ′nm a)2 − β 2 = jβ

(

1 − fcTE mn f

)

2

(11.23)

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11.1.2.3. Wavelength The wavelength of TE mn waves in a circular waveguide can be obtained as

λg =

2π

βg

2π

=

=

ω με− (χ ′nm a )

2

2

(

λ

1− f

TE mn c

(11.24)

)

2

f

11.1.2.4. Wave Impedance Substituting the phase constant from (11.17) into (10.43) the wave impedance for the TE mn waves in the circular waveguides can be expressed as Z TE mn =

ωμ 2 ω με− (χ ′nm a ) 2

=

(

1− f

η TE mn c

f

(11.25)

)

2

Where η = μ ε . 11.1.2.5. Phase Velocity and Group Velocity Using (8.41) and (8.116) along with (11.17) the phase velocity υ p and group velocity υ
of the TE mn mode waves in a circular waveguide can be obtained respectively as

υp =

υ
=

ω

βg

=

ω

β − (χ′nm a ) 2

2

=

(

1− f

υ TE mn c

(

f

)

(11.26)

2

dω 2 = ωυ 2 β 2 − (χ ′nm a ) = υ 1 − f cTE mn f dβ


)

2

(11.27)

11.1.2.6. Poynting Vector and Power Using (10.31) and putting in mind that for TE mn mode E zg = 0 , Z TE mn H φg = E ρg , and Z TE mn H ρg = −Eφg , the instantaneous Poynting vector for TE mn waves in a circular waveguide can be obtained as

⎛ E 2+E 2 ⎞ 1 ρg φg ⎟ ⎜ T = az ⎜ + H z∗g (a ρ Eφg − aφ E ρg ) ⎟ ⎜ 2Z TE mn ⎟ 2 ⎝ ⎠

(11.28)

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Note that as in the rectangular waveguide case, the Poynting vector in the direction of propagation is always real. The time-average power PaTE mn through the waveguide can be found from (10.32) and (10.33), taking into consideration that in cylindrical coordinates dS = a z ρ dρ dφ , then TE mn a

P

∫∫ (Eρ

2π a

1

) )

2

ρ dρ dφ 2Z TE mn 0 0 2 Z TE mn 2π a 2 = H + H ρ dρ dφ ρ s φ s 2 ∫0 ∫0 =

2

s

+ Eφs

(

(11.29)

Substituting Eρs and Eφs respectively from (11.18a) and (11.18b) into (11.29), the average power becomes TE mn a

P

2 2π a

⎛ ωμ ⎞ ⎜⎜ 2 m H mn ⎟⎟ = 2 Z TE mn ⎝ βc ⎠ 1 ⎛ ωμ χ ′mn ⎜ + H mn 2 Z TE mn ⎜⎝ βc2 a 1

∫ ∫ J (χ ′ 2 m

mn

ρ a ) sin 2 [m(φ − φo )]

0 0 2 2π a

⎞ ⎟⎟ ⎠

∫ ∫ J ′ (χ′ 2 m

mn

1

ρ

dρ dφ

ρ a ) cos2 [m(φ − φo )] ρ dρ dφ

0 0

(11.30) Evaluating the integrals in (11.30) with respect to φ , yields TE mn a

P

=

π 2Z TE mn

⎛ ωμ ⎜⎜ 2 H mn ⎝ βc

⎞ ⎟⎟ ⎠

2 ⎡ m2 2 ⎤ ⎛ χ ′mn ⎞ 2 ∫0 ⎢⎢ ρ J m (χ ′mn ρ a ) + ρ ⎜⎝ a ⎟⎠ J m′ (χ ′mn ρ a )⎥⎥ dρ ⎣ ⎦ (11.31)

2a

Using the substitution u = χ′mn ρ a , (11.31) becomes TE mn a

P

π ⎛ ωμ ⎜ = H mn 2Z TE mn ⎜⎝ βc2

⎞ ⎟ ⎟ ⎠

2 χ′

mn

∫ 0

⎡ m2 2 ⎤ 2 ′ ( ) ( ) J u u J u + m m ⎢ ⎥ du ⎣ u ⎦

(11.32)

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Make use of J m′ (u ) = − J m+1 (u ) + (m u )J m (u ) in (11.32), the transmitted average power of the TE mode fields through a circular waveguide becomes TE mn a

P

π ⎛ ωμ ⎜ = H mn 2Z TE mn ⎝⎜ βc2

⎞ ⎟ ⎟ ⎠

2 χ′

mn

∫ 0

⎡ 2m 2 2 ⎤ J m (u ) + uJ m2 +1 (u ) − 2mJ m (u )J m+1 (u )⎥ du ⎢ ⎣ u ⎦ (11.33)

The integral in (11.33), which depends only on χ′mn and the mode indices m and n , can be computed analytically or numerically. Letting the value of this integral be I mn (χ′mn ), then (11.33) can be written in a reduced form as TE mn a

P

π ⎛ ωμ ⎜ = H mn 2Z TE mn ⎜⎝ βc2

2

⎞ ⎟ I mn (χ ′mn ) ⎟ ⎠

(11.34)

Where χ ′mn

I mn (χ ′mn ) =

∫ 0

⎤ ⎡ 2m 2 2 J m (u ) + uJ m2 +1 (u ) − 2mJ m (u )J m+1 (u )⎥ du ⎢ ⎦ ⎣ u

(11.35)

I mn (χ ′mn ) can be evaluated analytically or numerically. Values for I mn (χ ′mn ) are computed numerically for 0 < m < 8 and 1 < n < 5 and listed in Table 11.2. An

analytical expression for I mn (χ′mn ) will be derived in the following analysis.

Using Bessel functions recurrence relations, (11.35) can be written in the form χ ′mn

χ ′mn

I mn (χ′mn ) = m ∫ [J m (u )J m −1 (u ) − J m +1 (u )J m (u )] du + ∫ uJ m2 +1 (u ) du 0 0

  

    I1

(11.36)

I2

Table 11.2. Values of I mn (χ ′mn ) for P andQ m=0

m=1

m=2

m=3

m=4

m=5

m=6

m=7

m=8

n=1

1.1908

0.4046

0.6305

0.8161

0.9804

1.1307

1.2710

1.4034

1.5295

n=2

2.2165

1.6428

2.0138

2.3417

2.6411

2.9199

3.1828

3.4328

3.6721

n=3

3.2267

2.6840

3.0952

3.4695

3.8171

4.1443

4.4550

4.7520

5.0375

n=4

4.2322

3.7021

4.1338

4.5340

4.9100

5.2666

5.6072

5.9342

6.2495

n=5

5.2356

4.7124

5.1568

5.5738

5.9688

6.3458

6.7075

7.0561

7.3932

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Using the integral formula (A.10b) in the Appendix A, the integral I1 can be obtained as χ ′mn

I1 =

∫ [J (u )J (u ) − J (u )J (u )] du m −1

m

m +1

m

0

m

m −1

k =1

k =1

= ∑ J 2 k (χ ′mn ) − ∑ J 2 k (χ ′mn ) = J 2 m (χ ′mn )

(11.37)

Using the substitution u = χ ′mn t for the integral I 2 in (11.36), it can be written in the form 1

I2 = χ′

2 mn

∫ tJ (χ ′ t )dt 2 m +1

(11.38)

mn

0

Using the following integral formula 1

2 ∫ tJ p (ct )dt = 0

[

]

c2 2 J p (c ) − J p −1 (c )J p +1 (c ) 2

(11.39)

and letting p = m + 1 and c = χ′mn , I 2 can be evaluated as 1

I 2 = ∫ tJ m2 +1 (χ ′mn t ) dt 0

χ′ = mn J m2 +1 (χ ′mn ) − J m (χ ′mn )J m + 2 (χ ′mn ) 2 2

[

(11.40)

]

Since χ′mn is a root of J m (u ) = 0, then using Bessel functions properties, we can write J m′ (χ′mn ) = J m (χ′mn ) − (χ′mn m) J m+1 (χ′mn ) = 0 , or J m +1 (χ ′mn ) =

m J m (χ ′mn ) χ ′mn

(11.41)

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In addition, from Bessel functions properties, we have 2(m + 1) J m +1 (χ ′mn ) − J m (χ ′mn ) χ ′mn

J m + 2 (χ ′mn ) =

(11.42)

From (11.41) and (11.42), we can express J m +1 (χ′mn ) in term of J m (χ′mn ) as

J m+ 2 (χ ′mn ) =

2m(m + 1) J m (χ ′mn ) − J m (χ ′mn ) 2 χ ′mn

(11.43)

Substituting J m +1 (χ′mn ) and J m+ 2 (χ′mn ) from (11.41) and (11.43) respectively into (11.40) with simplification, the integral I 2 reduces to

I2 =

(

)

1 2 χ ′mn − m 2 − 2m J m2 (χ ′mn ) 2

(11.44)

Substituting I1 and I 2 from (11.37) and (11.44) respectively, into (11.36), yields

I mn (χ ′mn ) =

(

)

1 2 χ ′mn − m 2 J m2 (χ ′mn ) 2

(11.45)

From (11.34) and (11.45) the time-average power transmitted through a rectangular waveguide supporting a TE mode wave becomes TE mn a

P

2

(

)

⎛ ωμ ⎞ ⎜⎜ 2 H mn ⎟⎟ χ′mn 2 − m2 J m2 (χ ′mn ) = 4Z TE mn ⎝ βc ⎠

π

(11.46)

Using (11.20) and (11.25), (11.46) can be written, after some manipulations, as TE mn a

P

ηπ a 2 H mn = 4

2

⎛ f ⎜⎜ TE mn ⎝ fc

⎞ ⎟⎟ ⎠

2

⎡ ⎛ m ⎞2 ⎤ ⎛ f TE mn ⎟⎟ ⎥ J m2 (χ ′mn ) 1 − ⎜⎜ c ⎢1 − ⎜⎜ ⎢⎣ ⎝ χ ′mn ⎠ ⎥⎦ ⎝ f

2

⎞ ⎟⎟ (11.47) ⎠

Example 11.1 A circular waveguide of radius a = 3 cm, operates at 2 GHz. If the waveguide is filled with polystyrene (εr = 2.56), determine, for the TE 11 mode, (a) The cutoff frequency.

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(b) The wavelength in the waveguide. (c) The phase velocity and group velocity. Solution (a) From Table 11.1, χ′11 = 1.8412 and a = 3 cm then the cutoff frequency can be computed using (11.20) as

f cTE11 =

′ χ11 2π με a

′ 1 χ11 3 × 108 1.8412 = −2 2π μ 0ε 0ε r a 2π 2.56 3 × 10 = 1.8315 GHz

1

=

(b) Since the operating frequency 2.0 GHz is greater than the cut-off frequency, then the wavelength of the TE 11 mode waves in the waveguide is

λg =

λ 1 − ( fc f )

2

3 × 108 1 = × = 23.34 cm 9 2 2 × 10 2.56 1 − (1.8315 2)

(c) The phase velocity for the TE 11 mode is

υp =

1 μ 0ε 0ε r 1 − ( fc f )

2

3 × 108 2.56

=

1 − (1.8315 2)

2

= 4.67 × 108 m sec

The group velocity for the TE 11 mode is

υg =

(

1 1 − f cTE mn f μ 0ε 0ε r

)

2

=

3 × 108 2 1 − (1.8315 2) = 1.927 × 108 m sec 2.56

Example 11.2 Find the average power for the modes TE 01 , TE 02 , and TE 03 in a circular waveguide using the (a) Formula (11.46). (b) Formula (11.34) with the data in Table 11.1.

Electromagnetics for Engineering Students

728

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Solution (a) The average power for TE0n can be obtained using (11.46) by substituting m = 0 , as TE 0 n a

P

⎛ ωμ ⎞ χ ′2 J 2 (χ ′ ) ⎜⎜ 2 F0 n ⎟⎟ 0 n 0 0n = 2ZTE 0 n ⎝ βc 2 ⎠ 2

π

For modes TE 01 , TE 02 , and TE 03 , n = 1 , n = 2 , and n = 3 respectively. For TE 01 mode: from Table 11.1, χ′01 = 3.8318 . From Bessel function tables J 0 (3.8318) ≈ −0.4028, then

⎞ (3.8318)2 (− 0.4028)2 π ⎛ ωμ ⎜ H 01 ⎟⎟ = 2Z TE 01 ⎜⎝ β c2 2 ⎠ 2 ⎞ π ⎛ ωμ ⎜⎜ 2 H 01 ⎟⎟ × 1.1909 = 2Z TE 01 ⎝ β c ⎠ 2

TE 01 a

P

For the TE 02 mode: from Table 11.1, χ′02 = 7.0156. From Bessel function tables J0 (7.0156) ≈ 0.3001, then TE 02 a

P

⎛ ωμ ⎞ (7.016)2 (0.3)2 ⎞ π ⎛ ωμ ⎜ ⎟ ⎜ ⎟⎟ × 2.216 = H = H 02 02 ⎟ ⎜ β2 Z 2ZTE 02 ⎜⎝ βc2 2 2 TE 02 ⎝ c ⎠ ⎠ 2

π

2

For the TE 03 mode: from Table 11.1, χ ′03 = 10.1735 , and from Bessel function tables J 0 (10.1735) ≈ −0.2497, then TE 03 a

P

⎛ ωμ ⎞ (10.174)2 (− 0.249)2 ⎞ π ⎛ ωμ ⎜⎜ 2 H 03 ⎟⎟ ⎜⎜ 2 H 03 ⎟⎟ × 3.227 = = 2ZTE 03 ⎝ βc 2 2ZTE 03 ⎝ βc ⎠ ⎠

π

2

2

(b) Using Table 11.2, the values of I 0n (χ′0n ) for n = 1 , n = 2 , and n = 3 can be obtained respectively, as

I01(χ′01 ) = 1.1908,

I02 (χ′02 ) = 2.2165,

I03 (χ′03 ) = 3.2267

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Substituting these values into (11.34), the average power for TE 01 , TE 02 , and TE 03 modes are obtained as follows TE 01 a

P

2

2

2

2

2

2

⎛ ωμ ⎞ ⎞ π ⎛ ωμ ⎜⎜ 2 H 01 ⎟⎟ I 01 (χ ′01 ) = ⎜⎜ 2 H 01 ⎟⎟ × 1.1908 = 2Z TE 01 ⎝ βc 2Z TE 01 ⎝ βc ⎠ ⎠

π

TE 02 a

⎛ ωμ ⎞ ⎞ π ⎛ ωμ ⎜⎜ 2 H 02 ⎟⎟ I 02 (χ ′02 ) = ⎜⎜ 2 H 02 ⎟⎟ × 2.2165 = 2ZTE 02 ⎝ βc 2Z TE 02 ⎝ βc ⎠ ⎠

TE 03 a

⎛ ωμ ⎞ ⎞ π ⎛ ωμ ⎜⎜ 2 H 03 ⎟⎟ I 03 (χ ′03 ) = ⎜⎜ 2 H 03 ⎟⎟ × 3.2267 = 2ZTE 03 ⎝ βc 2ZTE 03 ⎝ βc ⎠ ⎠

P

P

π

π

Results using formula (11.46) for the average power are identical to those obtained using the data from Table 11.2 with formula (11.34). 11.1.3. TM Mode Fields in the Circular Waveguide Following the same analysis as in the TE mode case outlined in Section 11.1.1, Ez , Et , and Ht for TM mode waves in the circular waveguide can be obtained starting from (10.19) and using (10.39) – (10.42). In cylindrical coordinate (10.19) can be written as follows

∂ 2 Ez 1 ∂Ez 1 ∂ 2 Ez + + 2 + βc2 Ez = 0 2 2 ∂ρ ρ ∂ρ ρ ∂φ

(11.48)

Applying the separation of variables method and following similar analysis as for the TE mode case, the solution of (11.48) can be written as

Ez (ρ ,φ ) = D′′J m (βc ρ )cos[m(φ − φo )]

(11.49)

Where m = 0,1, 2,  and D′′ and φo are constants. As for the TE mode case, the electric field on the walls of the waveguide is zero. Hence, the spatial field satisfies Ezs (a,φ , z ) = 0 , which implies that Ez (ρ = a) = 0 . Imposing this boundary conditions on (11.49), yields

J m (βc ρ ) ρ =a = 0

(11.50)

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There are an infinite number of roots that satisfy (11.50). Let the root n for J m (βc a ) be χ mn , then

βc = χ mn a

(11.51)

Substituting β c from (11.51) into (10.44), the phase constant of TM mode waves in the circular waveguide are found as

β g = β 2 − (χ mn a )2

(11.52)

Note that β g for TE mode given in (11.17) depends on the roots of the derivative of Bessel function χ ′mn , while that for TM mode depends on the roots of the Bessel function χ mn . Substituting β c from (11.51) into (11.49) and letting D ′′ = E mn , yields ⎛χ ⎞ Ez (ρ ,φ ) = Emn J n ⎜ mn ρ ⎟ cos[m(φ − φo )] ⎝ a ⎠

(11.53)

Using (11.48), we can write

⎛ ωε ⎞ ⎛ 1 ∂Ez ⎞ ∂E a ρ H ρ + aφ Hφ = j⎜⎜ 2 ⎟⎟ ⎜⎜ − aφ z + a ρ ⎟ ρ ∂φ ⎟⎠ ∂ρ ⎝ βc ⎠ ⎝

(11.54)

Substituting Ez from (11.53) into (11.54) and equating the quantities in the direction of ρ and φ in both sides, the components of the transverse quantity Ht can be obtained as

⎛ ωε ⎞ m H ρ = − j⎜⎜ 2 ⎟⎟ Emn J m (βc ρ ) sin[m(φ − φo )] ⎝ βc ⎠ ρ ⎛ ωε ⎞ ∂J (β ρ ) Hφ = − j⎜⎜ 2 ⎟⎟ Emn m c cos[m(φ − φo )] ∂ρ ⎝ βc ⎠

(11.55a) (11.55b)

Similarly, from (10.50), we can write

⎛β ⎞ a ρ Eρ + aφ Eφ = −⎜⎜
⎟⎟ (aφ H ρ − a ρ Hφ ) ⎝ ωε ⎠

(11.56)

Circular Waveguides

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Substituting H ρ and H φ from (11.55a) and (11.55b) respectively into (11.56), and equating the quantities in the ρ and φ directions on both sides, we get

⎛β ⎞ ∂J (β ρ ) Eρ = − j⎜⎜ g2 ⎟⎟ Emn m c cos[m(φ − φo )] ∂ρ ⎝ βc ⎠ ⎛β ⎞ m Eφ = j⎜⎜ g2 ⎟⎟ Emn J m (βc ρ ) sin[m(φ − φo )] ⎝ βc ⎠ ρ

(11.57a) (11.57b)

Substituting, Ez from (11.53), and the components of Ht , and Et from (11.55a), (11.55b), (11.57a), and (11.57b) respectively into (11.2a) and (11.2b), the expressions of the spatial TM mode waves components in a circular waveguide are obtained as

βg χ mn − jβ Emn J m′ (χ mn ρ a ) cos[m(φ − φo )] e 2 βc a β m − jβ z Eφs = j g2 Enm J m (χ nm ρ a ) sin[m(φ − φo )] e βc ρ

Eρs = − j

gz

g

Ezs = Enm J m (χ nm ρ a ) cos[m(φ − φo )] e
ωε m − jβ z H ρs = − j 2 Emn J m (χ mn ρ a ) sin[m(φ − φo )] e g βc ρ ωε χ − jβ z Hφs = − j 2 mn Emn J m′ (χ mn ρ a ) cos[m(φ − φo )] e g βc a H zs = 0 − jβ z

(11.58a) (11.58b) (11.58c) (11.58d) (11.58e) (11.58f)

As in the rectangular waveguides, the constant Enm is in ( V m ). The corresponding instantaneous components can be obtained by substituting Ez and the components of Et , and Ht in (10.1) and (10.2), as follows

βg χ mn E J ′ (χ ρ a ) cos[m(φ − φo )] sin (ω t − βg z ) βc2 a mn m mn β m Eφg = − g2 Enm J m (χ nm ρ a ) sin[m(φ − φo )] sin (ω t − βg z ) βc ρ

E ρg =

(11.59a) (11.59b)

732

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E z
= Enm J m (χ nm ρ a ) cos[m(φ − φo )] cos(ω t − β
z ) ωε m H ρg = 2 Emn J m (χ mn ρ a ) sin[m(φ − φo )] sin (ω t − β g z )

βc ρ

H φg =

ωε χ mn E J ′ (χ ρ a ) cos[m(φ − φo )] sin (ω t − β g z ) β c2 a mn m mn H z
= 0

(11.59c) (11.59d) (11.59e) (11.59f)

Note that J m′ (u ) = ∂J m (u ) ∂u , where u = χ mn ρ a . The roots χ mn of J m (χ mn ) = 0 are listed in Table 11.3 for some values of m and n . Table 11.3. The roots χ mn
of J m (x ) = 0 form andn m=0

m=1

m=2

m=3

m=4

m=5

m=6

m=7

m=8

n=1

2.4049

3.8318

5.1357

6.3802

7.5884

8.7715

9.9361

11.0864

12.2251

n=2

5.5201

7.0156

8.4173

9.7610

11.0647

12.3386

13.5893

14.8213

16.0378

n=3

8.6537

10.1735

11.6199

13.0152

14.3726

15.7002

17.0038

18.2876

19.5545

n=4

11.7915

13.3237

14.7960

16.2235

17.6160

18.9801

20.3208

21.6415

22.9452

n=5

14.9309

16.4706

17.9598

19.4094

20.8269

22.2178

23.5861

24.9349

26.2668

11.1.4. Characteristics of the TM Mode Fields in the Waveguide 11.1.4.1. Cut-off Frequency The cut-off frequency of T M mn mode waves in a circular waveguide can be obtained by letting β g = 0 in (11.52) and solving for f = f cTM mn . Doing this, the cut-off frequency is obtained as

f cTM mn =

1

χ nm 2π με a

(11.60)

and the corresponding wavelength is

λc =

2π a χ mn

The values of χ mn for 0 < m < 8 and 1 < n < 5 , are listed in Table 11.3.

(11.61)

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11.1.4.2. Phase and Propagation Constants From (11.52) the phase constant in a circular waveguide supporting a T M mn mode waves can be written as

βg = β 2 − (χ mn a )2 = β 1 − ( f cTM

f

)

(11.62)

1 − f cTM mn f

)

(11.63)

mn

2

Using
= jβ
, the propagation constant is

(

(χ mn a)2 − β 2 = jβ


=

2

11.1.4.3. Wavelength The wavelength of TM mn mode waves in a circular waveguide can be expressed using (11.62), as

λg =

2π

βg

2π

=

=

ω2με− (χ mn a )

2

(

λ

1− f

TM mn c

f

(11.64)

)

2

11.1.4.4. Wave Impedance The wave impedance for TMmn mode waves in a circular waveguide can be obtained using (10.52) and (11.62) as Z TM mn =

βg ωε

ω2με − (χ mn a ) = η 1 − f cTM mn f ωε 2

=

(

)

2

(11.65)

11.1.4.5. The Phase Velocity and Group Velocity The phase velocity υ p and group velocity υ
in a circular waveguide that supporting TM modes are similar in form to that of the TE mode case and are given by

υp = υg =

ω

=

f

)

υ dω = dβg 1 − fcTM mn f

)

βg

(

υ

1− f

(

TM mn c

2

2

(11.66)

(11.67)

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11.1.4.6. Poynting Vector and Power Following the same analysis as for the TE case, the instantaneous Poynting vector for the TM mode waves in a circular waveguide can be obtained as

⎛ E 2+ E ρ
φ
T = a z ⎜⎜ ⎜ 2ZTM mn ⎝

2

⎞ 1 ⎟− E a H ∗ −a H ∗ ⎟⎟ 2 z
ρ φ
φ ρ
⎠

(

)

(11.68)

Using (10.32) and (10.33), the average power through the waveguide becomes

1

∫ ∫ (Eρ

2π a

) )

2

ρ dρ dφ 2Z TM mn 0 0 2 Z TM mn 2π a = H ρs 2 + Hφs ρ dρ dφ ∫ ∫ 2 00 =

PaTM mn

2

s

(

+ Eφs

(11.69)

Substituting Eρs and Eφs from (11.58a) and (11.58b) respectively into (11.69) and evaluating the integrals with respect φ , yields TM mn a

P

⎛ β
⎜ = Emn 2Z TM mn ⎜⎝ βc2

π

⎞ ⎟⎟ ⎠

2 ⎡ m2 2 ⎤ ⎛ χ mn ⎞ ′2 ( ) ( ) J χ ρ a ρ J χ ρ a + ⎜ ⎟ ⎢ ⎥ dρ m mn m mn ∫0 ⎢ ρ ⎝ a ⎠ ⎥⎦ ⎣ (11.70)

2a

Letting u = χ′mn ρ a , (10.70) becomes TM mn a

P

⎛ β
⎞ ⎜⎜ 2 Emn ⎟⎟ = 2Z TM mn ⎝ βc ⎠

π

2 χ mn

∫ 0

⎡ m2 2 ⎤ 2 ⎢ u J m (u ) + u J m′ (u )⎥ du ⎣ ⎦

(11.71)

Using J m′ (u ) = − J m+1 (u ) + (m u )J m (u ), the average power through the circular waveguide supporting TM mode can be expressed as 2

TM mn a

P

⎞ π ⎛ βg ⎜⎜ 2 Emn ⎟⎟ I mn (χ mn ) = 2Z TM mn ⎝ βc ⎠ 2 2 π a Emn 2 ( = f fcTM ) 1 − ( fcTM 2 2η χ mn mn

[

mn

f

) ] I (χ ) 2

mn

mn

(11.72)

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Where I mn ( χ mn ) is given by

I mn ( χ mn ) =

χ mn

∫ 0

⎡ 2m 2 2 ⎤ 2 ⎢ u J m (u ) + uJ m +1 (u ) − 2 mJ m (u ) J m +1 (u )⎥ du (11.73) ⎣ ⎦

Following the procedure for the integral I mn (χ ′mn ) in Section 11.1.2, I mn (χ mn ) can be obtained from (11.73) as

I mn (χ mn ) = mJ m2 (χ mn ) +

[

]

2 χ mn J m2 +1 (χ mn ) − J m (χ mn ) J m + 2 (χ mn ) 2

(11.74)

Table 11.4. I mn ( χ mn ) values for 0 < m < 8 and 1 < n < 5 . m=0

m=1

m=2

m=3

m=4

m=5

m=6

m=7

m=8

n=1

0.7794

1.1909

1.5215

1.8108

2.0736

2.3173

2.5466

2.7643

2.9723

n=2

1.7640

2.2165

2.6090

2.9635

3.2912

3.5986

3.8897

4.1676

4.4342

n=3

2.7591

3.2267

3.6476

4.0355

4.3986

4.7419

5.0691

5.3828

5.6849

n=4

3.7567

4.2322

4.6695

5.0781

5.4640

5.8313

6.1832

6.5218

6.8489

n=5

4.7553

5.2355

5.6836

6.1063

6.5084

6.8932

7.2632

7.6204

7.9664

Values of I mn ( χ mn ) for 0 < m < 8 and 1 < n < 5 , are listed in Table 11.4. Since

χ mn satisfies J m (χ mn ) = 0 , then from Bessel functions properties it can be shown

that J m+1 (χ mn ) = J m−1 (χ mn ) = J m′ (χ mn ) and (11.74) reduces to

I mn (χ mn ) =

χ 2mn 2 χ2 χ2 J m +1 (χ mn ) = mn J m2 −1 (χ mn ) = mn J m′2 (χ mn ) 2 2 2

(11.75)

Substituting I mn (χ mn ) from (11.75) into (11.72), the average power PaTM mn becomes 2

TM mn a

P

⎞ π ⎛ βg ⎜⎜ 2 Emn ⎟⎟ χ 2mn J m2 −1 (χ mn ) = 4 Z TM mn ⎝ β c ⎠ 2 ⎞ 2 2 π ⎛ βg ⎜ E ⎟ χ mn J m +1 (χ mn ) = 4 Z TM mn ⎜⎝ β c2 mn ⎟⎠

(11.76)

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Substituting β
, β
, and Z TM mn from (11.51), (11.62), and (11.65) respectively, into (11.76) with some algebraic manipulations, the average power through a circular waveguide supporting TM mode waves can be expressed as TM mn a

P

π a 2 Emn = 4η

2

(f

f cTM mn

) [1 − ( f 2

TM mn c

f

) ] J (χ ) 2

2 m −1

(11.77)

mn

11.2. DOMINANT MODE IN THE CIRCULAR WAVEGUIDE We know that the dominant mode is the mode of the lowest cut-off frequency. The lowest possible modes in a circular waveguide are the TE 01 , TE 11, TE 21 ,

TM 01 , and TM11 modes and their corresponding frequencies are f cTE , f cTE , f cTE , f cTM , and f cTM11 respectively. Using (11.20) and (11.60) along with 01

12

11

01

Tables 11.3 and 11.4, the order of the above modes from lower to higher is TE 11, TM 01 , TE 21 , TE 01 , and TM11 . Therefore, the lowest cut-off frequency in circular waveguides is f cTE11 which corresponds to the TE 11 mode. Thus, the dominant mode in a circular waveguide is the TE 11 mode and the next higher mode is TM 01 mode. The instantaneous electromagnetic fields components of the dominant mode can be obtained by putting m = 1 and n = 1 in (59a) – (59f). The lines of electric and magnetic fields in orthogonal planes in the circular waveguide are shown in Fig. (11.2) for the dominant TE 11. Note that the dominant mode fields characterized by only one electric field component in the y direction, and one transverse magnetic component in the x direction, in addition to the magnetic field component in the direction of propagation. Consider the fields in (11.59a) – (11.59b) at the instant t = 0 sec with φo = 0 , on the orthogonal planes A and B within the circular waveguide of Fig. (11.2). The planes are Plane A: z = λ
4 , 0 ≤ ρ ≤ a , 0 ≤ φ ≤ 2π and Plane B: φ = 0, π ,

0 ≤ ρ ≤ a , − λ
4 ≤ z ≤ λ
4 . The electric and magnetic fields at the instant t = 0 sec are as follows: Plane A:

E zg = H zg = 0

(11.78a)

Circular Waveguides

Electromagnetics for Engineering Students

(ωμ =−

βc2 )

E1n J1 (χ1′n ρ a ) sin φ ρ Eφg = −(ωμ βc2 )E1n J1′(χ1′n ρ a ) cosφ

E ρg

H ρg = (βg βc2 )E1n J1′(χ1′n ρ a ) cosφ

H φg = −

(β

g

ρ

β c2 )

E1n J1 (χ1′n ρ a ) sin φ

737

(11.78b) (11.78c) (11.78d) (11.78e)

Plane B (φ = 0, z = λg /4):

E ρg = E zg = H φg = 0

(11.79a)

Eφg = −(ωμ βc2 )E1n J1′(χ1′n ρ a )

(11.79b)

H ρg = +(βg βc2 )E1n J1′(χ1′n ρ a )

(11.79c)

E ρg = Eφg = E zg = H ρg = H φg = 0

(11.80a)

H zg = Fmn J m (χ ′mn ρ a ) cos(β g z )

(11.80b)

E ρg = E zg = H φg = 0

(11.81a)

Eφg = +(ωμ βc2 )F1n J1′(χ1′n ρ a )

(11.81b)

H ρg = −(βg βc2 )F1n J1′(χ1′n ρ a )

(11.81c)

E ρg = Eφg = E zg = H ρg = H φg = 0

(11.82a)

H zg = − Fmn J m (χ ′mn ρ a ) cos(β g z )

(11.82b)

Plane B (φ = 0, z = λg /2):

Plane B (φ = π, z = λg /4):

Plane B (φ = π, z = λg /2):

Using the above equations, the field lines on the planes A and B, can be constructed as shown in Fig. (11.2).

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λg /4

y

-λg /4                   

       

z

a

x

E H

x

Fig. (11.2). Electric and magnetic fields lines in a circular waveguide supporting TE11 mode waves.

The characteristics of the dominant mode are obtained by letting m = 1 and n = 0 in the all relevant TE mn mode characteristics and the results are summarized in Table 11.5. Table 11.5. The characteristics of the dominant mode in a circular waveguide. Parameter

Mathematical Expression

Parameter

Cut-off Frequency f cTE11

′ χ 11 2π με a

Wave Impedance Z TE

Cut-off Wavelength λ


2π a ′ χ 11

Phase Constants β


β 1 − f cTE

Propagation Constants


jβ 1 − f cTE

Wavelength λ


1

Mathematical Expression

η

(

1 − f cTE

11

f

)

2

11

(

λ

Average Power Pa11

11

(

11

(

11

1 − f cTE

ηπ a 2 H 11

2

16.76

(f

TE11 c

f

)

f

)

2

Phase Velocity υ


υ 1 − f cTE

f

)

Group Velocity υ


υ

f

)

f

)

2

2

(

1 − f cTE11 f

(

(

11

1 − f cTE

11

f

)

)

2

2

2

2

Note: β = 2π f με , λ = 2π β , η = μ ε , and

υ = 1 με

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Example 11.3 A circular waveguide filled with a lossless dielectric medium of dielectric constant 4, must operate in a single dominant mode over a bandwidth of 1 GHz, find the (a) Cross-section radius of the waveguide. (b) Lower, center, and upper frequencies of the bandwidth. (c) Transmitted power through the waveguide at the center frequency. Solution ′ = 1.8412, (a) The dominant mode is TE11 and the next higher mode is TM 01 . χ11 and χ 01 = 2.4049 , then the lowest and highest frequencies from (11.20) are

fL = f

TE11 c

′ ′ χ11 1 χ11 3 × 108 1.8412 = = = 4π a 2π με a 2π μ 0ε 0ε r a 1

f H = f cTM 01 =

χ 01 1 χ 01 3 × 108 2.4049 = = 4π a 2π με a 2π μ 0ε 0ε r a 1

Since the bandwidth is 1.0 GHz , then

fH − fL =

3 × 10 8 2.4049 3 × 10 8 1.8412 − = 1 × 10 9 4π a 4π a

Solving for a then the radius is

a=

3 × 108 (2.4049 − 1.8412) = 0.13457 m 4π × 109

(b) Substituting in the above equations, the lower frequency can be computed as 3 × 108 1.8412 f L = f cTE11 = = 3.2664 GHz 4π 0.13457 and the higher frequency as

f H = f cTE11 =

3 × 108 2.4049 = 4.2664 GHz 4π 0.13457

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The center frequency is

fo =

fL + fH 2

3.2664 + 4.2664 2 = 3.7664 GHz =

(c) At f o = 3.7664 GHz , the mode in the waveguide is the TE11 mode. Hence, from Table 11.5, the transmitted power through the waveguide at the center frequency fo = 3.7664 GHz is 2

TE 11 a

P

a 2 E11 = 13.24η

(

[

)

]

⎡ 1 − f TE11 f 2 ⎤ (0.013457)2 1 − (3.2664 3.7664)2 2 c = E ⎥ ⎢ 11 2 2 TE 11 f ⎥⎦ 13.24 × 60π (3.2664 3.7664) ⎣⎢ f c

(

)

2

= 2.4 × 10 −8 E11 W 11.3. POWER LOSS AND ATTENUATION 11.3.1. Power Loss in the Circular Waveguide In this case, a n = a ρ and the cylindrical surface ρ = a , φo ≤ φ ≤ 2π + φo represents the wall of the waveguide. Using (10.146), the surface current density on the waveguide wall can be obtained as K ρ = a = a ρ × (a ρ H ρs + aφ Hφs + a z H zs ) = a z Hφs − aφ H zs

(11.83)

Following the same analysis as in the case of the rectangular waveguide and taking into consideration that dl = a dφ , the average power loss in the wall of the circular waveguide becomes a Pc = 2δσ c

2π

⎛ H 2 + H 2⎞ zs ⎠ ∫ ⎝ ϕs 0

dϕ

(11.84)

ρ =a

11.3.2. Waveguides Supporting TE Mode Fields Using the TE mode electromagnetic fields components (11.18a) – (11.18b), the power loss per unit length PcTE mn in a circular waveguide can be determined.

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Substituting H φs and H zs from (11.18e) and (11.18f) into (11.84), the power loss per unit length of the circular waveguide with TE modes fields can be written as

a[ H mn J m (χ ′mn )] = 2 δσ c

⎡⎛ β m ⎞ 2 ⎤ g 2 2 ⎜ ⎟ ⎢ ⎥ dφ ( ) ( ) [ ] [ ] − + − sin m φ φ cos m φ φ o o ∫0 ⎢⎜⎝ βc2 a ⎟⎠ ⎥⎦ ⎣ (11.85)

2 2π

TE mn c

P

This can be simplified to TE mn c

P

2 2 ⎤ aπ H mn ⎡⎛ β g m ⎞ ⎢⎜⎜ 2 ⎟⎟ + 1⎥ J m2 (χ ′mn ) = 2 δ σ c ⎢⎝ β c a ⎠ ⎥⎦ ⎣

(11.86)

Substituting β c and β g from (11.16) and (11.22), and carrying out some manipulations, (11.86) can be expressed in the following form 2

TE mn c

P

(

aπ H mn f f cTE mn = 2 δσ c

) {(m χ′ ) + [1 − (m χ′ ) ] ( f 2

2

2

mn

mn

TE mn c

f

) } J (χ′ ) 2

2 m

mn

(11.87) Substituting the average power of the TE mode obtained in (11.47), and PcTE mn from (11.87), into (10.140), the attenuation constant for the circular waveguide supporting TE mode can be obtained as mn = αTE c

1

η δσ c a 1 − ( f cTE

mn

f

)

2

⎡ 2⎤ 1 TE mn + f f ⎢ ⎥ c 2 ⎣ (χ ′mn m) − 1 ⎦

(

)

(11.88)

11.3.3. Waveguides Supporting TM Mode Fields Following the same analysis as for the TE mode case, the power loss per unit length of the circular waveguide that supporting TM modes PcTM mn can be obtained. Substituting H φs from (11.58e) into (11.84) with H zs = 0 , yields TM mn c

P

aπ = 2 δσ c

⎡ ωε χ mn ⎤ H mn J m′ (χ mn )⎥ ⎢ 2 ⎣ βc a ⎦

2

(11.89)

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Substituting β c from (11.51) into (11.89), and taking into consideration that

J m′ (χ mn ) = J m−1 (χ mn ) , (11.89) becomes

aπ H mn J m2 −1 (χ mn ) = f f cTM mn 2 2η δ σ c 2

TM mn c

P

(

)

2

(11.90)

The average power of the TM mode in a circular waveguide given by (11.77). Hence, using (11.77) and (11.90) along with (10.140) the attenuation constant for a circular waveguide supporting TM mode waves becomes

α cTM mn =

PcTM mn 1 1 = TM mn η δ σ c a 1 − f TM mn f 2 Pa c

(

)

2

(11.91)

Example 11.4 An air-filled circular waveguide has a cross-section radius of 5 cm. If the waveguide is made of copper σ c = 5.8 × 10 7 S m , find the conduction attenuation

(

)

factor in dB/m for the TM 01 mode at 30 GHz. Find the length of the waveguide that dissipates 10% of the input power. (Ignore the attenuation due to the dielectric loss). Solution Since a = 5 cm the cut-off frequency for the TM 01 mode is

f cTM 01 =

2.4049 χ 01 = × 3 × 108 = 22.95 GHz 2π a μ 0ε 0 2π × 0.05

The skin depth at 30 GHz is

δσ c =

1 5.8 × 107 = = 22.129 S π × 30 × 109 × 4π × 10− 7 π f μ0 σ c

01 can be obtained by substituting m = 0 , n = 1 in (11.91) as α TM c

01 α TM = c

1 1 η δ σ c a 1 − f TM 01 f c

(

)

2

1 1 120π × 22.129 × 0.005 1 − (22.95 30 )2 = 0.03721 Np m =

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In dB

(

)

mn ⇒ α TE = 20 log e −0.03721 = −0.3232 dB m c

Let the length of which Pd = 0.1Pa , is l , where Pd and P are the dissipated and input power respectively. Using (10.141), then

(

Pd = Pa 1 − e−2(α +α
)l

(

mn −2 α TE c

)

)

l ≈ Pa 1 − e TE mn P −P P ⇒ e −2 α c l = a d = 1 − d = 1 − 0.1 = 0.9 Pa Pa ln(0.9) ln(0.9) ⇒ l = TE mn = − = 1.416 m 2αc 2 × 0.03721

SOLVED PROBLEMS Solved Problem 11.1 Design an air-filled circular waveguide such that the difference between the cutoff frequencies of the dominant mode and the next higher-order mode is 1 GHz. Determine the cut-off frequencies of the dominant mode and the next mode. Solution The dominant mode propagating in the circular waveguides is the TE11 and the next higher mode is the TM 01 , then their cut-off frequencies are

f cT M 01

χ11′

1.8412 2π a μ 0ε 0 2π a μ 0ε 0 χ 01 2.4049 = = 2π a μ 0ε 0 2π a μ 0ε 0

f cTE11 =

=

The difference between the frequencies is

f cT M 01 − f cTE11 = 1 × 109 =

1 2π a μ 0ε 0

(2.4049 − 1.8412)

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It is required that f cTM 01 − f cTE11 = 1 GHz

⇒ a=

3 × 108 × (2.4049 − 1.8412) = 0.026896 m = 2.6896 cm 2π × 1 × 109

The cut-off frequencies are

1.8412 × 3 × 108 = 3.266 GHz 2π × 0.026896 2.4049 = × 3 × 108 = 4.266 GHz 2π × 0.026896

f cTE11 = f cT M 01 Solved Problem 11.2

If the radius of an air-filled circular waveguide is 1.2 cm, for the TM11mode, find: (a) (b) (c) (d)

The cut-off frequency. The guide wavelength at the operating frequency of 10 GHz. The phase constant. The wave impedance.

Solution (a) The cut-off frequency for TE11 mode

f cTE11 =

1.8412 χ11′ = × 3 × 108 = 8.79 GHz 2π a μ 0ε 0 2π × 0.012

(b) Since the operating frequency is 10 GHz > f cTE 11 , the wave will propagate in the waveguide with wavelength

λ
=

λ 1 − ( fc f )

2

3 × 108 1 = × = 6.29 cm 9 2 2 × 10 1 − (8.79 10 )

(c) The phase constant is

β
=

2π

λ


=

2π = 1.0 rad cm 6.29

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(d) The wave impedance is

Z TE11 =

η 1 − ( fc f )

2

=

120π

1 − (8.79 10 ) 120π = = 790.23 Ω 0.4768 2

Solved Problem 11.3 Find all TE mn modes that may exist in an air-filled circular waveguide having a cross-section radius of 1.089 cm, and operates at 12.5 GHz. Solution The lowest cut-off frequency for the circular waveguides is f cTE 11 where

f cTE11 =

1.8412 χ11′ = × 3 × 108 = 8.075 GHz 2π a μ 0ε 0 2π × 0.01089

The next higher modes cut-off frequencies are

f cTE01 = f cTE 21 =

χ 01′ 2.405 = × 3 × 108 = 10.55 GHz 2π a μ 0ε 0 2π × 0.01089 ′ χ 21 2π a μ 0 ε 0

=

3.054 × 3 × 108 = 13.4 GHz 2π × 0.01089

Since f cTE11 < f cTE 01 < 12.5 GHz < f cTE 21 , then only the TE11 and TE 01 modes can exist in the waveguide at 12.5 GHz.  Solved Problem 11.4 A circular waveguide of radius a = 3 cm, is used at a frequency of 2 GHz. For the TE11 mode, determine the cutoff frequency and the bandwidth over single-mode operation if the waveguide is filled with polystyrene (εr = 2.56).

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Solution The dominant mode is the TE11 mode, then the cut-off frequency is TE 11

f

χ11′ 1.8412 3 × 108 = = × = 1.8315 GHz 2π a μ 0ε 0ε r 2π × 0.03 2.56

The next mode to the dominant TE11 mode is the TM 01 mode and its cut-off frequency is

f

TM 01

χ01 2.4049 3 × 108 = = × = 2.3922 GHz 2π a μ 0ε 0ε r 2π × 0.03 2.56

The bandwidth over single-mode operation (The bandwidth that permits the propagation of only the dominant mode) is

f 01TM − f11TE = 2.3922 − 1.8315 = 560 .7 MHz .

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PROBLEMS 11.1. In an air-filled circular waveguide of radius a = 2cm, the z-component is given by

E zg = 10 J 0 (0.882ρ ) cos(12π ×109 t − βg z )

Vm

Find the (a) (b) (c) (d)

Mode of operation. Propagation constant. Wave impedance. Average power transmitted through the waveguide.

11.2. An air-filled circular waveguide of radius 3 cm, operating at 6 GHz has the spatial magnetic field

Hφs =

1 − jβ z J1 (0.6137 ρ ) sin φ e g A m 10 ρ

determine the (a) (b) (c) (d)

Mode of operation Cut-off frequency Phase velocity Average power transmitted through the waveguide.

11.3. An air-filld circular waveguide of radius a = 1.5 cm , is used at a frequency of 4 GHz. If the waveguide operates at the TE 01 mode, find the (a) (b) (c) (d)

Cut-off frequency. Phase constant in the waveguide. Wavelength in the waveguide. Phase velocity and group velocity.

11.4. A circular waveguide filled with a lossless dielectric medium of dielectric constant 2.56, to be operated in a single dominant mode over a bandwidth of 1.5 GHz, find the (a) Cross-section radius of the waveguide. (b) Lower, center, and upper frequencies of the bandwidth.

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(c) Transmitted power through the waveguide at the center frequency. 11.5. An air-filled circular waveguide has a radius a = 2.0 cm . If the waveguide operates at 9 GHz, find for the dominant mode, the (a) (b) (c) (d)

Cut-off frequency. Phase constant. Phase and group velocities. Wave impedance.

11.6. The radius of a circular waveguide is a . Compute the cut-off frequencies of the modes: TE 01 , TE11 , TE 21 , TM11, TM12 , and TM 22 normalized to the cut-off frequency of the dominant mode. 11.7. Find the time-average power for the lowest five modes in a circular waveguide normalized to the time-average power of the dominant mode. 11.8. A circular waveguide of the cross-section with radius a = 3.0 cm is filled with polystyrene (εr = 2.56). If the waveguide is operated at 2 GHz , for the dominant mode, find the (a) (b) (c) (d)

Cut-off frequency. Phase constant in the waveguide. Wavelength in the waveguide. Wave impedance.

11.9. An air-filled circular waveguide of radius a = 1.0 cm that is operated at 4 GHz . Find the (a) Cut-off frequency of the dominant mode. (b) Cut-off frequency of the next higher mode to the dominant mode. (c) Bandwidth over single mode operation. 11.10. A circular waveguide has a cross-section with radius a = 4.0 mm . The waveguide is filled with a dielectric with . At the center of the waveguide, the amplitude of the longitudinal electric field Ez is equal to 6000 V/m. (a) Determine the cut-off frequencies of the two lowest TM modes for this waveguide.

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(b) Consider a TM11 wave traveling down the waveguide with a single frequency that is half-way between the frequencies determined in part (a). Calculate the time-averaged power transmitted down the waveguide. 11.11. An air-filled circular waveguide of radius a operates in the TE11 mode and the operating frequency is higher than the cut-off frequency by 25%. If the conductivity of the waveguide is σ c  show that

σc a3

11 αTE = 0.095 c

If the waveguide is filled with a medium of dielectric constant ε r , then 1

αcTE11 = 0.095 ε r 4

σc a3

11.12. An air-filled circular waveguide of radius a operates in the TM mn mode and the operating frequency is higher than the cut-off frequency by a IDFWRUr . If the conductivity of the waveguide is σ c show that mn = αTM c

χ mn σ c 2η0 a3r 1 − r 2

(

)

11.13. An air-filled circular waveguide of radius a = 2.0 cm , operating in its dominant mode. Find the length of the waveguide required to introduce a phase shift of 150 in a signal propagating through it at 6 GHz . 11.14. A circular waveguide of radius a = 1.15 cm to be operated in its dominant mode only over a bandwidth of 1.75 GHz . Find the dielectric constant of the medium that fills the waveguide. 11.15. A circular waveguide of radius a = 3.0 cm , is filled with a dielectric material of relative permittivity ε r = 2.56 and operating in its dominant mode. Find the length of the waveguide required to introduce a delay of 2.0 μs in a signal propagating through it at 3.75 GHz .

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11.16. The operation frequency of a rectangular waveguide of dimensions 2.286 cm ×1.016 cm , is 9 GHz. If the waveguide is made of copper

(σ

)

and filled with polystyrene = 5.8 × 107 S m (ε r = 2.56, σ d ωε r ε0 = 3 × 10−4 ), find the attenuation constant and the total c

power loss in 50 m along the waveguide for the dominant mode.

(

)

(

)

11.17. A rectangular waveguide is made of copper σ c = 5.8 × 10 7 S m , and filled

(

)

with polystyrene ε r = 2.56, σ d = 10 −17 , has dimensions a = 4.2 cm and b = 2.6 cm . If the waveguide is used to transmit a power of 1.2 kW at 30 GHz, compute the dissipated power in a 60 m of the guide length at the mode TE10 . 11.18. A circular waveguide is made of aluminum σ c = 3.5 × 10 7 S m and filled

(

)

with Teflon ε r = 2.6, σ d = 10 −15 , operates at the TE11 mode. If the radius of the waveguide is a = 2.5 cm , calculate attenuation constant and the total power loss in in dB over 1.5 m of the waveguide at 4 GHz.

(

)

11.19. A circular waveguide is made of copper σ c = 5.8 × 10 7 S m and filled

(

with Teflon ε r = 2.6, σ d = 10

−15

). A wave of 9 GHz is propagating in the

waveguide at the TE10 mode. If the radius of the waveguide is a = 2.2 cm , calculate the total power loss in a length of 40 m of the waveguide. 11.20. A circular waveguide of the cross-section with radius a = 2 2 cm , is filled with a dielectric of relative permittivity ε r = 16 1 − j10 −4 . If the

(

)

waveguide is operated at the TM11 mode and the operating frequency is higher than the cut-off frequency by 15%. Find the attenuation constant and the distance that the wave travels down the waveguide before its amplitude is reduced by 4%.

Electromagnetics for Engineering Students, 2017, 751-784

751

CHAPTER 12

Resonant Cavities Abstract: The resonant cavities are structures used to store the electromagnetic energy at high frequencies. Cavities may be rectangular, cylindrical, or spherical in geometry. This chapter is devoted to discuss the rectangular and cylindrical cavities and their characteristics. The electromagnetic fields in the rectangular and cylindrical cavities are obtained by considering the cavities as shorted waveguides at their ends with two additional walls. The most important parameters that characterize the cavity are its resonant frequency and quality factor. The quality factor is a parameter that gives a measure of the ratio of the stored energy in the cavity to the dissipated power at a certain frequency. A detailed derivation for the electromagnetic fields, resonant frequency, and the quality factor are presented for both the rectangular and cylindrical cavities. The topics of the chapter are supported by a number of illustrative examples, figures, and solved problems. Homework problems are included at the end of the chapter.

Keywords: Cylindrical cavity electric energy magnetic energy power loss quality factor rectangular cavity resonance frequency resonant cavity surface current density INTRODUCTION The resonant cavities are structures used to store the electromagnetic energy at high frequencies. They are different from the RLC resonant circuits in their construction and efficiency. Beyond certain frequency, the RLC resonant circuits become inefficient because their power dissipation increases with frequency, while resonant cavities are characterized by low losses with high Q-factor at high frequencies. Moreover, resonant cavities are characterized by a simple construction than that of RLC resonant circuits. These merits of the resonant cavities make them ideal resonators to store electromagnetic energy efficiently at high frequencies. More information on the resonant cavities is available in [50, 92, 98, 111-117]. Cavities may be rectangular, cylindrical, or spherical in geometry. Only the rectangular and cylindrical cavities will be discussed in this chapter. The rectangular or cylindrical cavities can be considered as a shorted metallic waveguide at the ends z = 0 and z = c by insertion of two conducting plates as shown in Fig. (12.1). Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

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z

z a

c

ε, μ

c

ε, μ x

x b a

y

y (a)

(b)

Fig. (12.1). Geometry of resonant cavities. (a) Rectangular. (b) Circular.

12.1. ELECTROMAGNETIC FIELDS The electromagnetic waves in the cavity can be analyzed by considering the cavity as a waveguide of length c , with two additional walls at z = 0 and z = c . The spatial electromagnetic fields in the cavity can be expressed as the same form as in the waveguides as in (10.1) – (10.6). Let us rewrite the spatial fields E s and

H s as Es (x, y, z ) = Ets + a z Ezs = E(x, y ) e

− jβ g z

Hs (x, y, z ) = Hts + a z H zs = H(x, y ) e

− jβ g z

(12.1) (12.2)

where E and H are complex vectors that are independent of z and can be written in terms of their transverse components Et and H t , and z directed components Ez and H z , as

E = Et + a z Ez H = Ht + a z H z

(12.3) (12.4)

The presence of the additional walls at z = 0 and z = c , makes the fields Ezs and H zs reflect continuously from the walls at z = c and z = 0 . The reflection from the wall at z = c results in a wave traveling in the

(− z )

direction with z

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dependence factor e+ jβ
z , while those from the wall at z = 0 results in a wave traveling in the (+ z ) direction with z dependence factor e− jβ
z . 12.1.1. TE Mode Fields in a Resonant Cavity For TE mode fields, we have Ez = 0 and H z ≠ 0 . The spatial waves traveling in the (+ z ) direction in the cavity, can be written as

E +s = K + Et e

− jβ
z

H +s = K + (H t + a z H z ) e

(12.5) − jβ
z

(12.6)

Note that the propagation constant is − jβ
along the negative z direction, then the spatial waves in the (− z ) direction become

E −s = K − Et e

+ jβ
z

H −s = K − (− H t + a z H z ) e

(12.7) + jβ
z

(12.8)

Superimposing the fields in the (+ z ) and (− z ) directions, the spatial electric field Esc and magnetic field H sc in the cavity can be obtained as

E sc = E+s + E−s

(

= Et K + e H sc = H +s + H −s

(

= Ht K +e

− jβ g z

− K −e

− jβ g z

+ jβ g z

+ K −e

+ jβ g z

)+ a H (K e

)

+ − jβ g z

z

z

(12.9)

+ K −e

+ jβ g z

)

(12.10)

The transverse electric field Et vanishes at z = 0 and z = c , then

Et

z =0 z =c

=0

(12.11)

Enforcing the boundary conditions given by (12.11) on (12.9), yields K + = − K − and e + j 2β
c = 1 . Consequently,

βg =



pπ , c

p = 0,1, 2,

(12.12)

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Substituting K + = − K − and β
into (12.9) and (12.10), the general expressions for TE mode electromagnetic fields in the cavity can be obtained as

E sc = 2 jK + E t sin ( pπ z c )

(12.13)

H sc = 2 K + [H t cos( pπ z c ) + a z jH z sin ( pπ z c )]

(12.14)

12.1.2. TM Mode Fields in a Resonant Cavity In this case, Ez ≠ 0 and H z = 0 . Hence, The TM mode spatial fields in the cavity in the (+ z ) direction can be written as

E +s = K + (Et + a z E z ) e H +s = K + H t e

− jβ
z

(12.15)

− jβ
z

(12.16)

Similarly, the TM mode spatial waves in the (− z ) direction can be written as

E −s = K − (− Et + a z E z ) e H −s = K − H t e

+ jβ
z

(12.17)

+ jβ
z

(12.18)

Superimposing the fields in (+ z ) and (− z ) directions, the spatial electric field Esc and magnetic field H sc in the cavity can be obtained as

E sc = E+s + E−s

(

− jβ g z

= Et K + e H sc

+ jβ z

)

(

− K −e g + a z Ez K + e = H +s + H −s

(

= Ht K +e

− jβ g z

+ K −e

− jβ g z

+ jβ g z

+ K −e

)

+ jβ g z

)

(12.19) (12.20)

The boundary conditions are the same as for TE mode case. Enforcing the boundary conditions on (12.19), and following the same analysis as in TE case, the general expressions for TM mode electromagnetic fields in the cavity can be obtained as

E sc = 2 K + [ jE t sin ( pπ z c ) + a z E z cos( pπ z c )] H sc = 2 K + H t cos( pπ z c )

(12.21) (12.22)

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12.2. QUALITY FACTOR The quality factor is an important parameter that gives a measure of the ratio of the stored energy in the cavity to the dissipated power. Mathematically, the quality factor Q for time harmonic fields in the cavity is defined as Q = ωr

stored energy W + Wm 2We 2Wm = ωr e = ωr = ωr dissipated powe Pc Pc Pc

(12.23)

where ωr is the resonance angular frequency. We is the time average stored electric energy and Wm is the time average stored magnetic energy. Assuming the cavity is filled with a lossless dielectric material then Pc represents only the power loss in the internal walls of the cavity. We and Wm are given respectively by 2 ε E s dv 4 ∫v 2 μ Wm = ∫ H s dv 4v

We =

(12.24a) (12.24b)

where v is the internal volume of the cavity. A similar analysis that is used to obtain the power loss in the waveguides can be used for cavities. Thus, the average power loss in the surface S of the cavity walls can be computed using

Pc =

1 2δσ c

∫∫ K

2

dS

(12.25)

S

where K is the surface current density on the internal surface of the walls of the cavity. 12.3. RECTANGULAR CROSS-SECTION CAVITIES 12.3.1. TE Mode Fields in the Rectangular Cavities The TE mode electromagnetic fields in a rectangular cavity can be obtained using (12.13) and (12.14), putting in mind that, the components of Et = a x E x + a y E y ,



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H t = a x H x + a y H y and H z are those obtained in Section 10.3.1 for the TE mode fields in a rectangular waveguide as follows

ωμ nπ H cos(mπ x a )sin (nπ y b ) β c2 b mn

(12.26a)

Ey = − j

ωμ mπ H sin (mπ x a )cos(nπ y b ) β c2 a mn

(12.26b)

Hx = j

βg mπ H sin (mπ x a )cos(nπ y b) βc2 a mn

(12.26c)

Hy = j

βg nπ H cos(mπ x a )sin (nπ y b ) βc2 b mn

(12.26d)

Ex = j

H = H mn cos(mπ x a )cos(nπ y b)

(12.26e)

Since the modes order are determined by (m, n, p), the mode in the cavity is denoted by TE mnp . Letting H mnp = 2K + H mn and substituting the components E x ,

E y  H x , H y , and H z from (12.26a) - (12.26e) respectively into (12.13) and (12.14), the TE mnp mode electromagnetic fields in the rectangular cavity are obtained as

ωμ nπ H mnp cos(mπ x a )sin (nπ y b )sin ( pπ z c ) βc2 b ωμ mπ E yc = 2 H mnp sin (mπ x a )cos(nπ y b )sin ( pπ z c ) βc a Ezc = 0

Exc = −

π mp H mnp sin (mπ x a )cos(nπ y b )cos( pπ z c ) βc2 a c π2 n p H yc = j 2 H mnp cos(mπ x a )sin (nπ y b )cos( pπ z c ) βc b c H zc = jH mnp cos (mπ x a )cos(nπ y b )sin ( pπ z c )

H xc = j

(12.27a) (12.27b) (12.27c)

2

(12.27d) (12.27e) (12.27f)

Note that when p = 0 the TE mnp mode fields vanish. Therefore, TE mn0 mode fields cannot be exist in the rectangular cavity.

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757

From (10.83), we have

βg = ω2με− (mπ a )2 − (nπ b)2

(12.28)

Comparing (12.12) and (12.28), yields

pπ 2 2 = ω2με− (mπ a ) − (nπ b ) c

(12.29)

From which we can write

(mπ a)2 + (nπ b)2 + ( pπ c)2

β=

(12.30)

where β = ω με is the phase constant of a plane wave in the medium of the cavity. 12.3.1.1. Resonant Frequency The resonant frequency for the TE mnp mode fields in the rectangular cavity can be obtained using (12.30) as TE mnp

fr

=

1 2 με

(m a )2 + (n b)2 + ( p c )2

(12.31a)

where

m = 0,1, 2, n = 0,1, 2,

(12.31b)

p = 0,1, 2, The dominant mode is the mode with a lowest resonant frequency. If c > a > b  the dominant mode is the TE101 mode and its resonant frequency is

f rTE101 =



1 a2 + c2 2ac με

(12.32)

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12.3.1.2. Quality Factor The quality factor depends on the total time average stored energy in the cavity and the average power loss on the internal surfaces of the cavity walls as evident from (12.23). Since the total time average stored energy is equal to 2We = 2Wm , we need to determine either We or Wm using (12.24) or (12.25) respectively. For TE mode case, it is better to determine the total stored energy using 2We by computing We from (12.24), since Ezs = 0 . Doing this we obtained

We =

2 ε E s dv ∫ 4v

(

)

2 ε 2 = ∫ E xc + E yc dv 4v

(12.33)

Substituting Exs and E ys from (12.27a) into (12.27b), with dv = dx dy dz , We at resonance frequency ωr = 2π f r becomes

⎞ ε ⎛ ω μ nπ We = ⎜⎜ r2 H mnp ⎟⎟ 4 ⎝ βc b ⎠

2c b a

⎞ ε ⎛ ω μ mπ + ⎜⎜ r2 H mnp ⎟⎟ 4 ⎝ βc a ⎠

⎛ mπ x ⎞ 2 ⎛ nπ y ⎞ 2 ⎛ pπ z ⎞ ⎟ sin ⎜ ⎟ sin ⎜ ⎟ dx dy dz a ⎠ ⎝ b ⎠ ⎝ c ⎠

∫∫∫ cos ⎜⎝ 2

0 0 0

2c b a

⎛ mπ x ⎞ 2 ⎛ nπ y ⎞ 2 ⎛ pπ z ⎞ ⎟ cos ⎜ ⎟ sin ⎜ ⎟ dx dy dz a ⎠ ⎝ b ⎠ ⎝ c ⎠

(12.34a)

∫∫∫ sin ⎜⎝ 2

0 0 0

Substituting f r from (12.31) into (12.34a) and evaluating the integrals for all combinations of m , n , and p , the TE mnp mode time-average energy stored in the cavity can be expressed as TE mnp e

W

2 ⎤ ζ m (n b )2 + ζ n (m a )2 2⎡ ( μ p c) = abc H mnp ⎢1 + , p ≠ 0 (12.34b) 2 2⎥ 32 (m a )2 + (n b)2 ⎣ (m a ) + (n b ) ⎦

where

⎧2, ζm = ⎨ ⎩1,

m=0 m≠0

(12.35a)

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n=0 n≠0

⎧2, ζn = ⎨ ⎩1,

759

(12.35b)

Following the same above analysis starting from (12.25), it can be shown that Wm = We . Note that when p = 0 , the TE mnp mode fields vanish as evident from (12.27a) – (12.27f). The power loss Pc is the total loss in the walls of the cavity. The rectangular cavity shown in Fig (12.1) has six walls parallel to x = 0 , x = a , y = 0 , y = b , z = 0 , and z = c . Note that the current densities on the surfaces of these walls satisfy K y =0 = −K y =b , K x = 0 = −K x = a , and K z = 0 = −K z = c . The surface current density on the walls parallel to x = 0 , y = 0 , and z = 0 are listed in Table 12.1. Table 12.1. Surface currents on the walls of the cavity. :DOO x = 0 , 0 ≤ y ≤ b, 0 ≤ z ≤ c y = 0, 0 ≤ x ≤ a , 0 ≤ z ≤ c z = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b

an

K = an × H

+ ax + ay + az

+ a z H yc − a y H zc

− a z H xc + a x H zc + a y H xc − a x H yc

Using the data in Table 12.1, the total power loss in the walls is obtained as Pc

=

1

∫ ∫ (H

(

) )

(

)

c b

2

+ H zc

2

δσ c 0 0 c a 1 2 + H xc 2 + H zc ∫ ∫ δσ c 0 0 b a 2 1 2 + H + H xc yc δ σ c ∫0 ∫0 yc

x =0

y =0

dydz dxdz

(12.36)

dxdy z =0

Substituting H xs , H ys , and H zs from (12.27d) – (12.27f) into (12.36), and evaluating the integrals, the power loss in the cavity at resonance becomes 2

TE mnp c

P



2 ⎫ H mnp ⎧ ⎪ (12.37) ⎪ ( p c ) (c + a ζ m ) ⎡ 2 2 (c + b ζ n ) ⎤ ( ) ( ) ( ) = + b n b a m a c a ζ b ζ + + ⎬ ⎨ m n ⎥ 2⎢ 2 2 (c + a ζ m )⎦ 4 δσ c ⎪ ⎪ ⎭ ⎩ (m a ) + (n b ) ⎣

[

]

760

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Substituting WeTE mnp and PcTE mnp from (12.34b) and (12.37) respectively into (12.23), the quality factor of a rectangular cavity that operates at the TE mnp mode can be obtained. The quality factor for the dominant TE101 mode (c > a > b) is obtained using (12.23), (12.31a), (12.34b), and (12.37) with m = 1 , n = 0 , and p = 1 , as

Q

TE 101

(

)

3

2WeTE101 πη δ σ c b c2 + a2 2 = 2π f r TE101 = Pc 2 a 3 (c + 2b ) + c3 (a + 2b )

(12.38)

12.3.2. TM Mode Fields in the Rectangular Cavities The modes in this case are denoted by TM mnp . In a similar fashion as in the TE mode cavity case, the TM mode electromagnetic fields in a rectangular cavity can be obtained by substituting Et = a x E x + a y E y , Ez and H t = a x H x + a y H y in (12.21) and (12.22), where E x , E y  Ez , H x , and H y are the TM mode components in a rectangular waveguide given by

Ex = − j

βg mπ E cos(mπ x a )sin (nπ y b ) βc2 a mn

(12.39a)

Ey = − j

βg nπ E sin (mπ x a )cos(nπ y b) βc2 b mn

(12.39b)

E zs = Emn sin (mπ x a )sin (nπ y b) e

Hx = j Hy = − j

− jβ
z

(12.39c)

ωε nπ E sin (mπ x a )cos(nπ y b ) β c2 b mn

(12.39d)

ωε mπ E cos(mπ x a )sin (nπ y b ) β c2 a mn

(12.39e)

Letting Emnp = 2K + Emn , and substituting E x , E y  Ez , H x , and H y from (12.39a) – (12.39e) into (12.21) and (12.22), the TM mode field components in the rectangular cavity are obtained as

Exc = − j

π2 m p Emnp cos(mπ x a )sin (nπ y b )sin ( pπ z c ) βc2 a c

(12.40a)

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π2 n p E yc = − j 2 Emnp sin (mπ x a )cos(nπ y b )sin ( pπ z c ) βc b c Ezc = Emnp sin (mπ x a )sin (nπ y b )cos( pπ z c ) ωε nπ H xc = j 2 Emnp sin (mπ x a )cos(nπ y b )cos( pπ z c )

βc b ωε mπ H yc = − j 2 Emnp cos(mπ x a )sin (nπ y b )cos( pπ z c ) βc a H zc = 0

761

(12.40b) (12.40c) (12.40d) (12.40e) (12.40f)

Note that when m = 0 or n = 0 the TM mnp mode fields vanish. Therefore, TM 0 np and TM m0 p mode fields cannot be exist in the rectangular cavity. 12.3.2.1. Resonant Frequency TM mnp

The resonant frequency for the TM mnp mode cavity f r TE mnp mode and is given by TM mnp

fr

=

1 2 με

is the same as that for

(m a )2 + (n b)2 + ( p c )2

(12.41a)

where

m = 1, 2, 3, 

n = 1, 2, 3, 

p = 0,1, 2, 

(12.41b)

12.3.2.2. Quality Factor The time-average stored magnetic field Wm in the cavity can be obtained by substituting the TM fields H xs , H ys from (12.40d) and (12.40e) into (12.25), then following the similar analysis as in the TE mode case, Wm can be obtained as TM mnp

Wm



=

2⎡ ( p c)2 ⎤ ζ m (n b)2 + ζ n (m a )2 ε abc Emnp ⎢1 + 2 2⎥ 2 2 32 ⎣ (m a ) + (n b ) ⎦ (m a ) + (n b)

(12.42)

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ζ m and ζ n are defined in (12.36a) and (12.36b) respectively. Using (12.37) and (12.40d) – (12.40f), and following the analysis as in the case of TE mode, the power loss at resonant frequency in the walls of the cavity that operates at TM modes can be expressed as TM mnp c

P

2 ⎡ ⎤ ⎡ b (m a ) (a + c ζ p )+ a (n b ) (b + c ζ p )⎤ ( p c) = 2 ⎥ ⎢1 + ⎥⎢ 4η δσ c ⎣ (m a )2 + (n b )2 ⎦ ⎣⎢ (m a )2 + (n b )2 ⎦⎥ (12.43a) 2

Emnp

2

2

where

⎧2, ζp = ⎨ ⎩1, TM

TM mnp

Once We mnp and Pc using (12.26).

p=0 p≠0

(12.43b)

are known, then the quality factor can be determined

Example 12.1 Compute the first five resonance frequencies of a rectangular cross section resonant cavity of dimensions a = 60 cm , b = 50 cm , and c = 40 cm , if the cavity is filled with air. Determine the dominant mode and the quality factor of the cavity at this mode assuming the conductivity of its walls is σ c = 1.57 × 10 7 S m . Solution The dimensions of the cavity are a = 0.6 m , b = 0.5 m , and c = 0.4 m . The resonant frequency for TE mnp and TM mnp modes are given by (12.31) and (12.41) respectively, which are similar in formulation. Since the cavity is air filled, the TE mnp mode resonant frequencies are TE mnp

fr

υ0 (m a )2 + (n b )2 + ( p c )2 2 2 2 2 = 1.5 × 108 (m 0.6 ) + (n 0.5) + ( p 0.4)

=

where

m = 0,1, 2,

n = 0,1, 2,

p =1, 2, 3,

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The TM mnp mode resonant frequencies are TM mnp

fr

υ0 (m a )2 + (n b )2 + ( p c )2 2 2 2 2 = 1.5 × 108 (m 0.6 ) + (n 0.5) + ( p 0.4)

=

where

m = 1, 2, 3, 

n = 1, 2, 3, 

p = 0,1, 2, 

The first five resonant frequencies for the given dimensions of the cavity are TM110 , TE101, TE 011 , TM111 , and TE111. Hence, using the above formulas the corresponding first five resonant frequencies are computed as

f rTM110 = 390.5MHz , f rTE111 = f rTE111 = 541.4 MHz

f rTE101 = 450.7 MHz ,

f rTE 011 = 480.2 MHz ,

The dominant mode is the TM110 , since it is of a lowest cut-off frequency. Substituting m = 1 , n = 1 , and p = 0 in (12.42) and (12.43a), then TM mnp

Wm

=

2 ε abc Emnp 32

E ⎡ b3 (a + 2c ) + a3 (b + 2c )⎤ = 2110 ⎢ ⎥ 4η δσ c ⎣ a 2 + b2 ⎦ 2

TM110 c

P



Then the quality factor can be obtained using (12.23) as Q TM110 = ωr

(

From (12.41a), the resonant frequency is

f rTM110 =



)

2WeTM110 πη 2 ε 0δσ f rTM110 abc b 2 + a 2 = PcTM110 2 b3 (a + 2c ) + a 3 (b + 2c ) 

1 2 μ 0ε 0

(1 a )2 + (1 b)2 

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⇒ Q

TM 110

=

πη0 δ σ c 4

Sameir M. Ali Hamed

(

)

3

c b2 + a 2 2 b3 (a + 2c ) + a 3 (b + 2c ) 

Substituting a = 0.6 m , b = 0.5 m , and c = 0.4 m

⇒ f rTM110 =

1 2 μ 0ε 0

(1 a )2 + (1 b)2 = 1.5 × 108 (10 6)2 + (10 5)2 = 3.905 × 108 Hz σc

⇒ δσ c =

πf rTM μ 0 110

1.57 × 10 7 = 100.87 S π × 3.905 × 108 × 4π × 10 − 7

=

30π 2 × 100.87 × 0.4(0.36 + 0.25)2 = 12498 = (0.5)3 (0.6 + 2 × 0.4) + (0.6)3 (0.5 + 2 × 0.4)  3

⇒

Q

TM110

Example 12.2 Determine the quality factor of an air-filled rectangular cavity having dimensions a × b × c , where a = c = 2b = 4 cm and operating in the TE101 mode. The

(

)

conductivity of the cavity walls is σ c = 1.57 × 10 7 S m . Solution For TE101 mode, m = 1 , n = 0 , and p = 1 , ⇒ βc = π a , ζ m = 1 and ζ n = 2 . Substituting these quantities in (12.38), the power loss can be expressed as PcTE101 =

[

]

2 F101 a 3 (c + 2b ) + c3 (a + 2b ) 2 4 δσ cc

From (12.23), the store electric energy is

WeTE101 =

a2 ⎤ μ 2 ⎡ + 1 abcF101 ⎢ c2 ⎥ 16 ⎣ ⎦

The quality factor can be obtained using (12.23) and (12.31a) as

Q

TE 101

(

)

3

2WeTE101 πη δ σ cb c2 + a2 2 = ωr TE = Pc 101 2 a 3 (c + 2b ) + c 3 (a + 2b )

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Since a = c = 2b  ⇒ Q TE101 =

πη0 δ σ c 4 2

This result can be obtained also directly from (12.39). The skin depth is obtained as

δ=

1

π f rTE μ 0 σ c 101

where σ c = 1.57 × 10 7 S m . Substituting a = c = 2b = 4 cm , m = 1 , n = 0 , p = 1 , the resonant frequency is from (12.31a) obtained as

f rTE101 =

υ0 2

(1 a )2 + (1 c )2 = 1.5 × 10

8

0.04

2 = 5.303 G Hz

Thus,

Q TE101 =

πη0 δ σ c 4 2

=

30π 2 σc TE 101 2 π fr μ0 σ c

=

30π 2 1.57 × 10 7 = 5736 9 −7 2 π × 5.3 × 10 × 4π × 10

Example 12.3 Design a cubical air-filled cavity to have a resonant frequency of 7 GHz and compute its quality factor at the dominant mode. Solution For a cubical cavity a = b = c , then the resonant frequency for an air-filled cavity is

f rTE nmp =



υ0 2

(m a )2 + (n b)2 + ( p c )2 =

1 n 2 + m2 + p 2 2a μ 0 ε 0

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Thus, the dominant mode may be TE 011 , TE101 , or TE 011

⇒ 7 × 109 =

3 × 108 2 2a

⇒ a = 3.03 cm

For mode TE 011 , m = 0 , n = 1 , and p = 1 , then the stored energy can be obtained from (12.35) as

WeTE 011 =

[

μ 2 2 abcF011 2 1 + (b c ) 32

]

The power loss in the walls of the cavity from (12.38) can be obtained as PcTE 011 =

{

}

2 F011 (c + 2a )b3 + c3 (2a + b) 2 4 δσ cc

Using (12.39), the quality factor can be obtained as

Q

π η δ σ c a (c 2 + b 2 )

=

TE 011

3 2

2 (2ab 3 + cb 3 + bc 3 + 2ac 3 ) π δσ c = 3 2

The skin depth at 7 GHz is

1

δ=

πf

TE 011 c

μσ c

= 1.518 μm

Thus,

Q TE 011 = σ c δ

πη 3 2

= 1.57 × 107 × 1.518 × 10 − 6 × π = 6658

Similarly, it can be shown that for this cavity

QTE 011 = QTE101 = QTE110 .

120π 3 2

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12.4. CYLINDRICAL CAVITIES 12.4.1. TE Mode Fields in the Cylindrical Cavities The TE mnp mode electromagnetic fields in a cylindrical cavity can be obtained from (12.13) and (12.14), where Et , H t and H z are the TE mode quantities in a cylindrical waveguide, the components of Et = a ρ Eρ + aφ Eφ , H t = a ρ H ρ + aφ H φ and H z are those derived in Section 11.1.1, which are given by

ωμ m

H J (χ ′ ρ a ) sin [m(φ − φo )] β c2 ρ mn m mn ωμ χ ′ Eφ = j 2 mn H mn J m′ (χ ′mn ρ a ) cos[m(φ − φo )] βc a β χ′ H ρ = − j g2 mn H mn J m′ (χ ′mn ρ a ) cos[m(φ − φo )] βc a β m Hφ = j g2 H mn J m (χ ′mn ρ a ) sin[m(φ − φo )] βc ρ H z = H mn J m (χ′mn ρ a ) cos[m(φ − φo )] Eρ = j

(12.44a) (12.44b) (12.44c) (12.44d) (12.44e)

Substituting the components from (12.44a) – (12.44e) into (12.13) and (12.14) and letting H mnp = 2K + H mn , the TE mnp mode electromagnetic field components in a cylindrical cavity are obtained as ωμ m H mnp J m (χ ′mn ρ a ) sin [m(φ − φo )]sin ( pπ z c ) βc2 ρ ωμ χ ′ Eφc = − 2 mn H mnp J m′ (χ ′mn ρ a ) cos[m(φ − φo )] sin ( pπ z c ) βc a E z
= 0 Eρc = −

(12.45a) (12.45b) (12.45c)

π p χ ′mn H mnp J m′ (χ ′mn ρ a ) cos[m(φ − φo )]cos( pπ z c ) (12.45d) βc2 c a π mp Hφc = j 2 H mnp J n (χ ′mn ρ a ) sin [m(φ − φo )] cos( pπ z c ) (12.45e) βc ρ c (12.45f) H zc = jH mnp J m (χ ′mn ρ a ) cos[m(φ − φo )]sin ( pπ z c )

H ρc = − j



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12.4.1.1. Resonant Frequency From (11.17) and (12.12), we can write

β
=

pπ 2 = ω2με− (χ ′mn a ) c

(12.46)

This equation can be written in the form

β=

(χ ′mn a )2 + ( pπ c )2

(12.47)

where β = 2π f μ ε . The resonant frequency for TE mnp mode fields in a cylindrical cavity is obtained as TE mnp

fr

=

1 2π με

(χ′mn a )2 + ( pπ c)2

(12.48)

The dominant mode is the mode of the lowest resonant frequency. For the TE modes in a cylindrical cavity, the dominant mode is the TE111mode. In this case ′ = 1.8412. It follows that the resonant frequency at m = n = p = 1 and χ11 dominant mode is

f rTE111 =

1 2π με

(1.8412 a )2 + (π c)2

(12.49)

12.4.1.2. Quality Factor For TE mode fields in a cylindrical cavity, Ezs = 0 and We can be obtained from

We =

(

)

2 2 2 ε ε E dv = E + E dv s c c ρ φ 4 ∫v 4 ∫v

(12.50)

Substituting Eρs and Eφs from (12.45a) and (12.45b) respectively into (12.50), with dv = ρ dρ dφ dz and ωr = 2π f r , after integration with respect to φ and z , We becomes TE mnp e

W

ε⎛ω μ = ⎜⎜ r2 H mnp 4 ⎝ βc

2 ⎤ ⎞ πc a ⎡ m 2 2 ⎛ χ ′mn ⎞ ′2 ′ ′ ⎟⎟ ( ) ( ) J χ ρ a ρ J χ ρ a + ⎜ ⎟ ⎢ ⎥ dρ m mn m mn ∫ 2 ρ a ⎝ ⎠ ⎢ ⎥⎦ ⎠ 0⎣ 2

(12.51)

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The integral in (12.51) has been evaluated in Section 11.1.2 numerically and analytically. Using the results of Section 11.1.2, the time-average energy stored in a cylindrical cavity that is operating at TE mnp mode can be written as 2⎡ ⎛ pπa ⎞ ⎤ επ c ζm 2 ⎟⎟ ⎥ I mn (χ ′mn ) η H mnp ⎢1 + ⎜⎜ = 8 ⎢⎣ ⎝ χ ′mn c ⎠ ⎥⎦ (12.52) 2 2 2⎡ ⎛ pπa ⎞ ⎤ ⎡ ⎛ m ⎞ ⎤ 2 ε 2 2 ⎟⎟ ⎥ J m (χ ′mn ) ⎟⎟ ⎥ ⎢1 − ⎜⎜ = η πa c ζ m H mnp ⎢1 + ⎜⎜ 16 ⎢⎣ ⎝ χ ′mn c ⎠ ⎥⎦ ⎢⎣ ⎝ χ ′mn ⎠ ⎥⎦ 2

TE mnp e

W

where I mn (χ′mn ) is given by (11.45) and listed in Table 11.2 for some values of m and n . ζ m is defined in (12.35a). The power loss Pc in the walls of the cylindrical cavity is the sum of the loss in the cylindrical surface and the loss in the plates at the ends of the cavity. Once the surface current density on the walls is known, then the power loss can be obtained from (12.25). The surface current density on the walls of the cylindrical cavity are obtained and listed in Table 12.2. Table 12.2. Surface currents on the walls of the cylindrical cavity. :DOO

an

K = an × H

ρ = a , φo ≤ φ ≤ 2π + φo , 0 ≤ z ≤ c

+ aρ

+ a z Hφc − aφ H zc

z = 0 , 0 ≤ ρ ≤ a , φo ≤ φ ≤ 2π + φo

+ az − az

+ aφ H ρc − a ρ Hφc

z = c , 0 ≤ ρ ≤ a , φo ≤ φ ≤ 2π + φo

− aφ H ρc + a ρ Hφc

Note that from Table 12.2, K z = 0 = −K z = c . Using (12.25) and the data from Table 12.2, the average power loss in the walls of the cylindrical cavity can be expressed in general as Pc

c 2π

(



(

)

1 2 ρ dφ dz Hφc 2 + H zc ∫ ∫ ρ =a 2 δσ c 0 0 2π a 2 1 + ρ dρ dφ H ρc 2 + Hφc ∫ ∫ δσ c 0 0 z =0 =

)

(12.53a)

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Substituting H ρc , H φc and H zc from (12.45e) and (12.45f) into (12.53), yields

⎫ 2 2 π a ζ m ⎧⎪⎛ aπ p ⎞ ⎪ 2 ′ ⎜ ⎟ ( )( ) = − + + c 2 a m χ 2 a c ⎨⎜ ⎬ H mnp J m (χ ′mn ) mn ⎟ 4 δ σ c ⎪⎝ χ ′mn c ⎠ ⎪ 2

TE mnp c

P

⎩

[

]

(12.53b)

⎭

The quality factor for the TE mode fields in a cylindrical cavity can be determined by substituting (12.52) and (12.53b) into (12.23), and expressed as

Q

TE mnp

η δ σ c pπ a = 2c

(χ ′mn c 1 2 1 − (m χ ′mn )

pπa ) + 1 2a c −1 + (χ ′mn c pπa )2 + 1 2

(12.54)

12.4.2. TM Mode Fields in the Cylindrical Cavities Following the same analysis as above, the TM mnp mode electromagnetic fields in a cylindrical cavity can be obtained from (12.21) and (12.22). The components of Et , Ez , and H t are the same as those derived in Section 11.1.3 and given by

βg χ mn E J ′ (χ ρ a ) cos[m(φ − φo )] βc2 a mn m mn β m Eφ = j g2 Enm J m (χ nm ρ a ) sin[m(φ − φo )] βc ρ Ez = Enm J m (χ nm ρ a ) cos[m(φ − φo )] ωε m H ρ = − j 2 Emn J m (χ mn ρ a ) sin [m(φ − φo )] βc ρ ωε χ Hφ = − j 2 mn Emn J m′ (χ mn ρ a ) cos[m(φ − φo )] βc a Eρ = − j

(12.55a) (12.55b) (12.55c) (12.55d) (12.55e)

Substituting the components of Et , H t , and Ez from (12.55a) – (12.55e) into (12.21) and (12.22) with Emnp = 2K + Emn , the TM mnp mode fields in a cylindrical cavity are obtained as

E ρc =

π p χ mn Emnp J m′ (χ mn ρ a ) cos[m(φ − φo )]sin ( pπ z c ) βc2 c a

(12.56a)

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π mp Enmp J m (χ nm ρ a ) sin[m(φ − φo )]sin ( pπ z c ) βc2 ρ c Ezc = Enmp J m (χ mn ρ a ) cos[m(φ − φo )]cos( pπ z c ) ωε m H ρc = − j 2 Emnp J m (χ mn ρ a ) sin[m(φ − φo )]cos( pπ z c ) βc ρ ωε χ Hφc = − j 2 mn Emnp J m′ (χ mn ρ a ) cos[m(φ − φo )]cos( pπ z c ) βc a H z
= 0 Eφc = −

771

(12.56b) (12.56c) (12.56d) (12.56e) (12.56f)

12.4.2.1. Resonant Frequency Comparing (11.62) and (12.12), we can write

pπ 2 = ω2με− (χ mn a ) c

(12.57)

This can be written in the form

β = ω με 2

⎛ χ ⎞ ⎛ pπ ⎞ = ⎜ mn ⎟ + ⎜ ⎟ ⎝ a ⎠ ⎝ c ⎠

2

(12.58)

Consequently, the resonant frequency for the TM mnp mode fields in a cylindrical cavity is TM mnp

fr

=

1 2π με

(χ mn a )2 + ( pπ c)2

(12.59)

The dominant mode For the TM mnp modes in a cylindrical cavity is the T M010 mode. In this case χ 01 = 2.4049  the corresponding resonant frequency becomes

f rTM 010 =



2.4049 2π a με

(12.60)

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12.4.2.2. Quality Factor For TM mode fields in a cylindrical cavity, H zs = 0 and Wm = We is obtained using

Wm

2 μ H dv s 4 ∫v 2 2 μ = ∫ H ρc + Hφc dv 4v

=

)

(

(12.61)

Following similar analysis as for the TE mode case, the TM mode time-average energy stored in the cylindrical cavity at resonance frequency ωr = 2π f r , can be obtained as 2

⎞ μ πc ⎛ ωr ε ⎜⎜ 2 Fmn ⎟⎟ ζ m ζ p I mn (χ mn ) = 8 ⎝ βc ⎠ 2 ⎞ μ πc ⎛ ωr ε 2 ⎜ Fmn ⎟⎟ ζ m ζ p χ mn J m2 +1 (χ mn ) = 16 ⎝⎜ βc2 ⎠

TM mnp m

W

(12.62)

where I mn (χ mn ) is given by (11.74).Values of I mn (χ mn ) for limited number of m and n are listed in Table 11.4. ζ m and ζ p are defined in (12.35a) and (12.43b) respectively. Using (12.51) and the TM mode field expressions in (12.54d) – (12.54f), the power loss in the walls of the cylindrical cavity for the TM fields at the resonant frequency can be obtained as 2

TM mnp c

P

⎤ ⎛c π ζ m ⎡ ωr ε ⎞ 2 2 = ⎢ 2 Fmnp ⎥ ⎜ ζ p + 2 ⎟ χ mn J m+1 (χ mn ) 4 δσ c ⎣ βc ⎠ ⎦ ⎝a

(12.63)

Using (12.23), (12.59), (12.62), and (12.63), the quality factor for the TM mode fields in a circular waveguide can be expressed as

Q

TM mnp

=

ac η δσ c (χ mn a )2 + ( pπ c)2 ζp 2 (c ζ p + 2a )

(12.64)

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Example 12.4 A cylindrical resonant cavity of radius a = 4.5 cm  is filled with a dielectric of relative permittivity εr = 4 . If the cavity is operating at TM 010 mode, calculate its quality factor. The cavity length is equal to its diameter ( c = 2a ), and made of a material of conductivity ( σ c = 1.57 × 10 7 S m ). Solution For TM 010 , m = p = 0 , n = 1 ⇒ χ 01 = 2.4049 , ζ m = ζ p = 2 , and c = 2a , then using (12.64), the quality factor for the TM 010 is

Q TM 010 =

η δ σ c χ 01 120π 2.4049 1.57 × 10 7 = TM 2 (1 + a c ) × 4π × 10 − 7 2 4 (1 + 1 2) π × f r 30.133 = × 10 7 TM fr 010

010

From (12.60), we have f rTM 010 =

2.4049 2π a μ 0ε 0ε r

2.4049 × 3 × 108 2π × 4.5 × 10 − 2 × 4 = 1.275 × 109 Hz =

Substituting the frequency in the above quality factor equation, yields

Q TM 010 =

30.133 f

TM 010 r

× 107 =

30.133 × 107 = 8438 10 0.1275 × 10

SOLVED PROBLEMS Solved Problem 12.1 If the rectangular cavity shown in Fig. (12.1) operates at its dominant mode, find the length c that will resonate the cavity at 10 GHz . Given that c > a > b , a = 2b = 2 cm and the cavity is air-filled.



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Solution Since c > a > b , the dominant mode of the rectangular cavity is the TE101 mode and its resonant frequency is given by

f rTE101 =

1 a2 + c2 2ac με

Given that f rTE101 = 10 GHz and a = 2 cm . Thus

f rTE101 =

1 2 με

(1 a )2 + (1 c)2 = 10 ×109

Solving for c , yields 2

2

⎛ 10 × 109 ⎞ ⎛ 1 ⎞ 1  ⎟ −⎜ = ⎜⎜ 8 ⎟ −2 ⎟ c ⎝ 1.5 × 10 ⎠ ⎝ 2 × 10 ⎠

⇒

c = 2.27 × 10−2 m = 2.27 cm

Solved Problem 12.2 Write the instantaneous TE101 mode electromagnetic fields in a rectangular cavity assuming the constant H101 is real. Solution The spatial TE101 mode fields are obtained by putting m = p = 1 and n = 0 in (12.27a) – (12.27b). Thus

Exc = Ezc = H yc = 0 ωμ π E yc = 2 H101 sin (π x a ) sin (π z c ) βc a

π2 1 H101 sin (π x a ) cos(π z c ) βc2 ac H zc = jH101 cos (π x a )sin (π z c )

H xc = j

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The corresponding instantaneous fields are

E yc = Re[E yc e j ω t ]

H xc

= Re[H

E xc = E zc = H yc = 0 ωμ π = 2 H101 sin (π x a) sin (π z c) cos(ω t ) βc a

] = − β1 πac H 2

xc

e

j ωt

2 c

101

sin (π x a) sin (π z c) sin(ω t )

H zc = Re[H zce j ω t ] = − H101 cos (π x a) sin (π z c) sin(ω t ) Solved Problem 12.3 A cylindrical cavity is filled with a medium of a dielectric constant ε r = 2.25 . The resonant frequency of the TM 011 mode is higher than that of the dominant mode by 16.62%. If the radius of the cavity is 5 cm , find the (a) Resonant frequency of the dominant mode. (b) Length of the cavity. (c) Quality factor of the cavity at the resonant frequency of the dominant mode. Assume the cavity made of a material of conductivity ( σ c = 1.57 × 10 7 S m ). Solution (a) The dominant mode of the cavity is the TM 010 mode and its resonant frequency is

f rTM 010 =

2.4049 2.4049 × 3 × 108 = 1.53 GHz = 2π a μ 0ε 0ε r 2π × 0.05 × 2.25

(b) Let the resonant frequency of the TM011be f rTM 011 and the length of the cavity is c. Given that f rTM 011 = 1.1662 f rTM 010 and a = 0.05 m . Thus

f rTM011 = 1.1662 f rTM010 =



1 2π μ 0 ε 0 ε r

(χ

0.09 ) + (π c ) 2

01

2

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2.4049 × 1.1662 = 0.05

(χ

0.05) + (π c ) 2

01

2

Solving for c , yields

c=

0.05 × π = 0.1089 m = 10.89 cm 0.6 × 2.4049

TM 010 , m = p = 0 , n = 1 ⇒ χ 01 = 2.4049 , f rTM 010 = 1.53 GHz , ζ m = ζ p = 2 , a = 5 cm and c = 10.89 cm . Using (12.64), the quality factor for the TM 010 mode is

(c) For

Q TM 010 =

η δ σ c χ 01 120π 2.4049 1.57 ×10 7 = 2 (1 + a c ) 2 2.25 (1 + 5 10.89 ) π ×1.53 ×109 × 4π ×10 −7 = 10559

Solved Problem 12.4 For the cylindrical cavity in Fig. (12.2), determine the dominant mode. z a

c

ε, μ

x y Fig. (12.2). The resonant cavities of Problem 12.2.

Solution The lowest TE mode of the cylindrical cavity is the TE111 mode, and its resonant frequency from (12.49) is

f rTE111 =

1 2π με

(1.8412 a) + (π c) 2

2

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For the TM modes, the lowest mode is the TM 010 mode, and its resonant frequency from (12.60) is

f rTM 010 =

2.4049 2π a με

Since f rTE 111 and f rTM 010 are the lowest possible frequencies in the cylindrical cavity, one of them must be the dominant mode of the cavity depending on the values of a and c . Both frequencies represent the dominant mode when f rTE111 = f rTM 010  TE111 and TM 010 are known as degenerate modes. The TM 010 mode will be the dominant mode if f rTE111 > f rTM 010 or when

1 2π με

(1.8412 a) + (π c) 2

2

>

2.4049 2π a με

It follows that from this inequality,

c a < 2.03 Thus, the TM 010 mode is the dominant mode in a cylindrical cavity when

c a < 2.03, whereas for c a > 2.03 the dominant mode is the TE111 mode. Solved Problem 12.5 The quality factors of rectangular and cylindrical cavities that are operating at resonant in their dominant modes TE101 and TM 010 respectively are equal. If the cross section of the rectangular cavity is a square of side 2d and the radius of the circular one is d . The length of both cavities is 2d . If both cavities are air-filled, compare the conductivities of their walls. Solution Let the conductivity and skin depth of rectangular cavity be σ rect and δ rect respectively, and those of the circular one be σ circ and δcirc respectively. The



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quality factor of the rectangular cavity at its dominant mode TE101 is obtained by putting a = b = c = 2d in (12.39) as

Q

TE 101

=

πη δ rect σ rect 2

(

)

3

b c2 + a2 2 πη δ rect σ rect = 3 3 a (c + 2b ) + c (a + 2b ) 3 2

Similarly, using (12.64) the quality factor of the cylindrical cavity operating in the TM 010 mode is obtained by putting 2a = c = 2d

Q TM 010 = η δcirc σ circ

1.20245 1.20245 2.4049 = η δcirc σ circ = η δcirc σ circ (1 + d c ) (1 + 0.5) 3

Since QTM 010 = QTE101 , then

πη δrect σ rect 3 2

= η δcirc σ circ

2.4049 3

⇒

δrect 2.4049 ⎛⎜ σ circ ⎞⎟ 2⎜ = ⎟ δcirc π ⎝ σ rect ⎠

We have

δ rect =

1

π f

TE 101 r

f rTE101 =

μ σ rect

2 , 4d με

,

δ circ = f rTM 010 =

1

π f

TM 010 r

μ σ circ

2.4049 2π d με

Then 2

TM 010 ⎛ ⎞ ⎛ ⎞ ⎜ δ rect ⎟ = f r TE σ circ = 2.4049 2 ⎜ σ circ ⎟ ⎜δ ⎟ ⎜σ ⎟ f r 101σ rect π ⎝ circ ⎠ ⎝ rect ⎠

Substituting the ratio δ rect δcirc = 2.4049 2 (σ circ σ rect ) π that we obtained above into the last equation, yields

σ rect 2.4049 = 2 = 1.0825 σ circ π

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Hence, for the two cavities to have the same quality factor under the given conditions the conductivity of the rectangular cavity must be more that of the circular one by

σ rect − σ circ = (1.082 − 1)× 100 = 8.25% . σ circ



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PROBLEMS 12.1. An air-filled, lossless, resonant cavity of dimensions a = 70 cm , b = 45 cm, and c = 30 cm . List the five resonant frequencies of the lowest modes. 12.2. For the resonant cavity in Problem 12.1, determine the quality factor assuming the conductivity of its walls is σ c = 1.57 × 10 7 S m . 12.3. An air-filled, lossless, resonant cavity of dimensions a = 6 cm , b = 4 cm , and c = 10 cm . (a) List the five resonant frequencies of the lowest modes. (b) Find the quality factor for the dominant mode if the cavity is made of brass (σ c = 1.4 × 10 7 S m ). 12.4. For the rectangular cavity shown in Fig. (12.3), find the surface current densities on the walls y = 0 and y = b , assuming perfectly conducting walls. z

c

ε, μ

x b a

y

Fig. (12.3). Problem 12.4.

12.5. Show that for a rectangular resonant cavity of cross section a × b and length c , where (c > a > b),the dominant mode is the TE101 mode. If c = a , show that the quality factor at the dominant mode of this cavity is given by

Q=

1 πη δσ c 2 2 (a 2b + 1)

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If b = a 2 , find the dimensions that resonate the cavity at 1 GHz, when the cavity is (a) Air-filled. (b) Filled with a medium with dielectric constant ε r = 2.56 . 12.6. The dominant mode of an air-filled rectangular resonant cavity of cross section a × b and length c , is the TE101 mode. If a = c = 2.286 cm and b = 1.016 cm , find the quality factor for the dominant mode. 12.7. Show that for a resonant cavity of cross section a × b , and length c  the quality factor for the TM mm0 mode is given by

Q

TM mm 0

=

ηπ δσ c 2

(

)

3

mc a2 + b2 2 b3 (a + 2c ) + a 3 (b + 2c )

If the cross section of the cavity is square, show that Q TM mm 0 reduces to

QTM mm 0 =

ηπ δσ c m c 2 (a + 2c )

12.8. A resonant cavity is in the shape of a cube with edges of length a. Consider the resonant mode with indices m = 1 , n = 1 , and p = 0 . (a) Determine resonant frequency. (b) Write down expressions for all the non-zero components of both the electric and the magnetic field in terms of the non-zero longitudinal field amplitude. 12.9. Determine the time averages of the Poynting vector and the total electromagnetic energy within the cavity in Problem 12.8. 12.10. An air-filled resonant cavity is in a shape of a cube. Find the dimensions of the cavity if the resonant frequency of the dominant mode in the cavity is 20 GHz .



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12.11. In a rectangular cavity determine the dominant mode when (a) b > a > c . (b) a > b > c . (c) a < b < c . 12.12. An air-filled resonant cavity in a shape of a cube operates in the TE 011 mode. If the side of the cavity is 10 cm and the amplitude of the electric field in the cavity is 100 kV/m, find the time-average energy stored 12.13. Show that if all dimensions of a rectangular cavity are doubled, the total frequency content increased by a factor of 8 . 12.14. An air-filled cubical cavity of edge length 3.2 cm is made of brass (σ c = 1.4 × 107 S m ). If the cavity is operating at TE101, find (a) The resonant frequency of the dominant mode. (b) The quality factor. 12.15. Design an air-filled cubical cavity to have its dominant resonant frequency at 2.4 GHz . 12.16. Determine the maximum possible energy stored in a cubical resonant airfilled cavity with side length 10 cm , at a resonant frequency correspond to TE101 mode. 12.17. Compare the quality factors of a cylindrical cavity operating in the TM 010 mode to those of a rectangular cavity. The dimensions of each are such that the cylindrical cavity is circumscribed by the rectangular cavity. 12.18. An air-filled resonant cylindrical cavity of radius a and length c . The walls of the cavity are made of a conducting material of conductivity σ c . Find an expression for the quality factor of the cavity at TM 010 mode. 12.19. For the resonant cavity in Problem 12.18, if a = c , determine a for the cavity to operate at a wavelength of 1.5 cm at the dominant mode. Assuming the conductivity of the cavity walls is σ c = 1.57 × 10 7 S m , compute its quality factor.

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12.20. The dominant frequency of an air-filled cylindrical cavity is 9 GHz . If the resonant frequency of the next higher-order mode is higher than the resonant frequency of the dominant mode by 50%, find the dimensions of the cavity. 12.21. If the cavity in Problem 12.20 is operating at the dominant mode, find the quality factor at the resonant frequency. 12.22. Show that the quality factor of a resonant cylindrical cavity with c = 2a and operating at the TE mnp mode, is given by

Q

TE mnp

=

η δσ c 2

[1 − (m χ′ ) ] χ′ 2

mn

mn

2

+ ( pπ 2)

2

12.23. A cylindrical resonant cavity with a radius of 4 cm is filled with a medium with a dielectric constant ε r = 2.56. If the resonant frequency of the TM 011 mode is higher than that of the dominant mode by 20%, find the (d) Length of the cavity. (e) Resonant frequency of the dominant mode. 12.24. An air-filled cylindrical resonant cavity has a radius of 30 cm . If the resonant frequency of the TM 011 mode is higher than that of the dominant mode by 40%, find the (a) Length of the cavity. (b) Resonant frequency of the dominant mode. (c) Dielectric constant of the dielectric material that must replace the air inside the cavity to reduce the resonant frequency of the dominant mode by a factor of 1.5. 12.25. A cylindrical cavity of length 10 cm is filled with a medium of a dielectric constant εr = 4 . If the cavity is operating in TM 010 mode under resonant conditions, find the (a) Time-average energy stored in the cavity. (b) The radius of the cavity, so that the cavity to be resonant at 1.15 GHz .



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12.26. Find the resonant frequency of a cylindrical cavity of radius 1.15 cm , that operates in TM 010 mode when it is (a) Air-filled. (b) Filled with a medium with a dielectric constant of εr = 2 . 12.27. Find the quality factor of the resonant cavity in Problem 12.24(c) at the dominant mode, if the conductivity of the cavity walls is σ c = 6.18 × 107 S m . 12.28. Compare the conductivity of the walls of a cylindrical cavity operating in the TM 010 mode to that of a cubical cavity, if both cavities are air-filled and have equal quality factors in their dominant mode. The dimensions of each are such that the cylindrical cavity is circumscribed by the cubical cavity.

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CHAPTER 13

Transmission Lines Abstract: Transmission media in the communications system may be unbounded such as free space or bounded. The bounded transmission media include waveguides, optical fiber cables, and transmission lines. Transmission lines are made from two or more conductors separated by a dielectric material and they can transmit signals in a wide range of frequencies including DC signals. However, beyond certain frequencies, the transmission lines become inefficient in the transmission of electromagnetic power. This chapter focuses on the transmission lines and their characteristics in the frequency domain. The per unit line parameters and characteristics for some widely used transmission lines, such as coaxial cables, parallel wires lines, parallel plates lines, and micro-strip lines, are presented. The detailed discussion on reflection coefficient and voltage standing wave for terminated transmission lines is also presented. Applications of Smith chart in transmission lines calculations is covered at the end of the chapter. The topics of the chapter are analyzed in details and supported by illustrative examples and figures in addition to solved problems. Homework problems are included at the end of the chapter.

Keywords: Characteristic impedance, coaxial cables, lossless line, microstrip lines, matched line, parallel plates lines, parallel wires lines, reflection coefficient, quarter-wave transformer, Smith chart, stub matching, transmission coefficient, twisted pair lines, voltage standing wave ratio. INTRODUCTION Transmission lines are the class of the bounded media that have a wide range of application in communication systems. Transmission lines are made from two or more conductors separated by a dielectric material and they can transmit signals in a wide range of frequencies including DC signals. Unlike waveguides, the transmission lines can support all modes of propagation. However, most of the transmission lines used in practical applications, support TEM mode of propagation. The attenuation in the transmission lines increases with the frequency due to the high ohmic and dielectric losses. Therefore, the transmission lines are not efficient for electromagnetic power transmission beyond certain frequencies and the waveguides are more efficient for the electromagnetic power transmission at these high frequencies. The transmission lines commonly used in practical applications are coaxial cables, parallel wires lines, twisted pair lines, parallel-plates lines, and micro-strip lines. The subject of transmission lines is covered in many references, e.g. [18, 19, 21, 24, 93, 120-123]. Sameir M. Ali Hamed All rights reserved-© 2017 Bentham Science Publishers

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This chapter focuses on the transmission lines and their characteristics in the frequency domain. The per unit line parameters for some widely used transmission lines are presented. The detailed discussion of the reflection coefficient and voltage standing wave for terminated transmission lines is also presented. Applications of Smith chart in transmission lines calculations is covered at the end of the chapter. 13.1. PROPERTIES OF TEM MODE TRANSMISSION LINES Most of the transmission lines used in practical applications, support TEM mode of propagation. However, in some transmission lines, the propagation modes deviate from this TEM mode as in the case of micro-strip lines. y z

Surface S

x

L CA

dL

Conductor at a potential VB

ε μ σd +Q

dl

VA

+I

σc

E l

σc

-I VB

-Q

CB

H Conductor at a potential VA Fig. (13.1). Transmission line of two arbitrary conductors.

Consider an arbitrary line of two conductors A and B as shown in Fig. (13.1). The conductors are made of a metal of conductivity σ c and the medium between them is characterized by constitutive parameters ε , μ , and σ d . The conductor A is at a potential VA , while B at VB < VA , and their respective charge and currents are + Q , − Q and + I , − I . The common characteristics of the TEM lines can be summarized as follows: 1. Since the mode of propagation is a TEM, then the TEM waves equations given in Table 10.1 in Chapter 10 are applicable also for the electromagnetic fields propagating through the TEM lines. Assuming the propagation is in the

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positive z-direction, then the transverse components Et and H t in the line are governed by

a z × Et = η Ht a z × Ht = − Et η ∇ t × Et = 0

(13.1a) (13.1b) (13.1c) (13.1d)

∇t × H t = J c

∇t ⋅ Et = 0 ∇t ⋅ H t = 0

(13.1e) (13.1f)

where a z is the unit vector in the direction of propagation and η the intrinsic impedance of the medium surrounding the line. 2. The transverse electric and magnetic fields in the TEM line satisfy the static distribution in the transverse plane as clear from (13.1a) and (13.1b), despite that the excitation of these fields is time-varying. 3. According to (13.1b), if the conductivities of the conductors of the line are finite the current J c = a z σ c Ezc induces an electric field Ezc in the direction of propagation and the waves cannot be considered strictly TEM. However, for metallic conductors σ c is very high. Hence, Ezc → 0 , and the mode of propagation can be considered as a TEM. 4. On the surface of the conductors, the electric field Et is normal to the surface of the conductor and the magnetic field H t is tangential to it. 5. The potential difference VAB between the two conductors is obtained by finding the line integral of Et along any path l from conductor A to conductor B . If the path is chosen along the direction of Et , then

VAB = VA − VB = ∫ Et ⋅ dl = ∫ Et dl = η ∫ Ht dl l

l

(13.2)

l

Note that the potential distribution V between the conductors satisfies

∇ t2V = 0

(13.3)

Et = −∇tV

(13.4)

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6. The current through the conductors are equal and opposite in directions. The current in the conductor A is obtained by applying Ampere’s law. Since the periphery of the conductor CA represents a magnetic field line, then choosing CA as a closed path around the conductor and applying Ampere’s law, then

I = ∫ Ht ⋅ dL = CA

∫H

t

dL =

CA

1

η C∫

Et dL

(13.5)

A

where dL is the differential length along the path CA as shown in Fig. (13.1). 7. The total charge on the surface of each conductor is obtained by application of Gauss’s law. Let the surface of the conductor S be the Gaussian surface that enclosing the conductor A . Since E is normal to the surface, the direction of dS is the same as E , then the total charge on the conductor is

Q = ∫ ε Et ⋅ dS = ε ∫ Et dS = εη ∫ Ht dS S

S

(13.6)

S

For a line of length Δz , dS = Δz dL ; hence (13.6) becomes

Q = εη Δz ∫ H t dL = εη ΔzI

(13.7)

CA

The capacitance per unit length of the line is

C=

Q I = εη VAB Δz VAB

(13.8)

8. The total magnetic flux between the conductors is obtained from

ψ = μ ∫ Ht ⋅ ds = μ ∫ Ht ds = μ Δz ∫ Ht dl s

s

(13.9)

l

where ds = Δz dl . Using (13.2), (13.9) can be written as

μ

μ

ψ = Δz ∫ E dl = ΔzVAB η l η

(13.10)

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The external inductance per unit length can be obtained as

Lex =

ψ IΔz

=

μ VAB η I

(13.11)

From (13.8) and (13.11), the relation between the capacitance and inductance for the TEM mode lines is obtained as

LC = με

(13.12)

The total inductance of the line is the sum of its internal and external inductances

L = Lex + Lin

(13.13)

At angular frequency ω , Lin can be written from (8.77) in Chapter 8, as

Lin =

R ω

(13.14)

At high frequencies Lex >> Lin and the inductance can be approximated by

L ≈ Lex . 9. The relation between the capacitance and conductance is given by (3.116) and can be written for TEM lines as

G σd = C ε

(13.15)

10. The components of the magnetic field tangential to the surface of the conductors will be continuous across the wires. This magnetic field will induce currents in the wires which lead to ohmic power losses Pc per unit length of the wires given by

Pc =

⎞ ⎞ 1 ⎜⎛ 1 ⎛⎜ 2 2 2 2 ⎟ H H H H + = + dL dL dL dL A ∫C B ⎟ 2 σcδ ⎜ C∫ A ∫C B ⎟⎟ (13.16) 2 σ c δ ⎜⎝ C∫A B B ⎠ ⎠ ⎝ A

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where H A and HB are the tangential magnetic fields on the surfaces of the conductors A and B respectively and δ is the skin depth of the material of the conductors. 2

11. This power loss is equivalent to I R 2 , where I is given by (13.5) and R is the resistance of the line per unit length. Therefore, the resistance of the two wire line per unit length can be obtained using (13.16), as

R=

⎛ ⎞ ⎜ H A 2 dL + H B 2 dL ⎟ 2 ∫ ∫ ⎟ σ c δ I ⎜⎝ C A CB ⎠ 1

(13.17)

12. The average power transmitted through a TEM mode line can be obtained using (10.33) as

Pa =

η 1 1 2 2 Re E × H∗ ⋅ dS = Et dS = ∫∫ Ht dS ∫∫ ∫∫ 2 S 2η S 2 S

[

]

(13.18a)

The power can also be obtained using

[

1 ∗ Pa = Re VAB I 2

]

(13.18b)

13. The attenuation constant α c of the line due to the ohmic losses can be obtained by substituting the power loss and the total power from (13.16) and (13.18) respectively into (10.140). The attenuation constant due to the dielectric loss α d can be written using (10.134) as

ω εr 1 α d ≈ ησ d = tanδ′ 2 2υ0

(13.19)

where δ ′ is the loss angle of the dielectric and υ0 is the velocity in free space. Hence, the total attenuation becomes α = αc + α d

(13.20)

Note that from (8.20) the attenuation in dB is α (dB m) = 8.686 α (Np m) .

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13.2. TRANSMISSION LINE PARAMETERS Consider an infinitely long transmission line shown in Fig. (13.2). Assuming that impedances and admittances are uniformly distributed along the line, hence the impedance and admittance per unit length respectively are

Z = R + jωL Y = G + j ωC

Ω /m S/m

(13.21) (13.22)

where the elements R , L , G , and C are respectively, the resistance, inductance, conductance, and capacitance per unit length of the line. These elements are known as the per-unit parameters of the line and they depend on the geometry and properties of the material of the line. The characteristics impedance of the transmission line is defined as

Z R + jωL = Y G + j ωC

Zo =

Ω

(13.23)

It will be shown in Section 13.3 that Z o is the ratio between the potential and current waves along a transmission line in a certain direction. For lossless transmission lines R = G = 0 ; hence the characteristic impedance in (13.23) reduces to

Zo = L C

Ω

(13.24)

The parameters for some widely used transmission lines are discussed in the following subsections. RΔz

CΔz

LΔz

GΔz

Δz Fig. (13.2). The model of an infinitely long transmission line.

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13.2.1. Parallel-Plates Lines The geometry of the parallel-plates line is shown in Fig. (13.3). It is assumed that the line is long, the separation between the plates is h and the width of each plate is w such that w >> h . The thickness of each plate is t . The plates are made of a material of conductivity σ c and the space between them is filled with a dielectric  plate is medium of permittivity ε , permeability μ , a nd conductivity σ d . One grounded and the other is maintained at a potential Vo . Plate sc h

V = V0

w/2 t

h

e, s d w

E H

y

Dielectric

V=0

Plate

x

Fig. (13.3). Geometry of the parallel-plates line.

It is appropriate to use rectangular coordinates. Assuming the axis of the line is parallel to the z-axis, then the wave travels in the z-direction. The operators ∇ t and ∇ t2 can be written as

∇t = a x

∂ ∂ , +ay ∂x ∂y

∇ t2 =

∂2 ∂2 + ∂x 2 ∂y 2

(13.25)

The electric potential V between the plates satisfies Laplace equation in (13.3). Assuming the planes of the plates are parallel to the xz plane, the Laplace equation can be written as

∂ 2V ∂ 2V ∇V = 2 + 2 =0 ∂x ∂y 2 t

(13.26)

Since w >> h , the edge fringing can be ignored. Hence, V is considered as constant with x , or ∂V ∂x = ∂ 2V ∂x 2 = 0 and (13.26) reduces to

∂ 2V =0 ∂y 2

(13.27)

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Integrating twice with respect to y , the solution is obtained as

V ( y ) = K1 y + K2

(13.28)

where K1 and K2 are constants to be determined from the boundary conditions. The boundary conditions are: at y = 0 , V = 0 and y = h , V = V0 . Applying these boundary conditions, then K1 = Vo h and K2 = 0 . Substituting K1 and K2 in the above solution, the potential between the plates can be obtained as

V (y) =

Vo y h

(13.29)

Consequently, Et can be determined as

Et = −∇tV = −a x

∂V ∂V V + ay = −a y o ∂x ∂y h

(13.30)

Using (13.1a), we get

Ht = −

1 Vo 1 Vo az × a y = ax η h η h

(13.31)

Applying Ampere’s law around the closed loop ABCDA in Fig. (13.4b), yields B

C

D

A

1 Vo 1 Vo 1 Vo 1 Vo ⋅ d l + ∫ ax ⋅ d l + ∫ ax ⋅ d l + ∫ ax ⋅dl I = ∫ H
⋅ d l = ∫ a x η h η h η h η h l A B C D (13.32) y dS = az dxdy

y

Ey Hx

D

h

A

Hx = 0 Hx

C B

x -w/2

0 (a)

w/2

h

x -w/2

0 (b)

w/2

Fig. (13.4). Parallel-plates line. (a) Electric and magnetic fields in the xy plane. (b) Application of Ampere’s law.

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Electromagnetics for Engineering Students

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Since the magnetic field outside the plates is zero, the last term will be zero. The integral along the path BC is equal and opposite in sign to the integral along the path CD, and they cancel each other. The only remaining integral is that in the first term in which d l = a x dx . Thus, the current in the upper and lower plates are I and − I respectively, where

1 Vo I= η h

+w 2

∫a

x

⋅ a x dx =

−w 2

1 Vo w η h

(13.33)

Using (13.31) and (13.33), the magnetic field in terms of the current is obtained as

Ht = a x I w

(13.34)

The total charge is

V Q = ε ∫ Et dS = ε o h S

Δz w

∫∫ E

t

dxdz =

0 0

ε V0 wΔz h

(13.35)

Hence the capacitance of the line per unit length is

C=

Q εw = V0 Δz h

(13.36)

The external inductance and conductance are obtained from (31.11) and (13.15) respectively, as

Lex = μ h w G = σd w h

(13.37) (13.38)

The magnitude of the magnetic field on the surface of each plate can be obtained from (13.34) as H A = H B = I w. Substituting H A = H B = I w in (13.17), the resistance of the line per unit length is obtained as 2 w ⎞ ⎛ ⎞ 1 ⎛⎜ I 2 2 ⎟ ⎜ ⎟ R= dL dL 2 dx ( 0 ) H H + = + A B 2 2 ∫ ∫C ⎟ ⎜ ∫0 w2 ⎟ I σ c δ I ⎜⎝ C A σ δ c B ⎠ ⎝ ⎠ (13.39) 2 = σcδ w

1

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Substituting δ = 1 π fμ σ c in (13.39), the resistance of the parallel-plates line per unit length becomes R=

2 π fμ w σc

(13.40)

The resistance in (13.40) is valid only at high frequencies or δ