Brief Lessons of Physics 9789389960839, 9522285558, 9522263336

Citation preview

BRIEF LESSONS OF

PHYSICS

I

© Copyright, 2020, Deepak Kumar Singh All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy recording, or any information storage or retrieval system, without permission in writing from the publisher. The opinions/ contents expressed in this book are solely of the author and do not represent the opinions/ standings/ thoughts of Publications Name. No responsibility or liability is assumed by the publisher for any injury damage or financial loss sustained to persons or property form the use of the information, personal or otherwise, either directly or indirectly. While every effort has been made to ensure reliability and accuracy of the information within, all liability, negligence or otherwise, form any use, misuses or abuse of the operation of any methods, strategies, instructions or ideas contained in the material herein is the sole responsibility of the reader. Any copyrights not held by publisher are owned by their respective authors. All information is generalized, presented informational purposes only and presented “as is” without warranty or guarantee of any kind. All trademarks and brands referred to in this book are for illustrative purposes only, are the property of their respective owners and not affiliated with this publication in any way. Any trademarks are being used without permission and the publication of the trademark is not authorized by associated with or sponsored by the trade mark owner. ISBN: 978-93-89960-83-9 Price: 385.00 Publishing Year 2020

Published and Printed by: Rudra Publications Office Address: Kapil Nagar, New Sarkanda, Bilaspur, Chhattisgarh – 495001 Phones: +91 9522285558 +91 9522263336 Email: [email protected] Website: www.rudrapublications.com Printed in India

II

BRIEF LESSONS OF

PHYSICS

By Deepak Kumar Singh M.Sc. Electronics & Communication (Double Gold Medalist)

III

IV

Dedicated To

My Grandparents, Late Mrs. & Mr. D. Singh And My Parents, Mrs. Savitri Devi & Mr. Arun Kumar Singh

V

VI

Acknowledgements

This book has grown out of the notes developed by me over years of teaching this discipline to class 12 students of various CBSE school and postgraduate students of Electronics and Communication of physics department at Ranchi University, Ranchi. I would like to thank all those who directly or indirectly contributed in making this venture a success. I express my deep regards for Dr. Arun Kumar, a renowned physics associate professor and author under whose utmost care I have grown up as a student, a teacher and as an author. Sincere and specific mention must be made of Afzal Ansari. Thanks are also due to my students, past and present. Finally, I would like to thank Mr. N. N. Srivastava, principal of D.A.V Model School CFRI, Dhanbad and Mr. Sukhdev Singh for building keen interest of physics in me and for guiding me in all aspects when I was in school.

Deepak Kumar Singh

VII

VIII

Contents: Unit I: Electrostatics 1.

Electric Charges and Fields

2.

Electrostatic Potential and Capacitance

Unit II: Current Electricity 3.

Current Electricity

Unit III: Magnetic Effects of Current & Magnetism 4.

Moving Charge and Magnetism

5.

Magnetism and Matter

Unit IV: Electromagnetic Induction and Alternating Current 6.

Electromagnetic Induction

7.

Alternating Current

Unit V: Electromagnetic Waves 8.

Electromagnetic Waves

Unit VI: Optics 9.

Ray Optics and Optical Instruments

10.

Wave Optics

Unit VII: Dual Nature of Matter and Radiation 11.

Dual Nature of Matter and Radiation

Unit VIII: Atoms and Nuclei 12.

Atoms

13.

Nuclei

Unit IX: Electronic Devices 14.

Electronic Devices

IX

X

CHAPTER 1 Electric charges and fields Charge: Intrinsic property of all the elementary particles, which give rise to electric force between various objects. or Electric charge is a physical quantity due to which electrical and other related effects are produced in the matter. S.I Unit: coulomb or C. Methods of Charging: (1) Charging by Friction: when two insulating bodies are rubbed against each other, there is a transfer of electron from one body to the other. The body which had a lower work function, loses electron and becomes positively charged, while the body which gains e- becomes negatively charged. +Fur → glass → silk → human body → cotton → wool → sealing wax → amber→ resin → Sulphur → rubber → ebonite- (which appears earlier will positive) (2) Charging by induction: The process of charging, uncharged body without bringing a charged body in actual contact with it is known as charging by induction.

The maximum value of induced charge is given by  1 q ' = −q 1 −   k (3) Charging by Conduction: when a charged body is brought in contact with an uncharged body, e- are transferred from one body to the other. The body which loses e becomes positively charged and the body which gains e- becomes negatively charged. Properties of Charge: (i) Charges are additive in nature i.e. they are scalar and can be added by algebraic method. (ii) Charges are conserved i.e. they can neither be created nor be destroyed. (iii) Quantization of electric charge: It is a property of an electric charge which states that any charged body will have an integral multiple of the basic charged on an e - i.e. 1.6x10-19 C Q = ± ne , n = 1,2,3…….. (iv) Like charges repel each other whereas unlike charges attract each other. (v) Charge can't exist without mass where mass can exist without net charge. (vi) Charges are invariant in nature i.e. independent of velocity of charged particle (vii) Accelerated charges radiate energy. Coulomb's law:

1

According to this law, the force between any two point charges at relative rest, is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance of separation between them. (i) F  q1q1 1 F 2 (ii) r qq F  1 22 r 1 q1q2 F= air 4 o r 2 where  o constant = permittivity of free space 1 k= = 9 109 Nm2c −2 (SI Unit) 4 o in c.g.s. k = 1 1 q1q2 Fm = medium 4 r 2  = absolute permittivity of the medium Permittivity: The permittivity of a medium is a measure of the extent of difficulty with which the medium allows electric field lines to pass through it.  o = 8.854 10−12 C 2 N −1m−2 Relative permittivity (  r ) of the medium is equal to the ratio of the force between two charged in vacuum to that in the medium. 1 q1q2 4 o r 2 F  r = = = 1 q1q2  o Fm 4 r 2

r =

 =k o

Significance of coulomb's law: (i) Coulomb’s law holds over large range distance (10 -14 to few km). (ii) Coulomb's law is based on physical observation and cann't be derived. (iii) It is a universal law. (iv) Coulomb force, like gravitational force is conservative. (v) The magnitude of coulomb force doesn't depend on the polarity of the charges. (vi) Coulomb's law is valid only when the separation between charges are greater than 1015m. For distance, 10-15m (less than it) nuclear forces come into play and dominate all other forces. Drawbacks of coulomb's law: (i) The distance between the two point charges must remain constant. (ii) The charges should not radiate energy i.e. they should not be accelerated. Coulomb's law in vector form:

2

Let the position vectors of two point charges q1 & q2 in the rectangular coordinates system be r1 & r2 respectively.

From fig r21 = r2 − r1 r12 = r1 − r2

(i) (ii)

r21 = −r12

(iii)

From eqn. (i) & (ii) Let F12 = force exerted on q1 by q2

F21 = force exerted on q2 by q1 From Coulomb’s law

&

F12 =

1 q1q2 rˆ12 4 o ( r12 )2

F12 =

1 q1q2 r12 4 o ( r12 )3

(iv)

F21 =

1 q1q2 r21 4 o ( r21 )3

(v)

1 q1q2 r12 4 o ( r12 )3

(vi)

Putting the value of eqn. (iii) in eqn. (v) F21 = −

from eqn.(iv) & (vi)

F12 = − F21 (vii) from eqn.(vii) it is clear that Coulomb force, therefore act in action reaction pairs. Principle of superposition: According to the principal of superposition, total force acting an a given charge due to number of charges around it is the vector sum of the individual forces acting on that charge due to all charges. F = F1 + F2 + F3 + ...... + Fn (i) According to Coulomb's law 1 qqi Fi = ri (ii) 4 o ri 3 From eqn. (i) & (ii) q n qi F=  ri 4 o i =1 ri 3 Linear charge density:

3

This is defined as the amount of charge per unit length of a body and is denoted by . Its unit is Cm-1. Q = l Q = total Charge l = length Surface charge density: This is defined as the amount of charge per unit surface area of a body and is denoted by ‘’. Its unit is Cm-2. Q = A Volume charge density: This is defined as the amount of charge per unit volume of a body and is denoted by  . Its unit is Cm-3. Q = V Electric field: The space around a charge within which its electrical effect can be felt is called electric field. Electric field intensity ( E ): The electric field intensity at any point in an electric field is defined as the force experienced by a unit positive charge placed at that point. F E= qo kqq rˆ E= 2 o r qo

E=

kq rˆ r2

E= E =

kq r2

S.I Unit: NC-1 or Vm-1 The electric field intensity E at a point is obtained by using the principle of superposition. E = E1 + E2 + E3 + ....... + En Electric field Lines: Electric field lines is defined as the path followed by a unit positive charge when it is free to move in an electric field. Or It is an imaginary line use to represent the electric field around a charge. It is a vector quantity. The magnitude of the field is represented by the density of field lines. Electric field is strong near the charge, so the density of field line is more near the charge and the lines are closer. There are two types of field

4

(i) Uniform field (ii) Non- Uniform field. Note: Field of an isolated charge is non-uniform. The field lines are radial and not parallel to one another.

Properties of electric field lines: (i) They start from a positive charge and end at a negative charge. (ii) The tangent drawn at any point on a line gives the direction of the electric field at that point. (iii) Two electric lines of force never intersect each other. (iv) Electric field lines never form closed loop i.e. they never start from and end at the same charge. (v) In a charge free region, electric field lines can be taken to be continuous curves without any breaks. Motion of a charged particle in an electric field: When charged particle initially at rest is placed in the uniform field (i) F = qE F qE a= = (ii) m m (i) Motion along direction electric field: If ‘u’ be its initial velocity ‘v’ be its final velocity after a time ‘t’,

v = u + at

 qE  v = u +  t  m

(a)

[From eqn. (ii)] The path of the particle will be straight Line. (ii) Motion perpendicular to direction of applied field: Suppose that the initial velocity of the particle is ‘u’ along the x-axis and that a uniform electric field exist along the y-axis. The acceleration of the particle along the x-axis will be zero, while is acceleration along the y-axis will be qE ay = = constant m 1 y = u yt + a yt 2 2 1  qE  2 y=  (a) t 2 m 

5

1 x = u x t + ax t 2 2

x = ut

(b)

From eqn. (a) & (b)

1  qE  x 2 (c) y=   2  m  u2 From eqn. (c) it is clear that the transverse deflection of a charged particle in an electric field is directly proportional to its charge & inversely proportional to its mass. q y m Note: Proton have higher mass than electron. Proton is 1837 time heavier than an electron. Mp = 1.6726x10-27 kg Me = 9.1x10-31 kg q f ,v  m so e- will move with higher frequency than proton. Electric flux: Electric flux through a given surface is defined as the dot product of electric field and area vector over the surface.  = E  d S = EdS cos

It is a scalar quantity. S.I. Unit: (volt x m) or Nm2/C For a closed body outward flux is taken to be positive while inward flux is taken to be negative. Gauss's theorem: 1 The total electric flux through any closed surface in free space is times the total

o

electric charge enclosed by the surface

= =

q

 E dS = 

o

q

o

Note: (i) If a closed surface doesn't enclose any charge then  = 0.

6

(ii) The electric flux linked with a closed surface is independent of its shape and size as well as of the position of the charge within the closed surface. Prove of Gauss’s theorem: Let us consider a charge ‘q’ placed at the Centre of spherical Gaussian surface.

From figure, the electric field at the point ‘p’ is given as 1 q E= rˆ 4 o r 2

(i)

Total flux

E =  E  ds =  Eds cos  =  Eds cos 0 1 q ds 4 o r 2

= = =

E =

1 q ds 4 o r 2  1

q 4 r 2 4 o r 2 q

o

Application of Gauss's theorem: (i) Electric field due to a uniformly charged spherical shell: Consider a Gaussian surface at radius ‘r’ inside the hollow sphere.

Here the Gaussian surface will not enclose any charge therefore from Gauss theorem.  =  E.d S (i)

7

=

q

o

(ii)

From eqn.(i) & (ii)

q

 E.d S = 

o

q

 EdS cos = 

o

q

 EdS cos 0 =  E  dS =

o

q

o

E (4 r 2 ) =

0

o

E=0

Thus there is no electric field inside a hollow sphere. Consider a Gaussian surface at radius ‘r’ outside the hollow sphere having uniform charge.

As Gauss theorem

 =  E.d S

=

(i)

q o

(ii)

From eqn.(i) & (ii)

q

 E.d S = 

o

q

 EdS cos = 

o

q

 EdS cos 0 = 

8

o

E  dS =

q

o

q 4 o r 2

E=

(iii)

As Surface charge density

q 4 R 2 q =  4 R 2

=

(iv)

From eqn.(iii) & (iv)

E=

 4 R 2 4 o r 2

E=

 R2 or 2

(ii) Electric field due to a charged cylinder or infinitely long straight Conductor: Consider a cylinder having charge density ‘’.

Electric flux through a Gaussian surface

 =  E.d S

 =  EdS cos  =  EdS cos 0 According to Gauss’s law

 = E  dS = E (2 rl )

=

q l = o o

From eqn.(i) & (ii)

9

(i) (ii)

E (2 rl ) = E=

l o

l 2 rl o

E=

 2 r o

In vector form

E=

 2 r o

nˆ

(iii) Electric field due to thin sheet of charge: Let ABCD be a plane sheet carrying same charge on both side. Let  be the surface charge density.

Electric flux over the edges is given as

 =2

(  E.d S )

 = 2 EdS cos  = 2 EdS cos 0

According to Gauss theorem

 = 2E  dS = 2EA

=

q

as  =

o

=

A o

From eqn.(i) & (ii) 2 EA = E=

A o

 2 o

As

10

(i)

q A (ii)

=

q ; q = A A

Note: (i) If a dipole is enclosed by a surface  = 0. (ii) If a closed body (not enclosing any charge) is placed in an electric field (uniform or non-uniform) total flux linked with it will be zero. (iii) If charge is kept at the Centre of cube Q Q Q Q total = ;  face = ; corner = ; edge = o 6 o 8 o 12 o Electric field intensity at an axial point of a charged ring: Let us consider a charged ring having charge ‘Q’. let us consider a point ‘p’ on axial line. let ‘x’ be the distance of point ‘P’ from the Centre.

From figure

EY =  dEy =  dE sin  = 0

(i)

& EX =

 dE

x

EX =  dE cos 

from figure

cos  =

(ii)

b x = 2 h R + x2

& dE =

1 dq 4 o ( R 2 + x 2 )

using above eqn. in eqn. (ii) we get

EX =  EX =

1 dq x 4 o ( R 2 + x 2 ) R 2 + x 2

1 4 o

EX =

x 3

( R + x2 ) 2 2

1 4 o

 dq

Q

(R

for large ‘x’, x >> R then x2 + R2  x2 then from eqn. (iii)

11

2

+x

3 2 2

)

x

(iii)

EX =

1 Q x 4 o x 3

EX =

1 Q 4 o x 2

i.e. for large distance the ring behaves as a point charge Electric dipole: System of two equal and opposite charges separated by a small fixed distance is called a dipole.

Dipole moment: It is a vector quantity associated with two equal charges of opposite sign separated by a specified distance having magnitude equal to the product of the charge (either) and distance between the charges. Direction → negative to positive P = q ( 2a ) S.I. Unit: coulomb-meter or Cm or debye (D) 1 debye = 3.3x10-30Cm Ex- H2O, Chloroform (CHCl3), Ammonia (NH3), HCl, CO, are Polar molecules because the centres of negative & positive do not coincide. Non-polar molecules: CO2 & CH4 because the centres of positive & negative charge lie at the same place dipole moment is zero. Electric field intensity at any point on the axial line of a dipole: Let AX be the axial line of a dipole AB.

Electric field due to dipole at any point ‘C’ on the axial line E = EA + EB kq EA = along CA 2 (r + a) EB =

kq

(r − a)

2

As EB > EA, So net electric field act along CX .

12

along CX

(i) (ii) (iii)

E = EB − E A E=

kq

(r − a)

−

2

kq

(r + a)

2

 1 1  E = kq  − 2 2  ( r − a ) ( r + a )    4ra  E = kq  2  ( r 2 − a2 )    2 KPr E= 2 ( r 2 − a2 ) E=

(r

2 KPr 2

− a2 )

2

along CX

For short dipole r >> a r 2 − a2  r 2 2 KPr r4 2 KP E= 3 r Electric field at a point on equatorial line of a dipole: Let a dipole AB have a point ‘C’ on equatorial line. E=

E = EA + EB E = EA cos  + EB cos 

EA =

Kq Kq & EB = 2 2 (r + a ) ( r + a2 ) 2

From eqn. (ii) & (iii)

13

(i) (ii) (iii)

E = 2 E A cos  2 Kq  r 2 + a2

E=

a

(r

2

1

+ a2 )2

2 Kqa

E=

(r

E=

E=

2

3

+ a2 )2 KP

(r

2

3

+ a2 )2

KP r3

For short dipole r 2 + a 2  r 2 Note: Eaxial = 2 Equatorial The angle between E & P is zero in case of axial line & 180 in the case of equatorial line. Torque on an electric dipole in a uniform electric field: Let a dipole AB be placed in uniform electric field E.

Torque = Force x ⊥s distance

 = qE  BC  = qE 2a sin   = PE sin 

 = P E Direction of  is given by Right handed screw rule. Direction of torque is perpendicular to the plane of the paper directed inward. Note: When P is parallel to E, the dipole has a net force in the direction of increasing field (high field intensity to low field intensity). When P is antiparallel to E, the dipoles has a net force in the direction of decreasing field (low field intensity to high field intensity). Work done in deflecting a dipole in a uniform electric field: When an electric dipole makes an angle ‘’ with the direction of a uniform electric field (i)  = PE sin  To increase the angle of deflection from  to +d, work has to be done against the restoring couple. This work is dw = torque x displacement (angular) dw =  d (ii) dw = PE sin  d the total work done in deflecting the dipole from 0 to ‘’

14





0

0



w =  dw =  PE sin  d = PE  sin  d 0

w = PE  − cos  0 

w = PE  − cos  + cos 0

w = PE 1 − cos   for maximum work done  = 180, then from eqn. (iii) w = PE 1 − cos180 = PE 1 − ( −1)  wmax = 2 PE

15

(iii)

CHAPTER 2 Electrostatic Potential and Capacitance Electrostatic potential: It is defined as work done in bringing a charge from infinity to a point in any electric field region without acceleration. It is represented from ‘V’ i.e. W V = ext (i) qo we know w = F  ds kqq = 2o r r kqqo w= (ii) r kq 1 q V= = r 4 o r

V=

1 q 4 o r

It is scalar quantity. It’s S.I. Unit is J/C = volt. Dimensional formula [ ML2T-3A-1] Potential difference: The potential difference between two points in an electric field is the amount of work done by an external agent in bringing a unit positive charge from one point to the other (without changing its K.E) A U −UB 1 VA − VB = A =  Fex  dr q qB Note: Electric potential of a body is a physical quantity, which determines the flow of charge from that body to another body. Electric potential at a distance ‘r’ from a point charge: Consider a charge ‘Q’ fixed at a print ‘O’. let us find the potential due to this point charge at the point ‘P’.

Let us find the small amount of work done by displacing the test positive charge from point A to A'. The repulsive force acting on the positive test charge at A i.e.

16

F= The work done

kQqo x2

(i)

dw = F  dx = Fdx cos180 = − Fdx kQq dw = − 2 o dx x

(ii)

r

w =  dw 

r

1 dx 2 x 

w = −kQqo 

r

 1 w = −kQqo  −   x  kQqo w= r

(iii)

We know that

V=

w qo

V=

KQ r

Principle of superposition of potentials: The net potential at any point in the field of a group of charge is given by the algebraic sum of their individual potentials at that very point. Electric potential at any point due to an electric dipole: Consider a dipole having charge ‘q’ and distance between them is 2a. The net potential at point ‘P’ is due to both the charged and it follow the principle of superposition. so,

V− = −

kq kq ; V+ = r1 r2

The net potential Vp = V− + V+

(i)

kq kq Vp = − r2 r1 1 1 Vp = kq  −   r2 r1 

17

(ii)

From figure In  AOC OC cos  = AO OC cos  = a OC = a cos  Now

In  BOD OD cos  = OB OD cos  = a OD = a cos  r1 = AP  PC = PO + OC

r1 = r + a cos  r2 = PO − OD r2 = r − a cos 

(iii)

(iv) Putting the value of r1 & r2 in eqn.(ii) 1 1   V p = kq  −  r − a cos  r + a cos    r + a cos  − r + a cos   V p = kq   r 2 − a 2 cos 2   kq 2a cos  Vp = 2 r − a 2 cos 2  Pk cos  Vp = 2 r − a 2 cos 2  When r >> a Pk cos  Vp = r2 When point ‘P’ lies on the perpendicular bisector of the dipole i.e.  = 90 Vp = 0 Electric Potential at a point due to a system of charges: As potential is a scalar quantity, the potential at a point due a number of point charges is the algebraic sum of the potentials at that point due to the individual point charges. Thus if we have a number of point charges Q1, Q2, Q3, …… Qn situated at distances r1, r2, r3, …… rn from the point P then total potential at P will be V p = V1 + V2 + V3 + V4 + ...... + Vn Vp = Vp =

Q  1  Q1 Q2 Q3 + + ....... + n   + 4 o  r1 r2 r3 rn  1

n

Qi

r

4 o i =1 i Relation between field intensity and potential: Consider two point A & B separated by a very small distance x from each other.

18

As the point ‘A’ is closer to the charge +Q than the point B, the potential at ‘A’ is greater than that at B VA − VB = ( v + v ) − v = v (i) According to definition VA is the amount of work done in bringing a unit positive charge from infinity to the point A. Similarly VB is the amount of work done in bringing a unit positive charge from infinity to point B. VA - VB = amount of work done (w) in bringing a unit positive charge from the point B to the point A. i.e. (ii) w = v (iii) v = − Ex The negative sign appears because E & x are oppositely directed. v E=− (iv) x If x → 0 then in limiting case dv E=− (v) dx dv Where = potential gradient. dx from eqn. (v) it is clear that the electric field intensity at point in the electric field is equal to the negative potential gradient at that point and its S.I unit is Vm-1. Note: (i) The direction of the electric field is always along the direction of decreasing potential. (ii) Although electric potential is a scalar quantity potential gradient is a vector quantity. Equipotential surface: An equipotential surface is surface throughout which the electric potential is the same. If a test charge qo is moved from one point to another on such a surface, the work done W or the change in its potential energy U (= qVo) is zero, since the potential is constant and hence, the change in potential between any points (= Vo) is zero. The electric field lines are perpendicular to the equipotential surfaces and are directed from higher potential to lower potential. Electrostatic potential Energy: Work done in bringing the given charge from infinity to a point in the electric field is known as potential energy of the charge. Potential can also be written as potential energy per unit charge i.e. W V= Q V=

U Q

Potential Energy of a system of two charges: Suppose keeping one charge fixed, the other charge was moved slowly towards the first by an external agent. The external agent would have to perform some mechanical work against the repulsive force between the two charges. This work would be stored in the system of charges. This work would be stored in the system of charges as their electrostatic potential energy.

19

Let us consider a point charge Q1 at point A, as shown in figure. The potential due to Q1 at the point ‘B’ which is at a distance ‘r’ kQ V= 1 (i) r The amount of work that would need to be done to bring a unit positive charge (Q 2) from infinity to the point B. W = U = v.Q2

kQ1Q2 (ii) r Note: Electrostatic potential energy may be positive or negative, where as gravitational potential energy is always negative, since a gravitational field is always attractive. Case 1: Let in consider a number of point charges Q1, Q2, Q3, ……Qn situated at distances r1, r2, r3, ……rn respectively from a point charge Q. By the principle of superposition, the potential energy of the point charges ‘Q’ due to all the other charge will be U =V Q U=

Q Q Q Q Q Q  U = K  1 + 2 + 3 + ...... r r r 1 2 3   Q Q Q  U = KQ  1 + 2 + 3 + ...... r r r 2 3  1  n

U = KQ  i =1

Qi ri

Case 2: Consider a system consisting of three points Charges Q1, Q2 & Q3.

KQ2Q3 KQ1Q3 KQ1Q2 ; U 23 = ; U13 = r12 r23 r13 U system = U12 + U13 + U 23

U12 =

Q Q Q Q Q Q  U system = K  1 2 + 1 3 + 2 3  r13 r23   r12 Potential energy of a dipole in an electric field: The torque acting on dipole is given by  = PE sin 

20

(i)

If dipole is rotated through small angle d then small work done given by dw =  d (ii) dw = PE sin  d Work done in rotating the dipole to an angle ‘’ from the initial position say perpendicular to the direction of the electric field. 







w =  PE sin  d = PE  sin  d 2

2

    U = PE  − cos   = PE  − cos  −  − cos   2  2   U = − PE cos   

Case 1: When  = 0 , dipole is parallel U = − PE

 Case2: When  = , dipole axis is perpendicular 2 U =0 Case 3: When  =  , when dipole axis is antiparallel U = + PE Insulators: Insulators are those materials in which valence electrons are bound very tightly to their parent atoms thus requiring very large electric field to remove them from the attraction of their nuclei.

In terms of energy bands, it means that insulators have 1. Full valence band. 2. An empty conduction band. 3. A large energy gap (of several eV) between them. Conductor: Conducting materials are those in which plenty of free electrons are available for electric conduction in terms of energy bands, it means that electrical conductors are those, which have overlapping valence and conduction band.

21

Note: The electron are free charge carries inside a metallic conductor while positive ions fixed in lattice are bound charge carrier. Variation of electric field intensity & p.d by graphical method: 1 1 Hence E  2 and v  i.e. magnitude of E decreases rapidly with increase in r and r r magnitude of V decrease slowly with increase in r.

Electric field inside of charged conductors: Consider a conductor neutral or charged. There may also be an external electrostatic field in the static situation, when there is no current inside or on the surface of the conductor, the electric field is zero everywhere inside the conductor. An electric field is switched on, electrons move against electric field and accumulate at one side of metal. Naturally, other side gets positively charged so, the result is an internal field due to accumulation of charges. The internal field has exact same strength as external field. But in case insulator there are no such electron for above phenomenon. Dielectrics: Dielectrics (also called insulators) are materials that have no free charge carriers, so the electrical conductivity of an ideal dielectric is zero. However, when an electric field is applied across a dielectric, its behavior changes and at a particular value of the electric field, it starts behaving like a conductor. If the properties of a dielectric material are identical in all directions, the dielectric is said to be isotropic otherwise It is called anisotropic. The molecules of substance may be polar or non-polar. In a non-polar molecule; the centers of positive and negative charges consider. The molecule then has no permanent (intrinsic) dipole moment. Ex O2, H2 because of their symmetry, have no dipole moment. On the other hand a polar molecule is one in which the centers of positive and negative charges are separated (even when there is no external field). Such molecules have a

22

permanent dipole moment. An ionic molecules such as HCL or a molecule of water (H 2O) are examples of polar molecules. Polarization: When a dielectric is placed in an external electric field, the center of positive & negative dipoles get separated (in non-polar dielectrics) or get farther away (in polar dielectric), so that molecules of dielectric gain a permanent electric dipole moment, this process is called polarization and the dipole is said to be polarized. The induced dipole moment developed per unit volume in an electric field is called polarization P =  E . Dielectric Constant: The dielectric constant is defined as the ratio of the field Eo in free space to that of the field E within the dielectric. It is represented by  r or k . we know that Forcein air k = r = Forcein medium

k=

K=

qo Eo qo E

Eo (in term of electric field) E

Or

K=

 (in term of permitivity) o

Electrostatic Shielding: Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence. The field inside a conductor is zero. This is known as electrostatic shielding. Uses of this property in actual practice: (i) Sensitive instrument are shielded from outside electrical influences by enclosing them in a hollow conductor. (ii) During lighting it is safest to sit inside a car rather than near a tree. The metallic body of a car becomes an electrostatic shielding from lightening. Corona discharge: If a conductor has a pointed shape like needle and a charge is given to it, the charge density at the pointed end will be very high due to which the electric field near the pointed end will be very high which may cause dielectric breakdown in air i.e. the charge may leak from the conductor to the air because of increased conductivity of air. An electric wind is thus set up which takes away the charge continuously from the pointed end. This phenomenon is known as corona discharge. Capacitor or Condenser: Apart from resistors and inductors, a capacitor is the other basic component commonly used in electronic circuit. It is a device which has the ability to store charge which neither a resistor nor an inductor can do. Or

23

A Capacitor is a part of two conductors of any shape, which are close to each other and have equal & opposite charge. Note: Any two conductors separated by an insulator (or vacuum) form a capacitor. Capacitance of a Conductor: When a charge ‘q’ is given to a conductor, it spreads over the outer surface of the conductor. This increase the potential of the conductor. This potential (v) is directly proportional to the charge ‘q’ i.e.

qv q = cv

(i)

q (ii) c= v Where ‘c’ is a constant, known as the capacitance of the conductor or electrical capacitance of the conductor. if q = 1C & v = 1volt, then from eqn.(ii) C = 1F Hence one farad is defined as the capacitance of a capacitor which require a charge of one coulomb to establish a potential difference of one volt between its plates S.I Unit: CV-1 or farad or F Dimension: [M-1L-2T4I2] or [M-1L-2T4A2] It is a scalar quantity. Factor affecting the capacitance of conductor: (i) Area the conductor: capacitance increases directly with increase in plate area (A). (ii) Separation between the plate: As plate separation (d) decreases capacitance increases and vice versa. (iii) Nature of medium / Dielectric: It depends on the relative permittivity  r (Dielectric Constant) of the dielectric medium used. Higher the value of  r , greater the value of capacitance c =  r co

 A c = r  o   d  (iv) Independent of the material of the plates. Force between the plates a capacitor: Consider a parallel plate capacitor with plate area ‘A’. Suppose a positive charge ‘q’ is given to one plate and a negative charge ‘-q’ to the other plate. The electric field on the negative plate due to positive charge is

 2 o q E= 2 A o The magnitude of force on the charge in negative plate is E=

(i)

F = qE

q2 2 A o This is the force with which both the plates attract each other thus F=

24

(ii)

F=

q2 2 A o

Principle of capacitor: When a positively charge conductor placed near a non-charge conductor then due to induction inner surface of uncharged conductor is negatively charged and outer surface get positive charge. When outer surface is earthed positive charge of the surface shifted to ground. The negative charge of second conductor decreases the potential of 1 st conductor to increase its potential extra charge can be supplied to conductor. Hence the capacity of conductor increases.

Note: The term charge on a capacitor means positive charge on the plate of the capacitor and not the total charge on the two plates of capacitor, which is always zero. Types of Capacitor: (i) Parallel plate capacitor (ii) Spherical capacitor (iii) Cylindrical capacitor The symbol used to represent a capacitor irrespective of the type.

Capacitance of spherical conductor: Consider a hollow spherical conductor of radius ‘R’. Let ‘Q’ be the charged spreads uniformly over its entire surface. Let the potential at the spherical surface of the conductor be ‘V’. Thus 1 Q V= (i) 4 o R We know that Q C= V Eqn.(i) & (ii) C = 4 o R Capacitance of earth = 711 F =711x10-6F As R = 6400 km Capacitance of a spherical conductor is directly proportional to its radius. Equivalent capacitance of capacitors in series: Let V1, V2 & V3 be the potential difference across three capacitors C1, C2 & C3 respectively.

25

Then total voltage

V = V1 + V2 + V3

(i)

Now

C1 =

q V1

V1 =

q C1

q V2 = C2 q C3 Charge through all capacitor will be same as it is connected in series. From eqn.(i) & (ii) q q q V= + + C1 C2 C3 Let C be the equivalent capacitances then from eqn. (ii) q q q q = + + C C1 C2 C3

(ii)

V3 =

(iii)

1 1 1 1 = + + C C1 C2 C3

If ‘n’ capacitors are connected in series then equivalent capacitance of the combination is given by 1 1 1 1 1 = + + + ..... + Ceqv C1 C2 C3 Cn Equivalent capacitance of capacitors in parallel: Let three capacitors of capacitances C1, C2 & C3 be connected in parallel as shown in figure.

26

All Capacitors will have same potential difference ‘V’ across them. In parallel combination, total charge ‘q’ is sum of the charges stored by each capacitor i.e. (i) q = q1 + q2 + q3 If ‘C’ is equivalent capacitance of parallel combination then q = CV , q1 = C1V , q2 = C2V , q3 = C3V

From eqn.(i)

CV = C1V + C2V + C3V C = C1 + C2 + C3

If ‘n’ capacitor are connected in parallel, then C = C1 + C2 + C3 + ...... + Cn Note: (i) In series combination potential difference and energy distributes in the reverse ratio of capacitance i.e. 1 1 V and U  C C  C2   C1  v1 =  v  v & v2 =  C + C  1  C1 + C2  2  (ii) If ‘n’ identical capacitors are connected in series. v c Ceq = ; V ' = n n (iii) In parallel combination charge and energy distributes in the ratio of capacitance. Q  C and U  C  C1   C2  Q2 =  Q1 =  Q Q  C1 + C2   C1 + C2  (iv) If ‘n’ identical capacitor are connected in parallel. Q Q' = & Ceq = nc n (v) If Cp is the effective capacity when ‘n’ identical capacitors are connected in parallel cp = n2 and Cs is their effective capacity when connected in series, then cs Capacitance of a parallel plate Capacitor: Consider two parallel plates of conducting material separated by distance ‘d’.

27

If a voltage ‘V’ is applied to the capacitor, an electric field ‘E o’ will set up i.e. v = Eo d

(i)

 q & Eo = where  is surface charge density o A Eo =

q A o

(ii)

V=

qd A o

(iii)

From eqn. (i) & (ii)

Capacitance of a capacitor is given by

C=

q qA o = V qd

C=

o A

d If a dielectric medium of dielectric constant, ‘k’ is kept in between the plates then  KA C= o d Capacitance with dielectric slab between the plates: Capacitance of a parallel plate capacitor with air as the medium between two plates.  A C= o d Let a dielectric slab if thickness ‘t’ be put between the plates. Due to polarization, electric field reduce from E o to E.

Potential difference across the capacitor V = Eo ( d − t ) + Et

(i)

(ii)

we know

Eo E Eo E= K

K=

From eqn.(ii) & (iii)

28

(iii)

V = Eo ( d − t ) + As

Eo =

Eo t K

(iv)

 q =  o A o

So eqn. (iv) can be written as

V=

q  t ( d − t ) +   A o  K

(v)

As

C= C=

C=

q V qA o t   q ( d − t ) +  K  A o d −t +

C=

t K

A o 1  d − t 1 −   K

Energy stored in a capacitor: Let a capacitor of capacitance ‘C’ be charged to potential ‘V’. If q is the charge on the plate then q C= V q V= (i) C Small amount of work done by battery in charging the capacitor to small charge ‘dq’ at constant voltage v is given by dw = Vdq

[from eqn.(i)]

q dw =   dq C  w =  dw

(ii)

q

q w =    dq C 0 q

q

w=

1 1  q2  qdq =    C0 C  2 0

1 q 2 1 ( cV ) 1 = = CV 2 2C 2 C 2 This work done is stored inside the capacitor as potential energy. 2

w=

29

1 U = CV 2 2 Energy density of a capacitor: We know 1 U = CV 2 2 1  o A  2 U=   ( Ed ) 2 d  1 U =  o E 2V 2 U 1 = oE2 V 2 1 energy density =  o E 2 2 Energy change in two charged capacitors connected in parallel: Consider two capacitors having capacitance C 1, C2 charges q1, q2 and potentials V1, V2 respectively. Let these capacitors be connected in parallel by thin metallic wire. Charge will continuously flow from higher potential to lower potential till potential of both the capacitor becomes equal i.e. ‘V’. Let q1' & q2' be the new charge on the capacitors then according to conservation of charge. q1 + q2 = q1' + q2' C1V1 + C2V2 = C1V + C2V V=

C1V1 + C2V2 C1 + C2

q1' C1 r1 = = q2' C2 r2

1 1 Potential energy before sharing U = C1V12 + C2V22 2 2 After sharing 1 1 U ' = C1V 2 + C2V 2 2 2 1 U ' = ( C1 + C2 ) V 2 2 Loss of energy U = U − U ' 1 C1C2 2 U = (V1 + V2 ) 2 ( C1 + C2 ) Important points:

30

1  (i) The capacitance of parallel plate capacitor depends on ‘A’ (C  A) and ‘d’  C   . It d  doesn't depend on the charge on the plates or potential between the plates. (ii) If a number of dielectric slab are inserted between the plates. o A C= t t  t d − ( t1 + t2 + .... + tn ) +  1 + 2 + ..... + n  kn   k1 k2 (iii) When a metallic slab is inserted between the plates.  A C= o (d − t ) (iv) Force between the plates of a parallel plate capacitor.  2 A Q2 CV 2 F = = = 2 o 2 o A 2d (v) If ‘n’ identical plates are arranged, they constitute (n - 1) capacitors in series. If each  A o A capacitor has capacitance o then Ceq = d ( n − 1) d

31

Chapter 3 Current Electricity Electric Current: The electric Current through any cross section of a conductor is defined as the rate of flow of electric charge through that cross section. It is a scalar quantity. dq l= dt S.I. unit = Cs-1 or ampere or A. Note: (i) The direction of current flow merely gives the direction of flow of positive charge. (ii) Electric current does not obey the triangle law of vector addition. (iii) Current is also defined as the flux (or surface integral) of current density ( J ). I =  J  ds 1 Ampere (1A): Current through a conductor is said to be one ampere if one coulomb of charge flows through any cross-section of the conductor in one second. Current Density (J): It is defined as the amount of current flowing per unit area of the conductor held perpendicular to the direction flow of current. dI J= dA Drift Velocity (vd): It is defined as the average velocity with which free electrons in a conductor get drifted in a direction opposite to the direction of the applied field. or The drift velocity is the average velocity acquired by an electron in a conductor when an electric field is applied across it. Mobility (): It is the ratio of the drift velocity (vd) of current carrier and the applied electric field (E). v = d E If E = 1  = vd Numerically Thus mobility of a charge carrier may be defined as the magnitude of the drift velocity acquired by it in unit electric field. S.I Unit: m2V-1s-1 Relaxation time or Mean free time (): The small interval of time between two successive collisions of electrons and ions in the lattice of a substance is called relaxation time. Expression for drift velocity: Consider a conductor under the influence of electric field E . The force experienced by an election in the electric field is given by

32

F = −eE negative sign shows that the direction of F & E are opposite to each other. The acceleration of the electron F a= m from eqn. (i) & (ii) −eE a= m The electron accelerates for an average time internal . Therefore, the drift velocity is given by vd = u + a

(i)

(ii)

(iii)

eE  m eE (iv) vd = −  m eE vd =  (v) m Relation between drift velocity and electric current or current density: Consider a conductor of length ‘l’ and uniform cross-sectional area A. let ‘v’ be the applied potential difference across the ends of the conductor. vd = 0 −

The total number free elections in the conductor = (n) x (volume of the conductor)

= nAl

(Where n = number of electron per unit volume) total charge Q = ( nAl ) e the time taken by the charge to cross the conductor length. l t= vd vd = drift velocity. We know that Q I= t From eqn. (ii) & (iii)

33

(i)

(ii) (iii)

I=

nAle l vd

I = vd Ane

(iv)

I  vd

From eqn. (iv)

I Ane J vd = ne

vd =

(v)

J  vd

From eqn. (v) J = nevd

 eE  J = ne   m   ne 2  J = E  m  J =E

J =  E in vector form ne 2 = reciprocal of resistivity or conductivity. = m

Ohm's law: According to this law, the potential difference (v) across the ends of the conductor is proportional to the current (I) flowing through it, provided the physical conditions (like temp. Pressure, strain etc) of the conductor remain unchanged. I V V I V = IR where R = Resistance of the conductor. Ohmic Conductor: Substances which obey Ohm's law are called ohmic or linear conductor Ex- metals. Graph between V and I for a metallic conductor is a straight line as shown. At difference temp. V-I curves are different.

34

V tan  = = R I

tan 1  tan  2 so R1  R2 T1  T2

Non-Ohmic Conductor: The devices or substances which don't obey Ohm's law eg. gases, crystal rectifiers, transistor, thermionic valve etc. are known as non- ohmic or non-linear conductors. For these V-I curve is not linear. V tan  = I

Limitations of ohm's law: (a) V ceases to be proportional to I.

(b) For same value of V (magnitude) I is different i.e. relation between V & I depends on the sign of V. If I is the current for a certain V, then reversing the direction of V keeping its magnitude fixed, doesn't produce a current of the same magnitude as I in the opposite direction.

35

(c) The relation between V & I is not unique i.e. there is more than one value of V for same content I. Ex- GaAs, Gunn diode.

Resistance: The property of a substance by virtue of which it opposes the flow of current through it, is known the resistance. A body that offers resistance to the flow of electrical current through it is known as a resistor. V R= I Unit: V / A or ohm () Dimension: [ML2T-3A-2] Dependence of Resistance: Resistance of a conductor depends upon the following factor. (i) Length of the conductor. Rl (i) (ii) Area of cross-section of conductor. 1 R (ii) A From eqn. (i) & (ii) l R A l R= A where  = resistivity (iii) Temperature: For a conductor Resistance  temperature R = Ro (1 + t )  = temperature coefficient of resistance.

36

Resistivity: Resistivity is the intrinsic property of the substance. It is independent of shape and size of the body. l R= A RA = l If l = 1m, A = 1m2, then R =  numerically Resistivity is numerically equal to the resistance of a substance having unit area of cross section and unit length. S.I. Unit: ohm-meter Dimension: [ML3T-3A-2] Dependence of Resistivity: (i) Resistivity depends on the temperature for metals resistivity increase with temperature.  = o (1 + t ) (ii) Resistivity increases with impurity & mechanical stress.

Conductivity: Reciprocal of resistivity is called conductivity () 1 =  Unit: mho / m Dimensions: [M-1L-3T3A2] Conductance: Reciprocal of resistance is known as conductance. 1 C= R Unit: -1 or Siemen Stretching of wire: (i) If length is given then

R  l2 R1  l1  =  R2  l2  (ii) If radius is given then

37

2

R

1 r4

R

1 D4

R1  r2  =  R2  r1 

4

R1  D2  =  R2  D1 

4

Note: (i) For compression same above relation is used. (ii) If wire is cut along length, then Rl Derivation of microscopic form of Ohm’s law Resistivity: We know that I = neAvd

(i)

eE m

(ii)

&

vd = From eqn. (i) & (ii)

On comparing R = 

 eE  I = neA   m  ne 2 AE I= m ne 2 AV  I= ml V ml = I nA e 2 ml R= nA e 2

(iii)

l A

=

m n e2

Heating effect of Electric current: When a source of emf is connected across the free ends of a conductor a potential difference is set up across the ends. This potential difference accelerates the free electrons in the conductor. As they move, the electrons collide with positive ions and lose some of their energy to the positive ions. This increase the internal energy of conductor & its temp rises. In other words, when a current flows through a conductor, a part of the electrical energy is converted into heat. This is known as the heating effect of current.

38

(i) H = I 2 Rt This relation was established by Joule, so the phenomenon is known as joule heating. Electric Power: The rate at which electrical energy is dissipated (or consumed) in an electric circuit known as electric power. H P= t I 2 Rt P= t P = I 2R

S.I. Unit: JS-1 or watt (W) 1 kWh = 3.6 x 106 WS = 3.6 x 106 J The domestic consumption of electricity is measured in Board of Trade (BOT) units or simply units. 1 BOT = 1 kWh. Note: (i) When resistors are connected in series. P  R & P  V as I is constant. (ii) When resistors are connected in parallel. 1 P  , as V is constant. R

PI

In case of single resistor, we use the parallel condition. Colour code of carbon Resistors: Carbon resistors are used in electronic party because of their small size carbon resistors have a power rating of 1W or even less. The carbon resistance has normally four coloured rings or bands say A, B, C & D.

Colour bands A & B: Indicate the first two significant figures of resistance in ohm. Band C: Indicates the decimal multiplier i.e. the number of zeros that follows the two significant figures A & B. Band D: Indicates the tolerance in present about the indicated value or in other words it represents the percentage accuracy of the indicated value. Note: The tolerance in the case of gold is ±5% and in silver ±10% if only three bands are marked on carbon resistance, then it indicates a tolerance of ±20%. R = AB 10C  D% , where D is tolerance. B Black 0 100 B Brown 1 101 R Red 2 102 O Orange 3 103 Y Yellow 4 104

39

G B V G W

Green Blue Violet Gray White

5 6 7 8 9

105 106 107 108 109

“BB ROY Great Britain very good wife”

R = (46 x 102 ±5%) Ω R = (4600 ±5%)  Resistors in parallel: Resistors are said to be connected in parallel if the potential difference across each individual resistor is the same. When a cell is connected between the points A & B, different currents flow through the different resistors.

If ‘V’ be the potential difference between the points A & B and ‘I’ be the total current supplied by the cell. I = I1 + I 2 + I 3 V V V V = + + R p R1 R2 R3 1 1 1 1 = + +  R p  R1 R2 R3 

For ‘n’ resistors 1 1 1 1 1 = + + + ......... + Rp R1 R2 R3 Rn

If Resistors are identical

1 n = Rp R Rp = i' =

i n

40

R n

across each resistor. Resistors in series: Resistors are said to be connected in series if the current through each individual resistor is the same.

If V be the potential difference across the combination. V = V1 + V2 + V3 IRs = IR1 + IR2 + IR3 Rs = R1 + R2 + R3

for ‘n’ resistors Rs = R1 + R2 + R3 + ....... + Rn

If resistors are identical

Rs = nR V '=

V n

across each resistor. Cell: The device which converts chemical energy into electrical energy is known as electric cell. Cell is a source of constant emf but not constant current. Emf of cell(E): The potential difference across the terminals of a cell when it is not supplying any current is called it's emf. Potential difference (V): The voltage across the terminal of a cell when it is supplying current to external resistance is called potential difference or terminal voltage. V = IR Internal resistance (r): In case of a cell the opposition of electrolyte to the flow of current through it is called internal resistance of the cell. The internal resistance of a cell depends on the (a) Distance between electrodes (r  d) (b) Area of electrodes (r  1/A) (c) Nature concentration (r  C) 1 (d) Temperature of electrolyte [ r  ] temp A cell is said to be ideal, if it has zero internal resistance. Cell in various positions: Closed Circuit:

41

Current given by the cell

E r+R potential difference across the resistance V = IR potential drop inside the cell = Ir Equation of cell I=

E = V + Ir (During discharging i.e. current given to the circuit) E = V − Ir (when the cell is being charged i.e. current given to the cell) Series grouping of Cells: (a)

Eeq = E1 + E2

Eeq = E1 − E2 ( E1  E2 )

req = r1 + r2 req = r1 + r2 (b) When cells are identical: Let there be ‘n’ identical cells connected in series. Each cell has emf ‘E’ & internal resistance ‘r’.

E1 = E2 = E3 = En = E total resistance = nr + R total emf = nE

r1 = r2 = ......... = rn = r (i) (ii)

nE (iii) nr + R (c) Wrong connection in series connection: If ‘n’ is the total number of cells out of which ‘x’ cells are wrongly connected (that is, if positive terminals or negative terminals are connected together) then current in the circuit, I =

42

Total e.m.f. = ( n − 2 x ) E I=

( n − 2x) E

R + nr (d) Maximum power theorem: A source (cell) will deliver maximum power to a resistance when the internal resistance of the cell is equal to the external resistance. Therefore for maximum output power, we have Total internal resistance of the source = total external resistance. Kirchhoff's laws: The Junction law: The sum of all the currents directed towards a point in a circuit is equal to the sum of all the current directed away from the point. It is also known as current law (KCL). It is based on as conservation of charge.

I1 + I3 = I 2 + I 4

The loop law: The algebraic sum of all the potential differences along a closed loop in a circuit in zero. It is also known as voltage law (KVL). It is based on conservation of energy.  IR −  E = 0 Wheatstone bridge: Wheatstone bridge in an arrangement of four resistances which can be used to measure one of them in term of rest.

The bridge is said to be balanced when defection in galvanometer is zero i.e. no current flows through the galvanometer or in other words VB = VD In loop ABDA − I1P + ( I − I1 ) R = 0

 I1P = ( I − I1 ) R In loop DBCD

− I1Q + ( I − I1 ) S = 0

43

(i)

 I1Q = ( I − I1 ) S

(ii)

Dividing eqn. (i) & (ii)

P R = Q S If the bridge is not balanced current will flow from ‘D’ to ‘B’ i.e. VD > VB which gives. PS  RQ Meter Bridge: It is a device which is use to measure the unknown resistance. It works on the principle of Wheatstone bridge. It consists of 1m long resistance wire. The resistance of wire is divided into two resistances ‘P’ & ‘Q’. ‘R’ is known resistance and ‘S’ is unknown resistance.

P R = Q S RQ S= P   (100 − l ) A  S = R  A l   100 − l  S = R  l  Potentiometer: Potentiometer is a device mainly used to measure emf of a given cell and to compare emf of cells. It is also used to measure internal resistance of a given cell. Comparison of E.M.F. of cells:

let l1 & l2 be the balancing length with the cells E1 & E2 respectively, then E1 = Kl1 E2 = Kl2

Form eqn. (i) & (ii)

44

(i) (ii)

E1 l1 = E2 l2 Let E1 > E2 are both be connected in series. If balancing length is l 1 when cells assist each other and it is l2 when they oppose each other.

E1 + E2 = Kl1 E1 − E2 = Kl2

(i) (ii)

E1 + E2 l1 = E1 − E2 l2 or

E1 l1 + l2 = E2 l1 − l2 Determination of Internal resistance of a cell:

When K1 is closed and point ‘J’ is located on the wire such that galvanometer shows no deflection. Now (i) E = Kl1 Now K2 is pressed such that current is drawn by resistance ‘R’ from the cell & emf falls to voltage ‘V’ then (ii) V = Kl2 Dividing eqn. (i) by eqn. (ii) E l1 (iii) = V l2

45



Ir + IR l1 = IR l2



r + R l1 = R l2

l  r =  1 − 1 R  l2  Or We know that E  r =  − 1 R V 

From eqn.(i) & (ii) l  r =  1 − 1 R  l2 

Superiority of a potentiometer to a voltmeter: Potentiometer is a null method device. At null point, it does not draw any current from the cell and thus there is no potential drop due to the internal resistance of the cell. It measures the potential difference in an open circuit, which is equal to the actual emf of the cell. On the other hand, a voltmeter draws a small current from the cell for its operation. So it measures the terminal p.d in a closed circuit which is less than the emf of a cell. That is why a potentiometer in preferred over a voltmeter for measuring the emf of a cell. Sensitivity of a potentiometer: A potentiometer is sensitive if (i) It is capable of measuring very small potential difference. (ii) It shows a significant change in balancing length for a small change in the potential difference being measured. The sensitivity of a potentiometer depend on the potential gradient along it wire smaller the potential gradient, greater will be the sensitivity of the potentiometer. The sensitivity of a potentiometer can be increased by reducing the potential gradient. This can be done in two ways. (i) For a given potential difference, the sensitivity can be increased by increasing the length of the potentiometer wire. (ii) For a potentiometer wire of fixed length, the potential gradient can be decreased by reducing the current in the circuit with the help of a rheostat. Important fact of potentiometer: (i) The emf of the auxiliary battery must be greater than the emf of the cell to be measured. (ii) The balance point cannot be obtained on the potentiometer if the fall of potential along the potentiometer wire due to auxiliary battery is less than the emf of the cell to be measured.

46

(iii) The positive terminals of the auxiliary battery and the cell whose emf is to be delaminated must be connected to the zero ends the potentiometer. (iv) As we increase the emf of the cell to be measured the value of balance point will also increase. for example: At 1.5V → 60cm. 2.0V → 80 cm.

47

CHAPTER 4 Moving Charges and Magnetism Magnet: A substance which can attract small pieces iron, steel, nickel, cobalt etc. and rest in the north south direction when freely suspended is called a magnet. Properties: Attractive, Directive. Coulombs law for magnetic force: The force between two magnetic poles of strength m1 and m2 lying at distance ‘r’ is directly proportional to the product of pole strengths and inversely proportional to the square of distance between their centers. The force between the poles can be attractive or repulsive according to the nature of the poles. mm km m F  1 2 2 , F = 12 2 r r k=

o

4

= 10−7 WbA−1m −1

Magnetic field: Magnetic field is the region surrounding the magnet in which magnetic needles experience a torque which sets the needle in a particular direction. Magnetic lines of force: Magnetic lines of force are imaginary lines in magnetic field which indicates the direction of magnetic field. Properties of magnetic line of force: (i) Magnetic lines of force always start from North Pole & forming curved path enter South Pole and travel to North Pole inside the magnet, forming closed loops. (ii) Two lines of force never intersect, since in that case at that point of intersection there shall be two tangents representing two direction of magnetic field which is not possible. (iii) Near the pole of magnet where magnetic field is stronger. (iv) In earth’s magnetic field Lines of force at any place are equidistance and parallel. Such field magnetic field is called uniform magnetic field. Oersted’s Experiment: Whenever an electric current flow through a conductor a magnetic field exist around it. This phenomenon is known as the magnetic effect of current. “It is well known fact that an electric charge whether stationary or in motion produces an electric field around it. If it is motion then in addition electric field it also produces a magnetic field.” Oersted's observation established a direct connect between two different branches of Physics- electricity & magnetism and led to the development of a new branch, now known as electromagnetism. In 1980 Dansih scientist Oersted discovered that a small magnetic needle placed near a current carrying wire is deflected. It shows that similar to magnetic field surrounding a magnet, there is also a magnetic field surrounding a current carrying wire. If current in the wire is withdrawn, the magnetic field disappears and needle shown no deflection. Ampere’s swimming rule:

48

If a man is supposed to swim parallel to the wire with current entering his feet and leaving his head then the north pole of the magnetic needle is always deflected toward his left hand. Magnetic field intensity: If a test pole of pole strength mo experience a force F at a point in a magnetic field then the intensity of the magnetic field at that point is given by F B= mo If mo = 1 then

B=F Thus the intensity of magnetic field at a point is numerically equal to the force experienced by a unit test pole placed at that point. Force on a moving charge in magnetic field: Consider a charged particle carrying a positive charge ‘q’ placed in an electric field E . The force acting on the charge is Fe = qE (i) The force on the charge depend only on the magnitude of the charge and not on the state of the charge i.e, whether it is at rest or in motion. However, when a charged particle moves in a magnetic field, the force acting on it depends not only on the magnitude of the charge but also on its velocity. The magnitude of the magnetic force ‘Fm’ acting on a moving charge ‘q’ depends on the following:

(i) It is directly proportional to the magnitude of the charge ‘q’ i.e, Fm  q

ii) It is directly proportional to the component of its velocity v in a direction normal to the direction of magnetic field B . Fm  v sin  (iii) It is directly proportional to the magnitude of the applied magnetic field B . Fm  B

Combining the three proportionalities

Fm  qvB sin  Fm = kqvB sin 

49

Fm = qvB sin 

(ii)

(

(iii)

In vector notation

Fm = q v  B

)

The force Fm is therefore perpendicular to both v & B or the plane containing v & B . when (i)  = 0o ,180o ,sin 0o = 0, Fm = 0 (ii)  = 90o ,sin 90 = 1, Fm = qVB = maximum (iii) v = 0, Fm = 0 charged at rest does not experience any force. Lorentz Force: If a charged particle moves in a space in which an electric field as well as a magnetic field act then the net force F acting on the charged particle will the vector sum of the electric force Fe & magnetic force Fm acting on it. F = Fe + Fm

( ) F = q  E + ( v  B )   F = qE + q v  B

(i)

Eqn.(i) is known as Lorentz relation and the force F is called Lorentz force. let us assume that the velocity v , the electric field E and the magnetic field B are along the x, y and z axes respectively. Then v = viˆ, E = Ejˆ & B = Bkˆ

(

)

(

magnetic force Fm = q viˆ  Bkˆ = qvB iˆ  kˆ

)

( )

Fm = qvB − ˆj

electric force

( ) () = qvB ( − ˆj ) + qE ( ˆj )

Fe = qE = q Ejˆ = qE ˆj F = Fm + Fe

(ii) (iii)

F = q ( E − vB ) ˆj (iv) If the strength of the uniform electric and magnetic fields be so adjusted that the net force F acting on the charged particle is zero then the particle will emerge through combined fields un-deviated. From eqn. (iv) E E − vB = 0, v = B Right hand screw rule: This rule state that if a right handed screw be rotated from v to B then the direction of the linear motion of the screw will give the direction of the force Fm . This rule is also known as Maxwell's Corkscrew rule.

50

Right hand rule: If the fingers of the right hand be curled from v to B then the direction of the thumb gives the direction of Fm . These rules are valid for positively charged particles if the charge carried by a particle is negative then the direction of Fm will be the opposite. Right hand palm rule: According to this rule if we stretch our right hand palm such that the thumb points in the direction of current ‘I’ in the conductor and the finger point in the direction of magnetic field B , then the direction of magnetic force F acting on the Current Carrying Conductor in the magnetic field is along the perpendicular drawn out from the palm. Fleming left hand rule: According to this rule when we stretch the forefinger, central finger and thumb of our left hand in such a way that the thumb is perpendicular to the plane containing the two fingers then if the forefinger points in the direction of the magnetic field and central finger points in the direction of current, then the thumb will point in the direction of magnetic force on the current carrying conductor. Force on a current carrying conductor in B : Consider a straight conductor PQ of length ‘l’ placed in a uniform magnetic field. The length of the conductor is perpendicular to the magnetic field B .

Current flowing through the conductor due to flow of free electron is I = vd Ane

magnitude of current element of length ‘l’ is Il = vd Anel

magnetic lorentz force on an electron

(

Fm = e vd  B

)

If total number of free electron in the conductor is ‘nAl’ then

(

)

Fm = nAle vd  B = Fm = Il  B Fm = IlB sin  Where l is known as the displacement vector. Its magnitude is equal to the length of conductor and its direction is along the direction of flow of positive charge through the conductor. Motion of a charged particle in a uniform magnetic field: Suppose a positively charged particle having charge +q enters a uniform magnetic field B at a point ‘O’ with velocity v perpendicular to the field B .

51

The charged particle in the magnetic field will experience a force whose magnitude is given by Fm = qvB sin 90 = qvB . This force is perpendicular to v . Hence it has zero component along v . The direction of magnetic force is given by Fleming's left hand rule. Since the force Fm is at right angle to the velocity v , it will not change the magnitude of v but change only its direction. Thus the particle will move under such a force whose magnitude is constant but whose direction is at right angle to the velocity of the particle. Therefore the charged particles will describe a circular path in anticlockwise direction with constant speed v and force Fm will provide the necessary centripetal force to the particle for describing circular path. If ‘m’ be the mass of the charged particle and ‘r’ the radius of its circular path then the magnitude of centripetal force will mv 2 = Fm r mv 2 = Bqv r mv r= (i) qB Period of Revolution: It is the time taken by the charged particle in completing one revolution of the circle. Distance covered in one revolution T= Speed of the particle

T=

Circumference of the circle Speed

2 r T= = v

 mv  2    qB  v 2 m T= qB

Frequency of revolution: It is the number of revolution made by the particle per second. 1 qB = = T 2 m

52

(ii)

(iii)

Relationship radius & K.E: r=

mv qB

We know that P = 2mk Where k = kinetic energy r=

2mk qB

r k

When the angle between v & B is not 0o, 90o, 180o: Let a charged particle enter a magnetic field with a velocity v at an angle  with B .

Resolving the velocity v

v = v cos  v⊥ = v sin 

The radius of helical path is

r= & time period = T =

mv⊥ mv sin  = qB qB

2 m qB

Pitch of helical path: It is the distance covered by the charged particle in one complete revolution in the direction of B and is denoted by P. 2 m P = v  T = v cos   qB

P=

2 mv cos  qB

Cyclotron: It is a device which is used for accelerating positively charged particles like protons, deuterons, -particles etc to very high energies. These energetic particles are used for artificial disintegration of various nuclei. Cyclotron was discovered by Lawrence in 1932. Principle: If a charged particle is time and again moved in strong electric field of high frequency as well as strong magnetic field, the particle acquires large energy & gets accelerated. Construction:

53

It consists of strong electromagnets and two D-shaped metallic chambers DA & DB . A source positively Charged particles is Kept in the gap between the dees. These dees are connected to a HF oscillator (10 6 cycle per second). The magnetic field is perpendicular to the plane of the dees.

Working: If a charged particle (say proton) is emitted from source ‘S’, at an instant when D B is negatively charged and DA is positively charged, it accelerated toward D B. Inside DB it moves at right angle to the magnetic field and describes a semi-circle. It again enters the gap when polarities of dees reversed. The proton now goes towards D A and completes semicircle. This process is repeated till proton is accelerate through the window to hit the target. For the dynamics of circular motion mv 2 Bqv = r qBr (i) v= m time taken for semicircular path r m (ii) t= = = a constant v qB If ‘T’ be the time period of the oscillator, then T 2 m t =  T = 2t = 2 qB 1 qB Cyclotron frequency = = T 2 m In order to operate cyclotron it is necessary that the frequency of revolution of charged particle in dees should be equal to the frequency of AC oscillator. This condition is known as resonance condition of cyclotron. Limitations: (i) A cyclotron cannot be used to accelerate uncharged particles like neutrons. (ii) It cannot be used to accelerate electron because electron being light, are accelerated to a great extent and move with a very high speed. As a result, they quickly go out of step with the oscillating electric field. (iii) The speed of charged particle cannot be increased beyond the limit set by the following eqn.

54

mo

m=

v2 c2 At the speed comparable to ‘C’,the mass of the particle increase to such an extent that the specific charge (q /m) no longer remains constant and eqn. of time period no longer valid 2 m T= qB 1−

1

qB qB  v 2  2 f = = 1− 2 m 2 mo  C 2 

Biot Savart law: The Biot-Savart law is an experimental law put forth by Biot, Savart and Laplace. Biot-savart's law gives the magnitude of the magnetic field induction at a point due to small part of a current carrying conductor i.e. due to current element.

There is no mathematical derivation of this law. On the basis of experimental study this law is stated as follows: (i) It is directly proportional to the electric current ‘I’ flowing through the conductor. B  I (i) (ii) It is directly proportional to the length l of the current carrying element. B  I (ii) (iii) It is directly proportional to the sine of the angle ‘’ between the current element I .l and position vector r of the point P. B  sin  (iii) (iv) It is inversely proportional to the square of the magnitude r . 1 B  2 (iv) r Combining all the above factors.

B 

I l sin  r2

B = k

I l sin  r2

Where k = proportional constant.

55

(v)

In S.I Systems of units  k= o 4 o = 4 10−7 N/A 2 = 4 10−7 TA −1m  k = 10−7

From eqn.(v) In the case of free space

 I .l sin   B = 10−7  2   r Eqn.(vi) is known as mathematical form of Biot savart law. In vector form  I  B = o  3   lr sin   4  r 

(vi)

o  I    l  r 4  r 3  Comparison between Biot savart law and coulomb’s Law: (i) Both are inverse square law. (ii) Both the fields are long range. (iii) The principle of superposition is valid for both. Magnetic field due to straight current carrying conductor: Consider an infinitely long conductor AB through which current ‘I’ flows. Let ‘P’ be any point at a distance ‘a’ from the centre of conductor. Consider ‘dl’ be the small current carrying element at point ‘C’ at a distance ‘r’ from point ‘P’. ‘’ be the angle between ‘r’ and ‘dl’.

(

B =

from Biot - Savart law

dB =

o Idl sin  r2

4

56

)

From fig,

a = cos  r a r= cos 

sin  =

Again tan  =

(ii)

l a dl = a sec2  d

Fron eqn (i), (ii) and (iii) we get

dB =

(iii)

2 o I ( a sec  d ) cos  2 4  a     cos    I cos  d dB = o 4 a

(iv)

total magnetic field due to straight current carrying conductor is

B =  dB =

2

o I cos  d a

 4

−

1

I  B = o sin  − 4 a o I B= sin 2 + sin 1  4 a 2

1

Case 1: If conductor having infinite length then

1 =  2 =

(v)



2 o I B= tesla 2 a Case 2: if P is very close to finite length  1 &  2 → 2 o I B= tesla 2 a Magnetic field at a point on the axis of a current carrying loop / Circular Coil / Conducting Ring or Loop: Let a coil of radius ‘a’, centre ‘O’ carrying current ‘I’ be symmetrically placed with its axis along x-axis. Let ‘P’ be a point on the axis of the coil.

57

According to Biot savart law, magnitude of dB is  Idl dB = o 2 4 r Since angle between dl & r is 90o

(i)

sin  = sin 90 = 1

The vertical components cancel one another. Effective component of magnetic field along x-axis. Magnetic field due to whole coil  Idl B =  dB sin  =  o 2 sin  4 r o I B= sin   dl 4 r 2 I B = o 2 sin  ( 2 a ) 4 r I a a B = o 2 ( 2 a ) sin  = , r = a 2 + x 2 4 r r r   o 2 Ia 2 o  2 Ia 2  B= = 3  4 r 3 4  2 2 2  a + x ( )  

Magnetic field at the centre of a current carrying coil: As per Boit Savart’s law the magnetic field at point ‘C’ due to current element AB is given by  Idl  rˆ  Idl sin  dB = o or o 4 a 2 4 a2

58

Since angle between dl & rˆ is 90o dB =

o Idl 4 a 2

B =  dB =

o I  I dl = o 2 ( 2 a ) 4 a 2  4 a I B= o 2a

If the coil has ‘n’ turns then B=

o nI 2a

2 angle B =

o nI 2a

o nI

1 2  nI     o angle B = o   2a  2   nI B= o  4 a

1 angle B = o

2a



Ampere's Circuital law: The line integral of the magnetic field B along a closed path is equal to o times the total current crossing the area bounded by the closed curve.

59

Total current I = I1 + I 2 + I3 − I 4 Mathematically

 B.dl

= o I

o I 2 r Magnetic field due to a current carrying solenoid: Consider a very long solenoid having ‘n’ turn per unit length. Let current ‘I’ be flowing through the solenoid. B=

Let ‘P’ be a point well within the solenoid. Consider any rectangular loop ABCD (Amperian loop) passing through ‘P’ as shown in the figure. Then  B.dl = line integral of magnetic field across the Loop ABCD

 B.dl = N  I

(i)

o

(N = no. of turns) B

C

D

A

A

B

C

D

 B.dl +  B.dl +  B.dl +  B.dl

= N o I

(ii)

B is ⊥ to path BC & AD. Angle b/w B & dl is 90o. So C

A

B

D

o  B.dl =  B.dl =  Bdl cos 90 = 0

(iii)

since path CD is outside the solenoid so B = 0 For Path AB the direction of dl & B is same  = 0o , so B

B

A

A

 Bdl cos = B  dl = Bl from eqn (ii), (iii) & (iv)

Bl = o NI

60

(iv)

B=

o NI l

B=

= o nI at the centre

o nI at the end 2

Magnetic field due to a toried: A toroid is a doughnut-shaped object with a coil wound around it that is used as an inductor in electronic devices.

As

 B.dl

= o I

 Bdl cos =  I o

magnetic field ‘B’ is tangent to dl i.e.  = 0.

 Bdl =  I o

B 2 r = o I B=

If N is total number of turns then B =

o I 2 r

o NI 2 r

Force between parallel Current carrying conductors: Let two very long conductor AB & XY be placed parallel to each other.

61

(i)

Here magnetic field of one conductor links with the other conductor. Magnetic field due to current I2 flowing in XY I B2 = o 2 (i) 2 r (Ampere’s circuital law) Force experienced by conductor AB is given by  I  FAB = B2 I1l1 =  o 2  I1  l1  2 r   I I  FAB =  o 1 2  l1  2 r   I I  FXY =  o 1 2  l2  2 r  similarly If l1 = l2 = l , then Force is given by  I I  F =  o 1 2 l  2 r  This force is attractive in nature. Torque on rectangular Coil in a magnetic field: As the current carrying conductor experiences a force when placed in a magnetic field, each side of a current carrying rectangular coil experiences a force in a magnetic field.

62

F = BIl = F '  = (Force) ⊥ distance  = BIlb sin   = BIA sin 

(i)

( )

 = IA  B  = M B Galvanometer: The galvanometer in an instrument that is used to detect the electric current flowing in a circuit. It may be of following two types (i) Moving coil-galvanometer (ii) Moving magnet galvanometer The working of both is based on the interaction between an electric current and a magnetic field. Moving Coil galvanometer: It is a device used to measure /detect small electric current flowing in the electric circuit. It was first devised by Kelvin and later modified by D'Arsonval. With suitable modification a moving coil galvanometer can be used to measure current and potential difference. Principle: Moving coil galvanometer (MCG) is based on the fact that when a current carrying loop or coil is placed in the uniform magnetic field, it experiences a torque. Construction: It consists of a coil wound on a non-metallic frame. The coil is suspended between two poles of a permanent magnet which are cylindrical in shape. The coil is freely suspended by mean of a fine phosphor bronze wire which acts as path for the current to the coil. The strip is finally connected to the terminal T 1 and other end is connected to terminal T2 . The spring exerts a very small restoring couple on the coil.

Working: It is based on the fact that when a current carrying loop is placed in the uniform magnetic field it experiences a torque. As the field is radial the angle between the pole pieces and cylindrical core is 90o, so torque in given by  = NBIA (i)

63

Where N = number of turns in the coil B = magnetic field I = Current in the coil A = area of the coil (l x b) Let the deflection in the spring is given by 

   = C

C = torsional content From eqn. (i) & (ii)

k=

(ii)

C = ( NBIA )  C  I =   NBA  I = k

(ii)

I 

(iv)

C = Galvanometer constant NBA

From eqn. (iv) it is clear that deflection of the coil is directly proportional to the current following through it. Current sensitivity: The current sensitivity of a galvanometer is defined as the deflection produced per unit current flowing through it.  NBA = I C Voltage sensitivity: Voltage sensitivity of a galvanometer is defined as the deflection produced per unit applied voltage.   1    1  NBA  = =  =   v IR R  I  R  C  Conversion of a Galvanometer into Ammeter: An ammeter is an instrument which is used to measure electric current in the electrical circuits (small current to large currents). A galvanometer is used to measure small current in circuits if a large current flow through it either the coil burn out or the spring may break. The resistance of ideal ammeter is zero. A galvanometer is converted into ammeter by connecting a shunt (small resistance) parallel to Galvanometer. Since the resistance of shunt is very small, a large part of main current passes through the shunt and only a small part of the current passes through the galvanometer.

64

As shunt and galvanometer are in parallel connection. So VG = VS

IgG = ( I − I g ) S

S=

GI g

( I − Ig )

VG = VS

or

I gG = ( I − I g ) S I g G = IS − I g S I g ( G + S ) = IS Ig =

IS G ( + S)

 S  I g =   I  (G + S )  Conversion of Galvanometer into voltmeter: A voltmeter is used to measure the potential difference between two points in a circuit. The resistance of an ideal voltmeter should be infinite. The resistance of the voltmeter should be as high as possible so that on connecting it the circuit across any two of its points a small part of the circuit current passes through the coil of the voltmeter and nearly the same potential difference is developed between the terminals of the voltmeter which is between those points.

Let ‘V’ be the voltage to be measured by voltmeter. V = I g R + I gG V = Ig ( R + G) Ig =

or

V

(R + G)

V = I g R + I gG I g R = V − I gG R=

V −G Ig

65

CHAPTER 5 Magnetism and matter Current loop as a magnetic dipole: We know that magnetic field at a point on the axis of a current carrying loop is given by  2 Ia 2 Bn = o (i) 4 ( r )3 We can write  2 IA Bn = o 4 ( r )3

(ii)

Where A=πa2 & we know that 1 2P (iii) E= 4o ( r )3 On comparing equation (ii) & (iii) we get that a circular current loop behaves as a magnetic dipole of magnetic moment m=IA in vector m =IA Thus the magnetic dipole moment of any current loop is equal to the product of the current and its loop area. Magnetic field of a bar magnet at a point on axial line: Consider a bar magnet N-S of length ‘2l’ and pole strength ‘m’.

Let ‘P’ be a point situated on the axis at a distance ‘r’ from its centre ‘O’. The magnetic field intensity at ‘P’ due to the north pole of the magnet is   m m Bn = o = o along NP 2 4 ( NP ) 4 ( r − l )2 The magnetic field at ‘P’ due to South Pole   m m Bs = o = o along PS 4 ( SP )2 4 ( r + l )2 As Bn & Bs are oppositely directed, the magnitude at ‘P’

66

Be = Bn − Bs =

o m  1 1  −   2 4  ( r − l ) ( r + l )2 

Be =

 o m  4rl   4  ( r − l )2 ( r + l )2 

Be =

o m  2Mr  along NP 4  ( r 2 − l 2 )2 

  Where M = 2ml = magnetic moment Magnetic dipole: Magnetic dipole consists of a pair of magnetic poles of equal and opposite strength separated by a small distance. Magnetic dipole moment: Magnetic dipole moment is defined as the product of the pole strength of either pole and distance between the poles. Distance between the poles is called magnetic length and it is represented by ‘2l’. It is vector quantity. Direction from South pole to North pole.

M = 2ml or M = NIA

Magnetic field intensity at Equatorial line of bar magnet: Consider a bar magnet of length ‘2l’ and pole strength ‘m’. Let ‘p’ be a point on its perpendicular bisector at a distance ‘r’.

Bn =

o m along NP 4 ( r 2 + l 2 )

Bs =

o m along PS 4 ( r 2 + l 2 )

Since Bn = Bs = B

67

Be = Bn cos  + Bs cos  Be = B cos  + B cos  Be = 2 B cos  Be = 2 

o m cos  4 ( r 2 + l 2 )

Be = 2 

o m  4 ( r 2 + l 2 )

Be =

o

4

2ml

(r

2

+l

3 2 2

)

=

l

(r

2

+ l2 )

o

4

M

(r

2

3

+ l 2 )2

In case of magnet of very small length then l 2  r 2 M Be = o 3 4 r Torque on a magnet suspended in uniform magnetic field:

Torque = force x perpendicular distance between two force  = ( mB ) 2l sin 

 = MB sin   = M B We know

 = NIAB sin 

On comparing eqn.(i) M=NIA Potential energy of a bar magnet placed in uniform magnetic field: We know that

 = MB sin 

Work done in rotation of bar magnet is

dw =  d

Since

68

(i) (ii)

U =  dw =   d 2

U =  MB sin  d 1

U =  − cos 2 MB 

1

U = − MB  cos  2 − cos 1  U = MB  cos 1 − cos  2 

Case 1 When 1 = 90o & 2 = o

Case 2 When 1 = 90o & 2 = 0o (stable state)

U = MB  cos 90 − cos   U = − MB cos  = − M .B

U = MB cos90 − cos 0 = −MB

Case 3 When 1 = 90o & 2 = 180o (unstable state) U = MB cos90 − cos180 = MB Gauss's law in magnetism: Gauss’s law in magnetism states that the surface integral of a magnetic field over a closed surface is always zero. (Because it form closed loop as it goes from N-S outside the magnet & S-N inside the magnet. For any given area the equal number of magnetic line enter and leave that area) Proof: The inverse square law of force holds for magnetic poles too, so the expressions.

o m1m2

qq rˆ F = k 1 2 2 r2 r o m kq B= rˆ E= 2 4 r 2 r So Gauss’s law in magnetism can be derived as  B.dS = o  m F=

Where

 B.dS

4

is the magnetic flux through the closed surface and

(i)

m

is the net pole

strength within the closed surface. Unlike electric charges, isolated magnetic poles do not exist. Hence, the net magnetic pole strength (  m ) enclosed by any closed surface is always zero. i.e. (ii) m = 0 From eqn. (i) & (ii)

69

 B.dS = 0 Bar magnet as an equivalent solenoid: Consider a current-carrying solenoid of radius ‘a’, length ‘2l’ and number of turns ‘N’ N carrying a current ‘I’. The number of turns per unit length of the solenoid is n = 2l

Consider a small element of the solenoid of thickness dx at a distance ‘x’ from its centre ‘O’. The magnitude of the magnetic field at the point ‘P’ on the axis of the solenoid due to this element is o ( ndx ) Ia 2 dB = 3 2 2 ( r − x ) + a 2  2   (Expression from current carrying loop) As a  r & 2l  r so ( r − x ) + a 2 = r 2 So eqn.(i) can be written as 2

dB =

o ( ndx ) Ia 2

2r 3  nIa  nIa 2 2l  NIa 2  2l B =  dB = o 3  dx = o 3 = o 2r − l 2r 2l  2r 3 2 l

B=

o NIa 2 2r 3

=

 2 NIA B= o 4 r 3  2M B= o 3 4 r

2 o 2 NI ( a ) 4 r3

(A =a ) 2

( M = NIA)

Magnetic field strength due to a bar magnet of magnetic moment ‘M’ at a distant axial point of the magnet. Magnetic dipole moment a an orbital electron: Electron revolve around the nucleus of atom in anticlockwise orbits of electrons may be considered as tiny current loops, hence it is associated with a magnetic dipole moment.

70

q t ne I= t e I= t

I=

as n=1 for an electron. We know that, magnetic moment of current loop M =IA e e M =  r2 =  r2 t  2 r     v  evr M= 2

M=

eL 2me

(i) (ii)

Where L= mevr From eqn. (ii)

M e = L 2me The ratio of the magnetic dipole moment and the angular momentum of an electron is called the gyro magnetic ratio. According to Bohr's theory or Bohr's quantization law, angular momentum of electron is given by nh L= (where n = 1, 2, 3, ….) 2 so equation Can be written as M=

e  nh    2me  2 

71

When n = 1 M least =

eh 1.6 10−19  6.6 10−34 = 4me 4  3.14  9.110−31

M least = 9.27 10−24 Am 2 = Bohr Magneton Bohr Magneton ( M least = M B = B ) can be defined as the orbital magnetic moment of an electron circulating in the inner most orbit. Eqn (ii) can be written as, in vector notation  eL  M = −   2me 

The negative sign used to indicate that M and L are oppositely directed. In the case of a positive charge M and L have the same direction. Magnetic Intensity / Magnetizing field strength/Intensity of Magnetizing field: The degree or extent to which the magnetizing field can magnetize a substance is known as the intensity of magnetizing field. The magnetic field which magnetizes a substance placed in it is called the magnetizing field It is denoted by H . It is vector quantity. Dimension: [MoL-1A] Unit: Am-1 Intensity of Magnetization of Magnetization ( I M ): The degree or extent to which a substance is magnetized when placed in the magnetizing field is called intensity of magnetization. or It is defined as the magnetic moment per unit volume. m ( 2l ) mo ( 2l ) I= o = V A ( 2l ) mo A Thus it may also be defined as the pole strength per unit area of cross section. I=

Dimension & unit same as H . Magnetic flux (  ): The total number of magnetic line of force passing through a surface is called magnetic flux. S.I. Unit: weber (Wb) Magnetic flux density (B): The total number of magnetic lines of force passing normally through a unit area of a substance is called magnetic flux density.

B=

 A

Unit:– Wbm-2 or T, tesla Magnetic permeability (  ): The degree or extent to which magnetic lines of force can enter a substance is known as magnetic permeability. B = H Unit – WbA-1m-1 or TmA-1

72

If  be the permeability of the material and o that of vacuum, the ratio the relative permeability.

r =

 is called o

 o

Magnetic susceptibility ( xm  ) : It is the property of a substance which shows how easily the substance can be magnetized when placed in the magnetizing field. or Magnetic susceptibility is equal to the ratio of intensity of magnetization (I) to the intensity of magnetizing field (H). IH

I = H

=

I H

Curie Law: Curie law states that susceptibility of a material varies inversely with its temperature in kelvin. 1  T C = T C = curie constant Explanation: Curie discovered experimentally that intensity of magnetization ‘I’ of a magnetic material is directly proportional to magnetic field ‘B’ and inversely proportional to the temperature ‘T’ in kelvin. i.e. B H I i.e I  T T I 1 1  , x H T T Classification of Magnetic materials: As a matter of fact, any substance when placed in a magnetic field, behaves in a particular way which is not the same for all the substance. In 1846 Faraday, on the basis of behavior of various substance in the external magnetic field divided the substances into three categories. All substances solids, liquids and gases fall into one of the other these categories. i) Paramagnetic materials: Those substance which when placed in a magnetic field are feebly/weakly (lacks strength) magnetized in the direction of the magnetizing field. Ex- Aluminum, Sodium, Copper etc. When a paramagnetic substance in placed inside an external magnetic field, the magnetic field inside the paramagnetic substance is found to be a slightly greater than the external magnetic field.

73

ii) Ferromagnetic materials: Those substance which when placed in a magnetic field are strongly magnetized in the direction of the magnetizing field, are called ferromagnetic substances. When a ferromagnetic, substance is placed inside a magnetic field, the field inside the ferromagnetic substance gets greatly enhanced.

iii) Diamagnetic materials: The substance which when placed in a magnetic field are feebly magnetized in a direction opposite to that of the magnetizing field are called diamagnetic materials. When a diamagnetic substance is placed inside a magnetic field, the magnetic field inside the diamagnetic is found to be slightly less than external magnetic field.

Hysteresis: (dictionary meaning – “coming late”) The intensity of magnetization does not become zero on making magnetizing field zero but does so a little late and this effect is called Hysteresis. In other word we can say that once a ferromagnetic material is magnetized to its fullest extent, the magnetic field B in it does not change in phase with magnetic intensity H , but lags behind H . Note: This phenomenon is only possible in ferromagnetic.

Explanation of Hysteresis Curve: When H is increased gradually, it is found that the magnetic field B in the specimen first increase sharply then more and more slowly and finally reaches a point (a), after

74

which the magnetization of the specimen does not increase further with increase in H . In other words, the magnetization reaches saturation. If H is then reduced gradually, it is found that the decrease in B does not follow the path ao, but the path ab such that when H becomes zero, B has a definite value ob, (retentivity). If H is then increased in the reverse direction, B decrease and reaches zero when H = OC = − H c which is the reverse field required to completely demagnetize the specimen and is known as its coercivity. After this, as H is taken from –Hd to +Ha, B follows the path ‘defa’ and finally reaches ‘a’. If H is repeatedly taken through several cycles of change, B changes repeatedly along ‘abcdefa’ but never goes back to ‘O’. The process of taking a ferromagnetic material through a cycle of magnetization involves an expenditure of energy that cannot be recovered. This loss of energy is known as hysteresis loss and appears as heat in the specimen. The area of the hysteresis loop is a measure of the energy loss per cycle of magnetization per unit volume of the specimen & depends on the nature of the material. Comparing soft iron and steel: The low retentively and low coercivity of soft iron and the fact that the energy loss per cycle of magnetization is small makes it suitable for making electromagnets cores of transformers, moving coil galvanometers, generators and motors. On the other hand, steel which has a high retentivity and high coercivity is suitable for making permanent magnets of various kinds.

Retentivity: The magnetism present in a specimen even when magnetizing field is reduced to zero is called residual magnetism of the material and this property of the magnetic material is known as Retentivity. Coercivity: It is the property of a magnetic material which depends upon the value of reverse magnetizing field required to reduce the residual magnetism to zero. Geographic axis: The straight line passing through the geographical north and south pole of the earth is called its geographic axis. or The axis of rotation of earth is called geographic axis.

75

Magnetic axis: The straight line passing through the magnetic north & south poles of the earth is called its magnetic axis. Magnetic meridian: The vertical plane passing through the magnetic axis of a freely suspended small magnet is called magnetic meridian. The earth’s magnetic field acts in the direction of the magnetic meridian. or A vertical plane passing through the magnetic axis of the earth is called magnetic meridian. Geographic meridian: The vertical plane passing through the geographic north & south poles is called geographic meridian. or The vertical plane passing through the geographic axis is called the geographic meridian. Equator: The great circle on the earth's surface perpendicular to the geographic axis called equator. Magnetic equator: It is the great circle on the earth perpendicular to the magnetic axis. Angle of declaration or magnetic declination: The angle between the geographic meridian & magnetic meridian. It is denoted by  . Angle of dip or magnetic inclination: The angle made by the earth's total magnetic field B with the horizontal direction in the magnetic meridian is called angle of dip (  ) at any place.

76

From figure

BH = B cos 

BV = B sin 

From eqn. (i) & (ii)

(i) (ii)

B sin  BV = B cos  BH

BV BH

tan  =

On squaring and adding eqn. (i) & (ii) B 2 ( sin 2  + cos 2  ) = BH2 + BV2

B = BH2 + BV2

Time period of oscillation of a magnet/magnetic needle or a freely suspended magnetic dipole in uniform magnetic field: We know that time period is given as displacement ( ) (i) T = 2 acceleration ( ) We know that, torque is given as

 = MB sin 

For very small displacement sin can be written as ‘’ As we know that On equating eqn. (ii) & (iii)

 = MB

(ii)

 = I

(iii)

I = MB MB = I

Where I = moment of inertia Putting the value of eqn. (iv) in eqn. (i) T = 2

I MB

T = 2

I MB

77

(iv)

CHAPTER 6 Electromagnetic Induction Magnetic flux: It is defined as number of magnetic line of force crossing unit area to a surface normal to it.  = BA cos 

 = B. A S.I. Unit: weber Sometime magnetic field is known as flux per unit area i.e. flux density. B=



A Magnetic flux through close surface is zero. Faraday's laws of Electromagnetic Induction: 1st law: Whenever there is change in magnetic flux linked with coil, the e.m.f. induced and hence induced current which last as long change in flux continues. 2nd law: The magnitude of e.m.f. induced in a coil is directly proportional to rate of change of magnetic flux linked with the coil. d e=− dt Lenz's Law: The induced emf opposes the cause which produces it. Motional E.M.F.: Induced e.m.f. produced by changing the area of a closed circuit by the movement of the circuit or part of it through a uniform magnetic field is known as motional e.m.f. Consider a straight conductor moving in a uniform magnetic field. Let us consider a rectangular conductor in which the conductor (straight) is free to move.

78

d dt d ( BA ) e=− dt d ( lx ) e = −B dt dx e = − Bl dt e = − Blv e = Blv emf e=−

I=

e Bvl = current R R

Where R = Resistance We know that

F = BIl

(i) (ii)

(iii)

Putting the value of eqn (ii) & (i) F=

We know that

P=

 t

B 2 vl 2 R

F .ds = Fv t P = Fv

(iv)

=

(v)

From eqn (iv) P=

B 2 v 2l 2 R

Rotational E.M.F: Let us consider the case of a rod that is fixed at one end and rotating in a uniform magnetic field directed perpendicular to the plane of the rotation if the rod turns by an angle  in time ‘t’.

Where

79

d dt d ( BA ) dA =− = −B dt dt 1 d r ( r ) B d = −B 2 = r2 dt 2 dt 2 Br  =− 2

e=−

sin  =

p r

p r p = r

=

Eddy Current: When magnetic flux linked with a coil changes, induced e.m.f. is produced in it and the induced current flows through the wire forming the coil. These currents look like eddies or whirl pools and like-wise is known as eddy currents. They are also known as ‘Foucault’s currents’. Thus, eddy currents are the currents induced in a conductor when placed in a changing magnetic field. Application of Eddy currents: (i) Dead beat galvanometer (ii) Inductance motors (iii) Induction furnace (iv) Speedometer (v) Electrical brake The two most important undesirable effects of eddy currents are: (i) Opposing relative motion (ii) Loss of electrical energy in form heat Effect of Eddy Current: Since the resistance of a metallic conductor is quite low, the magnitude of eddy currents produced is quite large. As such, considerable amount of heat is produced in the conductor due to Joule's heating effect, if the large eddy currents are allowed to produce in the core of a choke coil, transformer, dynamo etc, it may produce undesirable effects. Self-Induction: The phenomenon according to which an opposing induced e.m.f. is produced in a coil as a result of change in current or magnetic flux linked with the coil is called self-induction.

80

I  = LI L=

 I

S.I. Unit: weber amp-1 or henry If I = 1 amp

=L Where L = coefficient of self-induction or self-inductance Self-Inductance (L): Self-Inductance (L) of a coil numerically equal to the magnetic flux linked with the coil, when a unit current flows through it. If ‘e’ is the induced e.m.f. then d e=− dt d ( LI ) e=− dt

e = −L

dI dt

e dI dt ‘L’ numerically equal to the induced e.m.f. produced in the coil, when the rate of change of current in the coil is unity. Inductors in Series: Consider two inductor of self-inductance L1 & L2 are connected in series. Let e1 & e2 are the induced e.m.f. produce in inductor. L =

As it is in series so net induced e.m.f. in series is given by es = e1 + e2 dI dI dI = − L1 − L2 dt dt dt Ls = L1 + L2

− Ls

Inductors in Parallel: Consider two inductor of self-inductances L1 & L2 are connected in parallel.Let I1 & I2 are the current flows through the inductor. As it is in parallel so net current is given by

81

I p = I1 + I 2 Differentiating both side w.r.t. ‘t’, we have dI p dI1 dI 2 = + dt dt dt ep e1 e2 − =− − Lp L1 L2

(i) (ii)

e e e = + L p L1 L2 1 1 1 = + L p L1 L2

As it is parallel, so e.m.f. will be same. Energy stored in an inductor: The source of e.m.f has to spend energy in sending current thorough the circuit against the induced e.m.f. The energy spent by the source of e.m.f. is stored in the form of magnetic field. we know that d p= dt d  = pdt = VIdt

dI Idt dt d = LdI .I

d = L

(i)

I

 =  d  =  LIdI 0

I

I2   = L   2 0 1  = LI 2 or 2 U=

1 2 LI 2

Mutual Induction: It is the phenomenon of inducing e.m.f. in a coil due to the rate of change of current in a nearby coil.

82

I  = MI If I = 1 amp. Then  = M where M= mutual inductance or coefficient of mutual induction. Mutual Inductance: Mutual inductance of the two coil is numerically equal to the magnetic flux linked with one coil, when a unit current flows through the neighbouring coil. If ‘e’ is the induced e.m.f. in secondary coil then d e=− dt d ( MI ) e=− dt dI dt The manner in which the two coils are oriented determines by the coefficient of coupling between them, which is given by M k= L1 L2 Self-Inductance of a Solenoid: We know that  = LI (i) magnetic field due to solenoid is given as  NI B= o (ii) l Magnetic flux linked with solenoid is given as  = NBA (iii) From eqn. (ii) & (iii) N 2 o AI (iv) = l on comparing eqn. (i) & (iv)  N2A L= o or L = o n 2lA l Where e = −M

83

Mutual Inductance of two long Solenoids:

N =n l N = nl

B1 = o

N1 I1 l

2 = N2 B1 A2 NN 2 = o 2 1 I1 A2 l

(i) (ii) (iii)

We know that

2 = M 2 I1 where M2 = mutual induction due to coil 1. on comparing (iii) & (iv) N N M 2 = o 2 1 A2 or M 2 = o n1n2lA2 l M1 =

o N 2 N1 l

(iv)

A1 or M 1 = o n1n2lA1

N1 N 2 N N =n = = , A1 = A2 = A & l l l l Then M1 = M 2 = M If

M = olAn 2 Coefficient of Coupling: The mutual induction between two coils depends upon the shape and size of the two coils, separation between them and the permeability of the material of the core (if any)on which the two coils are wound. Apart from this factor, the mutual induction between the two coils depends upon the manner in which the two coils are oriented relative to each other. The manner in which the two coils are oriented determines the coefficient of coupling between them, which is given by M k= L1 L2 A.C. Generator (A.C. Dynamo): An electric generator is a device used to convert mechanical energy into electrical energy. Principle: As per Faraday’s law of electromagnetic induction when a coil is rotated in uniform magnetic field an induced e.m.f. is produced in it. Construction:

84

It consists of large number of turns of conducting wire wound on iron core This arrangement is known as armature (A). N-S are strong field magnets, ‘B’ and ‘S’ are the brush and slip ring arrangement.

Working: When coil rotates in magnetic field, flux link with it change and according to Faraday's law whenever their is change in the magnetic flux an e.m.f. is induced so when the coil rotates the magnetic flux changes and an e.m.f. is induced in the coil. we know  = nBA cos t ( = t ) (i) Let E = induced e.m.f.

d dt d ( nBA cos t ) E=− dt E = nBA sin t

E=−

When sin wt = 1

Eo = nBA (maximum e.m.f.)

From eqn. (ii) & (iii)

E = Eo sin t

E Eo = sin t R R I = I o sin t

I=

85

(ii) (iii) (iv)

Chapter 7 Alternating Current Alternating Current: An electric current, magnitude of which changes with time and polarity reverse periodically is called alternating current (a.c.).

Mean or Average value (Iav) of a.c.: The average value of an a.c. is the average of all the instantaneous value during one alternation. Since the current increases from zero to peak value and decreases back to zero during one alternation, the average value must be some value between those two limits. or Average value of a.c. is that value of steady current, which sends the same amount of charge through a circuit in a certain time interval as is sent by an a.c. through the same circuit in half cycle. Iav = 0.637 Io Root Mean Square (RMS) value of a.c.: RMS value of a.c. is defined as that steady current which produces the same amount of heat in a conductor in a certain time interval as is produced by the a.c. in the same conductor during the time period ‘T’ (i.e. full cycle). I I rms = o 2 T

I2

onecycle

=

T

I

2

dt =

0

T

I

2 o

sin 2 tdt

0

T

 dt

 dt

0

0

T

I

2 onecycle

=I

 sin

2 0 o

2

T

 sin

tdt =I

T

 dt

2 0 o

0

I2

onecycle

2

tdt

T

 dt 0

1 I2 = I o2  = o 2 2 I2

onecycle

I rms = I rms =

I2

Io 2

=

I o2 2

onecycle

; Vrms =

86

(i)

Vo 2

(ii)

Note T

(a)

2  sin tdt 0

T

T

=

 cos

2

tdt

0

 dt 0

T

 dt

=

1 2

0

T

(b)

 sin t cos tdt 0

T

=0

 dt 0

A.C. through Resistor: Let us consider a circuit which contains a resistor of resistance ‘R’ connected with an a.c. source having voltage (i) V = Vo sin t

According to Ohm’s law

V R Vo sin t I= R I=

For maximum value

(ii)

I = I o , sin t = 1 from eqn. (ii)

Io =

Vo R

(iii)

from eqn. (ii) & (iii) I = I o sin t

Since, it is clear that voltage & current are in same phase in resistor.

A.C. through Inductor: Let us consider a circuit which contains an inductor of inductance ‘L’ connected with an a.c. source having voltage (i) V = Vo sin t

87

As we know that

V =L dI =

dI =

dI dt

V dt L

(ii)

Vo sin tdt (from eqn. (i)) L

Integrating both sides Vo

 dI =  L sin tdt Vo sin tdt L V I = o ( − cos t ) L Vo   I= sin  t −  XL 2  I=

Where X L =  L For maximum value

(iii)

  I = I o , sin  t −  = 1 from eqn. (iii) 2  V Io = o (iv) XL where X L = inductive reactance, S.I. Unit: ohm () from eqn. (iv) & (iii)   I = I o sin  t −  (v) 2  From eqn. (v) it is clear that in case of inductor current lag behind the voltage by (/2) or we say that voltage leads the current by (/2).

88

A.C. through Capacitor: Let us consider a circuit which contain a capacitor of capacitance ‘C’ connected with an a.c. source having voltage (i) V = Vo sin t

As we know that

q = CV

From eqn.(i)

q = CVo sin t Differentiating both side eqn (ii) we get dq d sin t = CVo dt dt I = CVo cos t

  I = CVo sin  t +  2  Vo   I= sin  t +  1 2  C V   I = o sin  t +  XC 2  Where X C =

(ii)

(iii)

1

C

for maximum value

  I = I o , sin  t +  = 1 2 

from eqn. (iii)

Io =

Vo XC

89

(iv)

from eqn. (iii) & (iv)

  I = I o sin  t +  2 

(v)

Where XC = capacitive reactance S.I. Unit of XC is ohm () from eqn (v) it is clear that in case of capacitor current leads the voltage by (/2).

Phasor: The complex quantities normally employed in a.c. circuit analysis, can be added and subtracted like coplanar vectors. Such coplanar vectors which represent sinusoidally time varying quantities are known as phasors. In this chapter current & voltage are phasors quantities. The different phasor diagram for different a.c. circuit is given below: (i) A.C. through Resistor: As V = Vo sin t; I = I o sin t;

so both quantities are in same phase. (ii) A.C. through Inductor:   As V = Vo sin t; I = I o sin  t −  ; 2 

Here voltage leads the current by /2. (iii) A.C. through Capacitor:   As V = Vo sin t; I = I o sin  t +  ; 2 

90

Here current leads the voltage by /2. Power across Resistor: We know that in case of resistor current & voltage are in same phase V = Vo sin t I = I o sin t the power of the given circuit is P = VI P = Vo I o sin 2 t the average power is

(i) (ii)

(iii)

t

Pav =

 Pdt

(iv)

0

t

 dt 0

From eqn. (iii) & (iv) t

V I

o o

Pav =

sin 2 tdt

0

t

 dt 0

t

Pav = Vo I o

 sin

2

tdt (v)

0

t

 dt 0

On solving t

 sin

2

tdt =

0

t

 dt 0

From eqn. (v) & (vi)

91

1 2

(vi)

1 2 Vo I o Pav = 2 2

Pav = Vo I o

Pav = Vrms I rms

Power across Inductor: We know that in case of inductor voltage leads the current by /2. V = Vo sin t

  I = I o sin  t −  2  The power of the given circuit is

(i) (ii)

P = VI

  P = Vo I o sin t sin  t −  2  P = −Vo I o sin t cos t

(iii)

the average power is T

Pav =

 Pdt

(iv)

0 T

 dt 0

From eqn.(iii) & (iv) T

 −V I

o o

Pav =

sin t cos tdt

0

T

 dt 0

T

Pav = −Vo I o

 sin t cos tdt 0

T

(v)

 dt 0

on solving T

 sin t cos tdt 0

T

=0

(vi)

 dt 0

from eqn. (v) & (vi) Pav = 0

Since Pav is zero, so the current is then stated wattles current. Power across Capacitor: We know that in case of capacitor current leads the voltage by /2. V = Vo sin t

92

(i)

  I = I o sin  t +  2  the power of the given circuit is

(ii)

P = VI

  P = Vo sin tI o sin  t +  2  P = Vo I o sin t cos t

(iii)

the average power is t

Pav =

 Pdt

(iv)

0

t

 dt 0

From eqn.(iii) & (iv) t

V I

sin t cos tdt

o o

Pav =

0

t

 dt 0

t

Pav = Vo I o

 sin t cos tdt 0

t

(v)

 dt 0

on solving t

 sin t cos tdt 0

t

=0

(vi)

 dt 0

from eqn. (v) & (vi) Pav = 0

Since Pav is zero, so the current is then stated wattles current. A.C. through L-C-R circuit in Series: Let us consider a inductor having inductance ‘L’, capacitor having capacitance ‘C’ and resistor having resistance ‘R’ connected to an a.c. source in series connection having an e.m.f. (i) V = Vo sin t

By plotting phasor diagram of given circuit

93

Equivalent phasor diagram of given

from diagram

V = Vo sin t

I = I o sin (t −  )

(ii)

By using Pythagoras theorem V 2 = (VL − VC ) + VR2 2

(iii) As it is series connection so current through each component will be same, then eqn.(iii) can be written as

( IZ )

2

= ( IX L − IX C ) + ( IR) 2 2

Z 2 = ( X L − X C ) + R2 2

1

Z = ( X L − X C ) + R 2  2   Where Z = total impedance (resistance) of the circuit. from diagram V −V tan  = L C VR 2

IX L − IX C IR X L − XC tan  = R

tan  =

 X L − XC   R  

 = tan −1  from diagam

94

VR V IR cos  = IZ R cos  = Z

cos  =

& cos = power factor It is also define as the ratio of true power and virtual power in a circuit. pure power cos  = virtual power Choke Coil: It is basically an iron cored wire wound inductor used to control current in an a.c. circuit without much loss of energy. or A choke coil is simply an inductor with large inductance which is used to reduce current in a.c. circuits without much loss of energy. R 0 Power factor: cos   WL i.e. average power dissipated by the coil is very small as Z = R 2 + W 2 L2 is large, so current is reduced without appreciable waste of power. Resonance condition of a series L-C-R circuit: LCR circuit is said to be in the resonance condition when the current through it has its maximum value & it is only possible when the impedance of the circuit is minimum. As

Z = R2 + ( X L − X C )

2

for Zmin X L − XC = 0

o L −

1

o C

=0

1 o C 1 2 o = LC 1 o = resonant angular frequency LC

o L =

1 LC 1 2 f o = LC

o =

fo =

1 resonant frequency 2 LC

95

The frequency at which the current amplitude I attain a peak value (Io) is called resonant frequency. Remark: The series resonant circuit is also called an acceptor circuit, when a number of frequencies are fed to it, it accepts only one frequency and rejects the other frequencies. The current is maximum at this frequency. Q-factor of Resonance Circuit: It defined as 2 times the ratio of the energy stored in the circuit to the energy dissipated in resistance per cycle of a.c. supply. ( 2π ) × ( energystored in circuit per cycle ) Q= energy dissipated per cycle or The ratio of the voltage drop across the inductance (or capacitance) at resonance to the applied voltage (voltage across resistor). V V Q= L Q= C VR VR IX L IR o L Q= R 1 L Q= LC R

Q=

Q=

1 L R C

IX C IR 1 Q= o CR

Q=

Q=

LC CR

Q=

1 L R C

Above grap show variation of current amplitude with frequency in an LCR Circuit. Transformer: It is a device which convert high voltage a.c. into low voltage a.c. and vice-versa. Principle: It is based on principle of mutual induction. There are two types of transformer: (i) Step up transformer: It is a transformer which convert low voltage into high voltage.

96

In this case the number of turn of primary coil is more than the number of turn of secondary coil. i.e. Np < Ns.

(ii) Step down transformer It is a transformer which convert high voltage into low voltage. In this case the number of turn of primary coil is more than the number of turn of secondary coil. i.e. Np>Ns.

Working: When an alternating current is passed through primary coil, the magnetic flux through iron core change which does two things, produce emf in primary coil and an induced emf is step in secondary coil. If we assume that there is no loss power i.e. power of primary equal to power of secondary PP = PS

eP I P = eS I S eP I S = eS I P If we assume that there is no leakage of magnetic flux then

97

(i)

d dt d eS = − N S dt

eP = − N P

On dividing eqn. (ii) by (iii)

eP N P = eS N S

From eqn. (i) & (iv)

eP N P I S = = eS N s I P The efficiency of a transformer is defined as power output = 100% power input The efficiency of real transformers is fairly high (90-99%) through not 100% Energy losses in transformer (i) Copper loss (ii) Eddy current loss (iii) Hysteresis loss (iv) Flux leakage (v) Humming loss

98

(ii) (iii)

(iv)

CHAPTER 8 Electromagnetic Waves Displacement Current: It is a current in a region where change of electric flux take place due to change in electric flux intensity. It is produced by displacement of electron caused by change in electric field. Hence the name displacement current. or It is due to time-varying electric field – d id =  o E dt Displacement current acts as a source of magnetic field in exactly the same way as conduction current. Expression for the displacement current: We know that, electric field inside capacitor  (i) E= o On differentiating both side w.r.t time

  d   o  dt  1   d  =      o   dt    q  d  1    A    =      o   dt   

dE = dt dE dt dE dt

dE  1   dq  =   dt   o A   dt  dq  dE  = ( o A)    dt  dt   dE  id = (  o A )     dt   d ( EA )  id =  o     dt 

dE dt Which is the expression of displacement current. Using Gauss’s theorem id =  o

99

(ii)

q E =    o  from equation (ii)

 q d   id =  o    o  dt    dq  id =    dt  id = ic

     

(iii) Where ic is conduction current. From equation (iii) it is clear that displacement current and conduction current are equal. Maxwell’s equations: Gauss law of electrostaticsq s E.ds =  o Gauss law of magnetism B.ds = 0 s

Faraday’s law of eletromagnetic induction-

 E.dl = − c

Modified form of Ampere’s circuital law-

 B.dl =  c

o

d B dt

d E    I c +  o dt 

Electromagnetic Waves: These waves propagates through spaceas coupled electric and magnetic fields, oscillating perpendicular to each other and to the direction of propagation of the wave.

1. Electromagnetic waves are produced only by charges that are accelerating, since acceleration is absolute, and not a relative phenomenon. 2. An electric charge oscillating harmonically with frequency, produces electromagnetic waves of the same frequency. 3. An electric dipole is a basic source of electromagnetic waves. 4. Electromagnetic waves with wavelength of the order of a few metres were first produced and detected in the laboratory by Hertz in 1887. He thus verified a basic prediction of Maxwell’s equations. 5. Electromagnetic waves are transverse in nature. 6. They do not require any material medium for their propagation.

100

Oscillation of Electric and Magnetic Fields: These oscillate sinusoidally in space and time in an electromagnetic wave. The oscillating electric and magnetic fields, E and B are perpendicular to each other and to the direction of propagation of the electromagnetic wave. For a wave of frequency ν , wavelength λ , propagating along z-direction

E = Ex ( t ) = Eo sin ( kz − t )  z   z t   = Eo sin  2  − vt   = Eo sin  2  −         T  B = By ( t ) = Bo sin ( kz − t )  z   z t   = Bo sin  2  − vt   = Bo sin  2  −         T  Eo =c Bo Relation between µo and o: The speed c of electromagnetic wave in vacuum is related to µo and o (the free space permeability and permittivity constants) as 1 C=

 o o

The value of c equals the speed of light obtained from optical measurements. Light is an electromagnetic wave; c is, therefore, also the speed of light. Electromagnetic waves other than light also have the same velocity c in free space. Speed of Light: The speed of light, or of electromagnetic waves in a material medium is 1 V=



Where µ is the permeability of the medium and  its permittivity. Electromagnetic waves carry energy as they travel through space and this energy is shared equally by the electric and magnetic fields. Energy Per Unit Volume: If in a region of space in which there exist electric and magnetic fields , there exists Energy Density (Energy per unit volume) associated with these fields is,  1 2 U = o E2 + B 2 2 o Where we are assuming that the concerned space consists of vacuum only. Electromagnetic waves transport momentum as well. When these waves strike a surface, a pressure is exerted on the surface. If total energy transferred to a surface in time t is U, total momentum delivered to this surface is U P= C .

101

Electromagnetic Spectrum: The orderly distribution of the electromagnetic waves in accordance with their wavelength or frequency into distinct groups having widely differing properties are called electromagnetic spectrum. The classification has more to do with the way these waves are produced and detected. Different Regions of Spectrum: Different regions are known by different names; У -rays, X-rays, ultraviolet rays, visible rays, infrared rays, microwaves and radio waves in order of increasing wavelength from 10-12 m to 106 m. (a) Radio Waves: (i) These are produced by accelerated motion of charges in wires. (ii) These are used in radio and television communication systems. (iii) These are generally in the frequency range from 500 kHz to about 1000 MHz or wavelength range 600 m to 0.1 m. (b) Microwaves: (i) These are short wavelength radio waves with frequency range 10 9 Hz to 1012 Hz or wavelength range 0.3 m to 10 -3 m. (ii) Due to their short wavelengths, they are suitable for radar systems used in aircraft navigation. (iii) Microwave ovens use them for cooking. (c) Infrared Waves: (i) Frequency range 1011 Hz to 5 x 1014 Hz or wavelength range 5 x 10 -3 m to 10-6 m. (ii) These are produced by hot bodies and molecules. (iii) They lie in the low frequency or long wavelength end of the visible spectrum. (iv) Treat muscular straw. (v) For taking photographs’ in fog or smoke. (vi) In green house to keep plants warm. (vii) In weather forecasting through infrared photography. (d) Visible Light: (i) The spectrum frequency runs from about 4 x 1014 Hz to about 7 x 1014 Hz or wavelength range . (ii) Our eyes are sensitive to this range of wavelengths. (e) Ultraviolet light: (i) It covers frequency range from 10 16 Hz to 1017 Hz or wavelengths range from 3.5 x 10-7 m to 1.5 x 10-7 m . (ii) The sun is an important source of UV rays. (iii) In the study of molecular structure. (iv) In sterilizing the surgical instruments. (f) X-rays: (i) It covers frequency range from 10 18 Hz to 1020 Hz or wavelengths range from 10-8m to 10-11 m . (ii) It is used in medical diagnosis. (iii) In detecting faults, cracks, flaws and holes in metal products. (iv) In the study of crystal structure. (v) For the detection of pearls in oysters. (g) Gamma Rays:

102

(i) (ii) (iii)

These lie in the upper frequency range (1018 Hz to 1022 Hz) of the spectrum, and have wavelengths in the range 10-14 m to 10-10 m . It is used in manufacture of polyethylene from ethylene. It is used for the study of nuclear structure.

103

CHAPTER 9 Ray Optics and Optical Instruments Optics: The branch of physics which deals with the study of nature, production and propagation of light. It consists of two branch (i) Ray optics / Geometrical optics (ii) Wave optics/ Physical optics Ray Optics: It concerns itself with the particle nature of light and is based on (i) The rectilinear propagation of light. (ii) The laws of reflection and refraction of light. In ray optics, the light is considered as a ray which travels in a straight line. It states that for each & every object there is an image. Wave Optics: It concerns itself with the wave nature of light and is based on the phenomena like (i) Interference (ii) Diffraction (iii) Polarization of light Wave optics treats light as a series of propagation electric and magnetic field oscillations. Reflection of Light: Bouncing back of light from polished/shining surface when is strike on it is known as reflection of light.

Laws of Reflection: Law of reflection has following two laws: (i) The incident ray, reflected ray and normal to the reflecting surface at the point of incidence all lie in the same plane. (ii) The angle of incidence (i) is equal to the angle of reflection (r) i.e. i = r

Note:

104

(i) When ray of light fall normal to the surface then i = r = 0o . (ii) Total number of images formed by two plane mirror inclined at an angle ‘’ with each other is given by 360 360 n= − 1 , if is even integer





360 360 n= , if is odd integer   Spherical Mirror: The reflecting surface of a spherical mirror forms a part of a sphere. The reflecting surface of a spherical mirror may be curved inwards or outwards. Spherical mirror are of two types: (i) Concave mirror (ii) Convex mirror Concave Mirror: A concave mirror is that spherical mirror in which reflection take place at the concave or inner curved surface.

Convex Mirror: A convex mirror is that spherical mirror in which reflection take place at the convex or bulging out surface.

The various terms associated with spherical mirror are: (i) Centre of Curvature: The centre of curvature of a spherical mirror is the central point of the hallow sphere of which the mirror is a part. (ii) Radius of Curvature: The radius of curvature of radius of a spherical mirror is the radius of the hallow sphere of which the mirror is a part. In other words, it is distance between the centre of curvature and pole of a mirror. It is represented by letter ‘R’. (iii) Pole:

105

The pole of spherical mirror is the centre or middle point of a spherical mirror. It is represented by the letter ‘P’. (iv) Principal axis: The line passing through the pole and the centre of curvature is called principal axis. (v) Aperture: It is the diameter (MM') of circular boundary of spherical mirror. (vi) Principal focus: When ray of light parallel to principal axis after reflection it pass through a point (in concave mirror) or appear to pass through a point (in convex mirror) on principal axis is called principal focus. (vii) Focal length: The distance between pole and focus is known as focal length.

F=

R 2

Real image: It is a kind of image which is formed by actual intersection of light rays after reflection. Virtual image: It is a kind of image which is formed by producing the reflected rays backward after reflection. Image: When light rays meet or appear to meet after reflection from a mirror, then it is called an image. Image formation by convex mirror: Whatever be the position of object in front of convex mirror, the image formed by a convex mirror is always behind the mirror, virtual, erect and smaller than object.

AB = Object A'B' = virtual erect Image formation by concave mirror: (i) Between the pole (P) and focus (F):

106

Position: behind the mirror AB = Object A'B' = image Nature: Virtual & erect (ii) At the focus (F):

Position of image: At infinity Nature: Real & inverted Size: highly magnified (iii) Between focus (F) and Centre of Curvature (C):

AB = Object A'B' = image Nature: Real & inverted Position: beyond the centre of curvature Size: larger than the object. (iv) At the centre of curvature:

AB= Object A'B' = image Nature: Real & inverted Position: at the centre of curvature Size: same size as the object

107

(v) Beyond the centre of curvature (C):

AB = object A'B' = image Nature: Real & inverted Position: between the focus and centre of curvature Size: diminished Uses of Convex Mirror: (i) Convex mirror are used as rear view mirrors in automobiles to see the traffic at back side as they give erect image and also highly diminished one giving the wide field view of traffic behind. (ii) Convex mirrors are used as ‘shop security mirrors’. Uses Concave Mirror: (i) Concave mirrors are used as shaving mirrors to see a large image of the face. (ii) Concave mirrors are used by dentists to see the large images of the teeth of patients. (iii) Concave mirrors are used as reflectors in torches, vehicle headlights. (iv) Concave dishes are used in TV dish antennas to receive TV signals from the distant communication satellites. Lateral inversion: The change of sides of an object and its mirror image is called lateral inversion. Sign Convention for Spherical Mirrors: (i) All the distances are measured from pole of the mirror. (ii) Distance measured in the same direction as that of incident light are taken as positive. (iii) Distance measured against the direction of incident light are taken as negative. (iv) Distance measured upward to the principal axis are taken as positive. (v) Distance measured downward to the principal axis are taken as negative. Mirror formula: A formula which gives the relationship between image distance (v), object distance (u) and focal length (f) of a spherical mirror is known as the mirror formula.

108

As LNF & A'B'F are similar

As ABC & A'B'C are similar

On equating eqn. (i) & (ii)

LN = A' B ' AB = A' B '

NF A' F NF A' F

AB AC = A' B ' A'C

NF AC = A ' F A 'C As P & N are very close to each other so we assume that they lie at same point. so eqn. (iii) can be written as PF AC = A ' F A 'C PF PA − PC = P ' A − PF PC − PA ' Using sign convention as PF = - f = focal length PA' = - v = image distance PA= - u = object distance PC = - r =- 2f = radius of curvature. Putting the above value in egn. (iv) ( − f ) = ( −u ) − ( −2 f ) ( −v ) − ( − f ) ( −2 f ) − ( −v )

(i)

(ii)

(iii)

(iv)

f u−2f = v− f 2 f −v 2 f 2 − fv = uv − 2 fv − fu + 2 f 2 2 fv − fv = uv − fu fv = uv − fu

On dividing above eqn. by uvf

1 1 1 = + f v u Linear Magnification: The ratio of the height of the image to that of the height of object is called linear magnification or transverse magnification. height of image hi m= = height of object ho

m=

hi −v = ho u

Note:

109

(i) Sign of the magnification indicate the nature of image & magnitude of the magnification indicate the size of image. (ii) if m = negative, Nature:Real image m = positive , Nature: virtual image Refraction of Light: When ray of light travel from one medium to another medium then it deviates from its original path. This phenomenon is known as refraction of light.

Note: (i) A medium in which speed of light is more is known as optically rarer medium and a medium in which speed of light is less is said to be optically denser medium. For example in air & water, air is rarer and water is denser medium. (ii) The bending of light or refraction occur due to change in the speed of light as it passes from one medium to another. Laws of Refraction: Two laws of refraction are given as below: (i) The incident ray, the refracted ray and the normal to the interface at point of incidence all lie in the same plane. (ii) The ratio of sine of the angle of incidence and the sine of the angle of refraction is constant for a given pair of media and for given colour of light. sin i = constant sin r This law is also known as Snell's law. Refractive index: It is ratio of speed of light in medium 1 (V1) to speed of light in medium 2 (V2). speed of light in medium 1 R.I = speed of light in medium 2

21 =

V1 V2

It has no unit. Types of refractive index (i) Absolute refractive index (ii) Relative refractive index Absolute Refractive Index: The refractive index of medium with respect to vaccum is called absolute refractive index. As

110

=

C V

=

vac med

=

vac med

i.e.  =

1



Where C = speed of light in vaccum V = speed of light in medium Relative Refractive Index: The refractive index of medium with respect to other than vaccum is called relative refractive index. V 21 = 1 V2 &

21 =

1 2

Factors on which the refractive index of a medium depends: (i) Nature of the medium (ii) Wavelength of the light used (iii) Temperature (iv) Nature of the surrounding medium. Relation between refractive index of different medium:

We know that

 AB =

VB VA

(i)

As C VA C B = VB

A =

from eqn. (i)

111

(ii) (iii)

VB C  VA C V C = B C VA

 AB =

 AB

(iv)

From eqn. (ii) & (iii)

 AB =

A B

(v)

As

From eqn. (v) & (vi)

sin i =  AB sin r

(vi)

sin i  A = sin r  B Refraction through rectangular glass slab:

Lateral shift or lateral displacement (x) The perpendicular distance between the emergent ray and incident ray, when the ray of light travel through glass slab is known as lateral shift. in BFC BF cos r = BC t cos r = BC t BC = (i) cos r in BCE CE sin ( i − r ) = BC CE = BC sin ( i − r ) from eqn. (i)

112

x=

t sin ( i − r ) cos r

Where t = thickness of glass slab Note: (i) The displacement of a emergent ray cannot exceed the thickness of the glass slab. (ii) The lateral shift increase with the increase in thickness of the glass slab. (iii) The lateral shift increases with the increase in refractive index of the glass slab. Real and apparent depth:

Here OP = real depth = t OP' = apparent depth PP' = Normal shift = d

=

real depth apparent depth

(i)

=

t apparent depth

(ii)

Normal shift (d): The height through which an object appear to be rise in denser medium is called normal shift. Nomal shift = Real depth - apparent depth t d =t−   t d = t 1 −   

Total internal reflection (TIR): The phenomenon in which a ray of light travelling at an angle of incidence greater than the critical angel from denser to a rarer medium is totally reflected back into denser medium is called total internal reflection (TIR).

113

Necessary condition for TIR (i) Light must travel from an optically denser medium to optically rarer medium. (ii) The angle of incidence in the denser medium must be greater than the critical angle for the two medium. Critical angle (ic): The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90o is called critical angle. It is denoted by ic. Relation between critical angle and refractive index: We know that sin i Vd (speed of light in denser medium) = sin r Vr (speed of light in rerer medium) As sin i r (i) = sin r d If r = 1 (for air) & i = ic then r = 90o d =  From eqn. (i) sin ic 1 = sin 90  sin ic =

1



Application of total internal reflection (TIR): (i) Sparkling of diamond: As the refractive index of diamond is very large, its critical angle is very small about 24.4o. (ii) Mirage: It is an optical illusion observed in deserts or over hot extended surfaces like a coal-tarred road, due to which a traveller sees a shivering pond of water some distance ahead of him and in which the surrounding objects like trees etc. appear inverted. (iii) Totally reflecting prism: A right-angled isosceles prism i.e, a 45 o- 90o- 45o prism is called a totally reflecting prism, critical angle of glass about 42 o. It is used in three ways: (a) To deviate a ray through 90o:

(b) To invest an image with deviation of rays through 180 o:

114

(c) To invert an image without deviation of prism (Erecting prism):

(iv) Optical fibres: An optical fibre is a hair-thin long strand of quality glass or quartz surrounded by a glass coating of slightly lower refractive index. It is used as a guided medium for transmitting an optical signal from one place to another.

Core: The central cylindrical core is made of high quality glass/silica of refractive index 1 and has a diameter about 10 to 100 m. Cladding: The core is surrounded by a glass/plastic jacket of refractive index µ2 (2E1) Note: we can find the expression for radii of the stationary orbits & the energy of the electron in different orbits by using Bohr model. Hydrogen Spectrum: The spectrum of light emitted from excited atoms consists of a series of isolated parallel bright line and it is termed as atomic spectrum. Hydrogen spectrum is an example of such type of atomic spectrum are called spectral lines. With the help of spectrometer the spectrum of hydrogen atom was first of all studied by Balmer in 1855. Let an electron jump from the mth orbit of energy Em to nth orbit (m>n) of energy En. The energy difference between the two orbits is emitted as a photon of radiation of frequency ν, given by E = Em − En

h = Em − En 1



=

( Em − En )

hC Substituting the value of Em & En , we get 1 1  1 = R 2 − 2   m  n

me4 = 1.1107 m − 1 8 o2 h3C (i) Lyman series: when electrons in hydrogen atoms jump to first orbit. n=1 , m= 2,3,4,5,.............................. 1 1 1  = R 2 − 2   1 m  (ii) Balmer series: when electron jump to second orbit. n= 2, m=3,4,5,............................. 1 1  1 = R 2 − 2   2 m  (iii) Paschen series: when electron jump to third orbit n=3, m=4,5,6,7,................................. 1 1  1 = R 2 − 2   3 m   (iv) Brackett series: when electron jump to fourth orbit n=4 , m=5,6,7,............................. Where R= Rydberg constant =

152

1  1 = R 2 − 2   4 m  (v) Pfund series: when electron jump to fifth orbit n= 5, m= 6,7,8,.......................................... 1 1  1 = R 2 − 2   5 m  1

Note: Number of spectral lines =

n ( n − 1) 2

153

Chapter 13 Nucleus Mass Number: The sum of the number of protons (Z) and the number of neutrons (N) is called mass number (A). Neutrons discovered by Chadwick. Size the nucleus: The greater the number of nucleons the larger is the volume of the nucleus. Assuming the nucleus to be a sphere of radius ‘R’, it has been found that the volume ‘V’ of the nucleus varies linearly with the mass number ‘A’ i.e. 4 V =  R3 3 VA 1

R  A3 1

R = Ro A 3 Where Ro is constant for all nuclei Ro = 1.2 fm Ro = 1.2x10-15 m Mass Defect: The difference between the sum of the masses of the nucleons and the mass of the nucleus is called the mass defect. m = Zmp + ( A − Z ) mn − M Where Z = atomic number A = mass number mp = mass of proton mn = mass of neutron M = mass of nucleus Binding Energy: The total energy required to disintegrate the nucleus into its constituent particles (nucleons) is called binding energy of the nucleus. Or Binding energy of the nucleus is define as the difference between the energy of constituent particles and the whole nucleus. We know that difference in mass is equivalent to energy according to Einstein mass energy relation (i) E = mc 2 Therefore the binding energy of nucleus. EB =  Zmp + ( A − Z ) mn − M  c 2 (ii) Average binding energy per nucleon

154

EB c 2 =  Zm p + ( A − Z ) mn − M  A A

(iii)

From graph, it is observed: E (i) B almost remain constant between A = 30 to 100 and decrease for small & large value A of A. (ii) For very light nuclei ‘A’ less than 30, there are rapid fluctuation shown by recurrence of peaks in the binding energy per nucleon for mass number, which are multiple of 4. For example 4He, 8B, 12C, 16O etc in which Z = N i.e. even-even nuclei. (iii) Less prominent peaks are observed at the value of Z & N equal to 20, 28, 50, 82 & 126 known as magic number. (iv) For very heavy nuclei greater than 180 binding energy per nucleon decrease with increase of A and for heaviest nuclei it is about 7.5 Mev per nucleon. (v) The appearance of peaks shows greater stability of the corresponding nuclei. Note: The binding energy of the nucleus is measure of its stability i.e. greater the binding energy more stable is the nucleus. Radioactivity: The phenomenon of spontaneous emission of highly penetrating particle and electromagnetic radiation from heavy element is called natural radioactivity. The element, which shows this property, are called radioactive elements. Law of Radioactive Decay (disintegration): 1. Radioactive decay is a spontaneous process and is not affected by the external conditions such as temperature pressure etc. 2. When a radioactive element decays to emitting an -particle, its position goes down by two places in the periodic table.  − particle A ⎯⎯⎯⎯ → Z − 2 X A− 4 Z X 3. When a radioactive element decays by emitting a -particle, its position is raised by one place in the periodic table.  − particle A ⎯⎯⎯⎯ → Z +1 X A Z X 4. When a radioactive element decays by emitting a -rays, its position remains the same in the periodic table. The radioactive element in the excited state come to its ground state by emitting the energy in the form of a photon or -ray.

155

*

 − ray X A ⎯⎯⎯ →ZXA 5. The rate of disintegration of a radioactive substance is directly proportional to the number of atoms remained un-decayed in the substance. The law is called radioactive decay law or disintegration law. Activity of Radioactive Substance: The activity of radioactive substance is defined as the rate of disintegration or decay of that substance. It is denoted by ‘R’. S.I. Unit: becquerel, curie, decay per second. dN R=− (i) dt we know that dN − = N (ii) dt From eqn. (i) & (ii) R =  N (as N = Noe− t ) Z

R =  N o e − t R = Ro e − t

1 curie = 3.7 x 1010 becquerel Where → Ro =  No  = radioactive decay constant Relation for Radioactive Decay (N = Noe- t): Let there be ‘N’ atoms of any radioactive substance at time ‘t’ and dN atom disintegrate at time dt. Then dN − N dt dN  = − N dt dN  = − dt N N t dN  = −  dt N No 0   log e N N = − t 0 N

t

o

  log e N − log e N o  = − t − 0  log e 

N = − t No

N = e − t No

 N = N o e − t

(negative sign indicates the reduction in number of atoms with time) a log e a − log b = log   b

156

Nuclear force: The nucleus of an atom contains protons & neutrons. Coulombian repulsive force acts between the protons, while gravitational force acts between all the nucleons. Gravitational force of attraction between the nucleons is about 10 -36 times smaller than repulsive force between the protons, so it can't hold the nucleons together against the force of repulsion. There must be some other strong force which overcomes the electrostatic repulsion between the protons and that is nuclear force which hold the protons together. It was the Japanese physicist Yukawa who proposed that a third force called nuclear force was responsible for the stability of the atom. Characteristics of Nuclear Force: (i) Nuclear force is independent of charge. (ii) It is short-range attractive force. (iii) The nuclear force between nucleons does not follow the inverse square law or any law based solely on the distance between them. (iv) It is not a centre force.

Mean Life or Average Life: The mean life of a radioactive element is defined as the ratio of the lives of all the radioactive atoms to the total number of such atoms in it.

157

Ta =  = Ta =

1

total lives of all atoms total number of atoms &



T1 = 0.6931Ta 2

Nuclear Fission: It is a process of splitting a heavy nucleus into two nuclei of comparable masses along with the emission of large amount of energy. It is a chain reaction. Ex- Atom bomb. 235 1 236 141 92 1 92U + 0 n → 92U → 56 Ba + 36 Kr + 3 0 n + Q Nuclear Fussion: It is a process in which two very light nuclei ( A  8 ) combine to form a nucleus with a large mass number along with simultaneous release of large amount of energy is called nuclear fussion. Ex-Hydrogen bomb. 2 2 4 1 H + 1 H → 2 He + 24MeV Half Life period: The half-life period (T1/2) of a radioactive substance is the time in which the amount of radioactive substance is reduces to half of its original value. We know that (i) N = N o e − t

t = T1 ; N = 2

No 2

from eqn (i) − T1 No = Noe 2 2 − T1 1 =e 2 2

T1

2=e 2 T1 = log e 2 2

T1 = 2

T1 = 2

log e 2



=

0.693



0.693



The determination of half-life period in very useful for geologists for estimating the age of mineral deposits, rocks & earth. t

 N   1  T1 Note:  =  2  No   2  Nuclear Reactor: Nuclear reactor is a device in which nuclear fission is maintained as a self-supporting yet controlled chain reaction.

158

Core: The core of the reactor is the site of nuclear fission. The core contains a moderator to slow down the neutrons. Reflector: The core is surrounded by a reflector to reduce leakage. Coolant: Coolant is a cooling material, which removes the heat generated due to fission in the reactor. Heat exchanger: This is a vessel the hot coolant gives up its heat to water at normal pressure.

159

CHAPTER 14 Electronic Devices Insulator: Insulators are those material in which valence electrons are bound very tightly to their parent atoms thus requiring very large electric field to remove them from the attraction of their nuclei. Conductors: Conducting materials are those in which plenty of free electrons are available for electric conduction. In terms of energy bands, it means that electric conductors are those which have overlapping valence and conduction bands. Semiconductors: A semiconductor material is one whose electrical properties lie in between those of insulators and good conductors. Ex- Germanium and Silicon. In terms of energy bands, semiconductor can be defined as those materials which at room temp have (i) Partially filled conduction band (ii) Partially filled valence band (iii) A very narrow energy gap (1 eV ) between them. Type of semiconductor: (i) Intrinsic or Pure semiconductor (ii) Extrinsic or Impure semiconductor (a) N- type semiconductor (b) P- type semiconductor Intrinsic semiconductor: An intrinsic semiconductor may be defined as one in which the number of conduction electrons in the conduction band are equal to the number of holes in the valence band in all temperatures, since the thermal excitation of an electron to the conduction band gives rise to a corresponding hole in the valence band. Extrinsic semiconductor: Those intrinsic semiconductors to which some suitable impurity or doping agent has been added in extremely small amount (1 part in 108) are called extrinsic or impurity semiconductors. Pentavelent doping atom is known as donar atom because it donates or contributes one electron to the conduction band of pure Germanium. Ex- Arsenic, Phosphorus. Trivalent atom, on the other hand is called acceptor atom because it accepts one electron from the Germanium atom. Ex- Gallium, Al, Boron etc. N- type Extrinsic semiconductor: This type of semiconductor is obtained when a pentavalent material like antimony(Sb), Phosphorus(P) are added to pure Germanium crystal. As shown in figure each Phosphorus atom forms covalent bonds with the surrounding four germanium atoms with the help of four of its five electons. The fifth electon is superfluous and is loosely bound to the Phosphorus atoms.

160

P-type Extrinsic semiconductor: When the impurity added or dopped into the basic crystal is a trivalent element such as Boron than the resulting crystal is called P- type semiconductor. Here the electron are the minority carrier and holes are majority carrier.

P-N junction & I-V characteristics: P-n junction diode is a two terminal device consisting of a P-n junction formed either in Ge or Si crystal. The P & N type regions are refered to as anode & cathode respectively as shown

The arrow head shows conventional direction of current flow when flow when forward biased. During the formation of a p-n junction, the following two phenomenon take place. (i) A thin depletion layer or region is set up on the both side of junction and is so called because it is depleted or devoid of the charge carriers. Its width is about 1µm (10-6 m). (ii) A junction or barrier potential Vb is developed across the junction. Vb= 0.3 for Ge Vb = 0.7 for Si

161

Working: The P-n junction diode is unidirected in its action. It offer low resistance when forward biased and behaves almost as an insulator when reversed bias. Hence such diode are used for rectification. V-I characteristic: (i) Forward bias characteristics: When the diode is forward biased and the applied voltage is increase from zero, there is hardly any current through the device in the beginning due to barrier potential. As soon as barrier potential neutralize, current through the diode increased rapidly and exponentially as shown in figure. (iii) Reverse bias characteristics: When the diode is reverse biased, majority carriers are blocked and only a small current due to minority carrier flows through the diode. As the reverse voltage is increased from zero, the reverse current quickly reaches its maximum or saturation value which is also known as leakage current as shown in figure.

Junction breakdown: If the reverse bias applied to a P-n junction is increased a point is reached when the junction breakdown and reverse current rises sharply to a value limited by the external resistance connected in series with the junction. This critical value of the voltage is known as breakdown voltage. The following two mechanisms are responsible for breakdown under increasing reverse voltage. (i) Zener breakdown: This form of breakdown occurs in junction which being heavily doped has narrow depletion layers. The breakdown voltage sets up a very strong electric field (about 108) across this narrow layer. This field is strong enough to break or rapture the covalent bonds thereby generating electron holes pairs. Even a small further increase in reverse voltage is capable of producing large number of current carriers. This is why the junction has very low resistance in the breakdown region. (ii) Avalanche breakdown: This form of breakdown occurs in junction which being lightly doped, have wide depletion layers where the electric field is not strong enough to produce Zener breakdown. Instead, the minority carriers accelerated by this field collide with the semiconductor atoms in the depletion region. Upon collision with valence electrons,

162

covalent bonds are broken and electron-hole pairs are generated. These newly generated charge carriers are also accelerated by the electric field resulting in more collision and hence further production of charge carriers. This leads to an avalanche of charge carriers.

P-N junction breakdown characteristics Optoelectronics Device: Optoelectronics devices are the products of a technology that combines optics with electronics. The few examples of optoelectronics devices are Photodiodes, LED, Solar Cell. Photodiodes: It is a reversed biased P-n junction, illuminated by radiation. When P-n junction is reversed biased with no current, a very small reverse saturated current flows across the junction called the dark current. When the junction is illuminated with light, electron-hole pairs are created at the junction, due to which additional current begins to flow across the junction. The current is only due to minority charge carriers.

Circuit diagram:

Photodiode is used to measure light intensity because reverse current increase with increase of intensity of light.

LED (Light Emitting Diode): A light emitting diode is simply a forward biased is simply a forward biased P-n junction which emits spontaneous light radiation. When forward biased is applied , the electron and holes at the junction recombine and energy released is emitted in the form of light. The advantages of LEDs are

163

(a) Low operational voltage & less power (b) Fast action with no warm up time. (c) They have long life. Symbolic representation:

V-I characteristics of LED:

Solar cell: A solar cell is a junction diode which convert light energy into electrical energy. A P-n junction solar cell consists of a large junction with no external biasing. The surface layer of P- region is made very thin so that the incident photons may easily penetrate to reach the junction which is the active region. It’s symbolic representation is:

Zener Diode: A zener diode is a specially designed heavily doped P-n junction, having a very thin depletion layer and having a very sharp breakdown voltage. It is always operated in breakdown region. Its breakdown voltage VZ is less than 6 V. At reverse voltage less than 6 V, Zener effect predominates whereas above 6 V, avalanche effects predominates. Strictly speaking, the first one should be called Zener diode and second one as avalanche diode. But the general practise is to call both types as Zener diodes. Zener breakdown occurs due to breaking of covalent bonds by the strong electric field set up in the depletion region by the reverse voltage.

164

Zener Diode as voltage regulator: It is a measure of a circuit’s ability to maintain a constant output voltage even when either input voltage or load current varies. A Zener diode when working in the breakdown region, can serve as a voltage regulator. In the figure Vin is the input DC voltage whose variation are to be regulated. The Zener diode is reverse connected across Vin. When potential difference across the diode is greater than Vz , it conducts and draws relatively large current through the series resistance ‘R s’. The load resistance RL across which a constant voltage Vout is required is connected in parallel with the diode. The total current I passing through ‘R s’ equals the sum of diode current and load current i.e. I = I Z + IL It will be seen that under all conditions, Vout =VZ

Rectifiers: The process of converting A.C into D.C is called rectification and device is known as rectifier. There are two types of rectifiers(i) Half wave rectifier (ii) Full wave rectifier Half wave rectifier: A simple Half Wave Rectifier is nothing more than a single pn junction diode connected in series to the load resistor. It is the device which convert half cycle of A.C only into D.C.

The half-wave rectifier circuit using a semiconductor diode (D) with a load resistance R L. The diode is connected in series with the secondary of the transformer and the load resistance RL. The primary of the transformer is being connected to the ac supply mains. The ac voltage across the secondary winding changes polarities after every half cycle of the input wave. During the positive half-cycles of the input ac voltage i.e. when the upper end of the secondary winding is positive w.r.t. its lower end, the diode is forward biased and therefore conducts current. If the forward resistance of the diode is assumed to be zero (in practice, however, a small resistance exists) the input voltage during the positive half-cycles is directly applied to the load resistance RL , making its upper-end positive w.r.t. its lower end. The waveforms of the output current and output voltage are of the same shape as that of the input ac voltage. During the negative half cycles of the input ac voltage i.e. when the lower end of the secondary winding is positive w.r.t. its upper end, the diode is reverse biased and so does not conduct. Thus during the negative half cycles of the input ac voltage, the current through and the voltage across the load remains zero. The reverse current, being very

165

small in magnitude, is neglected. Thus for the negative half cycles, no power is delivered to the load. It is obvious from the figure that the output is not a steady dc, but only a pulsating dc wave. To make the output wave smooth and useful in a DC power supply, we have to use a filter across the load. Since only half-cycles of the input wave are used, it is called a half wave rectifier. Advantages and Disadvantages of Half wave rectifier: A half wave rectifier is rarely used in practice. It is never preferred as the power supply of an audio circuit because of the very high ripple factor. High ripple factor will result in noises in the input audio signal, which in turn will affect audio quality. The advantage of a half wave rectifier is only that its cheap, simple and easy to construct. It is cheap because of the low number of components involved. Simple because of the straight forwardness in circuit design. Apart from this, a half wave rectifier has more number of disadvantages than advantages. Full wave rectifier: It is the device which convert complete cycle of A.C into D.C. A half wave rectifier can be converted into full wave rectifier by using two diodes and little bit modification in circuit.

The full wave rectifier circuit consists of two power diodes connected to a single load resistance (RL) with each diode taking it in turn to supply current to the load. When point A of the secondary transformer is positive with respect to point C, diode D1 conducts in the forward direction as indicated by the arrows. When point B is positive (in the negative half of the cycle) with respect to point C, diode D2 conducts in the forward direction and the current flowing through resistor R is in the same direction for both half-cycles. As the output voltage across the resistor R is the phasor sum of the two waveforms combined, this type of full wave rectifier circuit is also known as a “bi-phase” circuit. Advantages of Full wave rectifier over half wave rectifier: Like the half wave circuit, a full wave rectifier circuit produces an output voltage or current which is purely DC or has some specified DC component. Full wave rectifiers have some fundamental advantages over their half wave rectifier counterparts. The average (DC) output voltage is higher than for half wave, the output of the full wave rectifier has much less ripple than that of the half wave rectifier producing a smoother output waveform.

166

SOME SELECTED QUESTIONS OF

NCERT

167

Chapter 1: Electric charges and Fields 1. What do electric lines represent? Ans: Electric lines mainly represent the direction of the electric field. 2. Why do electrostatic field lines not form closed loop? Ans: The field lines starts from the positive charge and terminate on negative charge. If the electric field lines form closed loop, these line must originate and terminate on the same charge which is not possible because electric field lines always moves from positive to negative. 3. Why do electric field lines never cross each other? Ans: Electric field lines of a single charge can only indicate electric field in one direction at one point. Intersection of two lines will result into two direction of the electric field at one point which is not possible. 4. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? Ans: There is no effect of change in radius on to electric flux as q = o 5. What is the force between two small charged sphere having charges of 2x10 -7 C and 3x10-7 C placed 30cm apart in air? Ans: 1 q1q2 F= 4 o r 2

( 9 10 )  ( 0.2 10 )  ( 0.3 10 ) −6

9

F=

−6

0.3  0.3 −3

F = 6 10 N 6. The electrostatic force on a small sphere of charge 0.4 c due to another small sphere of charge -0.8 c in air is 0.2 N. a) What is the distance between the two spheres? b) What is the force on the second sphere due to the first? Ans: a) Force on charge 1 due to charge 2 is given by

168

F=

q1q2 4 o r 2

r2 =

1 q1q2 4 o F

1

( 9 10 )  ( 0.4 10 )  ( 0.8 10 ) −6

9

r2 =

−6

0.2

r 2 = 9 16 10−4 r = 12 10−12 m = 12 cm

b) F21 = F12 = attractive 7. Why can one ignore quantization of electric charge when dealing with macroscopic level? Ans: At macroscopic level, the quantization of charge has no practical importance because the charge at macroscopic level is very large as compared to elementary charge. For example, a small charge of 1 c has about 1013 electronic charges. In such cases the charge may be treated as continuous and not quantized. 8. Four point charges qA = 2 c, qB = -5 c, qC = 2 c and qD = -5 c are located at the corners of a square ABCD of sides 10cm. What is the force on a charge of 1 c placed at the centre of the square? Ans: The symmetry of the figure clearly indicates that 1 c charge will face equal and opposite forces due to equal charges of 2 c placed at ‘A’ & ‘C’. Similarly 1 c charges will face equal & opposite force due to -5 c charge placed at D & B. So net force on charge 1c is zero. 9. Two point chages qA = 3 c and qB = -3 c, are located 20cm apart in vacuum. a) What is the electric field at the midpoint ‘O’ of the line AB joining the two charges? b) If a negative test charge of magnitude 1.5x10-9 C is placed at this point, what is the force experienced by the test charge? Ans: a) Electric field at the mid-point of the separation between two equal and opposite charges given by E = E1 + E2

i.e.

E = E1 + E 2 E = 2 E1

as (E1 = E2)

169

E = 2

1

q 4 o r 2

( 9 10 )  ( 3 10 ) (10 10 ) −6

9

E = 2

−10 2

E = 5.4 106 NC-1 along OB

b) Force experienced by test charge due to charge 3 c = qoE = (1.5 x 10-9) x (5.4 x 106) = 8.1 x 10-3 N 10. A system has two charges qA = 2.5x10-7 C and qB = -2.5x10-7 C located at point A: (0, 0, -15)cm and B: (0, 0, +15)cm, respectively. What are the total charge and electric dipole moment of the system? Ans: It is clear that given points are lying on Z-axis. Distance between charges = 15 + 15 = 30cm = 0.3m Total charge = (2.5x10-7) + (-2.5x10-7) = 0 Dipole moment =q2a = (2.5x 10-7) x (0:3) = 7.5 x 10-8 C 11. An electric dipole with dipole moment 4x10-9 Cm is aligned at 30o with the direction of a uniform electric field of magnitude 5x10 4 NC-1. Calculate the magnitude of the torque acting on the dipole. Ans: As  = PE sin 

 = ( 4 10−9 )( 5 104 ) sin 30 1 2

 = 2 10−4  = 10−4 Nm 12. A point charge of 2 c is at the centre of a cubic Gaussian 9 cm on edge. What is the net electric flux through the surface? Ans: q =

o

=

2 10−6 = 2.26 105 Nm2C −1 8.854 10−12

13. A point charge +10 c is a distance 5 cm directly above the centre of a square of side 10cm, as shown in the given figure. What is the magnitude of the electric flux through the square?

170

Ans: Think of the square as one face of a cube with edge 10 cm. q Using  = , the flux through one square surface= o

1 q 1 10−5 = = 1.88 105 Nm2C −1 6  o 6 8.854 10−12 14. Consider a uniform electric field E = 3 103 iˆ N/C a) What is the flux of this field through square of 10 cm on a side whose plane is parallel to the Y-Z plane? b) What is the flux through the same square if the normal to its plane makes a 60o angle with X-axis? Ans: a)  = E.d S

1 ˆ  10 = 3 103 iˆ .   i  = 30 Nm2C −1  100 100  b)  = E.d S = EdS cos 

(

)

 10 10  = ( 3 103 )    cos 60  100 100  1 = ( 3 103 )(10−2 )  = 15 Nm 2C −1 2 15. An infinite line charge produces a field of 9x104 N/C at a distance of 2 cm. Calculate the linear charge density Ans: As 1  E= 2 o r Er 2 ( 9 104 )  ( 0.02 )

 = 2 o Er = 4 o =

1  9 109

2

= 10−7 Cm −1

171

Chapter 2: Electrostatic potential and capacitance 1. Does the charge given to a metallic sphere depends on whether it is hollow of solid? Give reason for your answer. Ans: In case of metallic sphere, charge given to it is mostly resides on its surface. Therefore, there is no difference whether the sphere is hollow or solid. As in both the cases, the charge that will reside will be same. 2. Two charges 5x10-8 C and -3x10-8 C are located 16 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Ans: Let the potential be zero at O, then VA + VB = 0 (i)



qB 1 qA 1 + =0 4 o x 4 o ( 0.16 − x )



qA qB =− x ( 0.16 − x )



5 10−8 3 10−8 = x ( 0.16 − x )

 5 ( 0.16 − x ) = 3 x  0.8 − 5 x = 3 x  8 x = 0.8  x = 0.1m  x = 10cm Electric potential is zero at a distance of 10 cm from the charge qA. 3. A regular hexagon of side 10 cm has a charge 5 C at each of its vertices. Calculate the potential at the centre of the hexagon. Ans: Total potential at O is given by  1 q V = 6   4 o r  = 6  ( 9 109 ) 

5 10−6 0.1

= 2.7 106 V

4. Two charges 2 C and -2 C are placed at points A and B, 6 cm apart. a) Identity an equipotential surface of the system

172

b) What is the direction of the electric field at very point on this surface? Ans: a) The equipotential surface is the plane perpendicular to the line AB joining the two charges and passing through the mid-point. On this plane, potential is zero everywhere. b) The direction of electric field is from positive to negative charge i.e. A to B which is infact perpendicular to the equipotential plane. 5. Three capacitors each of capacitance 9 pF are connected in series. a) What is the total capacitance of the combinations? b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply? Ans: a) 1 1 1 1 1 1 1 1 = + + = + + = 3 Ceq C1 C2 C3 C C C C 1 1 = 3 Ceq 9 10−12 Ceq = 3 10−12 F = 30 pF

b) Since three equal capacitors are across 120 V, so potential difference across each capacitor. 120 = = 40V 3 6. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. a) What is the total capacitance of the combination? b) Determine the charge on each capacitor if the combination is connected to a 100 V supply. Ans: a) Ceq = C1 + C2 + C3 = 2 + 3 + 4 = 9 pF q b) Using C = , we get q = CV V q1 = C1V = 2 10−12 100 = 2 10−10 C

q2 = C2V = 3 10−12 100 = 3 10−10 C q3 = C3V = 4 10−12 100 = 4 10−10 C 7. A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor? Ans: 1 1 E = CV 2 = 12 10−12  50  50 = 1.5 10−8 J 2 2 8. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? Ans:

173

1 1 E = CV 2 =  600 10−12  200  200 = 12 10−6 J 2 2 When connected to another capacitor of same capacity the voltage will be equally shared i.e. halved, but the capacitance will be doubled, so the energy is halved. Energy lost = 6x10-6 J 9. A long charged cylinder of linear charged density  is surrounded by a hollow coaxial conductive cylinder. What is the electric field in the space between the two cylinders? Ans:  E= 2 o r 10. A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2 .Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire), no matter what the charge q2 on the sell is. Ans: Charge resides on the outer surface of a conductor. So the charge on inner sphere will flow towards the shell through the conducting wire. The potential of inner sphere 1  q1 q2  V1 =  +  4 o  r1 r2  Potential of shell, 1  q2 + q1  V2 =   4 o  r2  Potential difference, V = V1 − V2 1  q1 q2   −  4 o  r1 r2  This is independent of q2 . So q2 does not matter in this case. If q1 is positive, potential difference is also positive =

11. How does the energy stored in a capacitor change if after disconnecting the battery, the plates of a charged capacitor are moved farther? 1 Ans: Capacitance C  , when plates of a capacitor are moved farther, the capacitance d decreases. After disconnecting the battery, the charge on capacitor remains constant, therefore the energy stored by capacitor.  q2  U =  increases.  2C  12. How does the energy stored in a capacitor change if the plates of a charged capacitor are moved father, the battery remaining connected? Ans: The capacitance of capacitor decrease on moving its plates farther. As the battery remains connected, the potential difference remains constant. Hence,

174

1 energy stored U = CV 2 , decreases 2 13. When a capacitor is charged by a battery, is the energy stored in the capacitor same as energy supplied by the battery? 1 1 Ans: No, Energy stored in capacitor CV 2 = qV while energy supplied by battery = 2 2 ; qV . The half of energy is dissipated as heat during charging. 14. For any charge configuration, equipotential surface through a point is a normal to the electric field justify. Ans: No work is done in moving the test charge from one point of an equipotential surface to the other B − A = 0 = − E.dl = E.dl = 0 Hence E ⊥ dl 15. What is the function of a dielectric in a capacitor? Ans: Dielectric reduces the effective potential on plates and hence increases the capacitance.

175

Chapter 3: Current Electricity 1. Is current density a scalar or a vector quantity? Ans: Current density a vector quantity. 2. What are the factors on which the resistances of a conductor depend? Give the corresponding relation. Ans: Keeping physical conditions like temperature, pressure, stress etc. constant, resistance depends directly upon resistivity ‘’ and length ‘l’ but indirectly upon area of cross-section ‘A’ of the conductor. l R= A 3. A wire of resistivity ‘’ is stretched to three times its length. What will be its new resistivity? Ans: Same 4. Why is a potentiometer preferred over a voltmeter for measuring the emf of a primary cell? Ans: Potentiometer uses null method so, no current is drawn by the galvanometer from the cell in the balanced condition of the potentiometer. It measure true emf of the cell. 5. Will the drift speed of free electrons in a metallic conductor increase or decrease with the increase in its temperature? Ans: Drift speed decreases with the increase in temperature. 6. If potential difference ‘V’ applied across a conductor is increased to 2 V, how will the drift velocity of the electrons change? Ans: eV vd =  ml Double the voltage means double the drift velocity 7. What is sensitivity of a potentiometer? Ans: The smallest potential difference that can be accurately measured by a potentiometer is called sensitivity of the potentiometer. 8. Which has greater resistance 1 kW electric heater or a 100W filament bulb both marked for 220V? Ans: Bulb has greater resistance V2 P= R 1 R P 9. The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.4, what is the maximum current that can be drawn from the battery?

176

Ans: V 12 I= = = 30 A R 0.4 10. A battery of emf 10V and internal resistance 3 is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resister? What is the terminal voltage of the battery when the circuit a closed? Ans: E = V + Ir E 10 R = −r = − 3 = 17 I 0.5 V = IR = 0.5 17 = 8.5V 11. a) Three resistor 1, 2 and 3 are combined in series. What is the total resistance of combination? b) If the combination is connected to a battery of emf 12V and negligible internal resistance, obtain the potential drop across each resistor. Ans: a) Req = R1 + R2 + R3 = 1 + 2 + 3 = 6 b) as V = IR V 12 I= = =2A Req 6 Voltage drop across R1 = IR1 = 2x1 = 2V Voltage drop across R2 = IR2 = 2x2 = 4V Voltage drop across R3 = IR3 = 2x3 = 6V 12. In a potentiometer arrangement, a cell of emf 1.25V gives a balance point at 35cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63cm, what is the emf of the second cell? Ans: As E1 l1 = E2 l2

E2 =

l2 63 E1 = 1.25 = 2.25V l1 35

13. Three identical cells, each of emf 2V and internal resistance of 0.2 are connected in series to an external resistor of 7.4. Calculate the current in the circuit. Ans: nE 3 2 6 I= = = = 0.75 A R + nr 7.4 + 3  0.2 8 14. Two bulbs whose resistances are in the ratio 1:2 are connected in parallel to a source of constant voltage. What will be the ratio of the power dissipation in these? Ans:

177

As V2 R V2 V2 P1 = & P2 = R 2R P1 2 = P2 1

P=

15. Determine the current in each branch of the network shown in Figure

Ans: Applying Kirchhoff's second law For ABDA −10 I1 − 5I 2 + 5 ( I − I1 ) = 0

 −10 I1 − 5I 2 + 5I − 5I1 = 0  I − I 2 − 3I1 = 0 For BC DB −5 ( I1 − I 2 ) + 10 ( I − I1 + I 2 ) + 5I 2 = 0  10 I − 15I1 + 20 I 2 = 0  2I − 3I1 + 4I 2 = 0 For ADCHG −5 ( I − I1 ) − 10 ( I − I1 + I 2 ) + 10 − 10I = 0

−5I + 3I1 − 2I 2 = −2 Multiplying eqs. (i) by 2 2I − 2I 2 − 6I1 = 0 Subtracting eqn. (iv) from eqn. (ii) 3I 2 + 6 I1 = 0

I1 + 2 I 2 = 0 I1 = −2 I 2 Subtracting eqn. (i) from eqn. (ii)

(i)

(ii)

(iii) (iv)

(v)

178

− I − 5I 2 = 0 I = −5I 2

(vi)

From eqn. (v), (vi) & (iii) −5 ( −5I 2 ) + 3 ( −2 I 2 ) − 2 I 2 = −2 25I 2 − 6 I 2 − 2 I 2 = −2 17 I 2 = −2 −2 I2 = A 17 From eqn. (v) −2 4 I1 = −2  = A 17 17 From eqn. (vi) −2 10 I = −5  = A 17 17

(vii)

4 A 17  4   −2  6 Current in BC = I1 − I 2 =   −   = A  17   17  17 4 Current in DC = I − I1 + I 2 = A 17 6 A Current in AD = I − I1 = 17 Current in AB = I1 =

179

Chapter 4: Moving Charges and Magnetism 1. Why do magnetic lines prefer to pass through iron than though air? Ans: Relative permeability of iron is high, therefore magnetic field lines prefer to pass through it. 2. What is the name given to the curve, the tangent to which at any point gives the direction of the magnetic field at that point? Ans: Magnetic field lines. 3. Write the expression in a vector from for the Lorentz magnetic force F due to a charge moving with velocity V in a magnetic field B . What is the direction of the magnetic force? Ans: F = q V  B

(

)

Direction: Perpendicular to the plane containing V& B

4. A beam of -particles projected along +X axis, experiences a force due to a magnetic field along the +Y Axis. What is the direction of the magnetic field? Ans: V = Viˆ, F = Fjˆ

( ) Fjˆ = q (Viˆ  B )

F = q V B

( )

The direction of magnetic field must be along -Z axis as B = B −kˆ

5. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.4 A. What is the magnitude of the magnetic field at the centre of the coil? Ans:  nI 4 10−7 100  0.4 B= o = =  10−4 T 2a 2  8 10−2 6. A long straight wire carries a current of 35A. What is the magnitude the field ‘B’ at a point 20cm from the wire? Ans:  I 4 10−7  35 B= o = = 3.5 10−5 T 2 a 2  20 10−2 7. A long straight wire in the horizontal plane carries a current of 50A in north to south direction. Give the magnitude and direction of B at a point 2.5m east of the wire. Ans:  I 4 10−7  50 B= o = = 4 10−6 T vertically up 2 a 2  2.5

180

8. An electron passes un-deviated through mutually perpendicular electric and magnetic fields. If the magnitudes of the electric & magnetic fields be 5kVm-1 and 0.1T and if the electron moves perpendicular to both the electric & magnetic fields. Calculate the speed of the electron. Ans: E 5 103 V= = = 50 km / s B 0.1 9. What is the magnitude of magnetic force per unit length. On a wire carrying a current of 8A and making an angle 30o with the direction of uniform magnetic field of 0.15T? Ans: F = BIl sin 

F 1 = BI sin  = 0.15  8  = 0.6 Nm−1 l 2 10. A 3cm wire carrying a current of 10A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27T. What is the magnetic force on the wire ? Ans:

F = BIl = 0.27 10  3 10−2 = 8.110−2 N 11. In a Chamber, a uniform magnetic field of 6.5G (1G = 10-4 T) is maintained. An electron is shot into the field with a speed 4.8x106 ms-1 normal to the field. Determine the radius of the circular orbit and the frequency of revolution. Ans: mv 9.110−31  4.8 106 r= = = 4.2 10−2 m Be 6.5 10−4 1.6 10−19 qB 1.6 10−19  6.5 10−4 f = = = 18.18 106 Hz 2 m 2  3.14  9.110−31 12. A part of a straight wire carrying of 12A is bent into a semiconductor arc of radius 2 cm. Find the magnitude of magnetic field at the centre. Ans:  I   As B = o  (i)  2a  2  Here  =  from eqn. (i) I  B= o  2a 2 o I 4 10−7 12 B= = = 1.89 10−4 T 4a 4  2 10−2 13. Two long and parallel straight wires ‘A’ & ‘B’ carrying currents of 8A and 5A in the same direction are separated by a distance of 4cm. Estimate the force on a 10cm section of wire ‘A’. Ans:

181

4 10  8  5  I I  F =  o 1 2 l = 10 10−2 2    4 10−2  2 r  F = 2 10−5 N attractive. −7

14. A square coil of side 10cm consists of 20 tums and carries a current of 12A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30o with the direction of a uniform horizontal magnetic field of magnitude of 0.8T. What is the magnitude of torque experienced by the coil? Ans: 1  = NIAB sin  = 20 12 100 10−4  0.8  2  = 0.96 Nm 15. A galvanometer can be converted into a voltmeter to measure up to i) V volts by connecting a resistance R1 in series with coil. ii) V/2 volts by connecting a resistance R2 in series with its coil. Ans: Here V = I ( G + R1 ) (i)

V = I ( G + R2 ) 2 V = 2I ( G + R2 )

&

A/Q 2V = I ( G + R ) equating eqn. (i) & (ii) I ( G + R1 ) = 2I ( G + R2 )

(ii) (iii)

G = R1 − 2R2 (iv) Putting the value of eqn. (i) in eqn. (iii) 2I ( G + R1 ) = I ( G + R ) G = R − 2R1 equating eqn. (iv) & (v) R − 2 R1 = R1 − 2 R2

(v)

R = 3R1 − 2 R2

182

Chapter 5: Magnetism and matter 1. Give one example of ferromagnetic, paramagnetic and diamagnetic. Ans: Iron, aluminium and zinc respectively. 2. What type of magnetic material is used in making permanent magnets? Ans: Magnetic materials with high coercivety, like steel, alnico etc are used for making permanent magnets. 3. Where on the surface of earth is the vertical component of earth’s magnetic field zero? Ans: At equator 4. A magnetic needle, free to rotate in vertical plane, orients itself with its axis vertical at a certain place on the earth. What is the value of a) horizontal components of earth’s magnetic field? b) angle of dip at this place? Ans: a) 0 b) 90 5. What is the angle of dip at a place where the horizontal and vertical components of earth’s magnetic fields are equal? Ans: It is 45o 6. Horizontal component of Earth’s magnetic field at a place is component. What is the value of angle of dip at this place? Ans: BH = 3BV

As tan  = tan  =

3 times the vertical

BV B 1 = V = BH 3BH 3

1 3

 = 30o 7. An iron bar magnet is heated to 1000 o C and then cooled in a magnetic field free space. Will it retain its magnetism? Ans: The Curie point for iron is about 770 o C. At 1000o C the magnet loses its magnetism. 8. What should be the orientation of a magnetic dipole in a uniform magnetic field so that its potential energy is maximum? Ans: When dipole moment is same as the direction of magnetic field. 9. Why should the materials used for making permanent magnets have high coercivity? Ans: A permanent magnet has to retain magnetism permanently so the reverse force required to vanish residual magnetism has to be very high.

183

10. A Short bar magnets placed with its axis at 30o with a uniform external magnetic field of 0.25T experiences a torque of magnitude 4.5x10-2 J. What is the magnitude of magnetic moment of the magnet? Ans:  = MB sin 

M=

 B sin 

=

4.5 10−2 2  4.5 10−2 = = 0.36 JT −1 0.25  sin 30 0.25

11. A closely wound solenoid 800 turns and area of cross section 2.5x10-4 m2 carry current of 3A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment? Ans: M = NIA

M = 800  3  2.5 10−4 = 0.60 JT −1 The sense is determined by the direction of the current. 12. The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s, magnetism. Why? Ans: Molten iron has a temperature more than Curie temperature so it is not ferromagnetic in nature. 13. The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion? Ans: Ionosphere of earth consists of charged ions. Motion of these ions causes a magnetic field which affects the magnetic field of earth at large distances from earth. The magnetic field due to ions depends upon extra-terrestrial disturbances like solar wind. 14. The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists known about the earth’s field in such distant past? Ans: During solidification of some rocks, the magnetic field of earth get weakly recorded in these rocks. Geomagnetic history of these rocks can be traced by making analysis of such rocks. 15. Interstellar space has an extremely weak magnetic field of the order of 10 -12 T. Can such a weak field be or any significant consequence? Explain. Ans: At very-very large distances like interstellar distances the small fields can significantly affect the charged particles like that of cosmic rays. For small distances, the deflections are not noticeable for small fields, but at very large distances the deflections are significant. mv As R = eB Clearly for small value of B gives a very large value of radius.

184

Chapter 6: Electromagnetic Induction 1. Why does acceleration of a magnet falling through a long solenoid decrease? Ans: Current induced in the solenoid offers retarding force to the falling magnet thus reducing its acceleration. 2. If the number of turns of a solenoid is doubled, keeping the other factors constants, how does the self-inductance of the solenoid change? Ans: as L  N2 L ' = 4L 3. A rectangular coil of N turns and area of cross-section A, is held in time varying magnetic field given by B = Bo sin t , with plane of coil normal to the magnetic field. Deduce an expression for the emf induced in the coil. Ans: as  = NBA cos 

 = N ( Bo sin t ) A cos 0o = NBo A sin t Induced emf d d sin t − = NBo A = NBo A cos t dt dt 4. If a rate of change of current of 4 A/s induces an emf of 20 mV in a solenoid, what is the self-inductance of the solenoid? Ans: as LdI e= dt e 20 10−3 L= = = 5 10−3 H dI 4 dt 5. A long solenoid with 15 turns per cm has a small loop of area 2 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2A to 4A in 0.1 sec, what is the induced emf in the loop while the current is changing? Ans: d dB e= = NA (i) dt dt B = o N ' I

B = ( 4 10−7 ) (1500 )( 4 − 2 ) B = ( 4 10−7 ) (1500 )( 2 )

e=

(ii)

( 2 10 )  ( 4 10 )  (1500 )  ( 2 ) −4

−7

0.1

e = 7.5110−6 V

185

6. A rectangular wire loop of sides 8cm and 2cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cms-1 in a direction normal to the a) Longer side b) Shorter side of the loop For how long does the induced voltage last in each case? Ans: a) As e = BlV

e = ( 0.3)  ( 8 10−2 )  (10−2 ) = 0.24 mV emf last long l 2 t = = = 2sec V 1 b) As e = BlV

e = ( 0.3)  ( 2 10−2 )  (10−2 ) = 0.06 mV emf last long l 8 t = = = 8sec V 1 7. A 1 m long metallic rod is rotated with an angular frequency of 400 rad S-1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field 0.5T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring Ans: As BlV e= 2 we know that v = l so eqn. (i) can be written as Bl 2 0.5 11 400 e= = = 100V 2 2 8. A horizontal straight wire 10m long extending from east to west is falling with a speed of 5 ms-1, at right angles to the horizontal component of the earth’s magnetic field, 0.30x10-4 Wbm-2 . a) What is the instantaneous value of the emf induced in the wire? b) What is the direction of the emf? c) Which end of the wire is at the higher electrical potential? Ans: a) As e = Bvl

e = ( 0.30 10−4 )  ( 5)  (10 ) = 1.5 10−3 V b) Using Fleming’s Right-hand rule, the direction of induced emf is West to East

186

c) Since the rod will act as a source, the Eastern end will be at higher electrical potential. 9. Current in a circuit falls from 5A to 0A in 1sec.If an average end of 200 V induced, give an estimate of the self-inductance of the circuit. Ans: as dI e =L dt 5 − 0 200 = L   0.1  200  0.1 L= = 4H 5 10. A pair of adjacent coils has a mutual inductance of 1.5H. If the current in one coil changes from 0 to 20A in 0.5 sec, what is the change of flux linkage with the other coil? Ans: as  = MI

d = MdI = 1.5  ( 20 − 0 ) = 30Wb 11. A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25m, if the earth magnetic field at the location has a magnitude of 5x10-4 T and the dip angle is 30o? Ans: The vertical component of earth’s magnetic field = B sin  = 5 10−4 sin 30 So induced emf e = ( B sin  ) vl = 5 10−4 sin 30  ( 500 )  25

(

)

1 e = 5 10−4   500  25 = 3.1V 2 12. A vertical metallic pole falls down through the plane of the magnetic meridian. Will any emf be produced between its ends? Give reasons for your answer. Ans: Flux linking with pole change so as per Faraday’s Law an emf may induce in it. 13. A bar magnet falls from height ‘h’ through a metal ring will its acceleration be equal to g? Give reasons for your answer. Ans: No, current induced in the ring offers retarding force to the falling magnet thus reducing its acceleration. 14. The electric current in a wire in the direction from B to A is decreasing. What is the direction induced current in the metallic loop kept above the wire as shown in figure?

187

Ans: Clockwise. 15. The current in wire PQ is increasing. In which direction does the induced current flow in the closed loop.

Ans: Clockwise

188

Chapter 7: Alternating Current 1. A Choke coil and a bulb are connected in series to a d.c. source. The bulb shines brightly. How does the brightness change when an iron core is inserted in the choke coil? Ans: Brightness of the lamp will not change because at steady d.c. the choke coil has no inductive reactance. 2. An electric lamp, connected in series with a capacitor and an a.c. source, is glowing with certain brightness. How does the brightness of lamp change on reducing the capacitance? Ans: On reducing capacitance, capacitive reactance increases which reduces brightness of lamp 1 As X c = c 3. Why can't transformer be used to step up d.c. voltage? Ans: D.C cannot produce varying field for secondary winding, therefore, induced emf cannot be produced in it. 4. The emf of an ac source is given by the expression E = 300sin 314t volts. Write the value of peak voltage and frequency of source. Ans: On comparing from standard expression E = Eo sin t Peak value = Eo = 300 V,  = 314, 2f = 314 314 f = = 50 Hz 2 5. The power factor of an AC circuit is 0.5. What is the phase difference between voltage and current of the circuit? Ans: As cos = power factor cos = 0.5 cos = cos 60  = 60 Phase difference between voltage and current of the circuit a 60o 6. A 100 resistor is connected to a 220V, 50 Hz ac supply. a) What is the rms value of current in the circuit? b) What is the net power consumed over a full cycle? Ans: V 220 = 2.2 A a) I rms = rms = R 100 b) Net power = VrmsxIrms = 220x2.2 = 484 W 7. a) The peak voltage of an ac supply is 300V. What is the rms voltage? b) The rms value of current in ac circuit is 10A. What is the peak current? V 300 Ans: a) Vrms = o = = 212.1V 2 2

189

b)

I rms =

Io 2

I o = 2 I rms = 2 10 = 14.1 A 8. A 44 mH inductor is connected to 220V, 50Hz ac supply. Determine the rms valve of the current in the circuit. Ans: X L = 2 fL = ( 2 )  ( 50 )  ( 44 10−3 )

I rms =

Vrms 220 = = 15.91 A X L 2  50  44 10−3

9. A 60 F capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. Ans: 1 1 XC = = 2 fC 2  60  60 10−6 V I rms = rms = (110 )  ( 2  60  60 10−6 ) = 2.49 A Xc 10. In exercise 8 & 9, what is net power absorbed by each circuit over a complete cycle. Explain your answer. Ans: In case of an idea inductor or capacitor there is no power loss. 11. Obtain the resonant frequency r of a series LCR circuit with L = 2 H, C = 32 F and R= 10. What is the Q-value of this circuit? Ans: 1 1 1 r = = = = 125 s −1 −3 −6 8  10 LC 2  32 10 Q=

X L r L 125  2 = = = 25 R R 10

12. Suppose the initial charge on the capacitor 30 F is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time? Ans: 1 Q 2 1 6 10−3  6 10−3 E= = = 0.6 J 2 C 2 30 10−6 Total energy is same for the later time. 13. A series LCR circuit with R = 20, L = 1.5H and C = 35 F is connected to a variable frequency 200V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? Ans: At natural frequency

190

X L = XC Z=R P=

V 2 200  200 = = 2000W R 20

14. A power transmission line feeds input power at 2300V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230V. Ans: V1 N1 = V2 N 2

N2 =

N1 4000  230 V2 = = 400 turns V1 2300

15. At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3s-1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms-2). Ans: Hydroelectric power = (pressure) x (volume of water flowing per second) =(hg)x(AV) = 300x9.8x103x100 Electric power = 60% of Hydroelectric power = (60/100)x300x9.8x103x100 = 176 MW

191

Chapter 8: Electromagnetic waves 1. What physical quantity is the same for x-rays of wavelength 10-10m, red light of wavelength 6800 Ao and radio waves wavelength 500m? Ans: Speed remains same in vacuum or given x-ray, red light and radiowave. 2. A plane electromagnetic wave travels in vacuum along Z-direction. What can you say about the direction of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? Ans: The electric field vector E and magnetic field vector B are in XY plane. They are normal to each other. we know that c 3 108 = = = 10 m v 30 106 3. A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? Ans: We know that c 3 108 = = = 40 m v 7.5 106 c 3 108 '= = = 25 m v ' 12 106 The corresponding wavelength band is 40 m to 25 m. 4. A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic wave produced by the oscillator? Ans: Frequency of the produced E.M. wave is same as the frequency of oscillating charged particles i.e. 109 Hz. 5. The amplitude of the magnetic field part of a harmonic electromagnet wave in vacuum is Bo = 510 nT. What is the electromagnetic waves produced by the oscillator? Ans: As Eo =c Bo

Eo = Bo c = ( 510 10−9 )  ( 3 108 ) = 153 NC −1

6. Suppose that the electric field amplitude of an electromagnetic wave is Eo = 120 N/C and that its frequency is v = 50 MHz. Determine Bo, , k and . Ans: i) Eo =C Bo Eo 120 = = 40 10−7 T C 3 108 ii)  = 2 v = 2  50 106 = 3.14 108 rad s −1 Bo =

192

2

2 = 1.05 rad m−1  6 c 3 108 = 6m iv)  = = v 50 106 iii) k =

=

7. Do electromagnetic waves carry energy momentum? Ans: Yes 8. In which direction do the electric and magnetic field vectors oscillate in an electromagnetic wave propagating along the x-axis? Ans: As we know that electric and magnetic field vector are always perpendicular to direction of propagation of light. Hence E is along Y-axis and B is along Z-axis. 9. How are radio waves produced? Ans: Radio waves produced by oscillating electric circuit having inductor and capacitor. 10. Which part of electromagnetic spectrum has largest penetrating power? Ans: -rays have highest frequency range and hence highest penetrating power

193

Chapter 9: Ray optics and optical instruments 1. A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to receive a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved? Ans: ho = 2.5cm, u = -27 cm, r = -36cm, f = -18cm We know that 1 1 1 + = v u f 1 1 1 1 1 1 = − = + =− v f u −18 27 54 v = −54 cm Image is formed at the distance of 54 cm from the mirror on the same side as the object. This is the position where a screen should be placed As v − ( −54 ) m=− = = −2 u −27 hi = m  ho = −2  2.5 = −5 cm The image is real, inverted and magnified. If the candle is brought closer to the mirror, the screen would have to be moved away from the mirror. If the distance of the candle from mirror is less than 18 cm, the image becomes and is not observable on the screen. 2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and magnification. Describe what happens as the needle is moved farther from the mirror Ans: u = -12cm = + 15cm, ho = 4.5 cm As we know that 1 1 1 + = v u f 1 1 1 1 1 4+5 9 = − = + = = v f u 15 12 60 60 60 v= = 6.7 cm 9 The image is formed at 6.7 cm behind the mirror. v 60 hi = mho = −  ho =  4.5 = 2.5 cm u 9 12 The image is virtual, erect and diminished. As we move the needle away from the mirror, the image goes on decreasing in size and moves towards the principal focus on the other side. 3. A tank is filled with water to a height of 12.5 cm. The apparent depth of the needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the

194

refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again? Ans: As Real depth  wa = Apparent depth

12.5 = 1.33 9.4 la = 1.63

 wa =

Real depth

12.5 = 7.66 cm 1.63 Distance through which the microscope has to be moved = 9.4 – 7.66 = 1.74 cm Apparent depth =

la

=

4. A small bulb is placed at the bottom of a tank in containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive of water is 1.33. Ans:

 = 1.33 4 3 we know that 1 sin C =  3 sin C = 4 3 tan C = 7

=

195

The light will emerge only through a circle of radius ‘r’ given by r = h tan C  3  r = 0.80     7 2

3   2 Area of circular surface of water =  r 2 = 3.14 0.80   = 2.6 m 7  5. Double convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm? Ans: we know that 1 1 1 = (  − 1)  −  f  R1 R2 

1  1  1  = (1.55 − 1)  −   20  R  − R  1 2 = 0.55  20 R R = 22 cm

6. An object of size 3 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved farther from the lens? Ans: We know that 1 1 1 = − f v u 1 1 1 1 1 1 1 5 = + = + =− − =− v f u −21 −14 21 14 42 42 v=− = −8.4 cm 5 v 42 hi = mho = ho = −  3 = 1.8 cm u 5  ( −14 ) The image is virtual, erect, diminished and is formed on the same side of the lens at a distance of 8.4 cm from the lens. If the object is moved away from the lens, the image move towards the principal focus and goes on decreasing in size. 7. What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system converging or a diverging lens? Ignore thickness of the lenses. Ans: f1 = 30 cm, f2 = -20 cm We know that 1 1 1 1 1 1 = + = − =− f f1 f 2 30 20 60

196

f = −60 cm Diverging lens.

8. A small pin fixed on table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? R.I of glass is 1.5. Does the answer depend upon the location of the slab? Ans: The lateral displacement  1 t = d 1 −    1   t = 15 1 −  = 5 cm  1.5  The answer does not depend upon the location of the slab. 9. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose? Ans: For a real, image, least distance between object and image should be four times the focal length. D 3 f = = = 0.75 cm 4 4 D = distance between object and image. 10. A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens. Ans: We know that

D 2 − x 2 ( 90 ) − ( 20 ) 7700 f = = = = 21.4 cm 4D 4  90 360 Where D = distance between object and image x = separation between two position of lens. 2

2

197

Chapter 10: Wave Optics 1. Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected and (b) reflected light? R.I. of water is 1.33 Ans: (a) For reflected light wavelength is unchanged i.e. 589 nm. Speed of light in air = 3x108 ms-1 c 3 108 v= = = 5.09 1014 Hz  589 10−9 (b) For refracted light, while moving from one medium to the other, frequency remains unchanged i.e. 5.09x1014 Hz  589 10−9 '= = = 443 10−9 m  1.33

v=

c



=

3 108 = 2.26 108 ms −1 1.33

2. In YDSE, the slits are separated by 0.28 mm and the screen is place 1.4 m away. The distance between the central bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment. Ans: D = 0.28x10-3 m, n =4, D = 1.4 m, Yn = 1.2x10-2 m as n D yn = d −2 −3 y d (1.2 10 )( 0.28 10 ) = n = nD 4 1.4  = 6 10−7 m 3. In YDSE using monochromatic light of wavelength, the intensity of light at a point on the screen where path difference is  is K units. What is the intensity of light at a point where path difference is

 ? 3

Ans: When path difference =  2  = 2 Phase difference,  =



Resultant intensity, I = 4 I o cos 2

 3 2  2 = Phase difference,  ' =  3 3

 2

= 4Io = k

When path difference =

198

(i)

I ' = 4 I o cos 2

'

2 2   I ' = k cos 2 = k cos 2 3 2 3 2

1 1 I'=k  = k 4 2

4. In YDSE, the distance between the slits and the screen is 2 m. when the distance between the slits is 5 mm, the fringe width is 0.1 mm. Determine the wavelength of the light used. Ans: D = d −4 −3 Bd (110 )  ( 5 10 ) = = = 2.5 10−7 m D 2 5. In YDSE, the interference fringes are formed on a screen 2 m away from the double slit of separation 0.1 mm. Determine the distance of the fifth dark band from the centre of the screen. The wavelength of light used is 6000x10-10 m Ans: ( 2n + 1)  D yn = 2 d  6000 10−10  2  y5 = ( 2  4 + 1)  = 0.054 m −3   2  0.110 

y5 = 5.4 cm 6. In YDSE, the separation between the slits is halved and the distance between the slits and the screen is doubled. What will be the new fringe width? Ans: D Fringe width  = d d d ' = , D ' = 2D 2  D '  2 D 4 D '= = = d' d d    2  

 ' = 4 7. In YDSE, the intensity ratio between the bright and dark fringes in the interference pattern is 9. Determine the ratio of the amplitude of two waves producing the interference pattern. Ans:

199

I max 9 = I min 1

( a1 + a2 ) = 9 2 ( a1 − a2 ) 1 ( a1 + a2 ) = 3 ( a1 − a2 ) 1 2

3a1 − 3a2 = a1 + a2 2a1 = 4a2 a1 2 = a2 1

8. Two polaroids P1, and P2 are placed with their pass axes perpendicular to each other. Unpolarised light of intensity Io is incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its pass axis makes an angle of 30o with that of P1. Determine the intensity of light transmitted through P 1, P2 & P3. Ans: As we know that after polarisation intensity of light becomes half I  Intensity through P1 = o 2 Intensity through 2

Io I 1 I cos 2 60 = o   = o 2 2 2 8 Intensity through P3 =

2

P2 =

Io I  3  3I o cos 2 30 = o   = 8 8  2  32

9. Can two independent sources of light be coherent? Ans: Ordinarily not. Word ‘never’ should not be used in the answer because laser is an exception. 10. If a slit is closed what will happen to the interference pattern? Ans: The interference pattern will disappear

200

Chapter 11: Dual Nature of matter and Radiation 1. Find the a) maximum frequency and b) minimum wavelength of x-rays produced by 30 kV electrons. Ans: 1.6 10−19  30000 eV = = 7.24 1018 Hz a) Vmax = h 6.63 10−34 −34 8 hc ( 6.63 10 )  ( 3 10 ) a) min = = = 0.04110−19 m eV (1.6 10−19 )(3 104 )

(

)

2 The photoelectric cut-off voltage in a certain experiment is 1.5V. What is the maximum kinetic energy of photoelectron emitted? Ans: Vo = 1.5V

( K .E )max = eVo = 1.6 10−19 1.5 = 2.4 10−19 J 3. The energy flux of sunlight reaching the surface of the earth is 1.388x103 W/m2. How many photons (nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm. Ans:  = 550 nm =550x10-9 m,  = 1.388x103 W/m2 hc 6.626 10−34  3 108 E= = = 3.611019 J  550 10−9  1.388 103 = 3.85 1021 no. of photons per square metre of earth = = E 3.6110−19 4. In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12x10-15 Vs. Calculate the value of Planck's constant. Ans: The slope of the graph in this case is h 4.12 10−15 = e h = 4.12 10−15  e = 4.12 10−15 1.6 10−19

h = 6.592 10−34 Js 5. A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light,which is incident on it. The wavelength of sodium light is 589 nm (a) what is the energy per Photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere? Ans: P =100 W,  = 589 nm = 589x10-9 m −34 3 108 hc 6.626 10 = = 3.38 10−19 J a) E =  589 10−9

(

)(

)

201

b) Rate at which the photons are delivered = P 100 = = 2.96 1020 photons / sec E 3.38 10−19 6. The threshold frequency for a certain metal is 3.3x1014 Hz. If light of frequency 8.2x1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission. Ans:

 o = 3.3 1014 Hz,  = 8.2 1014 Hz using eVo = hv − h o hv − h o h = ( v − o ) e e 6.626 10−34 Vo = = ( 8.2 1014 − 3.3 1014 ) = 2.03V 1.6 10−19

Vo =

7. The work function for a certain metal is 4.2eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm? Ans: W = 4.2 eV = 4.2x1.6x10-19 = 6.72x10-19 J As W 6.72 10−19 o = = = 1.011015 Hz h 6.626 10−34 C 3 108 o = = = 2.97 10−7 m = 297 nm  o 1.011015 Since o   Or  o   , the photoelectric emission for this radiation can not take place. 8. Light of frequency 7.21x1014 Hz is incident on a metal surface. Electron with a maximum speed 6x105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electron? Ans: 1 2 mvmax = hv − hvo 2 1m 2  o = − vmax 2h −31 5 1 9.110 ( 6 10 ) = 7.211014 −  = 4.74 1014 Hz 2 6.626 10−34 2

9. Calculate the a) Momentum and b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V. Ans: V = 56 V

202

a) P = 2mVe = 2  9.110−31  56 1.6 10−19 = 4.04 10−24 kgms −1 12.27 o 12.27 A = = 1.64 Ao b)  = V 56 10. What is photoelectric effect? Why it cannot be explained on the basis of wave nature of light? Ans: Photoelectric effect is a process of ejection of electrons from a metal surface when light radiations of sufficiently high frequency are incident on it. When light radiations strike a metal surface, they give energy to the electron which is thus liberated. This process can be explained better by treating it to be collision of particles, which is not possible on the basis of wave nature of light.

203

Chapter 12: Atoms 1. Suppose you are given a chance to repeat the alpha particle scattering experiment using a thin sheet of solid hydrogen in place of gold foil (hydrogen is a solid at temperature below 14K). What results do you except? Ans: As we know size of hydrogen nucleus = 1.2 x 10-15 m. Electrostatic potential energy of -particle at nuclear surface

Ve = Ve =

1 4 o

( 2e )( e ) = 9 109  2  (1.6 10−19 ) 1.2 10−15

r

9 109  2  (1.6 10−19 )

2

J

2

(1.2 10−15 )(1.6 10−19 )

= 2.4 106 eV = 2.4MeV

This energy is much less than incident energy of -particle (5.5 MeV). Therefore, the scattering of -particle by hydrogen nuclei will be quite different than the scattering of particle by gold nuclei. In this case the impact parameter and the distance of closest approach will be quite small. It is possible that -particle might penetrate the nucleus and no scattering will be observed. Size of hydrogen atom will not be assessed by this method. 2. What is the shortest wavelength present in the Paschen series of spectral lines? Ans: we know that 1 1  1 = R 2 − 2   n m   For Paschen series 1 1  1 = R 2 − 2   3 m  1 1 1  = R − 2   9 m  For shorter wavelength m =  1 R =  9 9  = = 9  911Ao = 8199 Ao R 1 As = 911Ao R 3. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when he atom transits from the upper level to the lower level? Ans: E = 2.3 eV = 2.3x1.6x10-19 = 3.68x10-19 J E 3.68 10−19 = 5.55 1014 Hz As  = = h 6.626 10−34 4. The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state?

204

Ans: We know that U − K .E = E = 2 U = 2 E = 2  −13.6 = −27.2 eV K .E = − E = +13.6 eV 5. A hydrogen atom initially in the ground level absorbs a photon which excites it to the n = 4 level. Determine the wavelength and frequency of photon. Ans: we know that −13.6 En = 2 eV n for ground level, n = 1 −13.6 E1 = 2 = −13.6 eV 1 For n = 4 −13.6 E4 = = −0.85 eV 42 E = E4 − E1 = −0.85 + 13.6 = 12.75 eV the hydrogen atom will go to n = 4 from n = 1 if the photon energy E = E

E = 12.75 eV = 12.75  1.6  10−19 E = 20.4 10−19 J h = 20.4 10−19 20.4 10−19 = 3.08 1015 Hz h C 3 108 = = = 9.74 10−8 m  3.08 1015

=

6. a) Using the Bohr's model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels. b) calculate the orbital period in each of these levels. Ans: a) we know that 1 c Vn =   137  n  When 1 3 108  = 2.18 106 ms −1 n =1 , V1 = 137 1 1 3 108  = 1.09 106 ms −1 n = 2, V2 = 137 2 1 3 108  = 0.73 106 ms −1 n = 3, V3 = 137 3 b) Orbital period of election in the nth orbit

205

2 rn 2 ao n 2 137  2 ao n3 = = 1 c Vn c  137 n Where ao = 0.53x10-10 m When n = 1 137  2  3.14  0.53 10−10 13 T1 = 3 108 −16 T1 = 1.52 10 sec when n = 2 Tn =

T2 = 23T1 = 8T1 = 8 1.52 10−16 = 1.22 10−15 sec When n = 3

T3 = 33 T1 = 27 1.52 10−16 = 4.110−15 sec 7. The radius of the innermost electron orbit of a hydrogen atom is 5.3x10-11 m. What are the radii of the n = 2 and n = 3 orbits? 2 Ans: rn = kn

4 o h 2 4 2 me2 for n =1 , r1 = k k = 5.3x10-11 m For n=2, r2 = 22k = 4k = 4x5.3x10-11 = 2.12x10-10 m For n=3 , r3 = 32k = 9k = 9x5.3x10-11 = 4.77x10-10 m where k =

8. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? Ans: hc E=  hc 6.625 10−34  3 108 = = = 0.994 10−7 = 99 nm E 12.5 1.6 10−19 On the excitation they may produce Balmer series also i.e. Lyman series. 9. The total energy of an electron in the first excited state of the hydrogen atom about 3.4 eV. a) What is the kinetic energy of a electrons in this state? b) What is the potential energy of the electron in this state? c) Which of the answer above would change if the choice of the zero of potential energy is changed? Ans: a) we know that U − K .E = E = 2 K .E = − E = +3.4 eV b) U = 2E = 2  ( −3.4 ) = −6.8 eV

206

c) K.E does not depend upon the choice of zero of potential energy. Therefore, its value remains unchanged. However, the potential energy gets charged with the change in the zero level potential energy. 10. Why does electron revolve round nucleus of an atom? Ans: So that the atom is stable. If election is not resolving but stationary in the orbit, it is pulled into the nucleus.

207

Chapter 13: Nucleus 1. Two stable isotopes of Lithium 36 Li and 37 Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u respectively. Find the atomic weight of lithium. Ans: We know that Average atomic mass = p m + p2 m2 = 1 1 p1 + p2 7.5  6.01512 + 92.5  7.01600 = 6.9409 u 7.5 + 92.5

=

2. Boron has two stable isotopes

10 5

B and 115B . Their respective masses are 10.01294 u and

11.00931 u and the atomic weight of boron is 10.811 u. Find the abundance of 11 5

10 5

B and

B.

Ans: Let x% and y% be the abundance of x + y = 100 (i) We know that Average atomic mass = p m + p2 m2 = 1 1 p1 + p2

10.811 =

10 5

B and 115B respectively

( x )(10.01294 ) + ( y )(11.00931)

x+ y 10.01294x + 11.00931y = 1081.1 Using eqn. (i)

(ii)

10.01294 (100 − y ) + 11.00931 y = 1081.1 y = 80.09% x = 100 − 80.09 = 19.91%

3. Write the nuclear equations for 226 a) the  decay 88 Ra b) the - decay +

32 15 11 6

P

c) the decay C Ans: 226 222 4 a) 88 Ra → 86 Rn + 2 He b)

32 15

P → 1632S + e− +

11 11 + c) 6 C → 5 B + e +

4. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce

208

a) 3.125% of its original value? b) 1% of its original value? Ans: a) We know that t

 N   1  T1  =  2  No   2  Given N 3.125 = No 100

(i)

N 1 = N o 32 5

N 1 =  No  2  on comparing with eqn. (i) we get t =5 T1

(ii)

2

t = 5T1 = 5T 2

b) Given  N   1   =   N o   100  on comparing with eqn. (i) we get t

 1   1  T1  =  2  100   2  t

 1   1 T  =   100   2  t log100 = log 2 T log100 2 t =T =T = 6.64 T log 2 0.3010 90

5. The half life of 38 Sr is 28 years. What is the disintegration rate of 15 mg of this isotope? Ans: T = 28 years = 28x365x24x3600 sec 0.6931 0.6931 = = s −1 T 28  365  24  3600 6.02 1023 15 103 90 Sr = =N No. of atoms in 15 mg 38 90 We know that Disintegration rate,

209

dN = N dt 0.6931 6.02  1023  15  10−3 = 28  365  24  3600  90 = 7.87 1010 Bq 6. Does the ratio of neutrons to protons in a nucleus increase, decrease or remain same after the emissions of - particles? Ans: Increases. 7. Why is heavy water used as moderator in a nuclear reactor? Ans: Heavy water is rich in protons and it does not absorb neutrons. 8. Among ,  and  radiations which get deflected by magnetic field? Ans:  and  radiations. 9. What is the difference between an electron and - particle? Ans: An electron resides outside the nucleus whereas  is an electron like particle of nuclear origin. 10. What is the effect of temperature and pressure on the radioactivity? Ans: Radioactivity is independent of temperature and pressure.

210

Chapter 14: Electronic Devices 1. In an n-type silicon, which of the following statement is true? a) Electrons are majority carriers and trivalent atoms are the dopants. b) Electrons are minority carriers and pentavalent atoms are the dopants. c) Holes are minority carriers and pentavalent atoms are the dopants. d) Holes are majority carriers and trivalent atoms are the dopants. Ans: (c) 2. Which of the statements given in above question is true for p-type semiconductor? Ans: (d) 3. In an unbiased p-n junction, holes diffuse from the p-region to n- region because a) Free electrons in the n-region attract them. b) They move across the junction by the potential difference. c) Hole concentration in p-region is more as compared to n-region. d) All the above Ans: (c) 4. When a forward bias is applied to a p-n junction, it a) raises the potential barrier. b) reduces the majority carrier current to zero. c) lowers the potential barrier d) none of the above. Ans: (c) 5. In half wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency? Ans: A half wave rectifier rectifies only one half cycle of input A.C.  Frequency of the output A.C = frequency of input AC = 50 Hz A full wave rectifier both halve cycles of the AC input frequency of output A.C = 2xFrequency of input A.C = 2x50 = 100 Hz 6. A p-n photodiode is fabricated from semiconductor with band gap of 2.8 eV. Can it detect a wavelength 6000 nm. Ans: −34 8 hc ( 6.62 10 )  ( 3 10 ) E= = = 3.3110−20 J −9  6000 10 3.3110−20 = = 0.2 eV 1.6 10−19 As the photon energy is much less than the band gap, hence it cannot detect the given wavelength 7. The number of silicon atoms per m 3 is 5x1028. This doped simultaneously with 5x1022 atoms per m3 of Arsenic and 5x1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Give that ni = 1.5x1016 per m3. Is he material n-type or p-type?

211

Ans: Number of electron, ne = 5x1022 – 5x1020 = 4.95x1022 m-3 ni = 1.5x1016 m-3 Number of holes, 16 ni2 (1.5 10 ) = = 4.54 109 m−3 ne 4.95 1022 2

nh =

nh = 4.54 109 m−3 Since ne >> nh, the semiconductor is n-type semiconductor.

8. Why is a semiconductor damaged with a strong current? Ans: Semiconductors are highly temperature dependent, strong current causes heating up of the semiconductor. Due to this reason covalent bonds the semiconductor break, resulting into ultimate breakdown of the semiconductor. 9. Give the ratio of number of holes and the number of conduction electrons in an intrinsic semiconductor. Ans: 1:1. 10. How does the width of the depletion layer of a p-n junction diode change with decrease in reverse bias? Ans: Width of depletion layer will decrease.

212