PHYSICS WORKBOOK-II FOR 11TH GRADE IBDP STUDENTS

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TED ANKARA COLLEGE FOUNDATION HIGH SCHOOL PHYSICS DEPARTMENT

NAME-SURNAME: _________________________ CLASS: _____________ SCHOOL NUMBER: _____________

PHYSICS WORKBOOK-II FOR 11TH GRADE IBDP STUDENTS 2020-2021 ACADEMIC YEAR

CONTENT

1. Circular Motion

2

Problems

3

Multiple Choice Items

12

Markscheme

16

2. Universal Gravitation

20

Problems

21

Markscheme

34

3. Electrostatics

42

Electric Force

43

Electric Field

44

Electric Potential and Potential Energy

45

Problems

51

Multiple Choice Items

59

Markscheme

65

4. Magnetism

72

Right Hand Rule Exercises

73

Magnetic Field Around Current Carrying Wire

76

Magnetic Force on Moving Charges

78

Problems

80

Multiple Choice Items

88

Markscheme

98

5. Magnetic Induction

102

Problems

103

Multiple Choice Items

110

Markscheme

120

6. Alternating Current

124

Problems

125

Multiple Choice Items

132

Markscheme

140

1 |TED Ankara College Foundation High School Physics Department

CIRCULAR MOTION

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Problems 1. Two bodies m1 and m2 ,shown in figure on right, are circling on the same plane with the periods of T1=3s and T2=5s. Show their positions after 12s on the figure on left. m2 T1 m1

T2 O

O

2. An object undergoes uniform circular motion with a period of 72 seconds. Find the time required for the object to move from P to Q y P

x 400

Q

3. Two objects A and B are rotating on the circular orbits on the same plane as shown. If the periods of A and B are 2 sec and 3 seconds respectively, what is the ratio of VA/VB? VB 3r

VA

2r

4. A mass of 1.5 kg moves in a horizontal circle of radius 0.25m at 2 rev/s. Calculate the magnitude of; (a) the linear speed.

(b) the acceleration

(c) centripetal force acting on the body.

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5. An object performs uniform circular motion with linear speed of 2m/s in a radius of 1m. If one revolution of object takes 4 sec, and object starts from point A, find y

V=2m/s

(a) centripetal acceleration?

A

x

r=1m

(b) change in velocity after 1 sec?

(c) average acceleration after 2 sec?

6. The equally divided rod in figure rotates around point S. If the centripetal forces acting on m 1 and m2 are equal, what is the ratio of m1/ m2? m2

m1 S

E

D

K

7. Two objects are attached to the midpoint and to the end of a rope and are being turned around horizontally as shown in figure below. The mass of inner object is twice of the mass of outer one. (a) Which one of the followings is equal for these masses? O

r

A

2m

B r

m

I

Period

II

Linear velocity

III

Angular velocity

IV

Centripetal acceleration

(b) What is the ratio of tensions in the ropes OA and AB?

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8. An object of mass 0.5 kg is attached to the end of a rope of 1 m. It performs uniform circular motion in vertical plane. If, it makes 2 revolutions per second; (a) Find the linear velocity of the object.

(b) Find the tension in the string at the top, bottom and side points of the circular path. top

C side

v

bottom

9. Mass m is attached to one end of a rope of length 1 m and is rotated around centre C, along a vertically aligned circular path. When mass m is at the lowest point on its path, the tension in the rope is measured to be “3mg”. What is the velocity of the mass m? P

C T v

10. A circular horizontal platform rotates around its central axis with a constant angular speed. A 0.2 kg mass is placed at the edge of the platform as shown. What is the maximum value of the linear velocity the platform can have, to keep mass on it, if the coefficient of friction between them is 0.5? m

0.8 m

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11. A small cart is released down a track, the bottom part of which has been bent into a loop as shown in figure below. K L

h

r

What must be the minimum initial height that the cart is released so that it can pass through the highest point of loop (point L) without falling down? Express the height (h) in terms of radius of circle (r).

12. An object is released from point A of the frictionless surface. What is the reaction force exerted by the rail to the object at point B? A B 30 r

2r

13. An object of mass, m=1 kg is let to swing from point B. It is attached to a rope of length, L neglecting the frictional effects, what is the tension in the string at point A? B

L A

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14. The object in figure is released to swing from point A. It passes through point B with a velocity of v and passes through point C with a velocity of v/2. It is known that the mass of object is 0.1kg and the length of string is 4m.

(a) What is the initial height (h) of object at point A? (Frictional effects are neglected.) A

C h

2.4m B

(b) What is the tension in the string at point A,

15. The objects with masses m1=2kg and m2=1kg tight together, are in equilibrium as in Figure. If the surface is frictionless, find the velocity with which object K move? m2=1kg r=2m

K

V=?

m1=2kg

16. An object is released from the top of a semicircular surface to slide down. At which height, the object would leave the surface?

r

h=?

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17. Figure below shows the London Eye.

(a) It has many capsules attached around the edge of a large vertical Wheel of radius 60m. The Wheel rotates about a horizontal axis such that each capsule has a constant speed of 0.26ms-1. (i)

Calculate the time taken for the Wheel to make one complete rotaion.

(ii)

Each capsule has a mass of 9.7x103kg. Calculate the centripetal force which must act on the capsule to make it rotate with the Wheel.

(b) Determine the direction and magnitude of force exerted by the Wheel on the capsule, when the capsule is (i)

at the lowest position on the Wheel.

(ii)

at the heighest position on the Wheel.

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(c) Figure below shows the drum of a spin-dryer as it rotates. A dry sock S is shown on the inside surface of the rotating drum. (i) Draw arrows on Figure to show the direction of centripertal force acting on S when it is at points A, B and C.

(ii)

State and explain at which position, the normal force between the sock and the drum will be The greatest:

The least:

18. This question is about circular motion. The diagram shows a car moving at a constant speed over a curved bridge. At the position shown, the top surface of the bridge has a radius of curvature of 50 m. car top surface of bridge of radius of curvature 50m

(a) Explain why the car is accelerating even though it is moving with a constant speed. [2]

(b) On the diagram, draw and label the vertical forces acting on the car in the position shown. [2]

(c) Calculate the maximum speed at which the car will stay in contact with the bridge. [3]

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19. This question is about the kinematics and dynamics of circular motion. (a)

A car goes round a curve in a road at constant speed. Explain why, although its speed is constant, it is accelerating. (2)

In the diagram below, a marble (small glass sphere) rolls down a track, the bottom part of which has been bent into a loop. The end A of the track, from which the marble is released, is at a height of 0.80 m above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is 0.35 m. A marble

C

0.80 m

0.35 m

ground

B

The mass of the marble is 0.050 kg. Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. The acceleration due to gravity, g = 10 ms–2. Consider the marble when it is at point C. (b)

(i)

(ii)

On the diagram opposite, draw an arrow to show the direction of the resultant force acting on the marble.

(1)

State the names of the two forces acting on the marble. (2)

(iii)

Deduce that the speed of the marble is 3.0 ms–1.

(3)

(iv)

Determine the resultant force acting on the marble and hence determine the reaction force of the track on the marble.

(4) (Total 12 marks)

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20.

Banked track A car travels at a steady speed v in a circular path of radius r on a circular track banked at an angle θ as shown in the plan and side views in the diagram.

The car’s speed is such that there is no sideways frictional force between the tyres and the track.

(a) Does the car have an acceleration? Explain why or why not. If you say yes, state its direction. [2]

(b) The car on the track is represented by a block in the figure below, moving perpendicular tothe page. Draw a force diagram, showing and labelling all the forces acting on the moving car. [2]

(c) Is there a resultant force on the moving car? If you say yes, explain why and state its direction. If you say no, explain why not. In either case, refer to your force diagram to support your answer. [2]

(d) The track is banked at an angle of 17" and the circular path of the car has radius 30 m.Calculate the speed at which the car must travel in order that there be no sideways frictional force between the tyres and the track. Show all working. [4]

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Multiple Choice Items 1. Two satellites of equal mass, S1 and S2, orbit the Earth. S1 is orbiting at a distance r from the Earth’s centre at speed v. S2 orbits at a distance 2r from the Earth’s centre at speed v . The ratio of the centripetal force on S1 to the centripetal force on S2 is 2 A.

1 . 8

B.

1 . 4

C. 4.

D. 8.

2. A point mass is moving in a horizontal circle with a velocity of constant magnitude v. At one particular time, the mass is at P. A short time later, the mass is at Q, as shown below.

v

Q

v

P

Which vector diagram correctly shows the change in velocity Δv of the mass during this time? A.

B. v

v

v

v

v

v

C.

D. v

v

v

v

v

v

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3. A particle travels at a constant speed around a circle of radius r with centripetal acceleration a. What is the time taken for ten complete rotations? A.

𝜋 𝑎 √ 5 𝑟

B.

𝜋 𝑟 √ 5 𝑎

𝑎

𝑟

C. 20𝜋√ 𝑟

D. 20𝜋√𝑎

4. Points P and Q are at distances R and 2R respectively from the centre X of a disc, as shown below.

Q P

2R R X

The disc is rotating about an axis through X, normal to the plane of the disc. Point P has linear speed v and centripetal acceleration a. Which one of the following is correct for point Q? Linear speed

Centripetal acceleration

A.

v

a

B.

v

2a

C.

2v

2a

D.

2v

4a

5. The centripetal force F acting on a particle of mass m that is travelling with linear speed v along the arc of a circle of radius r is given by A. F =

v2 mr

.

B. F = mv2r.

C. F = mr2v.

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D. F =

mv 2 r

.

6. For a particle moving in a circle with uniform speed, which one of the following statements is correct? A. B. C. D.

The kinetic energy of the particle is constant. The force on the particle is in the same direction as the direction of motion of the particle. The momentum of the particle is constant. The displacement of the particle is in the direction of the force.

7. A brick is placed on the surface of a flat horizontal disc as shown in the diagram below. The disc is rotating at constant speed about a vertical axis through its centre. The brick does not move relative to the disc.

Which of the diagrams below correctly represents the horizontal force or forces acting on the brick?

8. The centripetal force that causes a car to go round a bend in the road is provided by A.

the force produced by the car engine acting on the wheels.

B.

the friction between the tyres and the road.

C.

the weight of the car.

D.

the force exerted by the driver on the steering wheel.

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9. A particle of mass m moves in a circle of radius r at uniform speed, taking time T for each revolution. What is the kinetic energy of the particle? A.

𝜋2 𝑚𝑟 𝑇2

B.

𝜋2 𝑚𝑟 2 𝑇2

C.

2𝜋2 𝑚𝑟 2 𝑇

D.

2𝜋2 𝑚𝑟 2 𝑇2

10. A particle P is moving in a circle with uniform speed. Which one of the following diagrams correctly shows the direction of the acceleration a and velocity v of the particle at one instant of time? a

A.

a

B.

v

v

P

P

v C.

D.

a

a

P

P

v

π

11. A model car moves in a circular path of radius 0.8 m at an angular speed of 6 rad s–1. What is its displacement from point P, 6 s after passing P? A zero B 1.6 m C 0.4 πm

D 1.6 πm

12. A mass on the end of a string is whirled round in a horizontal circle at increasing speed until the string breaks. The subsequent path taken by the mass is A a straight line along a radius of the circle. B a horizontal circle. C a parabola in a horizontal plane. D a parabola in a vertical plane. 15 |TED Ankara College Foundation High School Physics Department

MARKSCHEME 1.

m1: at the same position

m2: rotated by 144 degrees

2.

28 s

3.

1

4.

(a) 3 ms-1 (b) 36 ms-2 (c) 54 N

5.

4m/s2 , 22m/s , 2m/s2

6.

2

7.

(a) I, III and IV

8.

a)12m/s, b) 67N, 77N, 72N

9.

25m/s

(b) 1/2

10. 2m/s 11. 2.5 r 12. mg/2 13. 3mg 14. 3.2 m ; 1.8 N 15.  40m/s 16.

ℎ=

2𝑟 3

17. (a) (i) 𝑉 = (ii) 𝐹 =

2𝜋𝑟 𝑇

; 0.26 =

𝑚𝑉 2 𝑟

;=

2𝜋×60 ; 𝑇

9700×0.262 60

T = 1450sec = 24min. ; = 11𝑁

(b) (i) Tlowest = Fc +mg = 11 + 9700 = 9711N ; upward force (ii) Thighest = Fc - mg = 9700 -11 = 9689N ; upward force

(c) (i)

(ii) The normal force is highest at C since mg is in opposite direction to centripetal force. And it is lowest at point C since mg is in the direction of centripertal force.

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18. (a) direction changing; velocity changing so accelerating;

2

(b) weight/gravitational force/mg/w/Fw/Fg and reaction/normal reaction/perpendicular contact force/N/R/FN/FR both labelled; (do not allow “gravity” for “weight”.) weight between wheels (in box) from centre of mass and reactions at both wheels / single reaction acting along same line of action as the weight; 2 (c) g = V2/r ; =(50x9.8)-1 ; 22ms-1; Allow [3] for a bald correct answer.

3 [7]

19. (a) Look for an answer on the following lines. the direction of the car is changing; hence the velocity of the car is changing; or since the direction of the car is changing; a force must be acting on it, hence it is accelerating; (b) (i) (ii)

arrow pointing vertically downwards;

1

weight; Do not penalize the candidate if they state “gravity”. normal reaction; Do not penalize the candidate if they state “push of the track on the marble”.

(iii)

2 max

2

loss in PE = 0.05 × 10 × (0.8 – 0.35); = gain in KE = 1 mv2; to give v = 3.0 m s–1; 2

or 2 gh to give v = 4.0 m s–1 at point B; and then use of v2 – u2 = 2gh with v = 4.0 m s–1 and h = 0.35 m;

use of v =

to get u = 3.0 m s–1;

3 max

Do not penalize the candidate if g = 9.8 m s–2 is used. (iv)

recognize that resultant force = =

mv 2 ; r

(0.05  9.0) mv 2 = 2.6 N; N = – mg; = 2.6 – 0.5 = 2.1 N; 0.175 r

4 [12]

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20. (a) Yes it has acceleration because its velocity direction is changing [1] – toward the centre of the track.

2

(b)] (Award [1] for weight downward, [1] for normal force perpendicular to track. Subtract marks for non-existent or wrong forces.) 2

(c) Yes, there is a resultant force: the car is not in equilibrium, but accelerating toward the centre of the circular track. [1] The resultant force on it is toward the centre, and in the force diagram it would be the resultant of the two forces shown. (d) Vertically there is equilibrium: N cosθ = mg Horizontally there is a net force and acceleration: Net force = N sinθ Mg.sin θ/cos θ = mg.tan θ N II:mg. Tanθ = mv2/r v2 = gr tanθ =10x30 tan17 = 91.5 v = 9.6 ms-1

2

4 [10]

Key for multiple choice items:

1. C

2. C

3. D

4. C

5. D

6. A

7. D

8. B

9. D

10. C

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11. B

12. D

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UNIVERSAL GRAVITATION

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Problems 1.

This question is about gravitation and orbital motion. (a)

Define gravitational field strength at a point in a gravitational field. (2)

The diagram below shows three points above a planet. The arrow represents the gravitational field strength at point A. C

B planet

(b)

A

Draw arrows to represent the gravitational field strength at point B and point C. (2) (Total 4 marks)

2.

This question is about gravitational fields. (a)

Define gravitational field strength. (2)

The gravitational field strength at the surface of Jupiter is 25 N kg–1 and the radius of Jupiter is 7.1 × 107 m. (b)

(i)

Derive an expression for the gravitational field strength at the surface of a planet in terms of its mass M, its radius R and the gravitational constant G. (2)

(ii)

Use your expression in (b)(i) above to estimate the mass of Jupiter. (2) (Total 6 marks)

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3.

This question is about a spacecraft. A spacecraft above Earth’s atmosphere is moving away from the Earth. The diagram below shows two positions of the spacecraft. Position A and position B are well above Earth’s atmosphere.

A

Earth

B

At position A, the rocket engine is switched off and the spacecraft begins coasting freely. At position A, the speed of the spacecraft is 5.37 x 103 m s–1 and at position B, 5.10 x 103 m s–1. The time to travel from position A to position B is 6.00 x 102 s. (a)

(i)

Explain why the speed is changing between positions A and B. (1)

(ii)

Calculate the average acceleration of the spacecraft between positions A and B. (2)

(iii)

Estimate the average gravitational field strength between positions A and B. Explain your working. (3)

(b)

The diagram below shows the variation with distance from Earth of the kinetic

energy Ek of the spacecraft. The radius of Earth is R. energy

Ek

0

R 0

distance

On the diagram above, draw the variation with distance from the surface of Earth of the gravitational potential energy Ep of the spacecraft. (2) (Total 8 marks)

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4.

This question is about gravitation. (a)

(i)

Define gravitational potential at a point in a gravitational field. (2)

(ii)

Explain why values of gravitational potential have negative values. (2)

The Earth and the Moon may be considered to be two isolated point masses. The masses of the Earth and the Moon are 5.98 x 1024kg and 7.35 x 1022 kg respectively and their separation is 3.84 x 108 m, as shown below. The diagram is not to scale. Earth mass = 5.98 1024kg

Moon mass = 7.35 1022kg

P

3.84 108m (b)

(i)

Deduce that, at point P, 3.46 x 108m from Earth, the gravitational field strength is approximately zero. (3)

(ii)

The gravitational potential at P is −1.28 x 106 J kg–1. Calculate the minimum speed of a space probe at P so that it can escape from the attraction of the Earth and the Moon.

(3) (Total 10 marks) 23 |TED Ankara College Foundation High School Physics Department

5.

This question is about gravitational potential. (a)

Define gravitational potential at a point in a gravitational field. (3)

(b)

A planet has mass M and radius R0. The magnitude g0 of the gravitational field strength at the surface of a planet is

g0 = G

M R0

2

where G is the gravitational constant.

Use this expression to deduce that the gravitational potential V0 at the surface of the planet is given by V0 = – g0R0. (2)

(c)

The graph below shows the variation with distance R from the centre of the planet of the gravitational potential V. The radius R of the planet = 5.0 x 106 m. 0

Values of V are not shown for R = R0. R / 10 7 m 0.5 0.0

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

–0.5 –1.0

V / 10 7 Jkg–1

–1.5 –2.0 –2.5 –3.0 –3.5 –4.0 –4.5 –5.0

Use the graph to determine the magnitude of the gravitational field strength at the surface of the planet. (3)

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(d)

A satellite of mass 3.2 x 103 kg is launched from the surface of the planet. Use the graph to estimate the minimum launch speed that the satellite must have in order to reach a height of 2.0 x 107 m above the surface of the planet. (You may assume that it reaches its maximum speed immediately after launch.) (4) (Total 12 marks)

6.

This question is about the gravitational field of Mars. (a)

Define gravitational potential energy of a mass at a point. (1)

(b)

The graph shows the variation with distance r from the centre of Mars of the gravitational potential V. R is the radius of Mars which is 3.3 Mm. (Values of V for r < R are not shown.) A rocket of mass 1.2 × 104 kg lifts off from the surface of Mars. Use the graph to (i) calculate the change in gravitational potential energy of the rocket at a distance 4R from the centre of Mars. (3)

(ii)

show that the magnitude of the gravitational field strength at a distance 4R from the centre of Mars is 0.23 N kg–1. (2)

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(c)

Use the answer to (b)(ii) to show that the magnitude of the gravitational field strength at the surface of Mars is 3.7 N kg–1. (2)

(d)

The gravitational potential at the surface of Earth is –63 MJ kg–1. Without any further calculation, compare the escape speed required to leave the surface of Earth with that of the escape speed required to leave the surface of Mars. (2) (Total 10 marks)

7.

This question is about gravitation. A space probe is launched from the equator in the direction of the north pole of the Earth. During the launch, the energy E given to the space probe of mass m is E=

3GMm 4Re

where G is the Gravitational constant and M and Re are, respectively, the mass and radius of the Earth. Work done in overcoming frictional forces is not to be considered. (a)

(i)

Explain what is meant by escape speed. (2)

(ii)

Deduce that the space probe will not be able to travel into deep space. (3)

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The space probe is launched into a circular polar orbit of radius R. (b)

Derive expressions, in terms of G, M, Re, m and R, for (i)

the change in gravitational potential energy of the space probe as a result of travelling from the Earth’s surface to its orbit. (1)

(ii)

the kinetic energy of the space probe when in its orbit. (2)

(c)

Using your answers in (b) and the total energy supplied to the space probe as given in (a), determine the height of the orbit above the Earth’s surface. (4)

A space probe in a low orbit round the Earth will experience friction due to the Earth’s atmosphere. (d)

(i)

Describe how friction with the air reduces the energy of the space probe. (2)

(ii)

Suggest why the rate of loss of energy of the space probe depends on the density of the air and also the speed of the space probe. (2)

(iii)

State what will happen to the height of the space probe above the Earth’s surface and to its speed as air resistance gradually reduces the total energy of the space probe. (2) (Total 18 marks)

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8.

This question is about escape speed and Kepler’s third law. Jupiter and Earth are two planets that orbit the Sun. The Earth has mass Me and diameter De. The escape speed from Earth is 11.2 km s–1. Data for Jupiter are given below. Mass:

1.90 × 1027 kg (318 Me)

Mean diameter: 1.38 × 105 km (10.8 De) (a)

(i)

State what is meant by escape speed. (1)

(ii)

Escape speed v is given by the expression

 2GM  v=  .  R  Determine the escape speed from Jupiter. (2)

(b)

(i)

State Kepler’s third law. (1)

(ii)

In 1610, the moon Ganymede was discovered orbiting Jupiter. Its orbit was found to have a radius of 15.0 R and period 7.15 days, where R is the radius of Jupiter. Another moon of Jupiter, Lysithea, was discovered in 1938 and its orbit was found to have a radius of 164 R and a period of 260 days. Show that these data are consistent with Kepler’s third law. (2) (Total 6 marks)

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9.

Motion of a satellite (a)

Define gravitational potential. (2)

(b)

A satellite of mass m is in a circular orbit around the Earth at height R from the Earth’s surface. The mass of the Earth may be considered to be a point mass concentrated at the Earth’s centre. The Earth has mass M and radius R.

orbit satellite mass m Earth mass M

R

(i)

R

Deduce that the kinetic energy EK of the satellite when in orbit of height R is

EK =

GMm . 4R (3)

(ii)

The kinetic energy of the satellite in this orbit is 1.5 x 1010 J. Calculate the total energy of the satellite. (3)

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(iii)

Explain how your answer to (b)(ii) indicates that the satellite will not escape the Earth’s gravitational field and state the minimum amount of energy that must be provided to this satellite so that it does escape. (3) (Total 11 marks)

10.

This question is about a satellite orbiting the Earth. A satellite S is in orbit round the Earth, a distance R = 4.2 × 107 m from the centre of the Earth.

S

Earth

R = 4.2 × 107 m

(a)

On the diagram above, for the satellite in the position shown, draw arrow(s) to represent the force(s) acting on the satellite. (1)

(b)

Deduce that the velocity v of the satellite is given by the expression

GM v2 = R

where M is the mass of the Earth. (1)

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(c)

Hence deduce that the period of orbit T of the satellite is given by the following expression. T2 =

4 2 R 3 GM

(3)

(d)

Use the following information to determine that the orbital period of the satellite is about 24 hours. GM = 10 ms−2, where 2 RE M is the mass of the Earth and RE is the radius of the Earth = 6.4 × 106 m. (2)

Acceleration due to gravity at the surface of the Earth g =

(e)

The satellite is moved

into an orbit that is closer to the Earth. State what happens to its (i)

potential energy. (1)

(ii)

kinetic energy. (1) (Total 9 marks)

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11.

Kepler’s third law. (a)

Kepler’s third law states that the period T of the orbit of a planet about the Sun is related to the average orbital radius R of the planet by the relationship T2 = KR3 where K is a constant. (i)

Suggest why the law specifies the average orbital radius. (1)

(ii)

State the name of the force that causes the acceleration of the planets orbiting the Sun. (1)

(iii)

State an expression for the magnitude F of the force in (ii) in terms of the mass MS, of the Sun, the mass m of the planet, the radius R of the orbit and the universal gravitational constant G. (1)

(iv)

Hence deduce, explaining your working, that the constant K is given by the 4 2 . expression: K = GM S (4)

(b)

Ganymede is one of the moons of Jupiter and the following data are available. Average orbital radius of Ganymede = 1.1 x 109 m

(i)

Orbital period of Ganymede

= 6.2x 105 s

Universal gravitational constant G

= 6.7 x 10−11 N m2 kg−2

Deduce that the gravitational field strength of Jupiter at the surface of Ganymede is approximately 0.1 N kg−1. (2)

(ii)

Estimate the mass of Jupiter. (3) (Total 12 marks)

32 |TED Ankara College Foundation High School Physics Department

12.

This question is about gravitation. A binary star consists of two stars that each follow circular orbits about a fixed point P as shown below.

star mass M1

P R1

star mass M2 R2

The stars have the same orbital period T. Each star may be considered to act as a point mass with its mass concentrated at its centre. The stars, of masses M1 and M2, orbit at distances R1 and R2 respectively from point P. (a)

State the name of the force that provides the centripetal force for the motion of the stars. (1)

(b)

By considering the force acting on one of the stars, deduce that the orbital period T is given by the expression

T2 =

4 2 2 R1 (R1 + R2 ) . GM 2 (3)

(c)

The star of mass M1 is closer to the point P than the star of mass M2. Using the answer in (b), state and explain which star has the larger mass.

(2) (Total 6 marks)

33 |TED Ankara College Foundation High School Physics Department

Markscheme

1.

(a)

(b)

F m F is the gravitational force; exerted on / experienced by a small / point / infinitesimal mass m;

2

Award [1] for each correct arrow. The one at B points in the same direction as that at A and is shorter. The one at C has the same length as that at A and points toward the centre of the planet.

2

g=

C

planet

A

B

[4]

2.

3.

(a)

the gravitational force exerted per unit mass; on a point / small mass;

(b)

(i)

use of g =

(ii)

M=

(i)

(deceleration due to) gravitational pull of Earth;

(ii)

a=

(a)

(iii) (b)

gR2 G

F M Mm and F = G 2 ; combine to get g = G 2 ; m R R

; substitute to get M = 1.9 × 1027 kg;

v 5100− 5370 = ; a = - 0.45ms-2; t 600

ecf from (ii): E =

F ; E = a; E = - 0.45Nk g-1; Accept ms-2 as correct units. m

general shape (1 / r); correct quadrant;

2 2

2 [6]

1 2

3 2

[8] 34 |TED Ankara College Foundation High School Physics Department

4.

(a)

(b)

(i)

work done per unit mass; in moving (small mass) from infinity to that point;

2

(ii)

gravitational forces are always attractive; work got out when moving from infinity; work done against field is negative;

2

(i)

(

)

2

ME 3.46 10 8 = 81.4 ; (ratio idea is essential) MM 3.84 10 8 − 3.46 10 8

(

)

2

= 82.9 ;

M should be same so QED; R2

some comment eg or: FEarth =

(G  5.98 10 ) (3.46 10 )

= 5.00 10 7 G ; (ratio idea is essential)

FMoon =

(G  7.35 10 ) (0.38 10 )

= 5.09 10 7 G ; some comment eg about same so

24

8 2

22

8 2

QED; or:

GM E GM M = ; (ratio idea is essential) 2 2 r 3.84 10 8 − r

(

81 .4 =

(ii)

)

r2

(3.84 10

8

−r

)

2

; r = 3.45 x 108 m;

3

indicates in some way that GPE is change in kinetic energy; 2 gravitational potential = 1 v ; v = 1.6 x 103 ms-1; 2

5.

(a)

(b)

the work done per unit mass; in bringing a small / point mass; from infinity to the point (in the gravitational field);

V0 = − G

3

M ; GM = g0 R02 to give V0 = - g0R0; R0

2

Do not award mark for data book expression V = − G

(c)

3 [10]

m . r

𝑉 3.9×107 from the graph V0 = 3.9 (±0.1) x107 J kg-1; 𝑔0 = 𝑟0 = 0.5×107 = 7.8(±0.2) Nkg-1; 0

35 |TED Ankara College Foundation High School Physics Department

3

(d)

2.0 x 107 m above surface is 2.5 x 107 m from centre; V between surface and 2.5 x 107m = (- 0.8 – (- 3.9)) x 107 = - 3.1 x 107 J kg-1; −∆𝑃𝐸 = ∆𝐾𝐸

1

− 𝑚∆𝑉 = 2 𝑚𝑣 2

𝑣 = √−2∆𝑉 = √−2 × (−3.1 × 107 ) = 7.9 × 103 𝑚𝑠 −1

4

Award [3 max] if the candidate forgets that the distances are from the centre, the candidate must show V. 6.

(a)

work done in moving mass from infinity to a point;

(b)

(i)

[12] 1

read offs –12.6 and –3.2; gain in gpe 1.2 × 104 × [12.6 – 3.2] or gain in g potential [12.6 × 106 – 3.2 × 106] = 1.13 ± 0.05 × 105 MJ or = 1.13 + 0.05 × 1011 J;

(ii)

3

use of gradient of graph to determine g;  6.7  10 6  ; values substituted from drawn gradient  typically 7  3.3  10 6   = 0.23 N kg–1 (allow answers in the range of 0.20 to 0.26 N kg–1)

Award [0] for solutions from

2

V . r

3 .7 (c) g at surface = 42 g at 4R; and = 16.1 or 16 × 0.23 = 3.7 N kg–1; 0.23

2

(d) escape speed for Earth > escape speed for Mars; potential less/more negative at Earth; 2 [10] 7.

(a)

(i)

speed of object at Earth’s surface; so that it will escape from the gravitational field / travel to infinity; 2

(ii)

gravitational potential energy at Earth’s surface = (–)

GMm ; Re

this must be provided for probe to escape; energy is less than this hence not escape; (b)

 1 1 1 1 −  ; Accept GMm  − (i) change = GMm   Re R   R Re (ii)

(c)

in orbit,

mv 2 GMm = 2 ; r r

1 2

mv 2 =

3

  . 

GMm ; 2R

idea of equating energies;

36 |TED Ankara College Foundation High School Physics Department

1

2

3GMm GMm GMm GMm = + − 4 Re 2R Re R height above surface = Re; (d)

(i)

1 1 = 4 Re 2 R

R = 2Re; 4 max

probe collides with air molecules; giving them kinetic energy and so losing energy itself;

2

Accept answers in terms of frictional forces. (ii)

greater density, more molecules of air with which to collide; higher speed, higher rebound speed for air molecules;

2

Accept answers in terms of magnitude of frictional force.

8.

(a)

(iii)

height becomes less; and speed increases;

(i)

speed with which object must leave (surface of planet) to (completely) escape its gravitational field;

(ii)

(b) (i) (ii)

v = 11.2 ×

2 [18]

318 or full substitution; = 60.8 km s–1; 10 .8

2

T2 ≈ R3 with symbols explained / in words; for Ganymede,

7.15 2 = 0.0151 15 3

1

1

for Lysthea,

260 2 = 0.0153; 164 3

Do not award the ratios if only 2 significant figure are given. This is not the significant figure mark. suitable comment based on candidate’s calculations;

9.

2 max [6]

Motion of a satellite (a)

(b)

the work done per unit mass; in bringing a small / point mass from infinity to a point (in the gravitational field); Ratio idea essential for first mark. (i)

2

mv 2 ; = equating gravitational force to centripetal GMm 2 r

r

2 GM ; to get the speed v =

r

and hence EK =

1 2

GMm = r

1 2

GMm = GMm ; 2R 4R

37 |TED Ankara College Foundation High School Physics Department

3

(ii)

potential energy in orbit is E P = −

GMm GMm =− ; r 2R

GMm − GMm = − GMm ; total energy is then E = 4R

2R

4R

ie E = -1.5 x 1010J;

3

Award [1 max] for bare answer without explanation. (iii)

at infinity the potential energy is zero and hence if satellite gets there its total energy will be E = 0; but the satellite has negative energy and hence it cannot escape to infinity / the satellite is bound to the Earth forever;

the minimum energy that must be supplied  the value must agree with the  is E = 1.5  1010 J; (or ECF from (b )(ii ))  candidates answer to (b) (ii). 3 [11] 10.

(a)

S F

F – towards centre of Earth;

1 max

Award [0] if any other forces are drawn.

Earth

R = 4.2 × 107 m

(b)

GM v 2 = R R2

(c)

2 πR v= ; T

(d)

T2 =

(e)

(i)

decreases;

1 max

(ii)

increases;

1 max [9]

GM so v2 = ; R 2 3 4π 2 R 2 GM 2 2 = 4π R ; = so v = ; so T R GM T2

4π 2 R 3 g 0 RE

2

; to give T = 85000 s = 24 hours;

38 |TED Ankara College Foundation High School Physics Department

1 max

3 max

2 max

11.

Kepler’s third law (a)

(i)

the orbits are elliptical / not circular; Do not accept “because R changes”.

1

(ii)

gravity / gravitational;

1

(iii)

F =G

M sm ; R2

1

Same symbols as in question must be used to receive the mark. (iv)

v2 let v = the speed of planet, the acceleration is then = R from Newton 2, F = G

v=

M s m mv 2  Some reference to Newton 2 = ; R is required to receive this mark. R2

M m 4 2 mR 2 2R ; therefore, = G s2 = ; R T T2

therefore, T 2 =

4 2 4 2 3 R ; so K = GM s GM s

4

Be aware of the many simple variants of this eg using a =  2 R =

(b)

(i)

40 1.110 9 v 2 4 2 R = ; = ; recognize that gravitational field strength = 2 R T2 6.2 10 5

(

= 0.1Nkg-1 (ii)

)

2

4 2 3 4 2 R 3 ; T = R therefore, M jup = GM s GT 2 2

4 2  (1.1) 10 27 3

M jup =

Or

4 2 R . T2

6.7 10

−11

(

 6.2 10

)

5 2

; 2.0 x 1027 kg;

(

GM 0.1 1.110 9 0.1R 2 0 .1 = 2 ; M = ;= R G 6.7 10 −11

)

2

= 2.0 = 1027 kg;

3 [12]

39 |TED Ankara College Foundation High School Physics Department

12.

(a)

gravitation / gravity;

(b)

gravitational force =

1 GM 1 M 2

; centripetal force = 2

(R1 + R2 )

gravitational force provides centripetal force

M 1 R1  4  2 ; T2

GM 1 M 2

(R1 + R2 )2

=

M 1 R1  4  2 ; T2

R1 (R1 + R2 )  4  2 GM 2 2

T2 =

(c)

from formula,

3

R1 is a constant; (so if R1 is smaller) then M2 is smaller / M1 is M2

larger;

2

Do not award second mark if no reasoning given or argument is fallacious. [6]

40 |TED Ankara College Foundation High School Physics Department

41 |TED Ankara College Foundation High School Physics Department

ELECTROSTATICS Electric Force, Electric Field, Electric Potential energy and Electric Potential

42 |TED Ankara College Foundation High School Physics Department

Electrostatic (Coulomb’s Force) 1. Find the net force on the charge placed at the center of the hexagon? +q

a

-q a

a +q

-q

+q a

a -q

a

+q

2. Three point charges are placed as in Figure. The net electric force acting on the charge +q is F. Find the ratio of charges q1/q2? q1 +q q2 F

3. Two charged spheres A and B are placed as shown in figure. A has a charge of 6q and a radius of 3r. B has a charge of q and a radius of r. Calculate the force of interaction between them in terms of k, q and r? Qa=+6q

Qb=+q

3r

r r

4. What should be the distance between 10C and 20C charges, so that the charge 20C is said to be in equilibrium? Q1=10C Q2=20C d=?

Q3=40C

18cm

43 |TED Ankara College Foundation High School Physics Department

Electric Field 5. Draw the electric field lines between the conductors given below. Neutral outer shell

+

-

+

-

+

-

+

-

-

(-)ly charged inner shell

-

(+)ly charged charged linear surface

(-)ly charged charged convex surface

6. Four point charges are located as in Figure. Calculate the magnitude of electric field at the center of a square with diagonals of 20 cm. q2=-2c

q1=-1c

q3=3c

q4=4c

7. Two concentric charged shells, one placed in other, is shown in figure. Outer shell is labeled as A, and the inner one is B. They have a common center and they have the net charges of QA=4x10-6 C, QB=1x10-6 C. Their radii are rA = 40 cm, rB = 10 cm. Find the electric field at a point 30 cm away from the common center O. QA=4x10-6C QB=1x10-6C O

8. Two point charges are located 2m apart. The charges have magnitude of 10-9C and 9x10-9C. Find the distance from the smaller charge to the point at which the net electric field is zero. Q2=-9x10-9 C

Q1=10-9 C 2m

44 |TED Ankara College Foundation High School Physics Department

9. A small charged body is suspended between two charged parallel plates. The rope makes an angle of 37 with the vertical. What is the electric field intensity between the plates if the mass of body is 4 gram and charge of the body is 1x10-8 Coulomb?

+

-

+

37

+

-

+

-

+

-

10. Two point charges q1 and q2 are placed as in Figure-I. The resultant electric field at point K is in the direction of the arrow. The direction of resultant electric field at point K is changed as shown in Figure-II when a third charge qx is added. What is the charge value of qx in terms of q? ( q1 = q) qx K

q1

E

K

q1 E

q2

q2

Figure-I

Figure-II

Electric potential and electric potential energy 11. The electric field due to charges q1 and q2 at point A is shown. Electric potential created at point B by q1 is V1, electric potential created at point B by q2 is V2. Find V1/V2.

q1

A B

E

q2

45 |TED Ankara College Foundation High School Physics Department

12. The net electric field intensity due to charges q and q’ at point O is E. Find the electric potential at point O in terms of k, q and r. q q' 600 r E r

13. The graph shows the magnitude of electric field of a hollow charged sphere against the distance r from its center. What is the electric potential on the surface of the sphere? E(N/C) 8

2

r(m)

0

3

6

14. K and L are two metallic shells. Excess charges of K and L are –4q and +q before touching each other. What will be the PE of the system when they are put in contact as shown in above figure K L 2R

R

15. The +q charge has an electric potential of 100 V at point Y. What is the potential difference between points X and Y? +q

a a/2

a X

Y

a

-q

46 |TED Ankara College Foundation High School Physics Department

16. A charged object +q is placed at point A in an uniform electric field. It is transferred from point A to point B by following two different routes as in Figure. Calculate the work done in the process in terms of q, E, d. Are they different? What are the signs of the works? d

Route (2)

B

d

+q

E

A

Route (1)

17. Two small charges are located at the corners of a rectangle as shown in Figure. Find the amount of work done if a charged particle +q is brought from A to B? 4a

B

Q1 =-12q

3a

Q2 =+12q

A

18. The excess charge on a spherical metallic shell is 6.10-9C. + + +

+

+ +

r=1m A

+

+ 1.5 m

+

+

+

B 2m

+

(a) What is the potential difference between A and B if the radius is 1 m?

(b) Calculate the work done by the electric force if the +2C is moved from A to B?

47 |TED Ankara College Foundation High School Physics Department

19. A triangle has charges of its vertices as shown in above Figure. How much work is required to bring a +2c charge from infinity to point A? Q3 =8x10-8C

5cm

5cm 4cm

Q1 =-6x10-8C

Q2 =2x10-8C A

20. A small charged particle 10-5 C is carried from A to B. Find the change in potential energy of particle to bring the charge from A to B? (It is known that E=105N/C, AB=2 cm, q=10-5 C)

B

530 A

21. The electric field between two parallel plates is 200 N/C. What is the potential difference between points A and B? 6cm A B 2cm

22. Three parallel conducting plates A, B and C are connected to two batteries as shown in Figure. A positively charged ion is released from plate A and it moves under the influence of the electric force along the x-axis. Discuss the motion of ion; draw a-t and v-t graphs. B

A d

C d

+x

+

V1=V

V2=2V

+ -

- +

48 |TED Ankara College Foundation High School Physics Department

23. Two parallel plates are connected to a power supply of 120V. If +q charge is released from point K as in Figure. Find the ratio of velocities at L and M, L/M=? KL=LM=MN (Ignore gravitational effects)

K

L +q

M

N

V=120V

24. The point charge q with mass m is in equilibrium as in figure. If the potential difference between the plates X and Y is halved; X d/2 d

(a)

q m

to which plate will the charge strike?

(b) with what velocity will the charge strike the plate? (Find the velocity of the charge in terms of q, V and m)

V +

Y

25. According to the given figure, (a)

4 cm K

What electrical force acts on +4C if it is at point A?

L

A

+q

B

+

-

V = 2000 V

(b) What is the change in KE of the charged particle from A to B if it starts from rest at point A? (Spaces are equally divided)

49 |TED Ankara College Foundation High School Physics Department

26. A negatively charged ion enters from small hole in the electric field, it slows down and stops somewhere between the plates. At what distance from the plate K does the ion stop? L

K 4cm

q=-2x10-8C V=0

KE=1,2.10-4J

- + V=8x103V

27. A charge of q = -2.10-6C is released from plate K. Find the final KE when it emerges from plate M. L

M

K 10 cm

10 cm KEf = ?

-q + -

- +

V2 = 300 V

V1 = 200 V

28. A charge of +2.10-9 C enters the system of plates from the hole in plate A with an initial kinetic energy of 8.10-6 J as shown. Will the charge be able to reach the plate C? B

A 6 cm

C 4 cm

(a) 2.10-9

q= C m = 2.10-2 kg

If yes what will be its velocity when it hits C?

(b) If no, what maximum distance away from B can it get to?

q

+ -

- +

V1 = 1000 V V2 = 3000 V

50 |TED Ankara College Foundation High School Physics Department

Problems 29. (a) State, in words, Coulomb’s law. [2] (b) The graph shows how the electric potential, V, varies with 1/r, where r is the distance from a point charge Q.

State what can be deduced from the graph about how V depends on r and explain why all the values of V on the graph are negative. [2]

(c) (i) Use data from the graph to show that the magnitude of Q is about 30 nC. [2]

(ii) A +60 nC charge is moved from a point where r= 0.20 m to a point where r = 0.50 m. Calculate the work done.

(iii) Calculate the electric field strength at the point where r= 0.40 m. [2]

51 |TED Ankara College Foundation High School Physics Department

30.

This question is about electric charge at rest. (a)

Define electric field strength at a point in an electric field. [2]

(b) Four point charges of equal magnitude, are held at the corners of a square as shown below. 2a +Q

+Q

2a

P

–Q

–Q

The length of each side of the square is 2a and the sign of the charges is as shown. The point P is at the center of the square. (i)

Deduce that the magnitude of the electric field strength at point P due to kQ one of the point charges is equal to . [2] 2a 2

(ii)

On the diagram above, draw an arrow to represent the direction of the resultant electric field at point P. [1]

(iii)

Determine, in terms of Q, a and k, the magnitude of the electric field strength at point P. [3]

52 |TED Ankara College Foundation High School Physics Department

31.

Fields and potential Electric fields and potential (a)

Define electric potential. [2]

(b) An isolated metal sphere of radius 50.0 cm has a positive charge. The electric potential at the surface of the sphere is 6.0 V.

50.0 cm metal sphere

(i)

On the diagram above, draw a line to represent an equipotential surface outside the sphere and draw additional lines to represent the electric field outside the sphere. [2]

(ii)

On the axes below, draw a sketch graph to show how the potential V outside the sphere varies with distance r from the surface of the sphere. V /V 6

4

2

0 0.0

(iii)

0.5

1.0

1.5 r / m

[4]

Explain how the graph drawn in (b) (ii) can be used to determine the magnitude of the electric field strength at the surface of the sphere. [2]

53 |TED Ankara College Foundation High School Physics Department

32.

This question is about electric fields and electric circuits. (a)

Two parallel, charged metal plates A and B are in a vacuum.

At a particular instant an electron is at point P. On the diagram, draw (i)

the electric field pattern due to the plates. [3]

(ii)

an arrow to represent the direction of the force on the electron at P. [1]

(b)

The acceleration of the electron at P is 8.8 × 1014 m s–2. Determine the magnitude of the electric field strength at the point P. [3]

(c)

The electric potential energy of the electron changes by 1.9 × 10–17 J as it moves from one plate to the other. Show that the potential difference between the plates is 120 V. [1]

54 |TED Ankara College Foundation High School Physics Department

33.

This question is about motion of a charged particle in an electric field. (a)

An α-particle of mass 4u and charge +2e is accelerated from rest in a vacuum through a potential difference of 2.4 kV. Show that the final speed of the αparticle is 4.8 × 105 m s–1. [2]

(b)

The α-particle is travelling in a direction parallel to and mid-way between two parallel metal plates.

The metal plates are of length 2.4 cm and their separation is 0.80 cm. The potential difference between the plates is 600 V. The electric field is uniform in the region between the plates and is zero outside this region.

(c)

(i)

Calculate the magnitude of the electric field between the plates. [2]

(ii)

Show that the magnitude of the acceleration of the α-particle by the electric field is 3.6 × 1012 m s–2. [2]

(i)

Calculate the time taken for the α-particle to travel a horizontal distance of 2.4 cm parallel to the plates. [1]

(ii)

Use your answers in (b)(ii) and (c)(i) to deduce whether, as the α-particle passes between the plates, it will hit one of the plates. [3]

55 |TED Ankara College Foundation High School Physics Department

34. The figure below shows a system that separates two minerals from the ore containing them using an electric field.

The crushed particles of the two different minerals gain opposite charges due to friction as they travel along the conveyor belt and through the hopper. When they leave the hopper they fall 4.5 metres between two parallel plates that are separated by 0.35 m.

(a) Assume that a particle has zero velocity when it leaves the hopper and enters the region between the plates. Calculate the time taken for this particle to fall between the plates. [2]

(b) A potential difference (pd) of 65 kV is applied between the plates. Show that when a particle of specific charge 1.2 × 10–6C kg–1is between the plates its horizontal acceleration is about 0.2 m s–2. [3]

(c) Calculate the total horizontal deflection of the particle that occurs when falling between the plates. [1]

(d) Explain why the time to fall vertically between the plates is independent of the mass of a particle. [2]

(e) State and explain two reasons, why the horizontal acceleration of a particle is different for each particle. [4]

56 |TED Ankara College Foundation High School Physics Department

35. Electrons are accelerated from rest from the cathode to the anode of a vacuum tube through a potential difference of 5000 V.

(a) Show that the speed v of an electron as it leaves the anode is approximately 4 × 107ms–1. [3]

(b) The emerging beam of electrons follows a parabolic path as it passes between a pair of horizontal parallel plates 5.0 cm apart with a potential difference of 1400 V between them.

(i) Calculate the strength E of the uniform electric field between the horizontal plates. [1]

(ii) Hence determine the force F exerted by this field on each electron. [1]

57 |TED Ankara College Foundation High School Physics Department

(c) An electron experiences an upward acceleration a as it travels between the plates. Its vertical 1

displacement h after a time t is given by ℎ = 2 𝑔𝑡 2 . Calculate the value of h as the electron leaves the plates. [4]

(d) (i) Add to Figure 2 the path that the electron beam would follow if the potential difference between the horizontal plates were decreased. Label this path A. [1] (ii) Add to Figure 2 the path that the electron beam would follow if the potential difference between the cathode and the anode were decreased. Label this path B. [1]

58 |TED Ankara College Foundation High School Physics Department

MULTIPLE CHOICE ITEMS 1.

A positively charged particle follows a circular path as shown below.

Which of the following electric fields could have caused the charged particle to follow the above path?

2.

Two isolated point charges, –7 μC and +2 μC, are at a fixed distance apart. At which point is it possible for the electric field strength to be zero?

(not to scale)

59 |TED Ankara College Foundation High School Physics Department

3.

The diagram below shows a uniform electric field of strength E. The field is in a vacuum.

An electron enters the field with a velocity v in the direction shown. The electron is moving in the plane of the paper. The path followed by the electron will be

4.

A.

parabolic.

B.

in the direction of E.

C.

in the direction of v.

D.

circular.

Three positive point charges of equal magnitude are held at the corners X, Y and Z of a right-angled triangle. The point P is at the midpoint of XY. Which of the arrows shows the direction of the electric field at point P?

60 |TED Ankara College Foundation High School Physics Department

5.

The diagram shows two parallel metal plates X and Y.

Plate X is at Earth potential (0 V) and the potential of plate Y is V0. Which of the following is correct in respect of the magnitude and the direction of the electric field between the plates? Magnitude

6.

Direction

A.

constant

X→Y

B.

increasing

Y→X

C.

constant

Y→X

D.

increasing

X→Y

The radius of a charged spherical conductor is R. Which of the following graphs best shows how the magnitude of the electrical field strength E varies with distance r from the centre of the sphere?

61 |TED Ankara College Foundation High School Physics Department

7.

Which of the following gives the acceleration of an electron of electric charge e and mass m in a uniform electric field of strength E? A. E

B. Ee

C.

Ee m

D.

m Ee

8.

Which arrangement of three point charges at the corner of an equilateral triangle will result in a zero electric field strength at the centre of the triangle, point P?

9.

Which diagram best represents the electric field due to a negatively charged conducting sphere?

62 |TED Ankara College Foundation High School Physics Department

10.

Which of the following diagrams illustrates the electric field pattern of a negatively charged sphere?

11.

Four point charges of magnitudes +q, +q, –q, and –q are held in place at the corners of a square of side r.

The Coulomb constant is k. Which of the following is the electrical potential at the centre of the square O? A. 0 12.

B.

4kq r

C.

4kq 2 r

D.

− 4kq 2 r2

The diagram below shows a particle with positive charge q accelerating between two conducting plates at potentials V1 and V2.

Which of the following is the kinetic energy gained by the charge in moving between the plates? A. V2q

B. V1q

C. (V1 – V2)q

D. (V2 – V1)q

63 |TED Ankara College Foundation High School Physics Department

13.

14.

A test charge is a A.

charged object with a very small mass.

B.

charged object with a very small charge.

C.

point charge which has no effect on the electric field in which it is placed.

D.

point charge which slightly changes the electric field in which it is placed.

The electric field strength at a point may be defined as A.

the force exerted on unit positive charge placed at that point.

B.

the force per unit positive charge on a small test charge placed at that point.

C.

the work done on unit positive charge to move the charge to that point from infinity.

D.

the work done per unit positive charge to move a small test charge to that point from infinity.

15. In the equation 𝑥 =

𝑎.𝑏 𝑟𝑛

, X represents a physical variable in an electric or a gravitational field, a is

a constant, b is either mass or charge and n is a number. Which line, A to D, in the table provides a consistent representation of X, a and b according to the value of n? The symbols E, g, V and r have their usual meanings.

64 |TED Ankara College Foundation High School Physics Department

MARKSCHEME 4kq2 a

q

8

6kq2

2. q1 = 9

3. 25𝑟2

4. 6cm

5. Suitable drawings

6. 45√2 × 105 𝑁/𝐶

7. 1 × 105 𝑁/𝐶

8. X=1cm to the left of q1

9. 3 × 106 𝑁/𝐶

10. 2

11. -1/2

12. −

13. 24V

14.

15. Zero

16. 4qEd, +

18. (a) 36V (b) 72 × 10−6 𝐽

19. 1.2 × 10−2 J

20. 1.2 × 10−2 J

22. (acc between BC) is equal to -2(acc between AB)

23.

1.

17.

2kq2 a

21. 4V 25. (a) 0.2N (b) 48 × 10−4 J

2

q

2kq2 3r

26. 1cm

1 √2

kq r

24. (a) plate Y (b) √

27. 2 × 10−4 J

28. 2 × 10−2 𝑚𝑠 −1

29. force between two (point) charges is proportional to product of charges ✓ inversely proportional to square of distance between the charges ✓ Mention of force is essential, otherwise no marks. Condone “proportional to charges”. Do not allow “square of radius” when radius is undefined. Award full credit for equation with all terms defined.

2

(b) Vis inversely proportional to r [or V ∝(−)1 / r ] ✓ (V has negative values) because charge is negative [or because force is attractive on + charge placed near it or because electric potential is + for + charge and − for − charge] ✓ potential is defined to be zero at infinity ✓ Allow V × r = constant for 1st mark.

max 2

65 |TED Ankara College Foundation High School Physics Department

𝑞𝑉 2𝑚

(c) (i) Q(= 4πɛ0 rV) = 4πɛ0× 0.125 × 2000 OR gradient = Q / 4πɛ0= 2000 / 8 ✓ (for example, using any pair of values from graph) ✓ = 28 (27.8) (± 1) (nC) ✓

(gives Q = 28 (27.8) ±1 (nC) ✓

2

(ii) at r= 0.20m V = −1250V and at r= 0.50m V = −500V so pd ΔV = −500 − (−1250) = 750 (V) ✓ work done ΔW (= QΔV) = 60 × 10−9× 750= 4.5(0) × 10−5(J) (45 μJ) ✓ (final answer could be between 3.9 and 5.1 × 10−5) Allow tolerance of ± 50V on graph readings. [Alternative for 1stmark: ΔV = (or similar substitution using 60 nC instead of 27.8 nC use of 60 nC gives ΔV = 1620V) ]

2

(iii) = ✓ = 1600 (1560) (V m−1) ✓ [or deduce E = by combining E = with V = ✓ from graph E = = 1600 (1560 ± 130) (V m−1) ✓ ] Use of Q = 30 nC gives 1690 (V m−1). Allow ecf from Q value in (i). If Q = 60 nC is used here, no marks to be awarded.

30.

(a)

2

[10]

the force exerted per unit charge; on a small positive (test) charge;

2

Accept either “small” or “test” or both. (b)

(i)

substitute for r = a 2 ; into E =

kQ kQ to get E = ; 2 2 2a 2 r

(ii) 1

(iii)

E for each component =

kQ kQ ; add vectorially; to get Etot =; 2 2 3 2 a a

Award [1] if not added vectorially ie Etot = 2

kQ a2

66 |TED Ankara College Foundation High School Physics Department

31.

Fields and potential (a)

(b)

energy / work per unit charge; in bringing a small positive test charge / positive point charge from infinity / positive test charge; Award [0] for quoting formula without definition of symbols. (i)

any roughly drawn circle drawn concentric to the sphere; Ignore any shape inside the sphere. Award [0] if other lines also shown.

2

1

(ii) curve starting from 6.0V; curves downwards; to zero potential at infinity; going through the point ( 0.5m, 3.0V ); 4

(iii) idea that E = ( ) potential gradient seen / implied; so draw tangent at r = 0 and find gradient;

32.

(a)

(i)

uniform field equal spacing of lines; edge effect; direction;

3

(ii)

1

as shown

(b)

combine F = qE and F = ma; to get E =

(c)

V=

33.

(a)

1.9 10 −17 1.6 10

−19

use of

ma ; E = 5.0 × 103 N C–1/V m–1 3 q

; = 120 V-1

1 2 mv = qV; 2

1 × 4 × 1.66 × 10–27 × v2 = 2 × 1.6 × 10–19 × 2400; 2 (v = 4.8 × 105 m s–1) (b)

(i)

E=

600

; 0.80  10 − 2 = 7.5 × 104 V m–1; (accept unit as N C–1

67 |TED Ankara College Foundation High School Physics Department

2

2

(ii)

force = (Eq =) 7.5 2×.4 10410× −214× 1.6 × 10–19; (= 2.4 × 10–14 N) − 27 acceleration = 4 1.66 10 ; (= 3.6 × 1012 m s–2)

Do not penalize twice for omission of 2 in charge of α-particle. (c)

(i) (ii)

 2.4  10 −2  –8    4.8  10 5 =  = 5.0 × 10 s  

for motion in direction of electric field 1 distance dropped in 50 ns = × 3.6 × 1012 × (5.0 × 10–8)2; 2 = 0.45 cm; α-particle starts 0.40 cm from above plate and so hits it / OWTTE

3 [10]

34. M5.(a) 4.5 = 1/2× 9.81 × t2 ✓

t= 0.96 s ✓

2

(b) Field strength = 186000V m–1 ✓ Acceleration = Eq / m or 186 000 × 1.2 × 10–6✓

0.22 m s–2✓

(c) 0.10(3)m (allow ecf from (i))✓

3 1

(d) Force on a particle = mg and acceleration = F / m so always = g✓ Time to fall (given distance) depends (only) on the distance and acceleration ✓ OR: g = GM / r ✓

2

Time to fall = √2s/g so no m in equations to determine time to fall✓

2

(e) Mass is not constant since particle mass will vary✓ Charge on a particle is not constant✓ Acceleration = Eq / m or (V / d) (q / m) or Vq / dm✓ E or V / d constant but charge and mass are ‘random’ variables so q / m will vary (or unlikely to be the same)✓

4

35. (a) Electron speed; Substitution of electronic charge and 5000V in eV (1) Substitution of electron mass in ½ mv2 (1) Correct answer [4.2 (4.19) × 107(ms–1), no ue] to at least 2 sf (1) 68 |TED Ankara College Foundation High School Physics Department

[Bald answer scores zero, reverse working can score 2/3 only] Example of answer:

(b) (i) Value of E; Correct answer [2.80 × 104Vm–1/N C–1 or 2.80 × 102Vcm–1] (1)

1

Example of answer: E = V/d = 1400 V / 5.0 × 10–2= 28 000 V m–1 (ii) Value of force F; Correct answer [4.5 × 10–15N, ecf for their E] (1)

1

Example of answer: F = Ee = 2.80 × 104V m–1× 1.6 × 10–19C= 4.48 × 10–15N (c) Calculation of h;

4

See substitution of a and t values [arrived at by above methods] into ½ at2 (1) Correct answer [h = 0.020 m – 0.022 m] (1) [Full ecf for their value of F if methods for a and t correct and their h ≤ 5.0cm] Example of answer: h = ½ a t2= ½ × 4.9 × 1015ms–2× (2.86 × 109s)2 = 2.0 × 10–2m (d) (i) Path A of electron beam Less curved than original (1)

1

(ii) Path B of electron beam More curved than original, curve starting as beam enters field [started by H of the Horizontal plate label] (1) 1

69 |TED Ankara College Foundation High School Physics Department

MULTIPLE CHOICE ITEMS KEY

1.

A

6.

D

11.

A

2.

D

7.

C

12.

C

3.

A

8.

A

13.

C

4.

B

9.

C

14.

B

5.

C

10.

A

15.

C

70 |TED Ankara College Foundation High School Physics Department

71

MAGNETISM

72

RIGHT HAND RULE EXERCISES

The magnetic force vector is shown for a positivelycharged particle moving through the magnetic field. What is the direction of velocity vector?

This is an end view (cross-section) of a current carrying wire in a uniform magnetic field. Magnetic field and current directions are shown. What is the direction of magnetic force?

What magnetic force direction will this positivelycharged particle experience?

The magnetic force and velocity vectors are shown for a charged particle moving through the magnetic field. What sign is the charge?

The external magnetic field is to the left. Is the particle’s charge positive or negative? What is the sign of charge?

73

The magnetic force vector direction is shown for a current-carrying wire in a magnetic field. What direction is the current?

What direction is the magnetic force on this currentcarrying wire?

The current direction is shown for a current-carrying wire in a magnetic field. What direction is the magnetic force?

This is an end view (cross-section) of a current carrying wire in a uniform magnetic field. Magnetic field and current directions are shown. What is the magnetic force direction?

V

What direction of magnetic force would this moving proton experience?

The magnetic force vector is shown for a charged particle moving through the magnetic field. What sign is the charge?

74

What is the magnetic field direction inside the current-carrying loop?

Which direction would a mapping compass point when placed above the middle of the solenoid?

Label the north and south poles of the selonoid.

What direction would an external magnetic field have to be to make the wire experience a magnetic force down the page?

Which wire would current enter to make a mapping compass point right when INSIDE the solenoid?

Draw the magnetic field lines around the wire.

75

Magnetic field around current carrying wire 1. Two long current carrying wires are placed on the plane of the page as shown. On which point ALOMG the x-axis, is the net magnetic field zero? 6 cm

x

I1=2A

I2=6A

2. Two long wires carrying currents of I1= 2A and I2 are placed perpendicular to the page parallel to each other as shown. If the net magnetic field is zero at point P a. Find the direction of I2?

A

2cm b. Find the magnitude of I2?

I1=2A

1cm P

3cm I2=?

c. Find the magnitude and direction of net magnetic field at point A?

3. A circular loop of wire and a long straight wire carry currents of I1 and I2, where I2=6I1. The loop and the straight wire are on the same plane. The net magnetic field at the center of the loop is zero. (Take π = 3) (a) Find the direction of current I1. (cw or ccw)

(b) Find the perpendicular distance H, expressing your answer in terms of r, the radius of the loop.

76

4. For the following solenoid having 2000 turn and carrying a current of 20 Amp, find the magnitude and direction of the magnetic field along its axis. 10cm

20A

5. A piece of wire of length 20cm is in a magnetic field of 4x10-4T. What is the magnitude and direction of the force acting on the wire?

R=10Ω

40 V

B = 4x10-4T x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

20cm

6. Two long wires are perpendicular to each other. If the net magnetic field is zero along the dotted line. I1

53

I2

a) Find the directions of the currents?

b) Find the ratio I1/ I2?

77

Magnetic force on moving charges 7. Determine the direction of magnetic field in each case given in Figures. F represents the force applied on charged particle or wire, V represents the velocity of charged particle and I represents the current in wire.

8. The drawing shows a top view of four interconnected chambers. A negative charge is fired into chamber 1. By turning on separate magnetic fields in each chamber, the charge can be made to exit from chamber 4 as shown. Describe how the magnetic field in each chamber should be directed?

9. Three particles have identical charges and masses. They enter a constant magnetic field and follow the path as shown in Figure.

Which particle is moving the fastest, and which is the slowest? What are the signs of charges? Justify your answer. 78

10. A thin, uniform rod which has a length of 0.40 m and a mass of 0.080 kg is shown in figüre below. This rod lies in the plane of the paper and is attached to the floor by a hinge. A uniform magnetic field of 0.30 T is directed perpendicularly into the plane of paper. There is a current of 4.0 A in the rod, which does not rotate clockwise or counterclockwise.

Find the value of cosθ. (Hint: The force maybe taken to act at the center of the gravity.) (g=10 m/s2)

11. A negatively charged particle moving at 4x10-2m/s has a mass of 1.6x10-29 kg and a charge of 1.6x10-19c. It enters a magnetic field of 2x10-2 T as shown. B=2x10-2 T

-q

V

(a) Draw the path of the particle. (b) What is the force acting on the particle in the field?

(c) What is the radius of the path?

79

Problems 12. This question is about the force between current-carrying wires. Diagram 1 below shows two long, parallel vertical wires each carrying equal currents in the same direction. The wires pass through a horizontal sheet of card. Diagram 2 shows a plan view of the wires looking down onto the card.

eye

sheet of card

diagram 1

(a)

(i)

diagram 2

Draw on diagram 1 the direction of the force acting on each wire. (1)

(ii)

Draw on diagram 2 the magnetic field pattern due to the currents in the wire. (3)

(b)

The card is removed and one of the two wires is free to move. Describe and explain the changes in the velocity and in acceleration of the moveable wire. (3)

80

13. (a) On the diagram below, draw the magnetic field pattern around a long straight currentcarrying conductor. (3) current-carrying wire

The diagram below shows a coil consisting of two loops of wire. The coil is suspended vertically.

0.20 cm

6.0 cm

Each loop has a diameter of 6.0 cm and the separation of the loops is 0.20 cm. The coil forms part of an electrical circuit so that a current may be passed through the coil. (b)

(i)

State and explain why, when the current is switched on in the coil, the distance between the two loops changes. (3)

When there is a current I in the coil, a mass of 0.10 g hung from the free end of the coil returns the separation of the loops to the original value of 0.20 cm. The circumference C of a circle of radius r is given by the expression C = 2πr. (ii) Calculate the current I in the coil. You may assume that each loop behaves as a long straight current-carrying wire. (5)

81

14.

This question is about magnetic fields. (a)

Using the diagram below, draw the magnetic field pattern of the Earth. (2)

North Earth

(b)

State what other object produces a magnetic field pattern similar to that of the Earth.

(c)

A long vertical wire passes through a sheet of cardboard that is held horizontal. A small compass is placed at the point P and the needle points in the direction shown.

cardboard sheet

direction of compass needle P

A current is passed through the wire and the compass needle now points in a direction that makes an angle of 30 to its original direction as shown below.

direction of compass needle with current in wire cardboard sheet

30 P

original direction of compass needle

82

(i)

Draw an arrow on the wire to show the direction of current in the wire. Explain why it is in the direction that you have drawn. (2)

(ii)

The magnetic field strength at point P due to the current in the wire is BW and the strength of the horizontal component of the Earth’s magnetic field is BE. Deduce, by drawing a suitable vector diagram, that BE = BW tan 600.

15.

(2)

This question is about motion of a charged particle in a magnetic field. A charged particle is projected from point X with speed v at right angles to a uniform magnetic field. The magnetic field is directed out of the plane of the page. The particle moves along a circle of radius R and centre C as shown in the diagram below. region of magnetic field out of plane of page

Y

v

R

(a)

C

X charged particle

On the diagram above, draw arrows to represent the magnetic force on the particle at position X and at position Y. (1)

(b)

State and explain whether (i)

the charge is positive or negative; (1)

(ii)

work is done by the magnetic force. (2)

83

(c)

A second identical charged particle is projected at position X with a speed

v in a 2

direction opposite to that of the first particle. On the diagram above, draw the path followed by this particle. (2)

16. This question is about forces on charged particles. (a)

(b)

A charged particle is situated in a field of force. Deduce the nature of the force-field (magnetic, electric or gravitational) when the force on the particle (i)

is along the direction of the field regardless of its charge and velocity;

(ii)

is independent of the velocity of the particle but depends on its charge;

(iii)

depends on the velocity of the particle and its charge.

An electron is accelerated from rest in a vacuum through a potential difference of 2.1kV. Deduce that the final speed of the electron is 2.7 × 107 m s–1.(3)

The electron in (b) then enters a region of uniform electric field between two conducting horizontal metal plates as shown below.

+95 V Path of electron

P 2.2 cm 7

2.7 × 10 m s

–1

0V 12 cm The electric field outside the region of the plates may be assumed to be zero. The potential difference between the plates is 95 V and their separation is 2.2 cm. As the electron enters the region of the electric field, it is travelling parallel to the plates.

84

(c)

(d)

(i)

On the diagram above, draw an arrow at P to show the direction of the force due to the electric field acting on the electron. (1)

(ii)

Calculate the force on the electron due to the electric field. (3)

The plates in the diagram above are of length 12 cm. Determine (i)

the time of flight between the plates. (1)

(ii)

the vertical distance moved by the electron during its passage between the plates (3)

(e)

Suggest why gravitational effects were not considered when calculating the deflection of the electron. (2)

(f)

In a mass spectrometer, electric and magnetic fields are used to select charged particles of one particular speed. A uniform magnetic field is applied in the region between the plates, such that the electron passes between the plates without being deviated. For this magnetic field,

(g)

(i)

state and explain its direction; (3)

(ii)

determine its magnitude. (2)

The electric and magnetic fields in (f) remain unchanged. Giving a brief explanation in each case, compare qualitatively the deflection of the electron in (f) with that of (i)

an electron travelling at a greater initial speed;

(ii)

a proton having the same speed;

(iii)

an alpha particle (α-particle) having the same speed. (7)

85

17.

The diagram above shows a double-caherged isotope that is projected into a vertical magnetic field of 0.28T, with the field directed upwards. The ion enters the field at a speed of 7.8x105ms-1. (a) State the initial direction of the magnetic force that acts on the ion.

(b) Describe the subsequent path of the ion as fully as you can. Your answer should include both qualitative description and an answer.

Mass of

ion=5.05x10-25kg.

(c) State the effect on the path in part (a) if the following changes are made separetely. (i) The strength of the magnetic field is doubled.

(ii) A singly-charged positive

ion replaces the original one.

86

18. The equation F=B.I.L, where the symbols have their usual meanings, gives the magnetic force that acts on a conductor in a magnetic field. (a) Give the unit of each quantities in the equation; F:__________________

B: ___________________

I:__________________

L: ___________________

(b) State the condition under which the equation applies.

(c) The diagram shows a horizontal copper bar of 25x25 mm square cross-section and length L carrying a current of 65A.

(i) Calculate the minimum value of the flux density of the magnetic field in which it should be placed if its weight is to be supported by the magnetic force that acts upon it. Density of copper = 8.9x103 kgm-3.

(ii) Draw an arrow on the diagram above to show the direction in which the magnetic field should be applied if your calculation in part (i) is to be valid. Label this arrow M.

87

MULTIPLE CHOICE ITEMS 1.

A strip of aluminium foil is held between the poles of a strong magnet, as shown below.

direction of current

magnet

aluminium foil When a current is passed through the aluminium foil in the direction shown, the foil is deflected. In which direction is this deflection?

2.

A.

Vertically downwards

B.

Vertically upwards

C.

Towards the North pole of the magnet

D.

Towards the South pole of the magnet

A current-carrying solenoid is placed with its axis pointing east-west as shown below. A small compass is situated near one end of the solenoid.

N W

E

axis of solenoid

S The axis of the needle of the compass is approximately 45° to the axis of the solenoid. The current in the solenoid is then doubled. Which of the following diagrams best shows the new position of the compass needle? A.

B. W

E

C.

W

E

W

E

D. W

E

88

3.

Two long, vertical wires X and Y carry currents in the same direction and pass through a horizontal sheet of card. X

Y

Iron filings are scattered on the card. Which one of the following diagrams best shows the pattern formed by the iron filings? (The dots show where the wires X and Y enter the card.)

4.

5.

A.

B.

C.

D.

A charged particle of mass m and charge q is travelling in a uniform magnetic field with speed v such that the magnetic force on the particle is F. The magnetic force on a particle of mass 2m, charge q and speed 2v travelling in the same direction in the magnetic field is A.

4F.

B.

2F.

C.

F.

D.

1 F. 2

A magnetic force acts on an electric charge in a magnetic field when A.

the charge is not moving.

B.

the charge moves in the direction of the magnetic field.

C.

the charge moves in the opposite direction to the magnetic field.

D.

the charge moves at right angles to the lines of the magnetic field. 89

6.

An electron is travelling in the direction shown and enters a region of uniform magnetic field. direction of travel of the electron

direction of magnetic field

e–

region of uniform magnetic field

On entering the field the direction of the force acting on the electron is

7.

A.

into the plane of the paper.

B.

out of the plane of the paper.

C.

towards the top of the page.

D.

towards the bottom of the page.

The currents in two parallel wires are I and 3I in the directions shown in the diagram below. wire 2 The magnetic force on wire 2 due to the current

wire 1

in wire 1 is F. The magnitude of the force on wire 1 due to the current in wire 2 is A.

F . 3

C. F. I

8.

B.

F . 2

D. 3F.

3I

The diagram below represents four long straight wires perpendicular to the plane of the paper.

A D

P

B

C The magnitude of the direct current in each wire is the same. Wires with have current into the plane of the paper and wires with have current out of the plane of the paper. Point P is the same distance from each wire. Which arrow shows the direction of the magnetic field at P? A.

A

B. B

C. C

D. D

90

9.

The diagram below shows a charged particle about to enter a region of uniform magnetic field directed into the page.

charged particle magnetic field

Which of the following correctly describes the change, if any, in the kinetic energy and the momentum of the particle in the magnetic field? Kinetic energy

10.

Momentum

A.

Changed

Changed

B.

Changed

Unchanged

C.

Unchanged

Changed

D.

Unchanged

Unchanged

The diagram below shows three parallel wires P, Q and R that are equally spaced.

I

I

wire P

wire Q

I

wire R

The currents in the wires are each of the same magnitude I and are in the directions shown. The resultant force on wire Q due to the current in wire P and in wire R is A.

perpendicular and into the plane of the paper.

B.

perpendicular and out of the plane of the paper.

C.

in the plane of the paper to the right.

D.

in the plane of the paper to the left.

91

11.

Two long, parallel, straight wires X and Y carry equal currents into the plane of the page as shown. The diagram shows arrows representing the magnetic field strength B at the position of each wire and the magnetic force F on each wire.

B Y (field strength at X due to Y)

X

F

F

Y

B X (field strength at Y due to X) The current in wire Y is doubled. Which diagram best represents the magnetic field strengths and forces? A.

BY X

F

2F

Y BX

B.

BY X

2F

2F

Y BX

C.

2B Y

X

2F

F

Y

2B X

D.

2B Y

X

2F

2F

Y BX

92

12.

An electron is moving in air at right angles to a uniform magnetic field. The diagram below shows the path of the electron. The electron is slowing down.

region of magnetic field

Which one of the following correctly gives the direction of motion of the electron and the direction of the magnetic field? Direction of motion

13.

Direction of magnetic field

A.

clockwise

into plane of paper

B.

clockwise

out of plane of paper

C.

anti-clockwise

into plane of paper

D.

anti-clockwise

out of plane of paper

A long, straight current-carrying wire is placed normal to the plane of the page. The current in the wire is into the plane of the page. Which of the following diagrams best represents the magnetic field around the wire? A.

B.

C.

D.

93

14.

The Earth’s magnetic field may be compared with that of a bar magnet. Which of the following diagrams correctly shows the orientation of the bar magnet in this model? A.

B.

geographical north pole

geographical north pole

N

S

N

C.

S

D.

geographical north pole

geographical north pole

S

N

N

S

15.

A straight conductor is in the plane of a uniform magnetic field as shown. current I magnetic field

The current in the conductor is I and the conductor is at an angle θ to the magnetic field. The force per unit length on the conductor due to the current in the magnetic field is P. Which is the correct expression for the magnitude of the magnetic field strength? A.

P sin  I

B.

P cos  I

C.

P I sin 

D.

P I cos  94

16.

A positively charged particle enters a region of uniform magnetic field. The direction of the particle’s velocity is parallel to the direction of the magnetic field as shown in the diagram below. region of uniform magnetic field charged particle

Which of the following diagrams correctly shows the path of the charged particle while in the region of magnetic field? A.

B.

C.

D.

17. A horizontal straight wire of length 40mm is in an eaast-west direction as shown in the diagram. A uniform magnetic field of 50mT is directed downwards into the plane of the diagram.

When a current of 5.0A passes through the wire from west to east, a horizontal force acts on the wire. Which line, A to D, in the table gives the magnitude and direction of this force? A. 2.0 mN towards north B. 10.0 mN towards north C. 2.0 mN towards south D. 10.0 mN towards south 95

18. A horizontal straight wire of length 0.30m carries a current of 0.2A perpendicular to a horizontal uniform magnetic field of flux density 5.0x10-2T. The wires ‘floats’ in equilibrium in the field.

What is the mass of wire? A. 8.0x10-4 kg

B. 8.0x10-3 kg

C. 3.0x10-4 kg

D. 3.0x10-3 kg

19. Which one of the following statements is correct? An electron follows a circular path when it is moving at right angles to A. a uniform magnetic field. B. a uniform electric field. C. Uniform electric and magnetic fields which are perpendicular. D. Uniform electric and magnetic fields which are in opposite directions.

20. The diagram shows a vertical square coil whose plane is at right angles to a horizontal uniform magnetic field B. A current, I, is passed through the coil, which is free to rotate about a vertical axis OO’.

Which one of the following statements is correct? A. The forces on the two vertical sides of the coil are equal and opposite. B. A couple acts on the coil. C. No forces act on the horizontal sides of the coil. D. The coil turns about OO’ axis. 96

21. The path followed by an electron of momentum p, carrying charge –e, which enters a magnetic field at right angles, is a circular arc of radius r. What would be the radius of the circular arc followed by an particle of momentum 2p, carrying charge +2e, which entered the same field at right angles? A. r/2 B. r C. 2r D. 4r 22. A beam of positive ions enters a region of uniform magnetic field, causing the beam to change direction as shown in the diagram.

What is the direction of the magnetic field? A. out of the page and perpendicular to it B. in to the page and perpendicular to it C. in the direction indicated by +y D. in the direction indicated by –y

97

MARKSCHEME

2. (a) outside 1. 1.5cm at right of wire 1

5. 3,2.10-4 cm 9. V3 ˃ V2 ˃ V1

12.

(a)

(i)

(b) 6A (c) 4x10-5 Tesla

6. i1: down I2: right Or i1: up I2: left 10.

3. (a) ccw (b) H = 2r

4. 0.48 T

7. (1) left (2) down

8. (1) in (2) out

(3) in (4) left (5) up

(3) out (4) in

11. (b) 1.28x10-22 N (c) 2x10-10m

3 5

They attract each other 1

(ii)

general shape: at least one circle around each wire and one loop around both wires; appropriate spacing of lines: increasing separation with distance from wires; correct direction of field; (b)

velocity increases; acceleration increases; because the force is getting larger the closer the wires get together;

3

3

Watch for ecf if force is drawn in wrong direction in (a) (i) ie velocity increases, acceleration decreases, force gets smaller. [7]

13.

(a)

concentric circles; separation increases (at least three circles required to see this); correct direction (anticlockwise); 3

(b)

(i)

(ii)

current in one turn produces magnetic field in region of the other turn; gives rise to a force on the wire; Newton’s third / idea of vice versa (gives rise to attraction) / idea of vice versa (gives rise to shortening); use of B =

0 I 2πr

3

(gives B = 1.0 × 10–4 I );

use of F = BIL; = 1.0 ×10–4 × I2 × 2π × 3.0 × 10–2; this force is equal to mg; hence I2 = 52.04, and I = 7.2 A;

5 98

[11] 14.

(a)

overall correct shape with no field lines touching; direction of field; (b)

bar magnet / solenoid; Do not accept just “magnet”.

(c)

(i)

2 1

upwards the direction of the compass needle is the resultant of two fields the field must be into the plane of the (exam) paper to produce a resultant field in the direction shown / OWTTE; Award [1] for “upwards because of the right-hand rule” / OWTTE.

2

(ii)

vector addition with correct values of two angles shown 300, 600 or 900; from diagrams, BE = BW ; tan 60 or BE =

BW ; tan 30

2 [7]

15.

(a)

(b)

(c)

(i)

two arrows directed towards the centre of the circular path, within 5.0 cm of the centre.

1

(i)

negative by stating any rule for the direction of the magnetic force;

1

(ii)

the work done is zero; since the force is at all times normal to the velocity;

2

a curved path starting at X and in the right direction ie counterclockwise; circular path of radius R ;

2

2

Allow diameter 3-4 cm and be generous with how round the circle is. [6] 16.

(a)

(i)

if independent of charge; must be gravitational;

(ii)

depends on charge so electric or magnetic; independent of velocity so electric;

(iii)

(depends on velocity and charge, so) magnetic;

5 max

99

(b)

idea of change in electric potential energy = gain in kinetic energy; qV = ½ mv2; (1.6 × 10–19 × 2.1 × 103) = (½ × 9.1 × 10–31 ×v2) / v = 7.38 × 1014;

3

Award [0] for v = 2.7 × 107 m s–1. (c)

(d)

(i)

correct force direction (upwards);

(ii)

force =

(i)

time to cross plates =

(ii)

(6.9  10 −16 ) vertical acceleration = (= 7.58 × 1014 m s–2); −31 (9.1  10 )

1

qV (1.6  10 −19  95) ;= = 6.9 × 10–16 N; −2 d (2.2  10 )

3

(12  10 −2 ) = 4.44 × 10–9 s; ( 2.7  10 7 )

1

distance = ½ × a × t2; = ½ × 7.58 × 1014 × (4.44 × 10–9)2 allow ecf; = 7.5 × 10–3 m; (e) (f)

gravitational force is very small; small in comparison with electric or magnetic force; (i)

(ii)

force due to B-field must be downwards; mention of Fleming’s left-hand rule / right-hand palm rule; hence field into paper;

2

3

Bqv = 6.9 × 10–16 N allow ecf B=

(g)

3

(6.9  10 −16 ) ; = 1.6 × 10–4 T; (1.6  10 −19  2.7  10 7 )

(i)

electric force unchanged; magnetic force is greater; hence deflection downwards;

(ii)

both forces reversed in direction; but not changed in magnitude; hence undeflected;

(iii)

undeflected;

2 max

7 max [30]

17.

100

18.

Multiple Choice Answers

1. B

12. D

2. B

13. A

3. A

14. D

4. B

15. A

5. D

16. C

6. B

17. B

7. C

18. C

8. C

19. A

9. D

20. A

10. B

21. B

11. D

22. A

101

ELECTROMAGNETIC INDUCTION

102

Problems 1. The magnetic flux through a coil of 50 turns is increasing at rate of 0.5 wb/s. What is the emf induced in the coil)

2. A rectangular loop with sides of 10 x20 cm is in uniform magnetic field of 0.04 T at right angle. Loop is rotated by 900 in 0.1 second as in figure. Find the induced emf in the loop

3.

It should in principle be possible to generate useful electric power by moving a conductor through the Earth’s magnetic field. In 1996 and 1992 there were (unsuccessful) attempts to do this using a satellite tethered to a space shuttle. The system consists of the satellite connected to the shuttle by a conducting cable which is insulated from the ionized gas through which it moves. (The attempts failed because the cable tangled and broke.)

(a) Suppose a cable of length l moves with a speed v perpendicular to a magnetic field B. By considering the area swept out by the wire in a time ∆t, write down an expression in terms of B, l and v, for the magnetic flux ∆Φ cut by the cable in time ∆t. 103

(b) Hence write an expression for the emf E induced across the ends of this cable.

(c) It was expected that the tethered cable would generate a maximum of 5000 V across its ends. If the length of the cable was 20.7 km, and the component of the Earth’s magnetic flux density perpendicular to it was 3.2 × 10–5T, calculate how fast the tether must have been moving.

4. (a) State Lenz’s law of electromagnetic induction.

(b) A bar magnet is dropped from rest through the center of a coil of wire which is connected to a resistor and data logger.

State the induced magnetic polarity on the top side of the coil as the magnet falls towards it.

104

(c) Add an arrow to the wire to show the direction of the induced current as the magnet falls towards the coil. (d) The graph shows the variation of induced current in the resistor with time as the magnet falls

Explain why the magnitude of I2 is greater than I1.

5.

Electromagnetic induction (a)

A long solenoid is connected in series with a battery and a switch S. Several loops of wire are wrapped around the solenoid close to its midpoint as shown below.

The ends of the wire are connected to a high resistance voltmeter V that has a centre zero scale (as shown in the inset diagram). The switch S is closed and it is observed that the needle on V moves to the right and then drops back to zero. Describe and explain, the deflection on the voltmeter when the switch S is re-opened. Description:

........................................................................................................... ...........................................................................................................

Explanation:

........................................................................................................... ........................................................................................................... 105

6. (a) Figure 1shows two coils, P and Q, linked by an iron bar. Coil P is connected to a battery through a variable resistor and a switch S. Coil Qis connected to a center-zero ammeter.

(i) Initially the variable resistor is set to its minimum resistance and S is open. Describe and explain what is observed on the ammeter when S is closed.

(ii) With S still closed, the resistance of the variable resistor is suddenly increased. Compare what is now observed on the ammeter with what was observed in part (i). Explain why this differs from what was observed in part (i).

(b) Figure 2shows a 40-turn coil of cross-sectional area 3.6 × 10–3m2 with its plane set at right angles to a uniform magnetic field of flux density 0.42 T.

(i)

Calculate the magnitude of the magnetic flux linkage for the coil. State an appropriate unit for your answer.

106

(ii)

7.

The coil is rotated through 90° in a time of 0.50 s. Determine the mean emf in the coil.

Electromagnetic induction A small circular coil of area of cross-section 1.7 x 10–4 m2 contains 250 turns of wire. The plane of the coil is placed parallel to, and a distance x from, the pole-piece of a magnet, as shown below.

coil P pole–piece of magnet

Q x

PQ is a line that is normal to the pole-piece. The variation with distance x along line PQ of the mean magnetic field strength B in the coil is shown below. 4.0

3.0 B / 10–2 T 2.0

1.0 5

10

15

x / cm

(a)

For the coil situated a distance 6.0 cm from the pole-piece of the magnet, (i)

state the average magnetic field strength in the coil;

(ii)

calculate the flux linkage through the coil.

107

(b)

(c)

The coil is moved along PQ so that the distance x changes from 6.0 cm to 12.0 cm in a time of 0.35 s. (i)

Deduce that the change in magnetic flux linkage through the coil is approximately 7.10–4 Wb.

(ii)

State Faraday’s law of electromagnetic induction and hence calculate the mean emf induced in the coil.

(i) State Lenz’s law

(ii) use Lenz’s law to explain why work has to be done to move the coil along the line PQ.

8.

Electrical conduction and induced currents (a)

A copper rod is placed on two parallel, horizontal conducting rails PQ and SR as shown below. B

B

copper rod

B

P conducting wire S

Q F R

The rails and the copper rod are in a region of uniform magnetic field of strength B. The magnetic field is normal to the plane of the conducting rods as shown in the diagram above. A conducting wire is connected between the ends P and S of the rails. A constant force, parallel to the rails, of magnitude F is applied to the copper rod in the direction shown. The copper rod moves along the rails with a decreasing acceleration. (i)

On the diagram, draw an arrow to show the direction of induced current in the copper rod. Label this arrow with the letter I. 108

(b)

(ii)

Explain, by reference to Lenz’s law, why the induced current is in the direction you have shown in (i).

(iii)

By considering the forces on the conduction electrons in the copper rod, explain why the acceleration of the copper rod decreases as it moves along the rails.

The copper rod in (b) eventually moves with constant speed v. The induced emf ɛ in the copper rod is given by the expression ɛ = Bvl where l is the length of copper rod in the region of uniform magnetic field. (i)

State Faraday’s law of electromagnetic induction.

(ii)

Deduce that the expression is consistent with Faraday’s law.

(iii)

The following data are available: F = 0.32 N l = 0.40 m B = 0.26 T resistance of copper rod = 0.15 Ω

Determine the induced current and the speed v of the copper rod. Induced current: ........................................................................................... ........................................................................................... Speed v:

........................................................................................... ........................................................................................... 109

MULTIPLE CHOICE ITEMS 1.

A magnetic field links a closed loop of metal wire. The magnetic field strength B varies with time t as shown.

0

t2

t1

B

t3

0

t

A current is induced in the loop during the time period

2.

A.

t1 only.

B.

t2 only.

C.

t2 and t3 only.

D.

t1 and t3 only.

The magnetic flux Φ through a coil having 500 turns varies with time t as shown below. 2.5

2.0

1.5 / 10–3 Wb 1.0

0.5

0.0

0

1

3 2 –3 t / 10 s

4

5

The magnitude of the emf induced in the coil is A. 0.25 V.

B. 0.50 V.

C. 250 V.

D. 1 000 V.

110

3.

A uniform magnetic field of strength B completely links a coil of area S. The field makes an angle Ø to the plane of the coil.

B

area S The magnetic flux linking the coil is

4.

A.

BS.

B.

BS cos Φ.

C.

BS sin Φ.

D.

BS tan Φ.

The north pole of a permanent bar magnet is pushed along the axis of a coil as shown below.

axis of coil

N

S

V

The pointer of the sensitive voltmeter connected to the coil moves to the right and gives a maximum reading of 8 units. The experiment is repeated but, on this occasion, the south pole of the magnet enters the coil at twice the previous speed. Which of the following gives the maximum deflection of the pointer of the voltmeter? A.

8 units to the right

B.

8 units to the left

C.

16 units to the right

D.

16 units to the left

111

5.

Two coils P and Q are arranged as shown below.

coil Q

V

coil P

Coil Q is connected to a sensitive voltmeter. The current I in coil P is varied as shown below. I

0

time

0

Which of the following graphs best shows the variation with time of the emf E induced in coil Q? A.

00

C.

B.

E

E

D.

E

00

00

time

time

time

E

00

time

112

6.

The diagram below shows two concentric loops lying in the same plane.

inner loop

outer loop

The current in the inner loop is clockwise and increases with time as shown in the graph below. current

0

0

time

The induced current in the outer loop is

7.

A.

constant in the clockwise direction.

B.

constant in the anticlockwise direction.

C.

variable in the clockwise direction.

D.

variable in the anticlockwise direction.

A thin copper ring encloses an area S. The area is linked by magnetic flux that is increasing. The rate of change of the magnetic flux from time t = 0 to time t = T is R. The emf induced in the copper ring during the time t = 0 to time t = T is A.

R.

B.

RS.

C.

RST.

D.

RS . T

113

8.

An electron is moving in air at right angles to a uniform magnetic field. The diagram below shows the path of the electron. The electron is slowing down.

Which of the following correctly gives the direction of motion of the electron and the direction of the magnetic field? Direction of motion A.

clockwise

into plane of paper

B.

clockwise

out of plane of paper

C.

anti-clockwise

into plane of paper

D.

anti-clockwise

out of plane of paper

A conductor in the shape of a solid square is moving with constant velocity in a region of magnetic field as shown in the diagram below. velocity

magnetic field into page conductor

+ + +

– – –

– – –

+ + +

–– –

+ + +

The direction of the field is into the plane of the page. Which of the following diagrams correctly represents the separation of the induced charges? B. D. A. C.

+ + +

9.

Direction of magnetic field

–– –

114

10.

A metal ring has its plane perpendicular to a magnetic field. metal ring

magnetic field

The magnetic flux through the ring increases at a constant rate by 4.0×10–5Wb in 5.0 s. During this change the e.m.f. induced in the ring

11.

A.

remains constant at 8µV.

B.

remains constant at 20µV.

C.

increases from zero to 8µV.

D.

increases from zero to 20µV.

The graph shows the variation with time t of the magnetic flux φ through a coil that is rotating in a uniform magnetic field.

The magnitude of the emf induced across the ends of the coil is maximum at time(s) A.

t1 and t3.

B.

t2 and t4.

C.

t3 only.

D.

t4 only.

115

12.

A copper sheet is suspended in a region of uniform magnetic field by an insulating wire connected to a horizontal support. The sheet is pulled to one side so that it is outside the region of the field, and then released.

The uniform magnetic field is directed into the plane of the paper. Which of the following is true for both the direction of the induced current in the sheet and the change in amplitude of the oscillations of the sheet with time? Direction of induced current

13.

Change in amplitude

A.

stays the same

no change

B.

changes

decreases

C.

stays the same

decreases

D.

changes

no change

A permanent bar magnet is moved towards a coil of conducting wire wrapped around a nonconducting cylinder. The ends of the coil, P and Q are joined by a straight piece of wire.

The induced current in the straight piece of wire is A.

alternating.

B.

zero.

C.

from P to Q.

D.

from Q to P.

116

14.

The magnetic flux Ø in a coil varies with time t as shown below.

t Which graph best represents the variation with time t of the emf E induced in the coil? A.

B. E

0

E

0

t

C.

t

D. E

0

E

t

0

t

117

15.

Why does the ammeter briefly show a non-zero reading? A.

The magnetic flux linkage in the coil increases then decreases.

B.

The magnetic flux linkage in the coil increases then becomes constant.

C.

The magnetic flux linkage in the coil decreases then increases.

D.

The magnetic flux linkage in the coil decreases then becomes constant.

16. The graph shows how the magnetic flux passing through a loop of wire changes with time.

What feature of the graph represents the magnitude of the emf induced in the coil? A.

The area enclosed between the graph line and the time axis.

B.

The area enclosed between the graph line and the magnetic flux axis.

C.

The inverse of the gradient of the graph.

D.

The gradient of the graph

118

17.

A magnetic field of strength B links a coil. The direction of the field is normal to the plane of the coil. The graph shows how B varies with time t.

Which of the following graphs shows how the induced emf ε in the coil varies with t?

119

MARKSCHEME 1. 25V 2. 8 × 10−3 V

3.

4.

5.

Electromagnetic induction (a)

description: on opening the switch, the reading on the voltmeter will deflect to the left and then drop to zero; explanation: when the switch is opened the field drops to zero so again a time changing flux; which will induce an emf in the opposite direction as the emf will now be such as to oppose the field falling to zero / Lenz’s law; when the current reaches zero, there will no longer be a flux change;

4 [5] 120

6.

7.

Electromagnetic induction (a)

(i)

3.3 x 10-2 T;

(ii)

flux linkage = 3.3 x 10-2 x 1.7 x 10-4 x 250;

1

= 1.4x 10-3Wb (turns);

2

Award [0] if answer given as flux in (a)(ii) but allow full credit in (b)(i).

121

(b)

(i)

new flux linkage = 7.23 x 10-4 Wb turns or B = 1.6 x 10-2 T; change = (1.4 - 0.7) x 10-3 or change = 1.6 x 10-2 x 1.7 x 10-4 x 250;

2

change = 7 x 10-4Wb turns (no mark for answer) (ii)

e.m.f. is proportional/equal to rate of change of flux  (do not allow  (linkage); "induced current")

(7 10 ) = 2 10 −4

emf = (c)

(i)

(ii)

0.35

−3

2

V;

emf / induced current acts in such a direction to (produce effects to) oppose the change causing it;

1

induced current produces a magnetic field in the coil / induced current is in field of magnet; this produces a force; (award only if the first marking point is correct) the force acts to oppose the motion of the coil;

3 [11]

8.

Electrical conduction and induced currents (a)

(i)

1 (ii)

(iii)

(b)

Lenz’s law says that the direction of the induced current is such as to oppose change; therefore, to produce a (magnetic) force that opposes F the current must be in direction shown / reference to left / right hand rule / OWTTE;

2

the force on the electrons is given by Bev; as v increases so does this force and therefore, so does the induced current; therefore, net force on rod decreases / OWTTE;

3

the induced emf is equal / proportional to the rate of change / cutting of (magnetic) flux; if the rod moves a distance x in time t then area swept out by rod = lx; flux = BlΔx; rate of change of flux =

Blx = Blv = ε ; t

4

122

(iii)

induced current:

I=

F ; substitute to give I = 3.1A; Bl

speed v: Ɛ = IR = 0.47; Ɛ = Bvl substitute to give v = 4.5 (4.4) ms-1;

4 [17]

Multiple Choice Answers 1.

D

2.

C

3.

C

4.

D

5.

D

6.

B

7.

A

8.

D

9.

A

10.

A

11.

B

12.

B

13.

C

14.

C

15.

B

16.

D

17.

D

123

ALTERNATING CURRENT (AC)

124

Problems: 1.

The diagram below shows the variation with time t of the current in a wire.

current

00

t

magnetic flux

00

t

00

t

e.m.f

(a)

Draw, on the axes provided, a sketch-graph to show the variation with time t of the magnetic flux in the coil.

(b) Construct, on the axes provided, a sketch-graph to show the variation with time t of the emf induced in the coil.

125

2.

This question is about alternating current. (a)

The graph shows the variation with time t of the output voltage V of an ac generator of negligible internal resistance.

A resistor of resistance 25 Ω is connected across the output of the generator. Calculate

(b)

(i)

the rms value of the current in the resistor. [2]

(ii)

the average power dissipated in the resistor. [1]

(iii)

the power dissipated in the resistor at 0.40 ms. [2]

The frequency of rotation of the generator coil is now doubled. Sketch, using the axes in (a), the variation with t of the new output voltage V. [2]

126

3.

The graph below shows how a sinusoidal alternating voltage varies with time when connected across a resistor, R.

(a) (i) State the peak voltage. [1]

(ii) State the rms voltage. [1]

(iii) Calculate the frequency of the alternating voltage. State an appropriate unit. [1]

(iv) Write down the equation of voltage that varies in the circuit. [1]

(b) On the graph above draw a line to show the dc voltage that gives the same rate of energy dissipation in R as produced by the alternating waveform. [2]

127

4. Domestic users in the United Kingdom are supplied with mains electricity at a root mean square voltage of 230V. (a)

State what is meant by root mean square voltage. [1]

(b)

(i)

Calculate the peak value of the supply voltage. [2]

(ii)

Calculate the average power dissipated in a lamp connected to the mains supply

when the rms current is 0.26 A. [1]

(d) The frequency of the voltage supply is 50 Hz. On the axes below draw the waveform of the supplied voltage labelling the axes with appropriate values. (e)

128

5. This question is about an ideal transformer. (a)

State Faraday’s law of electromagnetic induction. [2]

(b)

The diagram below shows an ideal transformer.

laminated core

primary coil

secondary coil

(i)

Use Faraday’s law to explain why, for normal operation of the transformer, the current in the primary coil must vary continuously. [2]

(ii)

Outline why the core is laminated. [2]

(iii)

The primary coil of an ideal transformer is connected to an alternating supply rated at 230V. The transformer is designed to provide power for a lamp rated as 12V, 42W and has 450 turns of wire on its secondary coil. Determine the number of turns of wire on the primary coil and the current from the supply for the lamp to operate at normal brightness. [3]

129

6. A hydroelectric power station has a power output of 2.0 MW when water passes through its turbines at a rate of 1.4 m3s–1.The water is supplied from a reservoir which is 750 m above the power station turbines, as shown in the diagram below. (density of water = 1000 kg m–3)

(a) Calculate (i) the mass of water passing through the turbines each second,

(ii) the loss of potential energy per second of the water flowing between the reservoir and the power station turbines,

(iii) the efficiency of the power station.

(b) The turbines drive generators that produce alternating current at an rms potential difference of 25 kV which is then stepped up to an rms potential difference of 275 kV by means of a transformer. (i) Calculate the rms current supplied by the generators to the transformer when the power output of the generators is 2.0 MW.

(ii) The transformer has an efficiency of 95%. Calculate the output current of the transformer.

130

7. (a) Explain what is meant by the term magnetic flux linkage. State its unit.

(b) Explain, in terms of electromagnetic induction, how a transformer may be used to step down voltage.

(c) A minidisc player is provided with a mains adapter. The adapter uses a transformer with a turn’s ratio of 15:1 to step down the mains voltage from 230 V. (i) Calculate the output voltage of the transformer.

(ii) State two reasons why the transformer may be less than 100% efficient.

131

MULTIPLE CHOICE ITEMS 1.

A sinusoidal ac power supply has rms voltage V and supplies rms current I. What is the maximum instantaneous power delivered? A.

2.

B.

2 VI

C.

VI

D.

VI 2

The rms current rating of an electric heater is 4 A. What direct current would produce the same power dissipation in the electric heater? A.

3.

2VI

4

A

2

B.

4A

C.

4 2A

D.

8A

An alternating current supply of negligible internal resistance is connected to two resistors that are in parallel.

The resistance of each resistor is R and the peak voltage of the ac supply is V0. Which of the following is the average power dissipated in the circuit? A.

2V0 R

B.

V0 R

2

2

C.

V0

2

2R 2

D.

V0 2R

132

4.

A rectangular loop of conducting wire rotates in a region of magnetic field. The graph shows the variation with time t of the induced emf in the loop during one cycle.

The resistance of the coil is 5.0 Ω. Which of the following is the average power dissipated in the loop? A.

B.

5.

6.

45 W 2

45 2

W

C.

45 W

D.

45 2 W

In order to reduce power losses in the transmission lines between a power station and a factory, two transformers are used. One is located at the power station and the other at the factory. Which of the following gives the correct types of transformer used? Power station

Factory

A.

step-up

step-up

B.

step-up

step-down

C.

step-down

step-up

D.

step-down

step-down

The maximum output voltage of a generator is V0. The frequency of rotation of the generator coil is doubled. What is the new maximum output voltage? A.

V0

B.

2 V0

C.

2 V0

D.

4 V0 133

7.

Raoul suggests that power losses in a transformer may be reduced by the following. I.

Constructing the core from a solid block of steel.

II.

Using large diameter wire in the coils.

III.

Using wire of low resistivity.

Which of the above suggestions would reduce power loss?

8.

A.

I only

B.

II only

C.

II and III only

D.

I, II and III

The variation with time t of the magnetic flux Φ through a coil is shown below.

0

0

t

Which of the following diagrams best shows the variation with time t of the emf E induced in the coil? A.

0

C.

B.

E

0

0

0

t

D.

E

0

t

E

0

t

0

t

E 0

(1)

134

9.

When a coil is rotated in a uniform magnetic field at a certain frequency, the variation with time t of the induced emf E is as shown below. E

0

0

t

The frequency of rotation of the coil is reduced to one half of its initial value. Which one of the following graphs correctly shows the new variation with time t of the induced emf E? A.

0

C.

B.

E

0

0

0

t

D.

E

0

t

E

0

t

0

t

E

0

135

10.

The maximum value of a sinusoidal alternating current in a resistor of resistance R is I0. The maximum current is increased to 2I0. Assuming that the resistance of the resistor remains constant, the average power dissipated in the resistor is now

11.

1 2

B.

I 0 R.

C.

2 I 0 R.

D.

4 I 0 R.

I 0 R. 2

2

2

A light bulb is connected to an ac supply. The variation with time of the current is sinusoidal having a maximum value of 0.50 A. The rms. current is

0.50 A. 2

A.

0.50

B.

12.

2

A.

A.

2

C.

0.50 A.

D.

0.50 2 A.

The graph below shows the variation with time t of the current I in a resistor. +I0 I

0

0

t

–I0

Which of the following is the root-mean-square value of the current I? A.

2I 0

B.

I0

C.

I0

D.

I0 2 136

13.

The diagram below shows the variation with time t of the emf E generated in a coil rotating in a uniform magnetic field. E e

0 0

T 2

T

3T 2

t

–e

What is the root-mean-square value Erms of the emf and also the frequency f of rotation of the coil? Erms

f

A.

e

2 T

B.

e

1 T

C.

D.

14.

e 2 e 2

2 T 1 T

An alternating supply of constant r.m.s. current and constant r.m.s. potential difference is connected to the primary coil of an ideal transformer. Which one of the following describes the effect, if any, on the r.m.s. current and on the r.m.s. power in the circuit of the secondary coil when the number of turns on the secondary coil is increased? r.m.s. current

r.m.s. power

A.

no change

increases

B.

no change

no change

C.

decreases

increases

D.

decreases

no change

137

15.

A resistor of resistance R is connected in series with a sinusoidal alternating supply having a maximum value of emf V0. The best estimate for the average power dissipated in the resistor during one cycle of the alternating current is 2

A.

2V0 . R 2

C.

V0 . R

2

B.

V0

2

2R

.

2

D.

16.

V0 . 2R

The rms voltages across the primary and secondary coils in an ideal transformer are Vp and Vs respectively. The currents in the primary and secondary coils are Ip and Is respectively. Which one of the following statements is always true?

17.

A.

Vs = Vp

B.

Is = Ip

C.

VsIs = VpIp

D.

Vs I s = . Vp I p

An ideal transformer has Np turns on the primary coil and Ns turns on the secondary coil. The input power of the primary coil is P. The output power at the secondary coil is A.

P.

B.

 Np  N  s

  P.  

C.

 Ns   Np 

  P.  

D.

 Ns 1 −  Np 

  P.   138

18.

19.

20.

139

MARKSCHEME

1.

(a)

sinusoidal and in phase with current;

(b)

sinusoidal and same frequency; with 90° phase difference to candidate’s graph for (a);

2.

(a)

(i)

100 = 4.0(A); 25 Imax = 2.8 A; I0 =

2

or Vmax = Imax =

100 2

= 70.7 V;

V rms 70.7 = = 2.8A; R 25

(ii)

200 W;

(iii)

V = 60(±2.0) V; V 2   60 2  P=  =  = 140(±10) W;  R   25 

1

2

(b)

peak voltage of 200 V; (allow 180 to 200 V) period = 0.50 ms;

2 [7]

140

3.(a) (i) 64 V (ii) Vrms= 64 / √2 =45.3 V (iii) frequency = 1 / 0.01 = 100Hz (iv) V= 64. Sin(200πt) (b) horizontal line through y = 45 (44 − 48) x =0 CE from (a)(iii) + / - half square straight line must extend to at least to 6.0 ms 4. (a) the square root of the mean of the squares of all the values of the voltage in one cycle (1) or the equivalent dc/steady/constant voltage that produces the same heating effect/power (1) (b) (i) peak voltage = 230 × √ 2 (1)

peak voltage = 325 V (or 324 V) (1)

(ii) average power = 230 × 0.26 = 60 W (1)

(c)

5.

(a)

(b)

e.m.f. induced proportional to/equal to; rate of change of flux (linkage) / rate of flux cutting; (i)

(ii)

(iii)

2

for e.m.f./current to be induced in secondary, flux must be changing in the core; changing flux is caused by varying current in primary;

2

induced currents in core are kept small; (do not allow reduced/ prevented) to reduce heating/energy losses; (do not allow mere “eddy current losses”)

2

use of

N S VS = ; to give NP = 8600 turns; N P VP

 42   = 180 mA; 3 and I P  =  230  141

6.

7. (a) product of flux and number of turns / Wb or equivalent (b) changing primary magnetic field due to alternating voltage (applied to primary) varying flux links with secondary induced emf ∑ rate of change of flux linkage NS NP so less voltage on secondary (c) (i) equation or correct substitution : 15.3 V (ii) not just “heating” or “heat loss”

Key for multiple choice questions: 1.

A

6.

C

11.

B

16.

C

2.

B

7.

C

12.

B

17.

A

3.

B

8.

A

13.

D

18. B

4.

A

9.

A

14.

D

19. B

5.

B

10.

C

15.

D

20. A

142