Chapter 10 Electromagnetic Induction

Worked solutions Chapter 10 Electromagnetic induction 10.1 Magnetic flux and induced currents 1

The flux is the field strength multiplied by the area of the loop, which is perpendicular to the field. In the initial set-up ( = 0°), this is given by: = BA = 2 × 10–3 × (4 × 10–2)2 = 3.2 × 10–6 Wb At other angles we multiply this result by cos to obtain: = 45°, = 2.3 × 10–6 Wb; = 60°, = 1.6 × 10-6 Wb; = 90°, = 0.

2

For For For For

3

a b c d

4

a b c d

5

There must be a non-zero rate of change of magnetic flux.

6

When S is closed, the change in flux produced by X will induce a momentary opposite (negative) current in Y as this current attempts to oppose the increasing flux. As S is held, there is no change of flux and so the current in Y will drop to zero. When S is opened, there will be a momentary positive current in Y as it attempts to oppose the loss of flux.

7

As the resistance is increased the current, and therefore the flux, decreases. This will result in a positive current in Y as it attempts to maintain the flux. As the resistance is decreased the opposite occurs, giving a negative current in Y.

8

a b

= BA = 2.0 × 10–3 × × 0.04² = 1.0 × 10–5 Wb Zero as there is no change in flux in the coil.

9

a b c

= 1.0 × 10–5 Wb / t = 1.0 × 10–5/1.0 × 10–3 = 0.01 Wb s–1 The same current (4.0 mA) will flow but in the opposite direction, i.e. from Y to X.

= 0° to 45°, the change of flux = 0° to 60°, the change of flux = 45° to 90°, the change of flux = 0° to 90°, the change of flux

= (3.2 – 2.3) × 10–6 = 0.9 × 10–6 Wb = (3.2 – 1.6) × 10–6 = 1.6 × 10–6 Wb = (2.3 – 0) × 10–6 = 2.3 × 10–6 Wb = (3.2 – 0) × 10–6 = 3.2 × 10–6 Wb

= (3.2 – 0) × 10–6 = 3.2 × 10–6 Wb = [3.2 – (–3.2 )] × 10–6 = 6.4 × 10–6 Wb = (3.2 – 6.4) × 10–6 = 3.2 × 10–6 Wb = (3.2 – 1.6) × 10–6 = 1.6 × 10–6 Wb No change of flux and hence no current. Opposite to the initial situation and so a negative current. Equivalent to the initial situation and so a positive current. Equivalent to situation in part b and so a negative current.

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10

a b c

Chapter 10 Electromagnetic induction

This change is equivalent to turning off the current but at twice the rate. This will result in twice the current, 8.0 mA, from X to Y. This coil will only have one-quarter of the area, and hence flux. The induced current will be 1.0 mA from X to Y. The current is proportional to the rate of change of flux. Therefore it will be half the previous current, 2.0 mA from X to Y.

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Chapter 10 Electromagnetic induction

10.2 Induced EMF: Faraday’s law 1

a b c d

2

a b c

3

a b

= BA = 2 × 10–3 × 6 × 10–4 = 1.2 × 10–6 Wb As the plane of the loop is in the same direction as the field there is no flux through the loop. 1.2 " 10!6 EMF = / t and thus EMF = = 3.0 × 10–5 V B 0.04 3 " 10!5 I = V/R = = 2 × 10–5 A = 2.0 × 10–5 A 1.5 = BA = 80 × 10–3 × 10 × 10–4 = 8 × 10–5 Wb 8 " 10!5 EMF = = 4.0 × 10–3 V B/ t and thus EMF = 0.02 In the 500 turns the total EMF will be 2000 mV = 2.0 V. = BA = 5 × 10–3 × (250 – 50) × 10–4 = 1.0 ! 10–4 Wb 30 " 1 " 10!4 EMF = N = 6.0 × 10–3 V B/ t and thus EMF = 0.5 = BA = 5 × 10–5 × × 4² = 2.5 × 10–3 Wb –3 EMF = B/ t and thus EMF = 2.5 × 10 × 8 = 0.02 V

4

a b

5

This problem could be done by finding the actual EMFs, but it is simpler to look at the ratios: EMF = N BX/ BY × AX/AY × tY/tX. Here B/ t = N BA/t and thus EMFX/EMFY = N X/NY × the field and the time are the same for each coil and so EMFX/EMFY = NX/NY × AX/AY = 20/10 × (4/8)² =

1 (1 : 2). Note that the area ratio is equal to the radius ratio squared. 2

6

D is the correct answer. As the loop enters the field there will be a current one way (clockwise). Next there will be a short period in which the flux does not change; hence no EMF. Then as the loop exits the field, the induced EMF will be in the opposite direction.

7

The EMF induced in a straight wire is EMF = vBl, where l = 4 cm in this case. Thus 5 × 10–3 = 2.5 × B × 4 × 10–2 and so B = 5.0 × 10–2 T.

8

a b c

9

In each case EMF = N B × 2 ! 10–3/t = 0.2 B/ t = N BA/t = 100 × a EMF = 0 as there is no change of flux b EMF = 0.2 B/t = 0.2 × 1 × 10–3/0.1 = 2.0 × 10–3 V c EMF = 0 as there is no change of flux d EMF = 0.2 B/t = 0.2 × 1 × 10–3/0.2 = 1.0 × 10–3 V

10

The period for one revolution is 60/33 = 1.82 s. We can picture a radius vector cutting the Earth’s flux at the rate of the area of the record in one turn, and thus inducing an EMF along the radius vector given by:

There is no change of flux and hence the current is zero. –4 –3 EMF = N B/ t and thus EMF = 10 × 0.1 × 8 × 10 /0.10 = 8.0 × 10 V By pulling the loop from between the poles of the magnet with a greater speed B/t

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Chapter 10 Electromagnetic induction

–5 EMF = × 0.26²/1.82 = 5.8 × 10–6 V B/ t = BA/t = 5 × 10 × (It’s not going to give him a shock!) Between opposite edges there will be no EMF as the diameter vector is turning in opposite directions on either side of the centre.

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Chapter 10 Electromagnetic induction

10.3 Direction of EMF: Lenz’s law 1

C. It is virtually a statement of Lenz’s law.

2

a b c

Out of the page in order to try to maintain the original flux Into the page in order to oppose the extra flux outwards Out of the page in order to try to maintain the original flux

3

a

The induced current must produce an upward flux in order to oppose the increasing downward flux. The right-hand screw rule tells us this current must be anticlockwise. The induced current will be the other direction (clockwise) as this is the opposite situation to that in part – or a similar argument to that in part a could be used. As the centre of the magnet passes through the ring there is no flux change for a moment. The strength of the magnet (B), the area of the ring (A), the speed of the magnet ( t) and the resistance of the ring and galvanometer circuit

b c d

4

Part a, the induced current will attempt to oppose the change of flux. As the coils are oriented the same way, this means that the induced current is in the same direction as that in coil X if the current (and hence flux) decreases and in the opposite direction if the current increases. Hence, in parts a and b the current is to the right (positive), while in part c it is to the left (negative).

5

a b c

6

a

b

We have a steady positive current induced in Q. This implies that the current in P must be either decreasing positive or increasing negative as Q attempts to maintain the original flux. This is shown in graph E. This situation is the opposite to part a, so the current must be increasing. This is shown in graph D. Zero current is produced when there is no change of flux, which will be the case in graphs A, B and C. The induced current will be from X to Y as the force on positive charges moving to the right will be downwards (from the right-hand palm rule). Another way to determine the current is to realise that the current must be such as to experience a force in the opposite direction to the motion. Left, as explained above

7

a b

The current will reverse as well – that is, from Y to X. The force must still oppose the motion – that is, it will be to the left. This is because both the field and current have reversed.

8

A. If this was not the case we would get perpetual motion!

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Chapter 10 Electromagnetic induction

10.4 Electric power generation 1

a b c d e f g

= BA cos , for 0.97 × 10–2 mWb 0.87 × 10–2 mWb 0.71 × 10–2 mWb 0.50 × 10–2 mWb 0.26 × 10–2 mWb 0 mWb

2

a

/

b c d e f g

= 0,

= 5 × 10–3 × 20 × 10–4 × 1 = 1 × 10–5 Wb = 1.0 × 10–2 mWb

(1 ! 0.97) " 10!2 = 0.3 mWb s–1 !3 10 1.0 mWb s–1 1.6 mWb s–1 2.1 mWb s–1 2.4 mWb s–1 2.6 mWb s–1 0 mWb s–1 t=

3

The rate of change of flux increases as the coil is rotated from 0° to 90°.

4

EMF = N B/ t a 0.03 V b 0.10 V c 0.16 V d 0.21 V e 0.24 V f 0.26 V

5

a b

6

B. The initial current is zero and then becomes positive and varies sinusoidally.

7

The peak EMF varies directly with the number of turns (N) the area that varies with R², the field (B) and the frequency (f). The period decreases with the frequency. a Double f gives graph C. b Double B and f gives graph D. c Half N, double B and f gives graph C. d One-quarter area, double B and f gives graph B. e Double N gives graph D.

8

The maximum EMF is equal to NBA2 f, so 8 × 103 = 1000 × B × B = 0.81 T.

The maximum rate of flux change occurs at 90°, where the flux is momentarily zero. The peak EMF will be close to 260 mV. (Actually it is NBA2 f = 100 × 5 × 10–3 × 20 × 10–4 × 2 × 42 = 0.263 V.)

× 0.1² × 2 ×

× 50 and so

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Chapter 10 Electromagnetic induction

10.5 Alternating voltage and current 1

D is correct by definition of what is meant by the RMS voltage.

2

Graph C. Our voltage is 240 V RMS at 50 Hz.

3

a b c d

4

Graph D shows a RMS current of 1 A and a period of 1/50 = 0.02 s.

5

a b c

6

The RMS current is I = P/V = 600/240 = 2.5 A a R = V/I = 240/2.5 = 96 Ω b Vp = 240 × √2 = 340 V (Well, 339 V on the calculator, but the 240 V figure is only held to about + or – 10 V.) c I = V/R = 340/96 = 3.53 A

7

a b c d

I = P/V = 60/24 = 2.5 A Ip = √2IRMS = 3.5 A Vpp = 2 × √2VRMS = 2 × √2 × 24 = 68 V R = V/I = 24/2.5 = 9.6 Ω (As this is an AC current in a magnetic device, this is better referred to as ‘impedance’ – a quantity that depends on the inductance in the coil and magnet system.)

8

a b c d

The RMS voltage is the equivalent DC voltage: VRMS = 170/√2 = 120 V PRMS = V²RMS/R = 120²/100 = 144 W (in parts i and ii, as that is the meaning of RMS) Peak power will occur twice each cycle – that is, 120 Hz. Peak power = Vp²/R = 170²/100 = 289 W

9

a b c

VRMS = Vp/√2 = 10/√2 = 7.1 V IRMS = Ip/√2 = 6/√2 = 4.2 A PRMS = VpIp/2 = 60/2 = 30 W

Vp = √2VRMS = 339 V 2Vp = 679 V Ip = Vp/R = 340/100 = 3.39 A I = V/R = 240/100 = 2.40 A

The period T = 20 ms = 0.02 s and f = 1/T = 1/0.02 = 50 Hz Vpp = 2 × 120 = 240 V VRMS = Vp/√2 = 120/√2 = 85 V

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Chapter 10 Electromagnetic induction

10.6 Transformers 1

a b c

2

Parts a and b have identical answers: A, B and D are all correct as, in an ideal transformer, the power in to the primary is equal to the power obtained in the secondary circuit.

3

a b

4

Power losses occur when electrical energy is converted into heat energy in the copper windings and in the iron core. Energy losses in the core are due to eddy currents.

5

a b c

Vp = Np B/ t Vs = Ns B/ t Vp = VsN1/N2

A is correct but the power from the secondary will be less. B and D are correct as they both represent the power used in the resistor.

B. No output for a steady input current. D. A steadily increasing current will generate a constantly increasing flux and hence a steady output voltage. A. The sinusoidal current will generate a sinusoidally varying flux and a sinusoidal output voltage.

6

a b c

Vp/Vs = Np/Ns and hence Vs = 8.0 × 200/20 = 80 V Is = VpIp/Vs = 8 × 2/80 = 0.20 A P = VI = 80 × 0.20 = 16 W

7

a b

Ns = 800 × 12/240 = 40 turns RMS power is P = VI = 12 × 2 = 24 W and so RMS current in primary is 24/240 = 0.10 A. Thus the peak current is √2 × 0.10 = 0.14 A. As in part b, P = 24 W for an ideal transformer.

c 8

No. A direct current cannot induce an EMF in the secondary coil because

B/

t = 0.

9

There will be no power consumed in the primary circuit of an ideal transformer. This is because the induced current in the primary coil is equal in magnitude but opposite in direction to the applied current. In a real transformer like this one there may be a loss of around, say, 2 W, which would add up to 1200 J in 10 min.

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Chapter 10 Electromagnetic induction

10.7 Using electrical energy 1

a

b

Power is transmitted at very high voltages because this means that the current used can be lower. (As P = VI, a large V means a low I for a certain amount of power.) The main source of power loss is the resistive loss in the power lines, which can be expressed as P = I²R. Because the power loss is proportional to the square of the current, reducing the current by a certain factor reduces the power loss by the square of that factor. At 500 kV the current is 2000 times less than at 250 V, thus the power loss is 2000² = 4 million times less! This means that thinner, and thus cheaper, cables can be used. The limit to the voltage is set by losses such as current leakage via insulators and the atmosphere.

2

a b

I = P/V = 500 × 106/250 × 103 = 2.00 × 103 A Twice the voltage will mean half the current: 1.00 × 103 A.

3

a

The current is 2000 A and thus the power loss is P = I²R = 2000² × 10 = 40 MW, which is 8.0% of the total output. At twice the voltage the current is half, and so the power loss (I²R) is only one-quarter 1 1 ( ² = ) or 2.0%. 2 4

b

4

This power line will have a resistance of half the other one as, although it is twice as long, it has four times the cross-section area. As the power loss is proportional to the resistance, the power loss in this cable will be half—that is, 1.0%.

5

a b c d

I = P/V = 5000/500 = 10 A Power loss = I²R = 100 × 4 = 400 W This is 400/5000 = 8.0% loss. The voltage drop along the cable will be the house is 460 V.

a b c

I = P/V = 5000/5000 = 1.0 A Power loss = 1² × 4 = 4 W or 4/5000 = 0.08% (Or, as we know the current is 1/10th, the power loss will be 1/100th.) The voltage drop is only 4 V, so the voltage available is 4996 V.

7

a b c d e

E = Pt = 1 × 2 = 2 kWh, so cost = 2 ×15 = 30c E = Pt = 0.08 × 0.5 = 0.04 kWh, so cost = 0.04 × 15 = 0.6c E = Pt = 0.25 × 12 = 3 kWh, so cost = 3 × 15 = 45c E = Pt = 0.006 × 7 × 24 = 1.008 kWh, so cost = 1.008 × 15 = 15.1c E = Pt = 0.003 × 365 × 24 = 26.28 kWh, so cost = 26.28 × 15 = $3.94

8

a

As P = VI, we find the current from I = P/V = 500 × 106/250 = 2 × 106 A or 2 MA (megaamps). The voltage drop along the line would be given by: V = IR = 2 × 106 × 2 = 4 MV, i.e. 4 million volts. This of course is completely impossible! This time I = P/V = 500 × 106/105 = 5000 A or 5 kA (kiloamps). The voltage drop along the line would now be given by: V = IR = 5000 × 2 = 10 000 V, or 10 kV. The voltage at the town would be 90 kV. While this is a significant loss of voltage (10%) it is at least feasible, unlike the previous scheme.

6

b

V = IR = 10 ! 4 = 40 V, and so the voltage at

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9

c

If the resistance is halved the voltage drop will halve also and so there will be a 5 kV drop leaving the town with 95 kV. The power loss in the lines can be found in several ways, but using P = VI gives us P = 5000 × 5000 = 25 × 106 = 25 MW. This is 5% of the 500 MW supplied by the power station.

a

Current from generator = P/V = 150 000/1000 = 150 A. The transformer increases the voltage by 10 and so decreases the current by 10 to 15 A. The voltage drop is given by V = IR = 15 × 2 = 30 V. The voltage at the town is therefore 9970 V. P = I²R = 15² × 2 = 450 W which is only 0.3% of the output and so not a problem.

b c 10

Chapter 10 Electromagnetic induction

a b c d

150 A Voltage drop = IR = 150 × 2 = 300 V and so input voltage at town = 1000 – 300 = 700 V Power lost = I²R (or VI) = 45 000 W = 45 kW. Power at town = VI = 700 × 150 = 105 kW. No it was not a good idea as 30% of the power output of the generator is being wasted.

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Chapter 10 Electromagnetic induction

Chapter review

c

I = EMF/R = / tR = 8 × 10–4 × 40 × 10–4/(1 × 10–3 × 1) = 3.2 mA clockwise (to oppose the increasing B) We could repeat a calculation as above, but we can see that this will produce the same rate of change of flux (from + to – in 2 ms is the same rate as from + to +2 in 1 ms) but in the opposite direction. So the answer is 3.2 mA anti-clockwise. This is half the rate, so the answer is 1.6 mA anti-clockwise.

2

a b

Emax = NBA2 f = 1 × 8 × 10–4 × 40 × 10–4 × 2 As R = 1 , I = 2.0 mA

3

We compare the situations given with the one just calculated in which the frequency was 100 Hz (T = 10 ms) and the field strength 8 ! 10–4 T. Remember that the maximum EMF (and therefore current) is proportional to both f and B. a Half the frequency will lead to half the current and twice the period, so graph A. b Double frequency and half B leads to same I but half T, so graph C. c Same frequency and half field leads to same T but half I, so graph B.

4

I = EMF/R = N / tR = 40 × 20 × 10–3 × × 0.04²/(0.1 × 2) = 0.020 A = 20 mA from Y to X. The induced current must create a field in the same direction as the original field.

5

A and C both increase the EMF, EMF = N BA/t, B decreases the EMF and D does not change the EMF.

6

The current would be opposite (that is, from X to Y) because in this situation we are reversing the action—more particularly, the induced current must oppose the increasing downward flux.

7

a

1

a b

b

× 100 = 2 × 10–3 = 2.0 mV

EMF = / t = Blv = 10 × 10–3 × 0.2 × 2 = 4.0 mV. As R = 1 Ω, the current is 4.0 mA from X to Y. The direction is determined either by considering the current required to replace the decreasing flux through the loop, or by considering the force on positive charges in the rod as the rod moves right in an outward field. F = IlB = 4 × 10–3 × 0.2 × 10 × 10–3 = 8.0 × 10–6 N. This tiny force would be to the left to oppose the motion (or use the right-hand palm rule).

8

There is no change in flux and hence no EMF or current. (And the switch is open.)

9

change of magnetic flux, and therefore the induced EMF, will be zero.

10

I = EMF/R =

11

a b c

Power out = power in, VSIS = VPIP, and so IS = 14 × 3/42 = 1.0 A Vp/Vs = Np/Ns and so Np = Vp/Vs × Ns = 14/42 × 30 = 10 turns P = VI = 42 × 1 = 42 W

12

a b c

Graph C. A steady voltage requires a steadily increasing primary current. Graph A. These spikes are due to opposite changes in the primary current. Graph B. A steady input current produces no output.

13

a

f = 1/T = 1/0.002 = 500 Hz

/ tR = B/t × A/R = 1 × 10–5 × 40/8 = 5.0 × 10–5 A = 50

A

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Chapter 10 Electromagnetic induction

b c

VRMS = Vp/√2 = 25/√2 = 17.7 V Vpp = 2Vp = 50 V

14

a b c

PRMS = ½Pp = ½ × VpIp = ½ × 25 × 15 = 188 W Pp = VpIp = 25 × 15 = 375 W R = V/I = 25/15 = 1.67 Ω (But remember that this is really impedance as it involves an AC circuit with magnetic induction.)

15

The peak-to-peak ‘power’ figure is 8 times the RMS figure (VppIpp = 2√2VRMS × 2√2IRMS = 8VRMSIRMS). The RMS figures are therefore 7.5 W, 30 W, 60 W, 15 W respectively. So C is consistent with the specification.

16

The power in this circuit is VI = 15 × 1 = 15 W, which is consistent with D in the table.

17

a b c d

18

a

b

c d

The peak-peak voltage is simply twice the peak voltage and so is 25.2 V. The RMS voltage is the peak voltage divided by √2 and so is 8.9 V. The output voltage is directly proportional to the rate of rotation and so it will also double, to give peak and RMS outputs of 25.2 V and 17.8 V, respectively. As the output is proportional to the frequency, the new frequency will be equal to the old one multiplied by the ratio of voltages, 16/8.9 = 1.8. So the new frequency will be 50 × 1.8 = 90 Hz. The output is proportional to the magnetic flux and so will reduce by a factor of 60/80 = ¾, giving a peak voltage ¾ of 12.6 = 9.5 V To operate correctly the machine will require I = P/V = 480/24 = 20 A. The input current to the transformer will therefore be 20/10 = 2 A. The voltage drop along the line will be V = IR = 4 V giving 236 V at the transformer. This is very close to the operating voltage and should cause no problems. If the machine was operating at its correct current of 20 A, there would be a voltage drop along the line of V = IR = 40 V, which is more than the supply of 24 V! Even with no resistance at the machine, the maximum current would be I = V/R = 24/2 = 12 A, clearly not enough to operate the machine. As there must be a 40 V drop (IR = 20 × 2) along the line, the output of the transformer will need to be 24 + 40 = 64 V. The turns ratio will therefore be 240/64 = 3.75:1. The power lost in the line will be P = VI = 40 × 20 = 800 W (or I²R = 20² × 2), much more than is being used by the machine!

19

Using this generator would be a very bad idea for two main reasons. First, many of the appliances in the house would require AC power because they have transformers in them. If DC is used, the transformers are likely to burn out because the current through them will be too high as there is no back EMF to reduce the current in the primary. In any case, they won’t work because there is no changing flux to produce a voltage in the secondary. Devices with purely resistive loads such as filament light globes or heaters will work normally as they do not need AC and the voltage will be correct. Second, if in fact the 2000 W load is entirely resistive, the current drawn would be I = P/V = 2000/240 = 8.3 A but this would create a IR = 8.3 × 8 = 66 V drop along the power line which would mean that the voltage at the house was only about 170 V.

20

He will need a step-down transformer with a turns ratio of 5:1. A 2000 W load will require a current of I = P/V = 2000/1200 = 1.7 A from the generator. This would result in a voltage drop of IR = 1.7 × 8 = 13 V along the power line. This would still mean that the voltage input to the transformer was 1187 V which will produce a secondary voltage of 1187/5 = 237 V. At no load Heinemann Physics 12(3e) Teacher’s Resource and Assessment Disk I S B N 9781442501263

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Chapter 10 Electromagnetic induction

the voltage would be 240 V (no voltage drop on the line) and at half load 238.5 V so the set-up will suit his purpose well.

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Chapter 10 Electromagnetic induction

Exam-style questions (Electric power) 1

a b c

A; both have A direction fields. B; left current has BC direction and right current has AB direction. G; C and A directions respectively cancel.

2

a b c

To the left, due to magnetic induction in the soft iron. More strongly to the left as the right end of the electromagnet is now a south pole. To the right as the right end of the electromagnet is now a north pole.

3

F = IlB = 100 × 1 × 1 × 10–5 = 1.0 mN

4

The right-hand force rule tells us that a current from west to east will experience an upward force.

5

The weight of 1 m of cable is mg = 0.05 × 9.8 = 0.49 N. For the magnetic force to equal this: I = F/B = 0.49/(1 × 10–5) = 4.9 × 104 A. (Not much chance of magnetic levitation for power cables!)

6

The change of force is from 1.0 mN up to 1.0 mN down – a change of 2.0 mN down.

7

B is the correct answer. The horizontal component of the current is now less and so there will be a smaller force per metre of cable.

8

F = IlB = 1 × 0.05 × 1 = 0.05 N = 50 mN

9

The right-hand force rule tells us that it is to the right.

10

F = IlB = 1 × 0.01 × 1 = 0.01 N = 10 mN

11

The direction of the force on side PQ is to the left.

12

a b

c

Sides AB and DC are parallel to the field. Hence there will be no force on them. (This is the case for parts i and ii.) i F = NIlB = 100 × 0.2 × 0.1 × 0.25 = 0.50 N. The force on AD is out of the page and on BC into the page. ii F = NIlB = 100 × 0.2 × 0.1 × 0.25 = 0.50 N. The force on AD is out of the page and on BC into the page. The coil will rotate through 90° until the plane of the loop is perpendicular to the field (and the page). It may swing back and forth until it settles in this position.

13

A, B and C. A will produce a greater total force, B will increase the current, and C will result in a stronger magnetic field through the coil. D would reduce the current.

14

a b c

Field is from N to S, so the right-hand rule shows that the force on side AB is upward and that on side CD is downward. In the position shown (with the coil horizontal), the direction of the forces on the sides AB and CD are at right angles to the radius and the torque is maximum. The torque becomes zero when the coil is in the vertical position. It continues to rotate for two reasons: (i) Its momentum will carry it past the true vertical position. (ii) At the Heinemann Physics 12(3e) Teacher’s Resource and Assessment Disk I S B N 9781442501263

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Worked solutions

Chapter 10 Electromagnetic induction

vertical position the commutator reverses the direction of the current through the coil and so the forces reverse, thus creating a torque that will rotate it another half turn—at which point the current reverses again. 15

a b c

40 = 4.0 A 100 ! 0.2 ! 0.5 The shorter side will halve the force, twice the current will double the force and half the turns halves the force. The net effect is to halve the force, so F = 20 N. The force increases with the field, so the new force is F = 40 ! 8/5 = 64 N. I = F/NlB =

16

The direction of the induced current is from Y to X. The direction is predicted by the right-hand force rule. The direction of the magnetic force on side XY must oppose the entry of the loop into the field.

17

EMF =

18

P = VI = 4 × 10–3 × 8 × 10–3 = 3.2 × 10–5 W (Note that I = V/R = 8 mA.)

19

The source of this power is the external force that is moving the loop towards the cube.

20

F = IlB = 8 × 10–3 × 0.20 × 0.4 = 6.4 × 10–4 mN

21

After 1.2 s, the loop will have moved 6 cm and so will be totally within the field. Thus there is no flux change and no EMF.

22

The direction of the induced current is from X to Y. The direction is predicted by the right-hand rule. The direction of the magnetic force on the side of the loop still in the field must oppose its exit from the field.

23

a b c d e

24

a b

B/

t = Blv = 0.4 × 0.2 × 5.0 × 10–2 = 4.0 mV

FB = BA cos , where is the angle between B and the normal to A. = 0 in this case. So FB = 1.0 × 10–3 × 5.0 × 10–3 = 5.0 Wb. No flux is threading the loop (as cos 90° = 0) and so FB = 0. EMF = B/ t and thus EMF = 5.0/2.0 = 2.5 mV I = EMF/R = 2.5/2.0 = 1.25 mA The electric current will quickly disappear as the electrical energy is rapidly converted into heat energy in the resistance, according to P = I2 R. The EMF, and hence the current, depends on the rate of change. If the rate is increased by 4, then the current will also increase by 4: I = 200 A. The EMF generated must have been EMF = IR = 50 × 10–6 × 600 = 3 × 10–2 V. As EMF = –2 –4 N R², and thus B/ t we can find B = 3 × 10 × 2 /100 = 6 × 10 Wb. The area is –4 the field B = 6 × 10 /( × 0.03²) = 0.21 T or 210 mT.

25

a b c

A sine wave with peak amplitude 0.9 V and period 0.01 s (10 ms) VRMS = VP/√2 = 0.64 V The output graph would have half the period and twice the amplitude. The RMS voltage will be 1.3 V.

26

Emax = NBA2 f = 200 × 0.5 × 100 × 10–4 × 2

27

VRMS = Vp/√2 = 314/√2 = 222 V

× 50 = 314 V

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Worked solutions

Chapter 10 Electromagnetic induction

28

B. This is twice the frequency and so the amplitude will be double and the period half.

29

D. This has twice the amplitude but the same period and so could be obtained by doubling B or A or N.

30

C. This time the frequency has doubled but the amplitude remains the same. Thus a combination of the other quantities must have halved (B has doubled, but N has reduced to one-quarter).

31

f = 1/T = ½ × 10–3 = 500 Hz

32

Vpp = 20 V

33

VRMS = Vp/√2 = 10/√2 = 7.1 V

34

Peak power is VpIp = 10 × 1 = 10 W. Therefore RMS power is half this: 5.0 W.

35

a b c d e

2.0 ! 600 = 0.40 A (RMS) 3000 Vpp = 2 × 3000 = 6000 V 1000 ! 600 Np = Ns × Vp/Vs = = 200 turns 3000 3000 Ps(RMS) = Vs(RMS) × Is(RMS) = = 850 W 2 ! 0.40 Peak power = 2 × RMS power = 1700 W (as Pp = √2VRMS√2IRMS = 2PRMS) Is = Ip × Vp/Vs =

36

C. The alternating current in the primary produces a changing magnetic flux, which induces an EMF in the secondary coils (as well as the primary coils).

37

C. The self-induced EMF is known as a back EMF and opposes the mains EMF.

38

With little or no current in the power line there will be almost no voltage drop. When the house appliances were used, there was a much higher current in the power line and hence a voltage drop. As the generator was supplying 4000 W at 250 V, the current in the line was I = 4000/250 = 16 A. The voltage drop along the line was therefore V = IR = 16 × 2 = 32 V and so the voltage at the house was only 218 V.

39

At the generator end a 1 : 20 step-up transformer is required (5 kV/250 V = 20). There will be 20 times as many turns in the secondary as the primary. At the house end a 20 : 1 step-down transformer is required. Both must be capable of handling 4 kVA. They could be the same type of transformers used the opposite way around.

40

a b c d

The current is given by I = P/V = 4000/5000 = 0.8 A. The voltage drop is V = IR = 0.8 × 2 = 1.6 V. The power loss is P = I²R = 0.8² × 2 = 1.3 W. 5000 ! 1.6 The voltage at the house will be given by = 249.9 V, which is almost the full 20 250 V. The power will be 4000 W – 1.3 W = 3999 W.

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Worked solutions

41

Chapter 10 Electromagnetic induction

The power loss in the first system was P = V × I = 32 × 16 = 512 W, which is about 13% of the power generated. In the second system the loss was 1.3 W, only 0.03% of the generated power. The reason is that the power loss in the power lines increases with the square of the current (P = I²R). Here the current was reduced by a factor of 20 and so the power loss decreased by a factor of 20² or 400.

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