139.Graphical Work With Electromagnetic Induction

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Number 139

Graphical work with Electromagnetic Induction To revise our ideas on electromagnetic induction look at Fig 1, showing a bar magnet and a coil connected to an oscilloscope (to display potential differences). As the coil is moved (at constant speed) horizontally to the left, a trace is displayed showing that an e.m.f. is induced in the coil. This e.m.f. is usually denoted by the letter ε . It arises due to the relative motion between the coil and the magnetic flux.

Fig 2 shows the trace on the oscilloscope as the coil moves from position 1 through to 5. Fig. 2

Fig 1 emf (ε) The coil is moved horizontally to the LEFT S

time

N

Qu 1. Now complete the trace as the coil passes completely over the magnet. Qu 2. On Fig 3, show the trace you would expect to obtain if the coil moved TWICE as fast across the magnet. (The original trace is shown to help you.) HINT - remember that if the speed is doubled then the rate of change of flux is doubled, what effect will this have on the induced e.m.f. ?

There are two ways to look at this :(i) as the coil moves through positions 1,2 and 3 it cuts the lines of flux. (ii) as the coil moves, it has a changing amount of flux threading the coil.

Fig. 3

It is this change in flux that is most important. You might have to use either of these two ideas. Faraday’s Law says ‘the induced e.m.f. is equal to the rate of change of flux’.

emf (ε)

∆Φ The equation used is ε = ∆t

time

where Φ is the amount of flux (measured in webers). Exam Hint:- You should be able to use Faraday’s law to explain how the size of the emf depends on the rate of change of flux. In a way you can say that the size of the e.m.f. depends on the relative speed between the magnet and coil providing that lines of flux are cut. Now imagine the coil at position 5. (i) Moving through position 5, the coil does not cut any flux. We conclude that there is no induced e.m.f. (at position 5) (ii) Or, there is flux threading the coil but this flux is not changing at position 5.

When the speed changes, the emf changes, the shape of the pulse changes, and the separation between the pulses changes.

Again we conclude that there is no induced e.m.f. (at position 5)

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139. Graphical work with Electromagnetic Induction Lenz’s Law tells us that the direction of the induced e.m.f. (and current) is always in such a direction as to oppose the change producing it.

Qu 4. Fig 6. shows the rotating coil of a generator between the poles of a magnet together with the emf output.

Fig 4

EMF output

V S

N I

S

N

N

S time

I

A north and south attracts, the motion is opposed

N

Two norths repel, again the motion is opposed

S rotation of coil at speed V

The speed of rotation is doubled from V to 2V. Which of the options A, B, C or D best shows the new output?

Look at Fig 4. It shows the induced current in the coil and how it makes the coil behave like a small bar magnet. In both cases the induced current is in a direction to oppose the motion producing it. This is the reason for the minus sign in the equation . ε = − ∆Φ ∆t It means that in the two questions you did in Fig 2 and Fig 3, the second ‘blip’ of e.m.f. is negative and should be below the horizontal line (time axis).

A

B

C

D

Exam Hint:- You should be able to use Lenz’s law to explain why two emfs similar to this are in opposite directions. Qu 3. Fig 5 shows a long bar magnet about to be dropped through a fixed coil. Think about how the magnet falls vertically and that it is ‘long’. Draw a sketch of the trace you would expect.

Fig 5.

Exam Hint:- There are two changes here. You may spot one but the examiner is looking for two.

N

Qu 5 Fig 7. shows a rotating axle whose speed of rotation is to be found. Four magnets are sunk into the axle as shown and each in its turn sweeps past a coil connected to an oscilloscope.The emf induced is recorded as shown in Fig 8.

S

Rotating axle

S

N

Fig 7

N

S

S

N

Coil 400 turns

S N

Fig 8

emf (ε) time

Time-base = 10ms cm-1 y amplifier = 5mV cm-1

(a) (i) From Fig 8, find the time for one quarter of a revolution and (ii) find the number of revolutions made in one minute.

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(b) (i) Use Faraday’s law to explain how the voltage pulses are produced. (ii) The coil has 400 turns. Calculate the maximum rate of change of flux through the coil. (c) The speed of rotation is now increased by 50%. On figure 9, sketch the new trace you would expect to see. (The original trace is shown to help you)

So, from O to A and B to C, the slope is constant and positive meaning that the emf is constant (and negative). From A to B, the slope is constant (and negative) giving a constant (positive) emf. This is shown in Fig 12.

Fig 9

emf ε

139. Graphical work with Electromagnetic Induction

Fig 12

t

Although you could be asked to do a calculation for frequency or maximum emf, you should concentrate on the shape of the waveform.

Questions 6 and 7: The diagrams show two different wave forms of flux φ against time t. For each one, decide which option A B C or D best shows the emf ε against time t.

Time-base = 10ms cm-1 y amplifier = 5mV cm-1

Exam Hint:- Examiners want you to understand graphs like this and make predictions when changes ( like speed ) are made.

Qu 6 φ

We don’t always need a permanent magnet moving relative to a coil. We can produce a magnetic field by sending a current through a coil (called a solenoid). Switching the current on and off will give the necessary change of flux to produce an emf .

t

The signal generator doesn’t just change the frequency, it can change the waveform as well.

emf ε

emf ε

Another way of changing the flux is to change the current in a solenoid as shown in Fig 10.

t

A emf ε

signal generator

solenoid

φt

B emf ε

Fig 10

t

t

C

e

t

D

Qu 7 φ

Fig 11 shows a triangular waveform for the flux φ. We have to find the emf ε in the small coil. A

C

t A

Always remember that on a φ-t graph, the slope (or gradient) is ∆φ ∆φ and because we know that ε = ∆t it means that the induced

C

∆t

emf is (-1) × the gradient on a φ - t graph.

3

t

B

emf ε

B

emf ε

t

t

emf ε

0

t

emf ε

Flux φ

D

t

139. Graphical work with Electromagnetic Induction Exam Hint:- Sometimes examiners will ask for the magnitude of the emf, that is the size of the emf with no regard to its sign.

Qu 2 emf ε

Fig 16 clearly shows the difference between the emf and the magnitude of the emf.

The coil moves at twice the speed so each pulse height is doubled. The time is halved so each pulse width is halved and two pulses take the same time as one of the ‘slower’ pulses.

Fig 16 variation of flux with time

t

Qu 3 variation of emf with time

emf ε

emf ε

t

variation of magnitude of emf with time

emf ε

The magnet accelerates from rest. The south enters the coil slowly and the north leaves more quickly. This makes the first ‘pulse’ of smaller height but a ‘longer’ base. (Because flux at each pole is the same in quantity, you may like to show that the area of each ‘pulse’ is the same.)

t

Qu 4 Option B is correct. Because the speed of rotation is doubled the maximum emf is doubled. Also, the frequency is doubled meaning the time period is halved. Two complete oscillations now occur in the time taken earlier by one.

Answers Qu 1

emf ε

Qu 5 (a) (i) From the graph, a quarter revolution takes 6cm at 10ms cm-1. So time for ¼ revolution = 60 ms 1

3

5

6

(ii) Time for one revolution is 60ms × 4 = 240ms = 0.24s. One minute is 60s, so the number of revolutions in one minute is 60/0.24 = 250 rpm (b) (ii) Maximum emf is 2cm at 5mV cm-1 ie 10mV. Coil has N = 400 turns so use ∆φ ∆φ ∆φ -3 -5 -1

7

ε=−N×

∆t

10 × 10 = 400 ×

(c) Because the coil is moving at the same speed, the pulse is of the same width and ‘height’. It is inverted because of the opposite polarity ( first a north then a south).

Acknowledgements: This Physics Factsheet was researched and written by Keith Cooper The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

Time-base = 10ms cm-1 y amplifier = 5mV cm-1

Qu 6: B Qu 7: C

4

∆t

∆t

= 2.5 × 10 Whs

The maximum emf increases by 50% (2cm to 3 cm). The time between pulses is reduced by one third from 6cm to 4cm. ( The width of each pulse is also reduced but difficult to display!!)