1. Coursebook A Level Physics

Chapter 28: Electromagnetic induction

Induced e.m.f. travel from right to left, making the left-hand side of the
conductor negative. What causes these electrons to move?
When a conductor is not part of a complete circuit, there Moving the conductor is equivalent to giving an electron
cannot be an induced current. Instead, negative charge within the conductor a velocity in the direction of this
will accumulate at one end of the conductor, leaving the motion. This electron is in an external magnetic field and
other end positively charged. We have induced an e.m.f. hence experiences a magnetic force Bev from right to left.
across the ends of the conductor. Check this out for yourself.

Is e.m.f. the right term? Should it be voltage? In this end is le with positive charge
Chapter 9 you saw the distinction between voltage and negative charge accumulates
e.m.f. The latter is the correct term here because, by at this end
pushing the wire through the magnetic field, work is done −
and this is transformed into electrical energy. Think of
this in another way. Since we could connect the ends of S
the conductor so that there is a current in some other
component, such as a lamp, which would light up, it must + induced
be an e.m.f. – a source of electrical energy. current

Figure 28.10 shows how the induced current gives rise movement
to an induced e.m.f. Notice that, within the conductor, of wire
conventional current is from negative to positive, in
the same way as inside a battery or any other source of Figure 28.10 An e.m.f. is induced across the ends of the
e.m.f. In reality, the free electrons within the conductor conductor.

QUESTIONS magnetic 439
B field lines
2 The coil in Figure 28.11 is rotating in a uniform
magnetic field. Deduce the direction of the induced C
current in sections AB and CD. State which terminal, X A
or Y, will become positive.
B
3 When an aircraft flies from east to west, its wings are XD
an electrical conductor cutting across the Earth’s
magnetic flux. In the northern hemisphere, which Y
wingtip will become positively charged? Why will this Figure 28.11 A coil rotating in a magnetic field.
wingtip be negative in the southern hemisphere?

Magnetic flux and magnetic flux linkage A. For a magnetic field normal to A, the magnetic flux Φ
So far in this chapter we have looked at the ideas of (Greek letter phi) must therefore be equal to the product of
magnetic flux density and the area A (Figure 28.12a).
electromagnetic induction in a descriptive way. Now we
will see how to calculate the value of the induced e.m.f. a b normal
and look at a general way of determining its direction.
B θB
In Chapter 26, we saw how magnetic flux density B is
defined by the equation

B = F
IL

Now we can go on to define magnetic flux as a quantity. area A area A
We picture magnetic flux density B as the number of
magnetic field lines passing through a region per unit Figure 28.12 a The magnetic flux is equal to BA when the field
area. Similarly, we can picture magnetic flux as the total is normal to the area. b The magnetic flux becomes BA cos θ
number of magnetic field lines passing through an area when the field is at an angle θ to the normal of the area.

Cambridge International A Level Physics

The magnetic flux Φ through area A is defined as: WORKED EXAMPLE
Φ = BA
1 Figure 28.13 shows a solenoid with a cross-sectional
where B is the component of the magnetic flux density area 0.10 m2. It is linked by a magnetic field of flux
perpendicular to the area. density 2.0 × 10−3 T and has 250 turns. Calculate the
magnetic flux and flux linkage for this solenoid.
How can we calculate the magnetic flux when B is not
perpendicular to A? You can easily see that when the Step 1 We have B = 2.0 × 10−3 T, A = 0.10 m2, θ = 0°
field is parallel to the plane of the area, the magnetic flux and N = 250 turns. Hence we can calculate the flux Φ.
through A is zero. To find the magnetic flux in general, we Φ = BA
need to find the component of the magnetic flux density Φ = 2.0 × 10−3 × 0.10 = 2.0 × 10−4 Wb
perpendicular to the area. Figure 28.12b shows a magnetic
field at an angle θ to the normal. In this case: Step 2 Now calculate the flux linkage.
magnetic flux linkage = NΦ
magnetic flux = (B cos θ) × A magnetic flux linkage = 2.0 × 10−4 × 250

or simply: = 5.0 × 10−2 Wb
A = 0.10 m2

magnetic flux = BA cos θ

(Note that, when θ = 90°, flux = 0 and when θ = 0°, B = 2.0 × 10–3 T N = 250 turns
flux = BA.)

For a coil with N turns, the magnetic flux linkage
is defined as the product of the magnetic flux and the
number of turns; that is:

440 magnetic flux linkage = NΦ Figure 28.13 A solenoid in a magnetic field.
or

magnetic flux linkage = BAN cos θ

The unit for magnetic flux or flux linkage is the weber QUESTIONS
(Wb). 4 Use the idea of magnetic flux linkage to explain

One weber (1 Wb) is the flux that passes through an area why, when a magnet is moved into a coil, the
of 1 m2 when the magnetic flux density is 1 T. e.m.f. induced depends on the strength of the
1 Wb = 1 T m2. magnet and the speed at which it is moved.
5 In an experiment to investigate the factors that
An e.m.f. is induced in a circuit whenever there is a change affect the magnitude of an induced e.m.f., a
in the magnetic flux linking the circuit. Since magnetic student moves a wire back and forth between two
flux is equal to BA cos θ, there are three ways an e.m.f. can magnets, as shown in Figure 28.14. Explain why
be induced: the e.m.f. generated in this way is much smaller
than if the wire is moved up and down in the field.
■■ changing the magnetic flux density B
■■ changing the area A of the circuit movement
■■ changing the angle θ. of wire

Now look at Worked example 1. S

Figure 28.14 A wire is moved horizontally in a
horizontal magnetic field. For Question 5.

Chapter 28: Electromagnetic induction

QUESTIONS 7 A bar magnet produces a uniform flux density
6 In the type of generator found in a power station of 0.15 T at the surface of its north pole. The pole
measures 1.0 cm × 1.5 cm. Calculate the magnetic flux
(Figure 28.15), a large electromagnet is made to rotate at this pole.
inside a fixed coil. An e.m.f. of 25 kV is generated; this
is an alternating voltage of frequency 50 Hz. What 8 A solenoid has diameter 5.0 cm and length 25 cm
factor determines the frequency? What factors do you (Figure 28.16). There are 200 turns of wire. A current
think would affect the magnitude of the e.m.f.? of 2.0 A creates a magnetic field of flux density 2.0  ×
10−5 T through the core of this solenoid. Calculate the
magnetic flux linkage for this solenoid.

25 cm

5.0 cm 200 turns

Figure 28.16  A solenoid. For Question 8.

Figure 28.15  The generators of this power station produce 9 A rectangular coil, 5.0 cm  × 7.5 cm, and having 120 turns,
electricity at an induced e.m.f. of 25 kV. For Question 6. is at right angles to a magnetic field of flux density 1.2 T.
Calculate the magnetic flux linkage for this coil.

Faraday’s law of 441
electromagnetic induction
Now look at Worked examples 2 and 3.
Earlier in this chapter, we saw that electromagnetic induction
occurs whenever a conductor cuts across lines of magnetic WORKED EXAMPLES
flux – for example, when a coil is rotated in a magnetic
field so that the magnetic flux linking the coil changes. We 2 A straight wire of length 0.20 m moves at a steady
can use Faraday’s law of electromagnetic induction to speed of 3.0 m s−1 at right angles to a magnetic field
determine the magnitude of the induced e.m.f. in a circuit: of flux density 0.10 T. Use Faraday’s law to determine
the e.m.f. induced across the ends of the wire.
The magnitude of the induced e.m.f. is proportional to
the rate of change of magnetic flux linkage. Step 1  With a single conductor, N = 1. To determine
the e.m.f. E, we need to find the rate of change
of magnetic flux; in other words, the change in
magnetic flux per second.

3.0 m s–1 B = 0.10 T

We can write this mathematically as: 0.20 m

E ∝ Δ(NΦ) 3.0 m
Δt Figure 28.17  A moving wire cuts across the
magnetic field.
where Δ(NΦ) is the change in the flux linkage in a time Δt.
Working in SI units, the constant of proportionality is Figure 28.17 shows that, in 1.0 s, the wire travels
equal to 1. Therefore: 3.0 m. Therefore:
change in magnetic flux = B × change in area
E = Δ(NΦ) change in magnetic flux = 0.10 × (3.0 × 0.20)
Δt = 6.0 × 10−2 Wb

The equation above is a mathematical statement of Faraday’s
law. Note that it allows us to calculate the magnitude of the
induced e.m.f.; its direction is given by Lenz's law, which is
explained in the next section on page 442.

Cambridge International A Level Physics

WORKED EXAMPLES (continued)

Step 2  Use Faraday’s law to determine the e.m.f. that the magnetic flux passes perpendicularly through
E =  Δ(ΔNtΦ)     (N = 1) the coil. The flux density of the field is 0.50 T. The coil is
pulled rapidly out of the field in a time of 0.10 s. What
ΔΦ = 6.0 × 10−2 Wb  and  Δt = 1.0 s average e.m.f. is induced across the ends of the coil?
Step 1  When the coil is pulled from the field, the flux
E =  6.0 × 10−2  = 0.06 V linking it falls to zero. We have to calculate the magnetic
1.0 flux linking the coil when it is in the field.

The induced e.m.f. across the ends of the wire is about To convert cm2 into m2, multiply by a factor of 10−4.
60 mV. Hence A = 1.2 × 10−4 m2.
magnetic flux linkage = NΦ = BAN
3 This example illustrates one way in which the flux = 0.50 × 1.2  × 10−4  × 2500
density of a magnetic field can be measured, shown in magnetic flux linkage = 0.15 Wb
Figure 28.18.

A search coil of wire having 2500 turns and of area Step 2  Now calculate the induced e.m.f. using Faraday’s
1.2 cm2 is placed between the poles of a magnet so law of electromagnetic induction.
search coil
to datalogger Δ(NΦ) = 0.15 Wb  and  Δt = 0.10 s

magnitude of induced e.m.f. = rate of change of flux
linkage
N
SE=  Δ(NΦ)  =  0.15  = 1.5 V
Δt 0.10
N
S Note that, in this example, we have assumed that the
flux linking the coil falls steadily to zero during the time
442 Figure 28.18  A search coil can be moved into and out interval of 0.10 s. Our answer is thus the average value of

of a magnetic field to detect magnetic flux. the e.m.f.

QUESTIONS 12 Figure 28.19 shows a search coil, having 2000 turns
and of area 1.2 cm2, placed between the poles of a
10 A conductor of length L moves at a steady speed v strong magnet. The ends of the coil are connected
at right angles to a uniform magnetic field of flux to a voltmeter. The coil is then pulled out of the
density B. Show that the e.m.f. E across the ends of magnetic field, and the voltmeter records an
the conductor is given by the equation: average e.m.f. of 0.40 V over a time interval of 0.20 s.
Calculate the magnetic flux density between the
E = BLv poles of the magnet.
(You can use Worked example 2 to guide you through 2000 turns
Question 11.)
11 A wire of length 10 cm is moved through a distance

of 2.0 cm in a direction at right angles to its length
in the space between the poles of a magnet, and
perpendicular to the magnetic field. The flux density
is 1.5 T. If this takes 0.50 s, calculate the average
e.m.f. induced across the ends of the wire.

voltmeter
Figure 28.19  Using a search coil to measure flux.

Chapter 28: Electromagnetic induction

Lenz’s law So, to summarise, there is an induced current
because the electrons are pushed by the motor effect.
We use Faraday’s law to calculate the magnitude of an Electromagnetic induction is simply a consequence of the
induced e.m.f. Now we can go on to think about the motor effect.
direction of the e.m.f. – in other words, which end of a
wire or coil moving in a magnetic field becomes positive, In Figure 28.20, electrons are found to accumulate at Y.
and which becomes negative. This end of the wire is thus the negative end of the e.m.f.
and X is positive. If the wire was connected to an external
Fleming’s right-hand rule gives the direction of an circuit, electrons would flow out of Y, round the circuit,
induced current. This is a particular case of a more general and back into X. Figure 28.21 shows how the moving wire
law, Lenz’s law, which will be explained in this section. is equivalent to a cell (or any other source of e.m.f.).
First, we will see how the motor effect and the generator
effect are related to each other. induced magnetic field lines
current current
The origin of electromagnetic induction through
cell
So far, we have not given an explanation of
electromagnetic induction. You have seen, from the + +
experiments at the beginning of this chapter, that it does
occur, and you know the factors that affect it. But what is movement
the origin of the induced current? of wire

Figure 28.20 gives an explanation. A straight wire XY Figure 28.21  A moving conductor in a magnetic field is a
is being pushed downwards through a horizontal magnetic source of e.m.f., equivalent to a cell.
field of flux density B. Now, think about the free electrons
in the wire. They are moving downwards, so they are in Forces and movement 443
effect an electric current. Of course, because electrons are
negatively charged, the conventional current is flowing Electromagnetic induction is how we generate most of
upwards. our electricity. We turn a coil in a magnetic field, and the
mechanical energy we put in is transferred to electrical
We now have a current flowing across a magnetic energy. By thinking about these energy transfers, we can
field, and the motor effect will therefore come into play. deduce the direction of the induced current.
Each electron experiences a force of magnitude Bev.
Using Fleming’s left-hand rule, we can find the direction Figure 28.22 shows one of the experiments from earlier
of the force on the electrons. The diagram shows that the in this chapter. The north pole of a magnet is being pushed
electrons will be pushed in the direction from X to Y. So a towards a coil of wire. There is an induced current in the
current has been induced to flow in the wire; the direction coil, but what is its direction? The diagram shows the two
of the conventional current is from Y to X. possibilities.

Now we can check that Fleming’s right-hand rule gives The current in the coil turns it into an electromagnet.
the correct directions for motion, field and current, which One end becomes the north pole, the other the south pole.
indeed it does. In Figure 28.22a, if the induced current is in this direction,
the coil end nearest the approaching north pole of the
conductor
pushed downwards a Incorrect
south
induced
Y current SN

movement X b Correct
B of electrons north

magnetic SN
field lines
Figure 28.20  Showing the direction of the induced current. Figure 28.22  Moving a magnet towards a coil: the direction of
the induced current is as shown in b, not a.

Cambridge International A Level Physics

magnet would be a south pole. These poles will attract one a Incorrect
another, and you could let go of the magnet and it would be force pushing
dragged into the coil. The magnet would accelerate into the wire downwards
coil, the induced current would increase further, and the
force of attraction between the two would also increase. induced
current
In this situation, we would be putting no energy into
the system, but the magnet would be gaining kinetic motor e ect
energy, and the current would be gaining electrical energy. force
A nice trick if you could do it, but against the principle of
conservation of energy! b Correct motor e ect
induced force
It follows that Figure 28.22b must show the correct current
situation. As the north pole of the magnet is pushed force pushing
towards the coil, the induced current makes the end of wire downwards
the coil nearest the magnet become a north pole. The two
poles repel one another, and you have to do work to push
the magnet into the coil. The energy transferred by your
work is transferred to electrical energy of the current. The
principle of energy conservation is not violated.

QUESTION Figure 28.23 Moving a conductor through a magnetic field:
13 Use these ideas to explain what happens if the direction of the induced current is as shown in b, not a.

a you stop pushing the magnet towards the coil, QUESTION
and b you pull the magnet away from the coil.
444

Figure 28.23 shows how we can apply the same 14 Draw a diagram to show the directions of the
reasoning to a straight wire being moved in a downward induced current and of the opposing force if you
direction through a magnetic field. There will be an now try to move the wire shown in Figure 28.23
induced current in the wire, but in which direction? Since upwards through the magnetic field.
this is a case of a current across a magnetic field, a force
will act on it (the motor effect), and we can use Fleming’s A general law for induced e.m.f.
left-hand rule to deduce its direction.
Lenz’s law summarises this general principle of energy
First we will consider what happens if the induced conservation. The direction of an induced current is such
current is in the wrong direction. This is shown in Figure that it always produces a force that opposes the motion
28.23a. The left-hand rule shows that the force that that is being used to produce it. If the direction of the
results would be downward – in the direction in which current were opposite to this, we would be getting energy
we are trying to move the wire. The wire would thus be for nothing. Here is a statement of Lenz’s law:
accelerated, the current would increase, and again we
would be getting both kinetic and electrical energy for no Any induced current or induced e.m.f. will be established
energy input. in a direction so as to produce effects which oppose the
change that is producing it.
The induced current must be as shown in Figure
28.23b. The force that acts on it due to the motor effect This law can be shown to be correct in any experimental
pushes against you as you try to move the wire through situation. For example, in Figure 28.3, a sensitive ammeter
the field. You have to do work to move the wire, and hence connected in the circuit shows the direction of the current
to generate electrical energy. Once again, the principle of as the magnet is moved in and out. If a battery is later
energy conservation is not violated. connected to the coil to make a larger and constant current

Chapter 28: Electromagnetic induction

in the same direction, a compass will show what the poles Using induction: eddy currents,
are at the end of the solenoid. If a north pole is moved into generators and transformers
the solenoid, then the solenoid itself will have a north pole
at that end. If a north pole is moved out of the solenoid, An induced e.m.f. can be generated in a variety of ways.
then the solenoid will have a south pole at that end. What they all have in common is that a conductor is
cutting across magnetic field lines (in some cases, the
QUESTIONS conductor moves; in others, the field lines move). The
alternative way to look at any change is to say that the flux
15 A bar magnet is dropped vertically downwards linking an area changes.
through a long solenoid, which is connected to
an oscilloscope (Figure 28.24). The oscilloscope Eddy currents
trace shows how the e.m.f. induced in the coil
varies as the magnet accelerates downwards. Induced e.m.f.s are formed in some unexpected places.
Consider the demonstration shown in Figure 28.25. A
a metal disc on the end of a rod swings freely between two
opposite magnetic poles.

metal disc

solenoid N
S
b E.m.f. 445
AB
CD

Time Figure 28.25  Demonstrating eddy current damping.

Figure 28.24  a A bar magnet falls through a long coil. Without the magnets, the disc oscillates from side to
b The oscilloscope trace shows how the induced e.m.f. side for a long time. This is because air resistance is small
varies with time. and it takes a long time for the energy of the disc to be lost.
When the magnets are present, the oscillation of the disc
a Explain why an e.m.f. is induced in the coil as dies away quickly. As the disc enters the magnetic field,
the magnet enters it (section AB of the trace). one side of the disc is cutting the magnetic field lines and
so an induced e.m.f. is created in that side but not in the
b Explain why no e.m.f. is induced while the side that has not yet entered. Since the disc is a conductor,
magnet is entirely inside the coil (section BC). the induced e.m.f. creates currents in the disc itself. These
currents are known as eddy currents. They flow in a
c Explain why section CD shows a negative circular fashion inside the disc. Lenz’s law predicts that
trace, why the peak e.m.f. is greater over this the induced currents that flow in the disc will produce
section, and why CD represents a shorter time a force that opposes the motion, just as in Figure 28.23.
interval than AB. Eddy currents, like other electrical currents, cause heating
and the energy of the oscillation dies away quickly. The
16 You can turn a bicycle dynamo by hand and oscillation is damped by the eddy currents.
cause the lamps to light up. Use the idea of Lenz’s
law to explain why it is easier to turn the dynamo This principle can be used in some types of
when the lamps are switched off than when they electromagnetic or eddy-current braking systems. For
are on. example, a large electromagnet suspended under a train
can cause eddy currents in the rails and slow the train
down. Better still, if the train has an electric motor, then

Cambridge International A Level Physics

the kinetic energy of the train can be used to turn the Figure 28.27 shows how the flux linkage varies with
electric motor to generate an induced e.m.f. With the time for a rotating coil. According to Faraday’s law, the
appropriate electronics the energy from the induced induced e.m.f. is equal to minus the gradient of the flux
current can be passed back to the power supply that runs linkage against time graph.
the train. This is an example of regenerative braking.
■■ When the flux linking the coil is maximum, the rate of
Generators change of flux is zero and hence the induced e.m.f. is zero.

We can generate electricity by spinning a coil in a ■■ When the flux linking the coil is zero, the rate of change
magnetic field. This is equivalent to using an electric of flux is maximum (the graph is steepest) and hence the
motor backwards. Figure 28.26 shows such a coil in induced e.m.f. is also maximum.
three different orientations as it spins. Notice that the
rate of change of flux linkage is maximum when the coil Hence, for a coil like this we get a varying e.m.f. – this
is moving through the horizontal position – one side is is how alternating current is generated. In practice,
cutting rapidly downwards through the field lines, the it is simpler to keep the large coil fixed and spin an
other is cutting rapidly upwards. In this position, we get a electromagnet inside it (Figure 28.28). A bicycle generator
large induced e.m.f. As the coil moves through the vertical (see Figure 28.7) is similar, but in this case a permanent
position, the rate of change of flux is zero – the sides of the magnet is made to spin inside a fixed coil. This makes for a
coil are moving parallel to the field lines, not cutting them, very robust device.
so that there is hardly any change in the flux linkage.
iron core (the rotor), wound iron outer shell (the stator),
field lines in alternating directions to with wire coil wound
produce electromagnet in alternating directions
poles as marked

rotation

446 S N

Induced e.m.f. Flux linkage X SN
N
SXY Y YX
Figure 28.26  A coil rotating in a magnetic field.

rate of flux output
change = 0
maximum rate of Figure 28.28  In a generator, an electromagnet rotates inside
flux change a coil.
0 Time
Transformers
e.m.f. = maximum e.m.f. = 0
0 Time Another use of electromagnetic induction is in
transformers. An alternating current is supplied to the
e.m.f. = –gradient of flux linkage against primary coil and produces a varying magnetic field in the
time graph soft iron core (Figure 28.30). The secondary coil is also
wound round this core, so the magnetic flux linking the
Figure 28.27  The magnetic flux linking a rotating coil secondary coil is constantly changing. Hence, according
as it changes. This gives rise to an alternating e.m.f. The to Faraday’s law, a varying e.m.f. is induced across the
orientation of the coil is shown above the graphs. secondary coil. The core is laminated – it is made up
of thin sheets of soft iron. Using soft iron in the core
increases the amount of the magnetic flux and, hopefully,
all of the magnetic flux from the primary coil passes to
the secondary coil. The thin sheets of iron in the core are

Chapter 28: Electromagnetic induction

separated by a non-conductor so eddy currents cannot flow so iron core
from one sheet to the next. This reduces the eddy currents
and the thermal energy that they create in the core. input output
voltage voltage
There is much more about transformers in Chapter 29,
where we will look at how they are used to change voltages
in circuits which make use of alternating currents.

primary secondary
(3 turns) (7 turns)

Figure 28.30  The construction of a transformer.

QUESTIONS 18 Does a bicycle generator (Figure 28.7) generate 447
17 Figure 28.29 represents a coil of wire ABCD being alternating or direct current? Justify your answer.

rotated in a uniform horizontal magnetic field. 19 The peak e.m.f. induced in a rotating coil in a
Copy and complete the diagram to show the magnetic field depends on four factors: magnetic
direction of the induced current in the coil, and flux density B, area of the coil A, number of turns
the directions of the forces on sides AB and CD N, and frequency f of rotation. Use Faraday’s law to
that oppose the rotation of the coil. explain why the e.m.f. must be proportional to each
of these quantities.
B
20 Explain why, if a transformer is connected to a
C steady (d.c.) supply, no e.m.f. is induced across the
A secondary coil.

B
D

Figure 28.29  A coil rotating in a magnetic field.

Summary ■■ Faraday’s law states that the magnitude of the
induced e.m.f. is equal to the rate of change of
■■ In a magnetic field of magnetic flux density B, the magnetic flux linkage:
magnetic flux passing through an area A is given
by Φ = BA. E  =  Δ(ΔNtΦ)

■■ The magnetic flux linking a coil of N turns is the ■■ Lenz’s law states that the induced current or e.m.f. is
magnetic flux linkage, NΦ. in a direction so as to produce effects which oppose
the change that is producing it.
■■ Flux and flux linkage are measured in webers (Wb).
1 Wb = 1 T m2. ■■ In an a.c. generator, an e.m.f. is induced because the
rotating coil changes the magnetic flux linking the coil.
■■ When a conductor moves so that it cuts across a
magnetic field, an e.m.f. is induced across its ends.
When the magnetic flux linking a coil changes, an
e.m.f. is induced in the coil.

Cambridge International A Level Physics

End-of-chapter questionsMagnetic flux density / T [3]
[2]
1 A student thinks that electrical current passes through the core in a transformer to the secondary coil. [3]
Describe how you might demonstrate that this is not true and explain how an electrical current is [3]
actually induced in the secondary coil. Use Faraday’s law in your explanation. [3]
[4]
2 A square coil of 100 turns of wire has sides of length 5.0cm. It is placed in a magnetic field of flux [2]
density 20mT, so that the flux is perpendicular to the plane of the coil.
a Calculate the flux through the coil. [2]
b The coil is now pulled from the magnetic field in a time of 0.10s. Calculate the average e.m.f. [2]
induced in it. [2]

3 An aircraft of wingspan 40m flies horizontally at a speed of 300ms−1 in an area where the vertical
component of the Earth’s magnetic field is 5.0×10−5T. Calculate the e.m.f. generated between the
aircraft’s wingtips.

4 What is an eddy current? State one example where eddy currents are useful and one where they
are a disadvantage.

5 Figure 28.27 shows the magnetic flux linkage and induced e.m.f. as a coil rotates. Explain why the
induced e.m.f. is a maximum when there is no flux linkage and the induced e.m.f. is zero when
the flux linkage is a maximum.

448 6 a Explain what is meant by a magnetic flux linkage of 1Wb.
b Figure 28.31 shows how the magnetic flux density through a 240 turn coil with a cross-sectional
area 1.2 × 10−4m2 varies with time.

0.60

0 0.8 Time / s
0 0.4

Figure 28.31 For End-of-chapter Question 6.

i Determine the maximum rate of change of flux in the coil.
ii Determine the induced e.m.f. in the coil.
iii Sketch a diagram to show the induced e.m.f. varies with time. Mark values on both the e.m.f.

and time axes.

Chapter 28: Electromagnetic induction

7 Figure 28.32 shows a square coil about to enter a region of uniform magnetic field of magnetic flux
density 0.30T. The magnetic field is at right angles to the plane of the coil. The coil has 150 turns and
each side is 2.0cm in length. The coil moves at a constant speed of 0.50ms−1.
2.0 cm

2.0 cm 0.50 m s–1

coil with B
150 turns

uniform magnetic field
(into plane of paper)
Figure 28.32 For End-of-chapter Question 7.

a i Calculate the time taken for the coil to completely enter the region of magnetic field. [1]
ii Determine the magnetic flux linkage through the coil when it is all within the region of magnetic field. [2]
[1]
b Explain why the induced e.m.f. is constant while the coil is entering the magnetic field. [4] 449
c Use your answer to a to determine the induced e.m.f. across the ends of the coil.
d What is the induced e.m.f. across the ends of the coil when it is completely within the magnetic field? [2]

Explain your answer. [2]
e Sketch a graph to show the variation of the induced e.m.f. with time from the instant that the coil [2]

enters the magnetic field. Your time axis should go from 0 to 0.08s.

8 a State Faraday’s law of electromagnetic induction.
b A circular coil of diameter 200mm has 600 turns (Figure 28.33). It is placed with its plane perpendicular
to a horizontal magnetic field of uniform flux density 50mT. The coil is then rotated through 90° about
a vertical axis in a time of 120ms.

axis

coil [2]
[2]
Figure 28.33 For End-of-chapter Question 8. [2]

Calculate:
i the magnetic flux passing through the coil before the rotation
ii the change of magnetic flux linkage produced by the rotation
iii the average e.m.f. induced in the coil during the rotation.

Cambridge International A Level Physics

9 a State Lenz’s law and explain how you would use a coil and a magnet to demonstrate the law. [4]
Make clear any other apparatus that you use.
[3]
b A vehicle brake consists of an aluminium disc attached to a car axle (Figure 28.34). Electromagnets [2]
cause an e.m.f. to be induced in the disc.
car wheel [2]
[2]
aluminium disc [2]
car axle [2]

electromagnets

Figure 28.34 For End-of-chapter Question 9.
i Explain how the induction of an e.m.f. causes the vehicle to slow down.
ii Explain why the braking effect increases when the speed of the car increases.
10 A bicycle wheel is mounted vertically on a metal axle in a horizontal magnetic field (Figure 28.35).
Sliding connections are made to the metal edge of the wheel and to the metal axle.
450

magnetic field
spoke

connections

Figure 28.35 For End-of-chapter Question 10.
a i Explain why an e.m.f. is induced when the wheel rotates.

ii State and explain two ways in which this e.m.f. can be increased.
b The wheel rotates five times per second and has a radius of 15cm. The magnetic flux density may be

assumed to be uniform and of value 5.0 × 10−3T.
Calculate:
i the area swept out each second by one spoke
ii the induced e.m.f. between the contacts.

Chapter 29: 451
Alternating
currents

Learning outcomes

You should be able to:

■■ measure frequency and voltage using a cathode-ray
oscilloscope

■■ describe an alternating current or voltage in terms of
period, frequency, peak value and r.m.s. value

■■ relate r.m.s. and peak values for sinusoidal currents

■■ solve problems involving transformers

■■ explain the benefits of transmission of electrical energy
at high voltages

■■ explain how diodes and capacitors can be used to
produce rectified, smoothed currents and voltages

Cambridge International A Level Physics

Describing alternating current Figure 29.1  Public electricity supplies made possible new
forms of street lighting and advertising.
In developed countries, mains electricity is a supply
of alternating current (a.c.). The first mains electricity
supplies were developed towards the end of the 19th
century; at that time, a great number of different
voltages and frequencies were used in different places.
In some places, the supply was direct current (d.c.).
Nowadays this has been standardised across much of
the world, with standard voltages of 110 V or 230 V (or
similar), and frequencies of 50 Hz or 60 Hz.

In this chapter we will look at some of the reasons
why a.c. has been chosen as standard. First, however,
we must take a close look at the nature of alternating
currents.

Sinusoidal current

An alternating current can be represented by a graph such QUESTION
as that shown in Figure 29.2. This shows that the current

varies cyclically. During half of the cycle, the current is 1 The following questions relate to the graph of
positive, and in the other half it is negative. This means Figure 29.2.

that the current flows alternately one way and then the a What is the value of the current I when time
452 other in the wires in which it is travelling. Whenever t = 5 ms? In which direction is it flowing?
you use a mains appliance, current flows backwards and
forwards in the wires between you and the power station b At what time does the current next have the
same value, but negative?
where it is being generated. At any instant in time, the
current has a particular magnitude and direction given by c What is the time T for one complete cycle?
the graph. d What is the frequency of the alternating

The graph has the same shape as the graphs used to current?

represent simple harmonic motion (see Chapter 19), and

it can be interpreted in the same way. The electrons in a An equation for a.c.
wire carrying a.c. thus move back and forth with s.h.m.

The current varies like a sine wave and so it is described As well as drawing a graph, we can write an equation to

as sinusoidal. (In principle, any current whose direction represent alternating current. This equation tells us the

changes between positive and negative can be described value of the current I at any time t:

as alternating, but we will only be concerned with those I = I0 sin ωt
which have a regular, sinusoidal pattern.) where ω is the angular frequency of the supply measured

/A 3 in rad s−1 (radians per second). This is related to the

2 frequency f in the same way as for s.h.m.:
ω = 2πf
1
0 5 10 15 20 25 30 35 t / ms and the frequency and period are related by:
–1
f = 1
T

–2 The quantity I0 is known as the peak value of the
alternating current, found from the highest point on the
–3 graph. (It is the amplitude of the varying current.)

Figure 29.2  A graph to represent a sinusoidal alternating
current.

Chapter 29: Alternating currents

QUESTIONS V
2 The following questions relate to the graph of V0

Figure 29.2. 0T T 3T t
a What are the values of I0 and ω? 2 2
b Write an equation to represent this current.
3 An alternating current (measured in amps, A) is –V0
represented by the equation:
I = 5.0 sin (120πt) Figure 29.4  An alternating voltage.
a For this current what are the values of I0, ω
QUESTION
and f ? What is the period T of the oscillation?
b Sketch a graph to represent the current. 4 An alternating voltage V (in V) is represented by
the equation:
Alternating voltages
V = 300 sin (100πt)
Alternating current is produced in power stations by large a What are the values of V0, ω and f for this
generators like those shown in Figure 29.3. voltage?
b What is the value of V when t = 0.002 s? (Recall
Figure 29.3  Generators in the generating hall of a large power that 100πt is in radians when you calculate this.)
station. c Sketch a graph to show two complete cycles of
this voltage.
In principle, a generator consists of a coil rotating in a
magnetic field. An e.m.f. is induced in the coil according 453
to the laws of electromagnetic induction. This e.m.f. V
varies sinusoidally, and so we can write an equation to Measuring frequency and voltage
represent it which has the same form as the equation for
alternating current: An oscilloscope can be used to measure the frequency and
voltage of an alternating current. Box 29.1 explains how to
V = V0 sin ωt do this. There are two types of oscilloscope. The traditional
where V0 is the peak value of the voltage. We can also cathode-ray oscilloscope (c.r.o.) uses an electron beam.
represent this graphically, as shown in Figure 29.4. The alternative is a digital oscilloscope, which is likely
to be much more compact and which can store data and
display the traces later.

BOX 29.1: Measurements using an oscilloscope

A c.r.o. is an electron beam tube, as shown in Figure
27.4 (page 424), but with an extra set of parallel
plates to produce a horizontal electric field at right
angles to the beam (Figure 29.5).

The principles of a c.r.o.

The signal into the c.r.o. is a repetitively varying
voltage. This is applied to the y-input, which deflects
the beam up and down using the parallel plates Y1
and Y2 shown in Figure 29.5. The time-base produces
a p.d. across the other set of parallel plates X1 and X2
to move the beam from left to right across the screen.

Cambridge International A Level Physics

BOX 29.1: Measurements using an oscilloscope (continued)

anode vacuum two controls that you must know about are the time-
base and the Y-gain, or Y-sensitivity.

X2 Y2 You can see in Figure 29.6 that the time-base control
X1 has units marked alongside. Let us suppose that this
reads 5 ms/cm, although it might be 5 ms/division. This

heated Y1 shows that 1 cm (or 1 division) on the x-axis represents
5 ms. Varying the time-base control alters the speed with
cathode electron beam which the spot moves across the screen. If the time-base
is changed to 1 ms/cm, then the spot moves faster and
electron gun screen each centimetre represents a smaller time.

Figure 29.5  The construction of a cathode-ray The Y-gain control has a unit marked in volts/cm,
oscilloscope. Cathode rays (beams of electrons) are or sometimes volts/division. If the actual marking is
produced in the electron gun and then deflected by 5 V/cm, then each centimetre on the y-axis represents
electric fields before they strike the screen. 5 V in the applied signal.

When the beam hits the screen of the c.r.o. it It is important to remember that the x-axis
produces a small spot of light. If you look at the screen represents time and the y-axis represents voltage.

and slow the movement down you can see the spot Determining amplitude and frequency

454 move from left to right, while the applied signal moves If you look at the c.r.o trace shown in Figure 29.7, you
the spot up and down. When the spot reaches the right can see that the amplitude of the waveform is 2 cm
side of the screen it flies back very quickly and waits and the distance along the x-axis for one complete wave
for the next cycle of the signal to start before moving to is 4 cm.
the right once again. In this way the signal is displayed
as a stationary trace on the screen. There may be many If the Y-gain or Y-sensitivity setting is 2 V/cm, then
controls on a c.r.o., even more than those shown on the the highest voltage is 2 × 2 = 4 V. If the time-base setting
c.r.o. illustrated in Figure 29.6. is 5 ms/cm, then the time for one wave (the period) is
4 × 5 = 20 ms.

The controls Since the x-axis measures time, the c.r.o. trace can

The X-shift and the Y-shift controls move the whole trace be used to measure frequency. In the above example,
in the x-direction and the y-direction, respectively. The
since

period = 1
frequency

frequency = 1 = 50 Hz
0.02

brightness focus
time-base Y-gain

X-shi Y-shi 1 cm
Y input on Figure 29.7  A c.r.o. trace when a sinusoidal alternating
o current is applied to the Y-plates.

Figure 29.6  The controls of a typical c.r.o.





























































Chapter 30: Quantum physics

End-of-chapter questions [2]

1 Calculate the energy of a photon of frequency 4.0 × 1018Hz. [3]

2 The microwave region of the electromagnetic spectrum is considered to have wavelengths ranging [1]
from 5mm to 50cm. Calculate the range of energy of microwave photons. [1]
[1]
3 In a microwave oven photons of energy 1.02 × 10−5eV are used to heat food.
a Express 1.02 × 10−5eV in joules. [1]
b Calculate the frequency of the photons.
c Calculate the wavelength of the photons. [1]
[1]
4 a Alpha-particles of energy 5MeV are emitted in the radioactive decay of radium. Express this energy [1]
in joules.
485
b Electrons in an cathode-ray tube are accelerated through a potential difference of 10kV. State [2]
their energy: [1]
i in electronvolts [2]
ii in joules.
[3]
c In a nuclear reactor, neutrons are slowed to energies of 6 × 10−21J. Express this in eV. [1]

5 A helium nucleus (charge = +3.2 × 10−19 C, mass = 6.8 × 10−27 kg) is accelerated through a potential [2]
difference of 7500 V.

Calculate:
a its kinetic energy in electronvolts
b its kinetic energy in joules
c its speed.

6 Ultraviolet light with photons of energy 2.5 × 10−18J is shone onto a zinc plate. The work function of
zinc is 4.3eV.
Calculate the maximum energy with which an electron can be emitted from the zinc plate.
Give your answer:
a in eV
b in J.

7 Calculate the minimum frequency of electromagnetic radiation that will cause the emission of
photoelectrons from the surface of gold.
(Work function for gold = 4.9eV.)

Cambridge International A Level Physics

8 Figure 30.28 shows five of the energy levels in a helium ion. The lowest energy level is known as the ground state.

–2.8 eV n=5

–3.4 eV n=4
–6.1 eV n=3

–13.6 eV n=2

–54.4 eV n = 1 (ground state)

Figure 30.28 For End-of-chapter Question 8.

a Calculate the energy, in joules, that is required to completely remove the remaining electron, [2]
originally in its ground state, from the helium ion. [2]
[2]
b Calculate the frequency of the radiation which is emitted when the electron drops from the level [3]
n = 3 to n = 2. State the region of the electromagnetic spectrum in which this radiation lies.
[1]
c Without further calculation, describe qualitatively how the frequency of the radiation emitted when [2]
486 the electron drops from the level n = 2 to n = 1 compares with the energy of the radiation emitted [1]

when it drops from n = 3 to n = 2.

9 The spectrum of sunlight has dark lines. These dark lines are due to the absorption of certain wavelengths
by the cooler gases in the atmosphere of the Sun.
a One particular dark spectral line has a wavelength of 590nm. Calculate the energy of a photon with
this wavelength.
b Figure 30.29 shows some of the energy levels of an isolated atom of helium.

Energy / 10–19 J

0

–1.6

–2.4
–3.0

–5.8

–7.6
Figure 30.29 For End-of-chapter Question 9.

i Explain the significance of the energy levels having negative values.
ii Explain, with reference to the energy level diagram above, how a dark line in the spectrum

may be due to the presence of helium in the atmosphere of the Sun.
iii All the light absorbed by the atoms in the Sun’s atmosphere is re-emitted. Suggest why a

dark spectral line of wavelength of 590nm is still observed from the Earth.

Chapter 30: Quantum physics

10 Figure 30.30 shows three of the energy levels in an isolated hydrogen atom. The lowest energy level
is known as the ground state.
Energy / 10–19 J

–2.4 n = 3
–5.4 n = 2

–21.8 n = 1 (ground state)

Figure 30.30 For End-of-chapter Question 10. [1]
[2]
a Explain what happens to an electron in the ground state when it absorbs the energy from a photon [3]
energy 21.8 × 10−19J. [2] 487
[2]
b i Explain why a photon is emitted when an electron makes a transition between energy levels [2]
n = 3 and n = 2.
[2]
ii Calculate the wavelength of electromagnetic radiation emitted when an electron makes a jump [3]
between energy levels n = 3 and n = 2. [2]

iii In the diagram, each energy level is labeled with its ‘principal quantum number’ n. Use the
energy level diagram to show that the energy E of an energy level is inversely proportional to n2.

11 a i Explain what is meant by the wave–particle duality of electromagnetic radiation.
ii Explain how the photoelectric effect gives evidence for this phenomenon.
Figure 30.31 shows the maximum kinetic energy of the emitted photoelectrons as the frequency
of the incident radiation on a sodium plate is varied.
2.5

2.0

E / eV 1.5

1.0

0.5

00 2 4 6 8 10
f / Hz × 1014

Figure 30.31 For End-of-chapter Question 11.

b Explain why there are no photoelectrons emitted when the frequency of the incident light is less
than 5.6 × 1014Hz.

c Calculate the work function for sodium.
d Use the graph to calculate the value of the Planck constant in Js.

Cambridge International A Level Physics

12 a Explain what is meant by the de Broglie wavelength of an electron. [2]
b Figure 30.32 shows the principles of an electron tube used to demonstrate electron diffraction.
[1]
hot cathode electron beam di racted electrons [3]
graphite film [2]
[3]
anode
[3]
θ [2]
[3]
5.0 kV [2]
[2]
Figure 30.32 For End-of-chapter Question 12.

i Calculate the kinetic energy E (in joules) of the electrons incident on the graphite film.
ii Show that the momentum of an electron is equal to 2Eme where me is the mass of an electron,

and hence calculate the momentum of an electron. (me = 9.11 × 10–31kg).
iii Calculate the de Broglie wavelength of the electrons.
c Discuss how the wavelengths of neutrons and electrons moving with the same energy would compare.

13 Blue light of wavelength 450nm is incident on a light-dependent resistor (LDR) made from a
488 semiconductor material. The energy gap between the valence and conduction bands of the material of

the LDR is 2.4eV.
a Explain what is meant by the valence band and the conduction band in a semiconductor. You may

include a diagram if you wish.
b Show that the energy of a photon of the blue light is about 3eV.
c Use band theory to explain why:

i the blue light causes the resistance of the LDR to decrease
ii the resistance of the LDR further decreases as the intensity of the blue light increases.
d Calculate the maximum wavelength of radiation which this LDR could detect.

14 Figure 30.33 shows the variation with temperature of the resistance of a semiconductor and of a metal.

Resistance metal

0 0 semiconductor [3]
Temperature [3]

Figure 30.33 For End-of-chapter Question 14.
In terms of the band theory of solids and the behaviour of free electrons, explain the changes in resistance of:
a the metal
b the semiconductor.