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20.1Electromagnetic induction
We know a current produces a magnetic field. Can a magnetic field produce a current? This is the question this chapter deals with.
Motional emf
A rod of length L is moved with velocity v in a region of a magnetic flux density B. B is directed into the plane of the page; the rod moves from left to right (Figure 20.1).

Figure 20.1: The rod is made to move normally to the magnetic field at constant speed. An emf develops between the ends of the rod.

The rod is conductingâthat is, it has many âfreeâ electrons. As it moves, the electrons within it also move from left to right. The magnetic field will exert a force on these moving electrons. The force on the electrons is directed downward (green arrow) and therefore the electrons are pushed downward. This means that the lower end of the rod has a net negative charge and the top end has an equal net positive charge. (The net charge of the rod is zero.) The flow of electrons towards the bottom end of the rod will stop when the electrons already there are numerous enough to push any new electrons back by electrostatic repulsion. There is, in other words, an electric field established in the rod whose direction is from top to bottom.
The value of this electric field E is given by:
E=ÎµL
where Îµ is the potential difference between the ends of the rod, known as the induced emf. The flow of electrons will stop when the electric force eE pushing the electrons back equals the magnetic force evB. Thus:
eE=evB
Dividing both sides by e and substituting for the electric field, this becomes:
Îµ=BvL
We have found the extraordinary result that a conducting rod of length L moving with speed v normally to a magnetic field B will have a potential difference BvL across its ends. This is called a motional emf, as it has been induced as a result of the motion of the conductor in the magnetic field.
We can check that the quantity BvL has the unit of potential difference (volt):
[BÃvÃL]=Â TeslaÂ ÃÂ metre per secondÂ ÃÂ metreÂ =TÃm sâ1Ãm
Substitute Tesla for NewtonAmpere metre:
(NA m)Ã(m sâ1)Ãm
Cancel out:
(NAÂ m)Ã(mÂ sâ1)Ãmâ(NA)Ãsâ1Ãm
Rearrange;
NmAsâ1
Substitute Amperes for Coulombs per second, and Newton metres for Joules, then cancel out:
JCsâ1Sâ1âJC
1Â Joule per CoulombÂ =1Â VoltÂ âJC=V
It is important to note that except for a very short interval of time initially, no current exists in the rod. But this example opens the way for generating an electric current out of magnetic fields.
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CHECK YOURSELF 1
Three loops enter a region of magnetic field with the same speed as shown in Figure 20.2.

Figure 20.2: For check yourself question 1.

Just as the loops enter the region of magnetic field, which loop has the larger induced emf and which has the smallest?
Suppose we modify things by letting the rod slide on two wires that are joined by resistor of resistance R, as shown in Figure 20.3. Now the moving rod behaves like a battery. There is a potential difference between the top and the bottom equal to BvL (this is the emf of the âbatteryâ) and a current equal to I=BvLR is established in the resistor and the moving rod, i.e. in the circuit on the left side of the diagram. This is because electrons in the bottom part of the circuit now have the opportunity to move up through the resistor; thus there is a counterclockwise current in the circuit.

Figure 20.3: As the rod is pushed along, a current is established in the circuit.

Notice also that now that we have a current, the rod needs to be pushed if it is to continue to move at constant speed. This is because the rod carries current I and is in a magnetic field, so it experiences a magnetic force F directed to the left (use the right-hand rule for force and the fact that the current is counterclockwise) given by:
F=BIL=BBvLRL=vB2L2R
For speed to remain constant a force of equal magnitude needs to act on the rod, directed to the right. The power P generated by this force is:
P=Fv=v2B2L2R
The power dissipated in the circuit as thermal energy in the resistor is:
P=Îµ2R=v2B2L2R
This is in perfect agreement with conservation of energy: the work done by the agent pushing the rod is dissipated in the resistor. Here, mechanical work (pushing the rod) is transferred into thermal energy.
EXAM TIP
The work done to move an electron from top to bottom in the wire is W = FL. The force is eE which equals evB and so the work done is W = evBL. The work done per unit charge is the emf, i.e. Îµ = BvL, as expected.
CHECK YOURSELF 2
A sheet of metal moves at constant speed in a region of uniform magnetic field that is directed into the plane of the page.
On a copy of Figure 20.4, show how charge separates in the sheet of metal.

Figure 20.4: For check yourself question 2.

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Magnetic flux and magnetic flux linkage
In 1831, Faraday experimented with coils of wire wrapped around an iron ring (Figure 20.5). He was hoping that somehow the current in the left circuit might induce a current in the right circuit. No such current was observed in the right circuit, but Faraday did notice that a small current was induced only during the opening and closing of the switch.

Figure 20.5: As the switch is closed a small current is registered by the galvanometer. While the switch remains closed a current exists in the circuit to the left but the current in the right circuit is zero. As the switch is opened another small current is established in the right circuit.

Similar results are obtained when a magnet is moved in or out of a coil of wire that is connected to a sensitive galvanometer (Figure 20.6). A current is induced.

Figure 20.6: As the magnet is allowed to enter the coil a current is induced in the coil and is registered by the galvanometer. If the magnet is then pulled out of the coil the induced current is opposite.

If the magnet is simply placed near the coil but does not move relative to it, nothing happens. The current is created as a result of the motion of the magnet relative to the coil. If we move the coil toward the magnet, we again find a current. This indicates that it is the relative motion of the coil and magnet that is responsible for the effect. If the magnet moves towards the coil faster, the reading on the galvanometer is greater. If a magnet of greater strength is used, the current produced is greater. If we try a coil with more turns of wire, we again find a greater current. We also observe that if the area of the loop is increased, the current also increases. But if the magnet is moved at an angle to the plane of the loop other than a right angle, the current decreases. To summarise, the observations are that the current registered by the galvanometer increases when:
âthe relative speed of the magnet and the coil increases
âthe strength of the magnet increases
âthe number of turns increases
âthe area of the loop increases
âthe magnet moves at right angles to the plane of the loop.
Faraday found that the common thread behind these observations is the concept of magnetic flux. Imagine a loop of wire, which for simplicity we take to be planar (i.e. the entire loop lies on one plane). If this loop is in a region of magnetic field whose magnitude and direction is constant, then we define magnetic flux as follows.
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KEY POINT
The magnetic flux Î¦ through the loop is:
Î¦=BAcosÎ¸
where B is the magnetic flux density, A is the area of the loop and Î¸ is the angle between the magnetic field direction and the direction normal to the loop area (Figure 20.7).

Figure 20.7: The definition of magnetic flux, Î¦ = BA cos Î¸.

EXAM TIP
Normal surfaces have two sides and so two normals: it does not matter which one we take.
If the loop has N turns of wire around it, the flux is given by:
Î¦=NBAcosÎ¸
The unit of magnetic flux is the weber (Wb). 1 Wb = 1 T m2.
KEY POINT
The magnetic flux multiplied by the number of turns in a loop is called the flux linkage.
This means that if the magnetic field is along the plane of the loop, then Î¸ = 90Â° and hence Î¦ = 0 (Figure 20.8a). The maximum flux through the loop occurs when Î¸ = 0Â°, when the magnetic field is normal to the loop area and its value is then BA (Figure 20.8b).

Figure 20.8: a The loop is not pierced by any magnetic field lines, so the flux through it is zero. b The magnetic field is normal to the loop, so the flux through it is the largest possible. c The flux here is BA cos Î¸.

The intuitive picture of magnetic flux is the number of magnetic field lines that pierce the loop area. Note that if the magnetic field went through only half the loop area, the other half being in a region of no magnetic field, then the flux would be Î¦=BA2. In other words, what counts is the part of the loop area that is pierced by magnetic field lines.
WORKED EXAMPLE 20.1
A loop of area 8.0 cm2 is in a constant magnetic field of B = 0.15 T. What is the magnetic flux through the loop when:
athe loop is perpendicular to the field
bthe loop is parallel to the field
cthe normal to the loop and the field have an angle of 60Â° between them?
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Answer
aIn this case Î¸ = 0Â° and cos 0Â° = 1. The area of the loop is 8.0 Ã 10â4 m2. Substituting in Î¦ = BA cos Î¸, the flux Î¦ is given by:
Î¦=0.15Ã8.0Ã10â4
Î¦=1.2Ã10â4Â Wb
bIn this case Î¸ = 90Â° and cos 90Â° = 0, so Î¦ = 0.
cIn this case Î¸ = 60Â°, so:
Î¦=0.15Ã8.0Ã10â4Ã0.5
Î¦=6.0Ã10â5Â Wb
Faradayâs law
What does magnetic flux have to do with how a magnetic field can create an electric current? The answer lies in a changing magnetic flux linkage. In Figure 20.6 we had a magnetic flux linkage through the coil, which was changing with time. As a magnet is brought closer to the loop area, the value of the magnetic field at the loop position is increasing and so is flux. If the magnet is held stationary near the loop, there is flux through the loop but it is not changingâso nothing happens. If the number of turns is increased, so is the flux linkage. Thus, we are led to consider the rate of change of magnetic flux linkage through the loop.
A changing flux creates an induced emf, not necessarily a current. There will only be a current if the loop is conducting; that is, if the resistance of the circuit is not infinite. For example, a loop containing an ideal voltmeter cannot let current through, but there will be an emf if the flux is changing.
KEY POINT
Faraday found that the induced emf is equal to the (negative) rate of change of magnetic flux linkage, that is:
Îµ=âNÎÎ¦Ît. This is known as Faradayâs law of electromagnetic induction.
The minus sign need not concern us too much, as mostly we will be dealing with the magnitude of the induced emf. However, if we use calculus, we need the minus sign:
Îµ=âNdÎ¦dt
CHECK YOURSELF 3
Figure 20.9 shows how the flux in a loop changes with time.

Figure 20.9: For check yourself question 3.

State a time at which the magnitude of the induced emf is a maximum.
WORKED EXAMPLE 20.2
The magnetic field through a single loop of area 0.20 m2 is changing at a rate of 4.0 T sâ1. What is the induced emf?
Answer
The magnetic flux through the loop is changing because of the changing magnetic field.
Using Î¦ = BA and Îµ=ÎÎ¦Ît:
Îµ=Î(BA)Ît=AÎBÎt=0.20Ã4.0=0.80Â V
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WORKED EXAMPLE 20.3
A pair of conducting rails is placed in a uniform magnetic field of magnitude 0.40 T directed downward, as shown in Figure 20.10. The rails are a distance L = 0.20 m apart. A rod is placed on the rails and pushed to the right at constant speed v = 0.60 m sâ1. What is the induced emf in the loop formed by the rod and the rails?

Figure 20.10: A rod on a pair of conducting rails.

Answer
We looked at this problem at the beginning of this section, but now we will solve it the âeasyâ way using the concept of a changing flux and Faradayâs law.
The flux in the loop is changing since the area of the loop is increasing. Therefore there will be an emf induced.
In a time interval ât the rod will move to the right a distance v ât and so the area will increase by âA = Lv ât (Figure 20.11).

Figure 20.11: As the rod moves along the rails, the area of the loop increases.

Using Îµ=ÎÎ¦Ît and Î¦ = BA, we see that Îµ=BÎAÎt and
so:
Îµ=BÃLvÎtÎt=BLv=0.40Ã0.20Ã0.60=48mV
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CHECK YOURSELF 4
A loop of conducting wire is in a region of uniform magnetic field directed out of the plane of the page. The loop can be rotated about any one of the three axes shown in Figure 20.12.

Figure 20.12: For check yourself question 4.

Rotation about which axis will give no change in magnetic flux?
Lenzâs law
Having seen that a changing magnetic flux will produce an emf and therefore a current in a conducting loop of wire, we now move to the interesting problem of determining the direction of this induced current. This is given by a law discovered by the German-Russian physicist Heinrich Lenz (1804â1865).
KEY POINT
Lenzâs law states that the induced emf will be such so as to oppose the change in the magnetic flux. It is equivalent to energy conservation.
What does this mean in practice? The flux in the loop is created by a magnetic field which we will call the external field. Now:
If the flux increases, the induced emf will create a current whose magnetic field is opposite to the external field.
If the flux decreases, the induced emf will create a current whose magnetic field is parallel to the external field.
This is a subtle and tricky formulation so let us try to understand it. Look at Figure 20.13. We know that the flux is increasing in the loop so an emf and a current will be induced.

Figure 20.13: a The rod is made to move to the right. The magnetic flux through the loop is increasing and a current will be established in the rod. b By Lenzâs law, the current must produce a magnetic field opposite to the existing magnetic field, i.e. out of the page. So the current must be counterclockwise.

Lenzâs law says we have to oppose the change in magnetic flux. This change is an increase in flux. Opposing the increase means we have to decrease the flux. One way of decreasing the flux is to reduce the magnetic field by creating another magnetic field inside the loop in the opposite direction, i.e. out of the page. This is the field in blue in Figure 20.13b. This is produced by the induced current. So now we ask: what should the direction of the induced current be, so that the field it produces is out of the page in the loop. By the right-hand rule, the current must be counterclockwise. (Alternatively, you can argue as follows: we want to oppose the increase in flux. A way of doing that is to prevent the rod from moving to the right. We can do that by having a force act on the rod, directed to the left. This force would be generated if a counterclockwise current flows in the loop.)
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Notice that Lenzâs law is just a statement of energy conservation: imagine that the current in the loop was clockwise. What would happen then? By the right-hand rule, the magnetic force on the rod would be directed towards the rightâin the direction of motion of the rod. The rod will therefore accelerate and its kinetic energy will increase. But there is nowhere that this energy can come from; energy conservation would be violated.
Let us look at another example. Consider the situation in which the current in the straight wire is decreasing. A loop of conducting wire is next to the wire, as shown in Figure 20.14a.

Figure 20.14: a A loop of wire near a straight wire in which the current is decreasing. b The current in the straight wire creates a magnetic field into the page at the position of the loop. The induced current in the loop produces a magnetic field in the same direction as to oppose the change in flux.

There will be an emf induced in the loop because the flux is changing; it is changing because the current is decreasing and so the magnetic field it produces decreases. What is the direction of the induced current? The change in the magnetic flux is a decrease in magnetic flux (the green field created by the green current in the wire is decreasing). We must oppose this decrease, i.e. we must create a magnetic field in the same direction as the green field, i.e. into the page inside the loop, Figure 20.14b. So the question now is: what is the direction of the current in the loop such that the field it produces is into the page inside the loop? From the right-hand rule, the current must be clockwise.
(Alternatively, you can argue as follows: we want to oppose the decrease in flux. A way of doing that is to make the loop come closer to the wire where the magnetic field is stronger. We can do that by having a force act on the loop, directed to the left. This force would be generated if a clockwise current flows in the loop.)
EXAM TIP
To work with Lenzâs law you must be very familiar with the rules for giving the direction of a magnetic field created by a current and the rule that gives the magnetic force on a current.
CHECK YOURSELF 5
A conducting loop enters and then leaves a region R of uniform magnetic field directed into the page, as shown in Figure 20.15. The loop moves with constant speed.

Figure 20.15: For check yourself question 5.

What is the direction of the induced current as the loop:
aenters region R
bleaves region R?
Let us make sure that we understand what is going on by looking at another example.
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WORKED EXAMPLE 20.4
A loop of wire has its plane horizontal, and a bar magnet is dropped from above so that it falls through the loop with the north pole first, as shown in Figure 20.16. Find the direction of the current induced in the loop.

Figure 20.16: A magnet is dropped into a loop of wire.

Answer
The flux in the loop is increasing because the magnetic field at the loop is getting larger as the magnet approaches. The induced current must then oppose the increase in the flux. This can be done if the induced current produces a magnetic field in the opposite direction to that of the bar magnet, as shown by the blue arrow in Figure 20.17a. Thus, the current will flow in a counterclockwise direction when looked at from above. This also means that there is force of repulsion between the coil and the magnet so the magnet slows down.
As the magnet leaves the loop from the other side, the flux is decreasing. So the current induced must produce a magnetic field in the same direction, i.e. down. This means the current is clockwise looked at from above, as shown in Figure 20.17b. (It follows that since the current changes from counterclockwise to clockwise, at some point it must be zero.) There is now a force of attraction between the coil and the magnet so the magnet again slows down.

Figure 20.17: Current induced by magnet.

Self induction
Suppose we connect a coil to a cell and a variable resistor R. By changing the resistance of R the current in the circuit will be changing; the magnetic flux linkage in the coil will therefore also be changing and an emf ÎµL will be induced in the coil. We call this phenomenon self induction. By Lenzâs law, the induced emf will oppose the change in the current. For example, suppose that the current in a coil is increasing, Figure 20.18a.

Figure 20.18: When the current is a increasing or b decreasing there is an induced emef in the coil.

The polarity of the induced emf is such as to drive current in the opposite direction to that of the current from the cell. In this way the induced emf opposes the increase in the current by decreasing the current in the circuit. If, instead, the current is decreasing, the polarity of the induced emf is shown in Figure 20.18b.
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TEST YOUR UNDERSTANDING
1The flux through a loop as a function of time is given by the graph in Figure 20.19. Sketch a graph of the emf induced in the loop as a function of time.

Figure 20.19: For question 1.

2The flux through a loop as a function of time is given by the graph in Figure 20.20. Sketch a graph of the emf induced in the loop as a function of time.

Figure 20.20: For question 2.

3Figure 20.21 shows the emf induced in a loop as a result of a changing flux in the loop.

Figure 20.21: For question 3.

aSketch a possible flux versus time graph that would give rise to such an emf.
bExplain why there isnât a unique answer.
4A loop of area 0.20 m2 has five turns of wire around it and finds itself in a region of uniform magnetic field 0.35 T, as shown in Figure 20.22. Initially the field is normal to the loop.

Figure 20.22: For question 4.

The loop is rotated by 180Â° about the axis shown in a time equal to 0.50 s. What is the average emf induced in the loop?
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5Two loops of wire enter a region of uniform magnetic field directed into the plane of the paper, as shown in Figure 20.23. The loops have the same area and speed.

Figure 20.23: For question 5.

Is the induced emf in the loops the same? Explain your answer.
6A current carrying wire is placed along the axis of a cylinder of length L and radius R, as shown in Figure 20.24.

Figure 20.24: For question 6.

The magnetic flux density created by the current on the surface of the cylinder is B. What is the magnetic flux through the cylinderâs surface area?
7Figure 20.25 shows a top view of three solenoids with their axes parallel. One of the small solenoids fits within the larger one. The third solenoid is outside the big solenoid. The bigger solenoid has a current flowing in the clockwise direction (looked at from above) and the current is increasing in magnitude.

Figure 20.25: For question 7.

Find the direction of the induced current in the smaller solenoids.
8A metallic ring is dropped from above a bar magnet as shown in Figure 20.26.

Figure 20.26: For question 8.

Determine the direction of the induced current in the ring as the ring falls over the magnet in each case, giving full explanations for your choices.
9A magnet is dropped from above into a metallic ring as shown in Figure 20.27.

Figure 20.27: For question 9.

Determine the direction of the current induced in the ring in each case.
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10For the diagram in question 9a, determine and justify the direction of the magnetic force on the ring as:
athe magnet approaches the ring.
bthe magnet moves away from the ring.
11A metallic rod of length L (shaded region) is dragged with constant velocity v in a region of magnetic field directed into the page, as shown in Figure 20.28.

Figure 20.28: For question 11.

By considering the force on electrons inside the rod, show that the ends of the rod will become oppositely charged. Determine the end that is positively charged.
12Find the direction of the current in the loop shown in Figure 20.29 as the loop is moved:
atowards to the wire
baway from the wire.

Figure 20.29: For question 12.

13An aluminium ring hangs from a string. A magnet is moved closer to the ring, as shown in Figure 20.30.

Figure 20.30: For question 13.

Determine which way the ring will move as the magnet approaches.
14A large coil has a smaller coil inserted inside it so that their axes are parallel. The smaller coil has 200 turns and a diameter of 2.0 cm. A changing current in the large coil causes the magnetic field to be increasing at a rate of 0.45 T sâ1. Calculate the emf induced in the smaller coil.
15Look at Figure 20.31. The conducting rod AB is free to move. The magnetic field is increasing.

Figure 20.31: For question 15.

Determine what will happen to the rod AB.
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16When lightning strikes, large magnetic fields are produced. This can be modelled by a long, straight, current-carrying wire vertical to the surface of the earth. In one strike, the magnetic field at the position of a rectangular loop of wire some distance from the strike point drops from 6.0 T to zero in a time of 8.0 Âµs. The loop has area 0.40 m2 and has 50 turns of wire around it.

Figure 20.32: For question 16.

What is the average induced emf in the loop if the plane of the loop is:
aparallel to the ground (Figure 20.32a)
bnormal to the ground (Figure 20.32b)?
17A loop of wire is placed in a region of magnetic field. In half the loop the magnetic field is out of the page and in other half the field is into the page. The area of the loop is A and the magnetic field in both directions has magnitude B. (See Figure 20.33.)

Figure 20.33: For question 17.

aWhat is the flux in the loop?
bThe magnitude of the magnetic field is increasing at the same rate in both directions. Is there an induced emf in the loop?
cThe magnitude of the field out of the page is increasing, and the field into the page is decreasing at the same rate. Is there an induced emf in the loop?
18A magnet is attached to a spring. The magnet oscillates in and out of a coil, as shown in Figure 20.34.

Figure 20.34: For question 18.

aDraw a sketch graph to show the variation with time of the displacement of the magnet when:
ithe switch is open
iithe switch is closed.
bExplain your sketches in part a.
19Two identical rings made out of conducting material are released from rest, from the same height above the ground and at the same time. One ring will fall through a region of a horizontal magnetic field. (See Figure 20.35.)

Figure 20.35: For question 19.

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State and explain which ring will reach the ground first.
20A clockwise current, when looked at from above, is established in a loop of wire X. (See Figure 20.36; note: connections to a battery are not shown.) A second conducting loop Y is placed underneath X so their planes are parallel.

Figure 20.36: For question 20.

aThe current in X is increasing.
iWhat is the direction of the induced current in Y?
iiWhat is the direction of the magnetic force on Y?
bThe current in X is now constant.
iIs there magnetic flux in Y?
iiIs there an induced emf in Y?
cThe current in X is now switched off. It takes 0.10 ms for the current to be reduced to zero.
iExplain why there will be an emf induced in Y during the 0.1 ms time interval.
iiWhat is the direction of the magnetic force on Y during this time?
21A loop of wire including a battery of emf 1.5 V is moved at speed 5.0 m sâ1 into a region of uniform magnetic field directed into the page (see Figure 20.37). The total resistance of the loop is 3.0 Î©. The magnitude of the magnetic field is 0.30 T. The length of the coloured wire is 0.20 m.

Figure 20.37: For question 21.

aWhat is the current in the loop before it enters the field?
bWhat is the current in the loop as it enters the region of magnetic field?
