Faraday's law — changing flux induces EMF — Physics

Before this

Internal resistance and e.m.f.
Electromagnets — current creates a magnetic field · coming soon
Electric fields and field lines

This leads to

A current with no battery

Connect a coil of wire to a sensitive meter. There is no battery anywhere in the circuit. Now push a bar magnet into the coil. The meter's needle flicks to one side. Hold the magnet still inside the coil, and the needle returns to zero. Pull the magnet out, and the needle flicks the other way.

This should surprise you. In the circuits lectures, every current had a source of e.m.f.: a battery, which gives each coulomb of charge some energy from its chemicals. Here there are no chemicals. And there is something stranger. When the magnet sits still inside the coil, its field is strongest there, yet nothing happens. The current flows only while the magnet moves.

So the magnet's field alone does not make the current. Something about change does. This lecture finds out exactly what changes, and how fast it must change to give a certain e.m.f. The answer is one of the most useful laws in physics. Almost every power station and every phone charger depends on it.

First, a number for the field: B

A magnet's field can be drawn with field lines, just as the lecture on electric fields drew them for charges. Where the lines are crowded, the field is strong. Where they spread out, it is weak.

Physics gives this crowding a number, the magnetic flux density, B. Its unit is the tesla, T. A later lecture defines the tesla exactly, by the force a field exerts on a current. For now, a sense of size is enough. The Earth's field at the surface is about 25 to 65 microtesla (µT), depending on where you are. A typical fridge magnet is a few millitesla. The magnet in a hospital MRI scanner is usually 1.5 T or 3 T. (Sources: Wikipedia, "Earth's magnetic field"; MRI values as in the resonance lecture.)

Some books call B the "magnetic field strength". We use magnetic flux density, because it tells you what B measures: how much field passes through each square metre.

Flux: how much field passes through the coil

Picture one flat turn of wire, with area A, in a uniform field B. If the field passes straight through the turn, at right angles to its surface, the amount of field passing through is B × A. This product is called the magnetic flux, Φ (the Greek letter phi). You can think of it as a count of the field lines that pass through the turn.

Now tilt the turn. Fewer lines pass through it, because it presents a smaller face to the field. To describe the tilt, we use the normal: a line at right angles to the surface of the turn. The angle θ is measured between the field and the normal. Tilt the coil below and watch how many lines pass through it.

Explore — tilt the coil, count the flux

The coil is seen edge-on, in a field of 0.50 T pointing to the right. Its area is 0.040 m². Change θ, the angle between the field and the normal. The darker lines are the ones passing through the coil.

Angle θ: 0°

This box shows how the flux through a coil of area 0.040 m² in a field of 0.50 T depends on the angle θ between the field and the normal.

The coil seen edge-on at θ = 60°. It presents a smaller face to the field, so fewer field lines pass through it than at θ = 0°.

θ	cos θ	Φ = BA cos θ (Wb)
0°	1	0.020
30°	≈ 0.866	≈ 0.0173
60°	0.5	0.010
90°	0	0

You should have seen the flux fall from its largest value at θ = 0° to zero at θ = 90°. The face the coil presents to the field shrinks by the factor cos θ, so the flux does too.

Magnetic flux

For a flat surface of area $A$ in a uniform field $B$ , where θ is the angle between the field and the normal to the surface:

Φ = B A cos θ

The unit of flux is the weber, Wb: 1 Wb = 1 T m².

A real coil has many turns, say $N$ . The same flux passes through every turn, so the coil "links" the flux $N$ times. The product $N Φ$ is called the flux linkage. Its unit is also the weber; some books write "weber-turns".

N Φ = NB A cos θ

Faraday's law: the change is what counts

Now return to the magnet and the coil. Pushing the magnet in makes the field inside the coil stronger, so the flux linkage rises. Holding it still keeps the flux linkage high but steady. Pulling it out makes the flux linkage fall. The meter flicked only while the flux linkage was changing.

Faraday found more. A faster push gave a bigger flick. A coil with more turns gave a bigger flick. Both point to one quantity: how fast the flux linkage changes.

Faraday's law of electromagnetic induction

The size of the e.m.f. induced in a circuit equals the rate of change of the flux linkage through it:

ε = \frac{Δ ( N Φ )}{Δ t}

Many books write a minus sign in front: $ε = - Δ (N Φ) /Δ t$ . The sign gives the direction of the induced e.m.f., and the next lecture, on Lenz's law, is about it. This lecture works with sizes only.

Check that the units agree. One weber per second should be one volt. A tesla is a newton per ampere-metre, N/(A m). So 1 Wb/s = 1 T m²/s = 1 N m/(A s). An ampere-second is a coulomb, and a newton-metre is a joule, so this is 1 J/C. That is exactly a volt, the energy given to each coulomb, as in the lecture on e.m.f. ✓

That is it. That is the whole idea. A flux that is large but steady induces nothing. A small flux that changes quickly can induce a large e.m.f. The e.m.f. depends on how fast the flux linkage changes, never on how big it is.

Stop and think

A very strong magnet, 1.2 T at the coil, rests inside a coil of 500 turns. A much weaker magnet is pushed quickly into an identical coil. Which coil has the larger induced e.m.f.?

Show the answer

The coil with the weak, moving magnet. The strong magnet gives a huge flux linkage, but it is not changing, so the rate of change is zero and the e.m.f. is zero. The weak magnet's flux linkage is small, but it is changing, so there is an e.m.f. Asking "how big is the flux?" is the wrong question. Ask "how fast is it changing?"

Reading the law from a graph

"Rate of change" has a picture. On a graph of flux linkage against time, the rate of change is the gradient. So the e.m.f. at any moment is the steepness of the $N Φ$ – $t$ graph at that moment. A flat part of the graph means zero e.m.f., however high it sits. Build a flux linkage graph below and watch the e.m.f. graph that comes with it.

Graph builder — flux linkage and its rate of change

Set the flux linkage at each time. The top graph joins your points. The bottom graph shows the gradient of each piece: the rate of change of NΦ, which is the size of the induced e.m.f. A negative value means the e.m.f. acts the other way.

This box shows a flux linkage graph and the rate of change of flux linkage that goes with it. A magnet is pushed slowly into a coil, held still, then pulled out fast.

Pushing in over 0.04 s: a gradient of 1.0 ÷ 0.04 = 25 V. Holding still: a flat graph and no e.m.f., even though the flux linkage is at its largest. Pulling out in 0.02 s: a gradient of −50 V, twice as large and the other way.

Two things should stand out. The e.m.f. was largest where the graph was steepest, not where it was highest. And a flux linkage that falls gives an e.m.f. the other way. This is how examiners test the law: they give a graph of flux linkage against time and ask you to sketch the e.m.f. Read the gradient of each piece, and you have the answer.

Three ways to change the flux

Since $N Φ = NB A cos θ$ , there are three ways to change the flux linkage of a coil, and each one induces an e.m.f.

Change B. Move a magnet nearer or further away, or switch an electromagnet on or off.
Change A. Change the area of the coil that is inside the field, for example by sliding the coil into or out of the field.
Change θ. Turn the coil in the field. This is how a generator works.

Worked examples

Worked Example 1 — Switching off a field

A coil of 200 turns and area 0.050 m² sits in a uniform field of 0.10 T, at right angles to the plane of the coil. The field is switched off steadily over 0.020 s. Find the induced e.m.f.

Think first. The field is at right angles to the plane, so it is along the normal: θ = 0° and cos θ = 1. Only B changes.

At the start, $N Φ_{1} = NB A = 200 \times 0.10 \times 0.050 = 1.0$ Wb. At the end, $B = 0$ , so $N Φ_{2} = 0$ . The change is 1.0 Wb.

ε = \frac{1.0}{0.020} = 50 V

Check. Units: Wb/s = V. ✓ Switching the field off twice as fast would halve Δt and double the e.m.f. to 100 V.

Worked Example 2 — Turning a coil

The same coil (200 turns, 0.050 m²) is in a steady field of 0.40 T. It is turned so that θ goes from 0° to 90° in 0.10 s. Find the average e.m.f.

At the start, $N Φ_{1} = 200 \times 0.40 \times 0.050 \times cos 0^{\circ} = 4.0$ Wb. At 90°, cos 90° = 0, so $N Φ_{2} = 0$ .

ε = \frac{4.0}{0.10} = 40 V

This is an average over the 0.10 s. The flux changes slowly at first and fastest near the end, because cos θ hardly changes near 0° and changes fastest near 90°. So, as the coil turns, the e.m.f. is close to zero just as the flux is at its largest, and it peaks just as the flux passes through zero. The lecture on the generator builds this into a full cycle.

Worked Example 3 — The angle to the plane, and a reversed field

A coil of 50 turns measures 4.0 cm by 5.0 cm. A field of 0.30 T makes an angle of 30° with the plane of the coil. In 0.050 s, the field is reversed: same size, opposite direction. Find the average e.m.f.

Think first. Two traps hide here. The angle in the formula is measured from the normal, which is at 90° to the plane, so θ = 90° − 30° = 60°. And the area must be in square metres.

The area is $A = 0.040 \times 0.050 = 2.0 \times 1 0^{- 3}$ m². So the flux linkage is $N Φ = 50 \times 0.30 \times 2.0 \times 1 0^{- 3} \times cos 6 0^{\circ} = 0.015$ Wb.

When the field reverses, the flux linkage goes from +0.015 Wb to −0.015 Wb. The flux passes through zero and keeps going, so the change is 0.030 Wb, not 0.015 Wb.

ε = \frac{0.030}{0.050} = 0.60 V

Check. Using cos 30° by mistake would give about 1.04 V. The field makes only a shallow angle with the plane, so less of it passes through the coil: the smaller answer is the sensible one. ✓

Worked Example 4 — Pulling a coil out of a field

A square coil of 20 turns and side 0.10 m is pulled at a steady 2.0 m/s. It moves out of a field of 0.50 T, which points into the page and fills the shaded region. Find the e.m.f. while the coil is leaving.

As the coil moves right, the part of it inside the field shrinks. In a time t, a strip of width vt leaves the field.

Think first. Here B and θ stay the same. What changes is the area of the coil inside the field.

The whole coil starts inside the field, so $N Φ_{1} = 20 \times 0.50 \times 0.1 0^{2} = 0.10$ Wb. It ends outside, so $N Φ_{2} = 0$ . The coil must move its own width, 0.10 m, to leave, which takes $0.10 \div 2.0 = 0.050$ s.

ε = \frac{0.10}{0.050} = 2.0 V

There is a neat second route. In each second, a strip of area $Lv$ leaves the field, where $L$ is the side of the coil. So the flux linkage falls by $NB Lv$ every second, and $ε = NB Lv = 20 \times 0.50 \times 0.10 \times 2.0 = 2.0$ V. ✓ This form, $ε = B Lv$ for one wire, describes any straight conductor moving across a field.

Three traps

Trap 1 — Big flux, so big e.m.f.

The e.m.f. depends on the rate of change of flux linkage, never on its size. A coil in a huge steady field has no induced e.m.f. On a graph, look at the gradient, not the height.

Trap 2 — The angle measured from the plane

In Φ = BA cos θ, θ is the angle between the field and the normal. If a question gives the angle to the plane of the coil, subtract it from 90° first. A quick check: when the field lies along the plane, the flux must be zero.

Trap 3 — Units and reversals

Areas must be in m²: 1 cm² = 10⁻⁴ m², so 20 cm² is 2.0 × 10⁻³ m². Remember the N. And when a field reverses, the change in flux linkage is twice the starting value, because the flux goes from positive, through zero, to negative.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A coil of 150 turns has an area of 3.0 cm². A field of 0.20 T passes through it at right angles to its plane. (a) Find the flux through one turn and the flux linkage. (b) The field falls steadily to zero in 0.15 s. Find the induced e.m.f.

Hint 1
Convert the area first: 3.0 cm² = 3.0 × 10⁻⁴ m².

Hint 2
"At right angles to its plane" means along the normal, so θ = 0°. Then ε = change in NΦ ÷ time.

Full answer
(a) $Φ = B A = 0.20 \times 3.0 \times 1 0^{- 4} = 6.0 \times 1 0^{- 5}$ Wb. $N Φ = 150 \times 6.0 \times 1 0^{- 5} = 9.0 \times 1 0^{- 3}$ Wb. (b) $ε = 9.0 \times 1 0^{- 3} \div 0.15 = 0.060$ V, or 60 mV. Check: units Wb/s = V ✓, and the answer is small because the coil is small. ✓
The flux linkage of a coil rises steadily from 0 to 0.60 Wb between t = 0 and t = 0.20 s. It stays at 0.60 Wb until t = 0.40 s. It then falls steadily to 0 by t = 0.50 s. Find the size of the induced e.m.f. in each of the three time intervals, and sketch the e.m.f.–time graph.

Hint 1
The e.m.f. in each interval is the gradient of that piece of the graph.

Hint 2
First piece: 0.60 ÷ 0.20. Last piece: 0.60 ÷ 0.10.

Full answer
0 to 0.20 s: $0.60 \div 0.20 = 3.0$ V. 0.20 to 0.40 s: the graph is flat, so 0 V. 0.40 to 0.50 s: $0.60 \div 0.10 = 6.0$ V, in the opposite direction to the first. The sketch is three flat steps: 3.0 V, then 0, then 6.0 V on the other side of the time axis. Check: the fall is twice as fast as the rise, so its e.m.f. is twice as large. ✓
A coil of 80 turns and area 25 cm² is in a field of 0.12 T. At first the field makes an angle of 40° with the plane of the coil. The coil is turned in 0.040 s until the field is at right angles to its plane. Find the average induced e.m.f.

Hint 1
An angle of 40° to the plane is 50° to the normal. "At right angles to the plane" means θ = 0°.

Hint 2
Find $N Φ$ at θ = 50° and at θ = 0°, and subtract. Use $A = 2.5 \times 1 0^{- 3}$ m².

Full answer
$NB A = 80 \times 0.12 \times 2.5 \times 1 0^{- 3} = 0.024$ Wb. At the start, $N Φ_{1} = 0.024 cos 5 0^{\circ} \approx 0.01543$ Wb. At the end, $N Φ_{2} = 0.024$ Wb. The change is about 0.00857 Wb, so $ε \approx 0.00857 \div 0.040 \approx 0.21$ V. Check: if you used 40° instead of 50°, you would get about 0.14 V. The flux must increase as the coil turns to face the field, and it does: 0.0154 Wb rises to 0.024 Wb. ✓
A stretch. An aircraft with a wingspan of 60 m flies level at 250 m/s. Where it flies, the vertical part of the Earth's field is about 40 µT (a typical value). (a) Use the idea of Worked Example 4 to find the e.m.f. between the wingtips. (b) A passenger suggests connecting a lamp between the wingtips with a long wire inside the plane. Explain why the lamp would not light.

Hint 1
The wing is a straight conductor sweeping across the vertical field. Each second it sweeps an area of wingspan × speed.

Hint 2
(b) The wire to the lamp travels with the plane too. Does the flux through the loop made by the wing and the wire change?

Full answer
(a) $ε = B Lv = 40 \times 1 0^{- 6} \times 60 \times 250 = 0.60$ V. Check the conversion: 40 µT = 4.0 × 10⁻⁵ T. ✓ (b) The wire to the lamp crosses the same field at the same speed, so the same e.m.f. is induced in it, and the two e.m.f.s cancel round the loop. Seen another way: the loop made by the wing and the wire moves as a whole through a nearly uniform field, so the flux through it does not change. With no change in flux linkage, there is no e.m.f. round the loop and no current. An e.m.f. exists between the wingtips, but you cannot use it from inside the plane.

How the law was found

Michael Faraday searched for a way to make electricity from magnetism through most of the 1820s. On 29 August 1831, in the basement laboratory of the Royal Institution in London, he wound two coils of wire on opposite sides of an iron ring. When he connected a battery to one coil, the meter on the other coil flicked, then fell back to zero. When he disconnected it, the meter flicked the other way. A steady current did nothing. Only switching it on or off did. On 17 October 1831 he got the same effect by pushing a bar magnet into a coil. The American physicist Joseph Henry found induction independently, and published in 1832. Faraday's ring still survives at the Royal Institution, which describes it as the first transformer. (Sources: Royal Institution, "Faraday's ring-coil apparatus"; H. Bence Jones, The Life and Letters of Faraday, 1870.)

The bigger idea

Faraday's law turns motion into electricity. Turn a coil in a magnetic field, and its flux linkage changes again and again, so an e.m.f. appears again and again. That is a generator, and it is how most of the world's electricity is made, whether the coil is turned by steam, falling water or wind. Change the current in one coil, and the changing field induces an e.m.f. in a second coil wound next to it. That is a transformer. Both are later lectures in this topic.

The law also raises a question. Where does the energy come from? There is no battery. The energy must come from whoever pushes the magnet or turns the coil, so pushing a magnet into a coil must be harder when a current flows. The next lecture, on Lenz's law, shows that this is true, and that it is the reason for the minus sign in Faraday's law.

Key concepts in this lesson

Magnetic flux density B, measured in tesla, describes how concentrated a magnetic field is.
The magnetic flux through a flat area is Φ = BA cos θ, in webers, where θ is the angle between the field and the normal to the surface.
For a coil of N turns, the flux linkage is NΦ, because the same flux passes through every turn.
Faraday's law states that the size of the induced e.m.f. equals the rate of change of flux linkage, ε = Δ(NΦ)/Δt.
A large but steady flux induces no e.m.f., and on a graph of flux linkage against time, the e.m.f. is the gradient.
Flux linkage can be changed by changing B, the area inside the field, or the angle θ, and any of these induces an e.m.f.