The generator — motion to electricity — Physics

Before this

This leads to

The transformer — stepping voltage up or down
Graphs of sin, cos and tan · coming soon

A machine that fights back

The Lenz's law lecture ended with a strange fact. A generator is hard to turn when it supplies a large current, because the induced current opposes the turning. So every time someone switches on a kettle, a machine somewhere becomes a little harder to turn, and something must push it harder.

That machine is almost always spinning. In 2024, about 7% of the world's electricity came from solar panels, which have no moving parts. Almost all the rest came from generators that turn: driven by steam from coal, gas or nuclear fuel, by falling water, or by wind (source: Ember, Global Electricity Review 2025).

The Faraday's law lecture showed that a turning coil's e.m.f. is close to zero just as the flux through it is largest. It promised that this lecture would build that into a full cycle. Here it is: the e.m.f. at every instant of the turn, why it keeps reversing, and why a power station must spin at one exact speed.

Build a generator from what you know

Take a rectangular coil of $N$ turns and area $A$ . Put it in a uniform magnetic field of flux density $B$ , and turn it about an axis through its middle at a steady angular velocity $ω$ , in radians per second. One full turn takes a time $T$ , and the number of turns per second is the frequency $f$ . As in the circular motion lectures, $ω = 2 π f = 2 π / T$ .

Measure the angle $θ$ between the field and the normal to the coil, as in the Faraday's law lecture. Start the clock when the coil faces the field, so that $θ = 0$ at $t = 0$ . After a time $t$ , the coil has turned through $θ = ω t$ . The flux linkage is then

N Φ = NB A cos ω t

This rises and falls like a cosine graph. It is largest when the coil faces the field, and zero when the coil lies along the field. Half a turn later it is most negative, because the field then passes through the coil from the other side.

How big is the e.m.f. at each instant?

Faraday's law says the e.m.f. is the rate of change of flux linkage. The Faraday's law lecture also gave a second route for a wire moving through a field, $ε = B Lv$ . That route gives the e.m.f. at any instant, so use it here. This is the hard part of the lecture. Spend time with the figure.

The coil seen end-on, turning anticlockwise about its axis. Each long side moves along a circle. Only the part of its velocity across the field lines, v sin θ, cuts them; the part along the lines cuts nothing. When θ = 0 the sides move along the lines; when θ = 90° they move straight across them.

Let the two long sides of the coil each have length $L$ , and let the coil be $w$ wide, so $A = L w$ . Each long side moves in a circle of radius $w /2$ around the axis, at speed $v = ω \times \frac{w}{2}$ .

A wire only cuts field lines with the part of its velocity that goes across them. The figure shows that this part is $v sin θ$ . The part along the lines cuts nothing. So each long side has an e.m.f. of $B Lv sin θ$ .

The two long sides move in opposite directions, one up and one down. So their e.m.f.s push charge the same way round the loop, and they add. The short ends move along their own length, so they add nothing. For one turn of wire, the e.m.f. is $2 B Lv sin θ$ . Now put in $v = ω w /2$ :

2 B L \times \frac{ω w}{2} \times sin θ = B (L w) ω sin θ

Since $L w = A$ , the e.m.f. of one turn is $B A ω sin θ$ .

With $N$ turns in series, the e.m.f.s of all the turns add.

The e.m.f. of a generator

A coil of $N$ turns and area $A$ turns at angular velocity $ω$ in a uniform field $B$ . Timing from when it faces the field, its e.m.f. is

$ε = NB A ω sin ω t$

Its largest value, the peak e.m.f., is $ε_{0} = NB A ω$ .

That is it. That is the whole idea. Every generator in every power station is this formula, made larger and stronger.

Check the units. $B A$ is a flux, in Wb, and $ω$ is in s⁻¹ (the radian has no unit). So $NB A ω$ is in Wb/s, which the Faraday's law lecture showed is the volt. ✓

Stop and think

Where is the coil when its e.m.f. is largest: when it faces the field and holds the most flux, or when it lies along the field and holds none?

Show the answer

When it lies along the field ( $θ = 9 0^{\circ}$ ), holding no flux at all. At that moment the long sides move straight across the field lines, so the flux is changing fastest. When the coil faces the field, the sides move along the lines for an instant. The flux is at its largest, but it is not changing, so the e.m.f. is zero. The e.m.f. depends on how fast the flux changes, not on how much flux there is.

Now the Faraday's law lecture's promise, as a full cycle. The graph below shows the flux linkage and the e.m.f. over one turn.

One turn of the coil, which takes a time T. The small lines at the top show the coil edge-on, with the field running left to right. When the coil faces the field (0, T/2, T), the flux linkage is largest and the e.m.f. is zero. When the coil lies along the field (T/4, 3T/4), the flux linkage is zero and the e.m.f. is largest. The e.m.f. is the gradient of the flux-linkage graph, with the sign from Lenz's law.

Read the graph as the Faraday's law lecture taught: the e.m.f. is the gradient of the flux-linkage graph. Where the dashed curve is flat, at its top and bottom, the e.m.f. is zero. Where the dashed curve crosses the axis most steeply, the e.m.f. is largest. And the sign is opposite to the gradient, as Lenz's law requires: when the flux linkage is falling, the e.m.f. is positive.

Notice also that the e.m.f. changes sign every half turn. So the current in the coil reverses twice in every turn. This is an alternating current, called a.c. A turning coil produces it naturally.

What makes the e.m.f. bigger?

The peak e.m.f. is $ε_{0} = NB A ω$ . Each factor makes physical sense. More turns give more wires cutting the field, and a stronger field gives more lines to cut. A larger coil sweeps through more of the field, and a faster turn cuts the lines sooner. Double any one, and the peak e.m.f. doubles.

But one of the four does something extra. Before you try it, make a prediction.

Predict — what happens when the coil turns twice as fast?

A generator's coil is made to turn twice as fast, with nothing else changed. What happens to the graph of its e.m.f. against time?

This box explains what happens to the e.m.f. of a generator when its coil turns twice as fast.

The peaks become twice as tall, because $ε_{0} = NB A ω$ and $ω$ has doubled. The peaks also come twice as often, because each turn now takes half the time, so the frequency of the a.c. doubles too.

Check your prediction on the generator below. It has 50 turns of area 4.0 × 10⁻³ m² in a field of 0.25 T, so $NB A = 0.050$ Wb. The motion is shown slowed down 50 times, but every number is for the real speed.

Simulation — a generator, slowed down

Press Play to turn the coil. Watch the e.m.f. graph grow as it turns. Then switch to 100 turns per second, and try the split-ring commutator.

This table shows the generator (NBA = 0.050 Wb) at 50 turns per second. Then ω = 100π ≈ 314 rad/s and ε₀ ≈ 15.7 V.

t (ms)	Angle turned	NΦ (Wb)	ε with slip rings (V)	Output with a commutator (V)
0	0°	0.050	0	0
2.5	45°	≈ 0.035	≈ 11.1	≈ 11.1
5	90°	0	≈ 15.7	≈ 15.7
10	180°	−0.050	0	0
15	270°	0	≈ −15.7	≈ 15.7
20	360°	0.050	0	0

At 100 turns per second, every peak is ≈ 31.4 V and one turn takes 10 ms, so the peaks come twice as often.

You should have seen three things. At 50 turns per second the peaks reach about 15.7 V, and one turn takes 20 ms. At 100 turns per second the peaks reach about 31.4 V and come twice as often. And with the commutator, the negative halves flip up, so the output never reverses.

Getting the current out: slip rings and commutators

Two different parts can join the turning coil to the still wires outside.

With slip rings, each end of the coil is fixed to its own metal ring on the axle. A fixed carbon contact called a brush presses on each ring. Each end of the coil always connects to the same outside wire, so the outside circuit receives the coil's e.m.f. exactly as it is: alternating.

With a split-ring commutator, both ends of the coil connect to one ring cut into two halves. Every half turn, each brush moves from one half to the other. This swaps the connections at the moment the e.m.f. reverses, so the outside circuit always receives the e.m.f. the same way round. The output is direct current, but it rises and falls from zero to the peak twice in each turn.

The history follows the same path. Michael Faraday built the first generator in 1831: a copper disc turning between the poles of a magnet. In 1832 the French instrument maker Hippolyte Pixii turned a magnet past fixed coils. His current reversed each time a different pole passed, so he added a commutator. Large power-station generators today work like Pixii's machine, not like the coil in this lecture. A spinning electromagnet turns inside fixed coils. Only the relative motion matters, and fixed coils are easier to connect to the very large current they carry.

Worked examples

Worked Example 1 — A generator at different instants

A coil of 50 turns and area 4.0 × 10⁻³ m² turns at 50 turns per second in a uniform field of 0.25 T. Find the peak e.m.f., and the e.m.f. 2.5 ms and 5.0 ms after the coil faces the field.

Think first. The formula needs $ω$ in rad/s, not the turns per second.

$ω = 2 π f = 2 π \times 50 = 100 π \approx 314$ rad/s. Then $NB A = 50 \times 0.25 \times 4.0 \times 1 0^{- 3} = 0.050$ Wb, so

So $ε_{0} = NB A ω = 0.050 \times 100 π$ .

ε_{0} \approx 15.7 V

At $t = 2.5$ ms, $ω t = 100 π \times 0.0025 = π /4$ rad, which is 45°. So $ε = 15.7 sin 4 5^{\circ} \approx 11.1$ V.

At $t = 5.0$ ms, $ω t = π /2$ rad, which is 90°. So $ε = 15.7 sin 9 0^{\circ} \approx 15.7$ V, the peak.

Check. One turn takes $1/50$ s = 20 ms, so 5.0 ms is a quarter turn. After a quarter turn the coil lies along the field, which is exactly where the e.m.f. should peak. ✓ Your calculator must be in radians for $sin (ω t)$ , or you must turn $ω t$ into degrees first.

Worked Example 2 — Average or instant?

For the same generator, find the average e.m.f. over the first quarter turn, using Faraday's law. Compare it with the peak.

Think first. Faraday's law with a change gives an average over the time of the change, as in the Faraday's law lecture.

In the first quarter turn, the flux linkage falls from 0.050 Wb to 0, a change of 0.050 Wb. The time is 5.0 ms. So the average e.m.f. is $0.050 \div 0.0050 = 10.0$ V.

Check. The average must lie between 0 and the peak of 15.7 V, because the e.m.f. rises from 0 to 15.7 V during this quarter turn. ✓ The ratio is $10.0 \div 15.7 \approx 0.64$ , which is $2/ π$ . The average over a quarter turn is always about 64% of the peak, whatever the generator.

Worked Example 3 — Why a power station cannot just spin faster

The same generator must give a peak e.m.f. of 31.4 V instead of 15.7 V. Which of $N$ , $B$ , $A$ and $ω$ should be doubled if the generator supplies mains electricity at 50 Hz?

Think first. Doubling any one factor doubles $ε_{0}$ . But one factor also changes the frequency.

Doubling $ω$ would give 31.4 V at 100 Hz, and the supply must stay at 50 Hz. So a power station keeps its speed fixed. It raises the e.m.f. through the other factors: more turns, a larger coil, or a stronger field. In a power station the field comes from an electromagnet, so it can be adjusted by changing the current in it.

Not every power-station generator spins at 50 turns per second, though. A rotor with one north and one south pole gives one cycle per turn, so it turns at 3000 turns per minute. Steam turbines usually drive generators like this. A rotor with more poles gives more cycles per turn. Water turbines turn slowly, sometimes at only about 100 turns per minute. At that speed, 50 Hz needs 30 north–south pairs, which is 60 poles.

Check. $\frac{100}{60}$ turns per second × 30 cycles per turn = 50 cycles per second. ✓

Why a generator is hard to turn

Now the Lenz's law lecture's promise. With nothing connected, no current flows, and the coil turns easily. When a lamp is connected, a current flows. Lenz's law says the current's magnetic effect opposes the change that causes it, so the field pushes back on the coil against its turning. The bigger the current, the bigger this push.

This is energy conservation, seen from inside the machine. Every joule the lamp receives must be supplied as work by whatever turns the coil.

A whole country's electricity system shows this on a grand scale. All the large generators in Great Britain turn together, locked to the 50 Hz of the grid. When demand rises suddenly, they become harder to turn, and they start to slow down. The frequency falls. Operators must add power at once, for example by letting more steam into turbines, to bring it back. The law sets limits: the frequency must stay between 49.5 Hz and 50.5 Hz (source: National Energy System Operator). The largest sudden rise from television, a "TV pickup", came on 4 July 1990, at the end of England and West Germany's World Cup semi-final. Millions of people switched on kettles at once, and demand jumped by 2800 MW (source: National Grid press release, 15 June 2016).

Three traps

Trap 1 — "Most flux means most e.m.f."

The e.m.f. depends on how fast the flux linkage changes, not on how large it is. When the coil faces the field, the flux is largest and the e.m.f. is zero. The e.m.f. peaks when the coil lies along the field and holds no flux at all.

Trap 2 — Turns per second in place of ω

$ε_{0} = NB A ω$ needs $ω$ in rad/s. Turn the frequency into $ω$ with $ω = 2 π f$ . If you are given turns per minute, divide by 60 first. Using 50 instead of 314 gives an answer about 6.3 times too small.

Trap 3 — "The generator makes electricity from nothing"

A generator neither creates charge nor creates energy. The charge is already in the wire, and the energy comes from the work done to turn the coil. This is why a generator becomes harder to turn as the current it supplies grows.

Try it yourself

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

A coil of 120 turns measures 5.0 cm by 4.0 cm. It turns at 25 turns per second in a field of 0.30 T. Find (a) the peak e.m.f. and (b) the e.m.f. one-eighth of a turn after the coil faces the field.

Hint 1
Convert the area to m² on its own line, and find $ω$ from the turns per second.

Hint 2
One-eighth of a turn is 45°, so $ε = ε_{0} sin 4 5^{\circ}$ .

Full answer
(a) $A = 0.050 \times 0.040 = 2.0 \times 1 0^{- 3}$ m². $ω = 2 π \times 25 = 50 π \approx 157$ rad/s. So $ε_{0} = 120 \times 0.30 \times 2.0 \times 1 0^{- 3} \times 50 π \approx 11.3$ V. (b) $ε = 11.3 \times sin 4 5^{\circ} \approx 8.0$ V. Check: 8.0 V is below the peak, and $sin 4 5^{\circ} \approx 0.71$ of it. ✓
A generator's e.m.f. graph has peaks of 6.0 V, and one cycle takes 20 ms. Its coil has 40 turns of area 3.0 × 10⁻³ m². (a) Find the flux density of the field. (b) Describe the graph if the coil turns at half the speed.

Hint 1
Find $f$ from the time for one cycle, then $ω$ .

Hint 2
Rearrange $ε_{0} = NB A ω$ for $B$ .

Full answer
(a) $f = 1 \div 0.020 = 50$ Hz, so $ω = 100 π \approx 314$ rad/s. Then $B = ε_{0} \div (N A ω) = 6.0 \div (40 \times 3.0 \times 1 0^{- 3} \times 314) \approx 0.16$ T. (b) Half the speed halves both: peaks of 3.0 V, and one cycle takes 40 ms. Check: $40 \times 0.159 \times 3.0 \times 1 0^{- 3} \times 314 \approx 6.0$ V. ✓
A water turbine drives a generator whose rotor has 48 poles (24 north–south pairs). (a) At what speed, in turns per minute, must it turn to supply 50 Hz? (b) The same machine is used in a country whose supply is 60 Hz. What speed is needed there?

Hint 1
Each north–south pair passing a coil gives one cycle of the e.m.f.

Hint 2
Cycles per second = pairs × turns per second.

Full answer
(a) Turns per second $= 50 \div 24 \approx 2.083$ , so turns per minute $= 2.083 \times 60 = 125$ . (b) $60 \div 24 = 2.5$ turns per second, which is 150 turns per minute. Check: $24 \times 125 \div 60 = 50$ Hz. ✓ The frequency is set by the speed and the number of poles together.
A stretch. On 4 July 1990, demand on Britain's grid rose by 2800 MW within minutes. (a) How many 3.0 kW kettles is that? (b) Explain, using Lenz's law, why such a rise makes the grid's frequency fall unless operators act. (c) A two-pole generator normally turns at 3000 turns per minute. At what speed is the frequency 49.5 Hz, the lowest the law allows?

Hint 1
(a) Divide the powers, with both in watts.

Hint 2
(b) More current in the coils means a bigger opposing push. What happens to a machine whose driving push stays the same? (c) A two-pole generator gives one cycle per turn.

Full answer
(a) $2800 \times 1 0^{6} \div 3000 \approx 933000$ kettles, nearly a million. (b) The extra kettles draw a larger current from every generator. By Lenz's law, the induced current's field opposes the turning, so each generator needs more work to keep its speed. If the steam or water supply does not increase at once, the generators slow down, and since the frequency is set by the speed, it falls. Operators prevent this by adding power quickly. (c) 49.5 cycles per second is 49.5 turns per second, which is $49.5 \times 60 = 2970$ turns per minute. Check: $2970 \div 3000 = 0.99$ , and $49.5 \div 50 = 0.99$ . ✓ A fall of just 1% in speed is the legal limit.

The bigger idea

A generator is a machine for moving energy, not making it. At one end, something turns a shaft: steam, falling water or the wind. At the other end, charge is pushed round a circuit. Between them sits one law, $ε = NB A ω sin ω t$ , and Lenz's law makes sure the account balances.

The same machine also runs backwards. Feed a current into the coil, and the field pushes it round: that is an electric motor. And the alternating e.m.f. this lecture produced is exactly what the next lecture needs. A transformer works only with a current that keeps changing, which is one reason the world's power grids use a.c.

Key concepts in this lesson

A coil of N turns and area A turning at angular velocity ω in a uniform field B has an e.m.f. ε = NBAω sin ωt, with a peak e.m.f. ε₀ = NBAω.
The e.m.f. is zero when the coil faces the field (most flux) and largest when the coil lies along the field (no flux), because the e.m.f. depends on how fast the flux linkage changes.
The e.m.f. reverses every half turn, so a turning coil produces alternating current.
Slip rings deliver the alternating e.m.f. unchanged; a split-ring commutator swaps the connections every half turn to give bumpy direct current.
Turning faster raises both the peak e.m.f. and the frequency, so power stations keep their speed fixed and the grid's frequency is set by that speed and the number of poles.
A generator moves energy from a turning shaft to a circuit, and by Lenz's law it becomes harder to turn as the current it supplies grows.