The AC generator

A lamp with no battery

A bicycle has a small lamp that runs from a dynamo. Pedal slowly and the lamp is dim. Pedal fast and it is bright. Stop, and it goes out. There is no battery anywhere. The energy comes from your legs.

How can a turning wheel make a voltage? The answer is Faraday's law. A coil that turns in a magnetic field has a changing flux, so it has an emf. A machine built on this idea is a generator.

The flux through a turning coil

Take a flat coil of $N$ turns and area $A$ in a uniform field $B$. Turn it at a steady rate. The angle $\theta$ between the normal to the coil and the field grows as $\theta = \omega t$, where $\omega$ is the angular speed in rad/s. We start the clock when the coil is at right angles to the field.

The flux through one turn is $\Phi = BA\cos\theta = BA\cos\omega t$. Faraday's law gives the emf of $N$ turns:

The largest value of the sine is $1$. So the peak emf is

Because $\omega = 2\pi f$, where $f$ is the number of turns each second, we can also write $e_0 = 2\pi f\,NBA$.

Look at one turn

Follow the coil round, and ask two questions at each place: how big is the flux, and how fast is it changing?

• Turned $0^\circ$. The coil is at right angles to the field. The flux is greatest. But the flux is at the top of a cosine curve, where it is changing most slowly. The emf is zero.
• Turned $90^\circ$. The plane of the coil lies along the field. No field line goes through it, so the flux is zero. But the cosine is falling fastest here. The emf is greatest.
• Turned $180^\circ$. The flux is greatest again, now with the other face towards the field. The emf is zero.
• Turned $270^\circ$. The flux is zero again. It is now rising fastest, so the emf is greatest in the other direction.

Many students expect the largest emf where the flux is largest. It is the other way round. The emf follows the slope of the flux, not the flux itself.

The emf changes direction twice in each turn, at the places where it passes through zero. A generator that turns $50$ times each second makes a current that reverses $100$ times each second.

A second look: the sides cut the field lines

We can find the same result from the motional emf. Let the sides parallel to the axle have length $l$, and let the other sides have length $b$, so $A = lb$. Each long side moves in a circle of radius $b/2$, so its speed is $v = \omega b/2$.

Only the part of the velocity at right angles to the field makes an emf. That part is $v\sin\omega t$. Each long side gives $Blv\sin\omega t$, and the two sides add up:

For $N$ turns, multiply by $N$. The two ways agree.

The parts of a generator

A real generator has a coil, wound on an iron core, that turns between the poles of a magnet. The two ends of the coil go to two metal slip rings on the axle. Fixed carbon brushes press on the rings and carry the current to the outside circuit. Because each ring stays joined to the same end of the coil, the outside current reverses with the coil. This is an alternating current, an AC.

A power station turns the coil with a turbine. Steam, falling water or wind drives the turbine. In a large machine the electromagnet turns and the coils are fixed, because it is easier to take a very large current from fixed coils. The emf is the same.

If we replace the two rings by a single ring cut in two, a split-ring commutator, the connections to the brushes are swapped every half turn. The outside current then keeps one direction, though it still rises and falls. This is a DC generator.

A worked example

A coil of $80$ turns has an area of $150\ \text{cm}^2$. It turns $1200$ times each minute in a field of $0.12$ T. Find the peak emf, and the emf $5.0$ ms after the coil was at right angles to the field.

1. The frequency. $f = \dfrac{1200}{60} = 20$ Hz, so $\omega = 2\pi f = 125.7$ rad/s.
2. The area. $150\ \text{cm}^2 = 150 \times 10^{-4}\ \text{m}^2 = 0.015\ \text{m}^2$.
3. The peak emf. $e_0 = NBA\omega = 80 \times 0.12\ \text{T} \times 0.015\ \text{m}^2 \times 125.7\ \text{rad/s} = 18.1\ \text{V}$.
4. At $5.0$ ms. The angle turned is $\omega t = 125.7\ \text{rad/s} \times 0.0050\ \text{s} = 0.628\ \text{rad}$, which is $36^\circ$. So $e = 18.1\ \text{V} \times \sin 36^\circ = 10.6\ \text{V}$.

Make the coil turn faster. Both the frequency and the height of the curve go up. Why does the height go up too? Because the flux changes faster, and the emf is the rate of change.

Where does the energy come from?

Connect a resistor $R$ to the brushes. A current $i = e/R$ flows round the coil. Now the coil carries a current in a magnetic field, so the field pushes on it. The torque on the coil is $\tau = NBiA\sin\omega t$, and it opposes the turning. This is Lenz's law once more: the effect of the induced current works against the cause.

So the driver must push against this torque. The power it supplies is torque times angular speed:

This is exactly the electrical power $ei$. Every joule that the load receives is a joule of work that the driver did. A generator makes no energy. It only turns mechanical work into electrical energy.

That explains the dynamo on the bicycle. With the lamp connected the dynamo is harder to turn. With no lamp, no current flows, there is no opposing torque, and it turns freely.

The steady value

The emf of a generator rises and falls, so we need a way to say how big it is. The power in a resistor is $e^2/R$. The average of $\sin^2\omega t$ over a turn is $\tfrac{1}{2}$. So the average power is $\dfrac{e_0^2}{2R}$.

A steady emf that heats a resistor equally would be $e_0/\sqrt{2} = 0.707\,e_0$. This is the rms value, the root mean square. The voltage that is quoted for a mains supply is the rms value, not the peak.

Practise with new numbers

Try it in Python

The program finds the peak emf in two ways. The first uses $NBA\omega$. The second finds the flux in tiny steps of time and takes its slope. It also finds when the emf is greatest.

Common mistakes

• Thinking the emf is greatest when the flux is greatest. The emf is greatest when the flux is changing fastest, which is when the coil lies along the field.
• Using $f$ in place of $\omega$. The peak emf needs $\omega = 2\pi f$.
• Leaving $N$ out. Every turn adds its own emf.
• Mixing cm² and m². $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$.
• Reading revolutions each minute as revolutions each second. Divide by $60$ first.
• Using degrees and radians together. If $\omega t$ is in radians, the calculator must be in radian mode.
• Thinking the load takes energy from nowhere. The driver does work against the torque, and the work is the energy that the load receives.

Remember these

• $\Phi = BA\cos\omega t$ for each turn. $e = NBA\omega\sin\omega t$, and the peak emf is $e_0 = NBA\omega$.
• $\omega = 2\pi f$. The emf reverses twice in each turn.
• The emf is zero when the flux is greatest, and greatest when the flux is zero.
• Slip rings give an alternating output. A split-ring commutator gives one direction.
• Mechanical power in $= \tau\omega = ei =$ electrical power out.
• The rms emf is $e_0/\sqrt{2}$, and the average power in a resistor is $e_0^2/2R$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A coil has $40$ turns of area $0.025\ \text{m}^2$. It turns $25$ times each second in a field of $0.30$ T. What is the peak emf?

2. Where is the coil, compared with the field, when the emf is zero? Where is it when the emf is greatest?

3. The driver of a generator doubles the speed of the coil. What happens to the frequency, and to the peak emf? What happens to the average power in the same resistor?

4. Why is it harder to turn a generator when it feeds a lamp than when nothing is connected?

5. A generator gives $e = 12 \sin(100\pi t)$, with $e$ in volt and $t$ in second. What are the peak emf, the frequency, and the emf at $t = 5.0$ ms?