Spin a coil in a steady magnetic field and its flux rises and falls over and over. Faraday's law turns that into an emf that swings back and forth: alternating current. Almost all of the world's electricity comes from machines built on this idea.
Key idea
A coil of $N$ turns, each of area $A$, turning at angular speed $\omega$ in a field $B$:
$$\Phi = NBA\cos\omega t, \qquad \varepsilon = NBA\omega\sin\omega t,$$
with peak emf $\varepsilon_0 = NBA\omega$ and $\omega = 2\pi f$.
Why it works
The angle between the coil's normal and the field grows steadily, $\theta = \omega t$, so the flux through each turn is $BA\cos\omega t$. Differentiating, $\varepsilon = -\dfrac{d}{dt}(NBA\cos\omega t) = NBA\omega\sin\omega t$. The emf is largest when $\sin\omega t = \pm 1$, which is when $\cos\omega t = 0$: the coil is edge-on to the field and the flux is zero, but changing fastest. Spinning faster raises both the frequency and the peak emf.
Worked examples
Worked example: Peak emf
A 100-turn coil of area $0.020\ \text{m}^2$ spins at $60\ \text{Hz}$ in a $0.25\ \text{T}$ field. Find the peak emf.
- $\omega = 2\pi(60) = 377\ \text{rad/s}$.
- $\varepsilon_0 = NBA\omega = (100)(0.25)(0.020)(377) = 188\ \text{V}$.
Answer: $\varepsilon_0 \approx 188\ \text{V}$
Worked example: At a given angle
What is that generator's emf when the coil's normal is $30^\circ$ from the field?
- $\varepsilon = \varepsilon_0\sin 30^\circ = (188)(0.50) = 94.2\ \text{V}$.
- At $0^\circ$ (field straight through) it would be zero; at $90^\circ$ (edge-on), the full $188\ \text{V}$.
Answer: $\varepsilon \approx 94.2\ \text{V}$
Worked example: From rpm
A coil turning at 1800 revolutions per minute: what is $\omega$?
- $1800\ \tfrac{\text{rev}}{\text{min}} \times \dfrac{2\pi\ \text{rad}}{1\ \text{rev}} \times \dfrac{1\ \text{min}}{60\ \text{s}} = 188\ \text{rad/s}$, that is $30\ \text{Hz}$.
Answer: $\omega \approx 188\ \text{rad/s}$
Common mistake
"The emf is largest when the most field passes through the coil." At that moment the flux is at its peak, so for an instant it is not changing, and the emf is zero. Faraday's law follows the slope of $\Phi(t)$, not its height. Check with the graphs: the emf curve crosses zero exactly where the flux curve is at a peak.
A motor is a generator run backward, and it is a generator at the same time: its spinning coil makes an emf that opposes the supply, called the back emf. That is why a motor draws its largest current when it is just starting, before it spins up.