In Generators and Induction you saw that moving a magnet near a coil makes a current. Faraday found the rule behind it in 1831, and it runs every generator and transformer. It is stated in terms of Magnetic Flux.
Key idea
A changing magnetic flux through a coil of $N$ turns makes an emf $$\varepsilon = -N\frac{\Delta\Phi}{\Delta t}.$$ The size is $N|\Delta\Phi|/\Delta t$. The minus sign is Lenz's law, which sets the direction (the next topic). A steady flux, however large, makes no emf.
- Change the field: push a magnet in, or turn up an electromagnet.
- Change the area: squash the loop or slide a rod along rails.
- Change the angle: turn the loop, as a generator does.
Why it works
Why the rate of change? An emf is energy per charge. Pushing a magnet in faster does the same change in less time, so the same work is done more quickly on the charges: a bigger emf for a shorter time. The emf times the time, $\varepsilon\,\Delta t = -N\Delta\Phi$, depends only on how much the flux changed.
Common mistake
Thinking a strong but steady field makes an emf. A magnet sitting still inside a coil makes no current at all. Only a change in flux counts. Check: if nothing in $\Phi = BA\cos\theta$ is changing, the answer is $0\ \text{V}$.
Worked example: A field that grows
A coil of $200$ turns with area $0.010\ \text{m}^2$ faces a field that grows from $0.10\ \text{T}$ to $0.50\ \text{T}$ in $0.20\ \text{s}$. Find the emf.
- Change in flux per turn: $$\Delta\Phi = \Delta B \cdot A = (0.40)(0.010) = 4.0 \times 10^{-3}\ \text{Wb}.$$
- $$\begin{aligned} |\varepsilon| &= N\frac{|\Delta\Phi|}{\Delta t} = 200 \cdot \frac{4.0 \times 10^{-3}}{0.20} \\ &= 4.0\ \text{V}. \end{aligned}$$
Answer: $4.0\ \text{V}$
Worked example: Turning a loop
A single loop of area $0.20\ \text{m}^2$ faces a $0.30\ \text{T}$ field. It is turned edge-on in $0.15\ \text{s}$. The loop has a resistance of $2.0\ \Omega$. Find the average emf and current.
- Start: $\Phi_i = BA = (0.30)(0.20) = 0.060\ \text{Wb}$. End: edge-on, so $\Phi_f = 0$.
- $|\varepsilon| = \dfrac{0.060}{0.15} = 0.40\ \text{V}$.
- $I = \dfrac{\varepsilon}{R} = \dfrac{0.40}{2.0} = 0.20\ \text{A}$.
Answer: $0.40\ \text{V}$ and $0.20\ \text{A}$