You know Faraday's law as "a changing flux makes an emf", with Lenz's law giving the direction. With flux as a function of time and the derivative, we can find the emf at any instant, not just an average over an interval.
Key idea
$$\varepsilon = -N\frac{d\Phi_B}{dt}.$$
The minus sign is Lenz's law: the induced current makes a field that opposes the change. For $\Phi = BA\cos\theta$, the product and chain rules give one term for each thing that changes.
Why it works
Faraday found that the emf depends on how fast the flux changes, not on how big it is. "How fast, at this instant" is a derivative. Dividing a change by a time interval, $\Delta\Phi/\Delta t$, only gives the average emf over that interval; letting the interval shrink to zero gives the emf at one moment. The sign is fixed by energy: if the induced current reinforced the change, it would feed itself without limit.
Worked examples
Worked example: A field that speeds up
A 50-turn coil of area $0.010\ \text{m}^2$ sits in a perpendicular field $B = 0.020t^2 + 0.10t$ (tesla, seconds). Find the emf at $t = 2.0\ \text{s}$.
- $\dfrac{dB}{dt} = 0.040t + 0.10$, which is $0.18\ \text{T/s}$ at $t = 2.0\ \text{s}$.
- $|\varepsilon| = NA\dfrac{dB}{dt} = (50)(0.010)(0.18) = 0.0900\ \text{V}$.
Answer: $|\varepsilon| = 0.0900\ \text{V}$
Worked example: A growing loop
A circular loop in a steady $0.50\ \text{T}$ field (perpendicular to it) grows: its radius is $10\ \text{cm}$ and increasing at $2.0\ \text{cm/s}$. Find the emf.
- $\Phi = B\pi r^2$, so $\dfrac{d\Phi}{dt} = B\cdot 2\pi r\dfrac{dr}{dt}$ by the chain rule.
- $|\varepsilon| = (0.50)(2\pi)(0.10)(0.020) = 6.28 \times 10^{-3}\ \text{V}$, about $6.28\ \text{mV}$.
Answer: $|\varepsilon| \approx 6.28\ \text{mV}$
Worked example: Average or instant?
A field through a loop rises from $0$ to $0.50\ \text{T}$ in $10\ \text{s}$, all of it in the last second. Compare the average emf with the emf during that last second.
- The average over all $10\ \text{s}$ uses $\Delta\Phi/(10\ \text{s})$.
- For the first $9\ \text{s}$ nothing changes, so the emf is zero. In the last second the flux changes ten times faster than the average rate, so the emf is ten times the average.
Answer: zero, then ten times the average
Common mistake
Using an average $\Delta\Phi/\Delta t$ over a long time when the question asks for the emf at one instant. Unless the flux changes at a steady rate, the average hides what happens at any particular moment. Write $\Phi(t)$, differentiate it, and put in the time you want. Check: if $\Phi(t)$ is not a straight line in time, the emf cannot be constant.