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The Timpson Lyceum · University Physics II (2026) – Electricity & Magnetism

Electromagnetic Induction – Lecture 12 (University Physics II)

Overview

This lecture takes Faraday's law from a qualitative rule to a working tool, then uses it to close Maxwell's equations. Magnetic flux can be changed in exactly three ways — by changing BB, by changing the area AA, or by changing the angle between them — and each route produces a device: the AC generator, the DC generator, and the slide-wire rod. Working through the slide wire in detail shows where the energy comes from and reveals that the induced emf is really the magnetic force separating charge inside the conductor. From there the focus shifts to induced electric fields, which turn out to be non-conservative, and finally to Maxwell's addition of the displacement-current term to Ampère's law — the step that makes electromagnetic waves fall out of the equations.

Key Concepts

Magnetic flux and Faraday's law

E=−dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}

AC and DC generators (changing ϕ\phi)

ΦB=ABcos⁡ωtE=−dΦBdt=ωABsin⁡ωt\begin{aligned} \Phi_B &= AB\cos\omega t \\ \mathcal{E} &= -\frac{d\Phi_B}{dt} = \omega AB\sin\omega t \end{aligned}

Slide-wire generator (changing AA)

E=−dΦBdt=BLvI=ER=BLvR\begin{aligned} \mathcal{E} &= -\frac{d\Phi_B}{dt} = BLv \\ I &= \frac{\mathcal{E}}{R} = \frac{BLv}{R} \end{aligned}

Motional emf, microscopically

qvB=qEE=vB\begin{aligned} qvB &= qE \\ E &= vB \end{aligned} E=∮(v⃗×B⃗)⋅dl⃗\mathcal{E} = \oint (\vec{v} \times \vec{B}) \cdot d\vec{l}

Faraday's paradox

E=∫0RvB dr=∫0RωrB dr=12ωBR2\begin{aligned} \mathcal{E} &= \int_0^R vB \, dr = \int_0^R \omega r B \, dr \\ &= \frac{1}{2}\omega B R^2 \end{aligned}

Induced electric fields

∮E⃗⋅dl⃗=−dΦBdt\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}

Eddy currents

Generalized Ampère's law and the displacement current

iD=dqdt=ϵdΦEdt∮B⃗⋅dl⃗=μ0(iC+ϵ0dΦEdt)\begin{aligned} i_D &= \frac{dq}{dt} = \epsilon \frac{d\Phi_E}{dt} \\ \oint \vec{B} \cdot d\vec{l} &= \mu_0 \left( i_C + \epsilon_0 \frac{d\Phi_E}{dt} \right) \end{aligned}

Electromagnetic waves and the spectrum

∮B⃗⋅dl⃗=μ0ϵ0dΦEdt∮E⃗⋅dl⃗=−dΦBdt\begin{aligned} \oint \vec{B} \cdot d\vec{l} &= \mu_0 \epsilon_0 \frac{d\Phi_E}{dt} \\ \oint \vec{E} \cdot d\vec{l} &= -\frac{d\Phi_B}{dt} \end{aligned}

Superconductors

Worked Example

Average emf of a DC generator. The DC generator's output is E=ωAB ∣sin⁡ωt∣\mathcal{E} = \omega AB\,|\sin\omega t| — a train of identical positive humps. Because every hump is the same, the long-run average equals the average over one of them, and the average height of a function is its area divided by the width of its base.

One hump runs from t=0t = 0 to t=π/ωt = \pi/\omega (at which point sin⁡ωt=sin⁡π=0\sin\omega t = \sin\pi = 0). Averaging the sin⁡ωt\sin\omega t factor over that interval:

⟨sin⁡ωt⟩=1π/ω∫0π/ωsin⁡ωt dt=ωπ[−cos⁡ωtω]0π/ω=1π(−cos⁡π+cos⁡0)=1π(1+1)=2π\begin{aligned} \langle \sin\omega t \rangle &= \frac{1}{\pi/\omega}\int_0^{\pi/\omega} \sin\omega t \, dt \\ &= \frac{\omega}{\pi} \left[ -\frac{\cos\omega t}{\omega} \right]_0^{\pi/\omega} \\ &= \frac{1}{\pi}\left( -\cos\pi + \cos 0 \right) \\ &= \frac{1}{\pi}(1 + 1) = \frac{2}{\pi} \end{aligned}

Averaging the sine alone is enough, since the amplitude ωAB\omega AB is a constant multiplier. So the average output is:

Eav=2π ωAB\mathcal{E}_{\text{av}} = \frac{2}{\pi}\,\omega A B

To design for a target voltage, solve for the required spin rate:

ω=π Eav2AB\omega = \frac{\pi \, \mathcal{E}_{\text{av}}}{2AB}

With NN coils the factor carries through as Eav=2πNωAB\mathcal{E}_{\text{av}} = \dfrac{2}{\pi}N\omega AB, so doubling the number of coils halves the spin rate needed for the same average voltage.

Summary