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Faraday's Law of Induction

Source lecture(s): SC134 Lec 11–13

Intuition

A changing magnetic flux drives an electric field around a loop. This is the moment the two halves of the course stop being separate subjects: electricity and magnetism become electromagnetism, and the door opens to light itself.

\[\mathcal{E} = -\frac{d\Phi_B}{dt}, \qquad \Phi_B = \int\mathbf{B}\cdot d\mathbf{A}\]

Three ways to change the flux

\(\Phi_B = BA\cos\theta\), so there are exactly three levers, and every generator, transformer and induction cooker pulls one of them:

Change Example
\(B\) transformer, induction hob, metal detector
\(A\) a loop being stretched or squashed
\(\theta\) a rotating coil — this is an electrical generator

The rotating case gives \(\Phi = BA\cos\omega t\), hence \(\mathcal{E} = BA\omega\sin\omega t\): a sinusoid. Alternating current is sinusoidal because rotation is sinusoidal, and the world's electricity grid is a direct consequence of \(\cos\theta\).

Lenz's law is the minus sign

The induced current opposes the change that produced it. This is not an extra empirical rule — it is energy conservation. If the induced effect reinforced the change, the change would grow without limit and you would have free energy.

You can feel it: drop a magnet down a copper pipe and it falls in slow motion. The induced eddy currents oppose the changing flux, and the magnet must do work against them the whole way. The energy does not vanish — it appears as resistive heating in the pipe.

Motional EMF: the same law twice

A bar of length \(L\) sliding at \(v\) across a field \(B\) develops \(\mathcal{E} = BLv\). There are two completely different-looking derivations:

  • In the lab frame: the flux through the circuit changes as the bar moves. Faraday's law.
  • In the bar's frame: the charges move through \(\mathbf{B}\) and feel the magnetic force \(q\mathbf{v}\times\mathbf{B}\), which pushes them along the bar. No changing flux at all.

Both give \(BLv\). That the same number arises from a "magnetic" argument in one frame and an "electric" argument in another is not a coincidence — it is a direct clue that \(\mathbf{E}\) and \(\mathbf{B}\) are frame-dependent aspects of one field. Einstein opens his 1905 relativity paper with precisely this observation.

The deeper form

Written locally, Faraday's law becomes

\[\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}\]

and this says something structurally new. In electrostatics \(\nabla\times\mathbf{E} = 0\), so \(\mathbf{E}\) is a gradient and a potential exists. With a changing \(\mathbf{B}\) the field is not conservative: \(\oint\mathbf{E}\cdot d\boldsymbol{\ell}\neq0\), and "the voltage between two points" stops being well defined — it depends on the path, which is why a voltmeter's reading around an induction loop depends on which way its leads are routed. This genuinely confuses people in the laboratory, and the resolution is exactly this equation.

Common mistakes

  • Flux, not field. A large but constant \(B\) induces nothing. Only \(d\Phi/dt\) matters.
  • Forgetting the area vector's orientation. The sign of \(\mathcal{E}\) is meaningless until you fix a direction for \(d\mathbf{A}\) and use the right-hand rule consistently.
  • Treating Lenz's law as separate. It is the minus sign, and it is energy conservation.
  • Assuming a potential still exists. With \(\partial\mathbf{B}/\partial t\neq0\) the electric field is non-conservative and no single-valued potential exists.

Knowledge graph position

Prerequisites: magnetic flux, magnetic field, electric field. Leads to: Maxwell's equations, electromagnetic waves, generators, transformers, inductance.

Quiz

Q1 (computational). A 200-turn coil of area 0.01 m² sits in a field rising from 0 to 0.5 T in 0.1 s. What is the average EMF?

Answer

\(\mathcal{E} = -N\,d\Phi/dt = -200\times(0.5\times0.01)/0.1 = -10\) V. The sign says the induced current opposes the increase; the magnitude is 10 V.

Q2 (conceptual). A magnet falls slowly through a copper pipe. Where does its lost potential energy go?

Answer

Into resistive heating of the pipe. The changing flux drives eddy currents; by Lenz's law they oppose the magnet's motion, so the magnet does work against that drag, and the eddy currents dissipate it as \(I^2R\). Energy is conserved — it has simply moved into the copper.

Q3 (MCQ). With \(\partial\mathbf{B}/\partial t \neq 0\), the electric field:

  • (a) is still conservative, with a well-defined potential
  • (b) is non-conservative: \(\oint\mathbf{E}\cdot d\boldsymbol{\ell} \neq 0\), so no single-valued potential exists
  • (c) vanishes
  • (d) is always uniform
Answer

(b). \(\nabla\times\mathbf{E} = -\partial\mathbf{B}/\partial t \neq 0\), so the field is not a gradient. "Voltage between two points" becomes path-dependent — which is a real and frequently confusing effect when probing induction circuits.