A changing flux linkage induces an EMF — Faraday's law

Electromagnetic Induction — AQA A-Level Physics

Key Definition

Faraday's law — The magnitude of the induced EMF is equal to the rate of change of flux linkage.

\begin{aligned} \varepsilon &= -N\frac{\Delta\Phi}{\Delta t} \\ &= -\frac{\Delta(N\Phi)}{\Delta t} \end{aligned}

$ε$: induced EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V). (V)
$N$: number of turns on the coil
$Φ$: magnetic fluxThe product of magnetic flux densityMass per unit volume of a material. Measured in kg m⁻³. and the area perpendicular to the field. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area. (Wb). through one turn (Wb)
$NΦ$: flux linkageThe product of magnetic flux and the number of turns of a coil. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area.-turns (Wb turns). (Wb turns)
$Δt$: time interval (s)

This is the central law of electromagnetic inductionThe generation of an EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V). (and hence currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). in a closed circuit) when the magnetic fluxThe product of magnetic flux densityMass per unit volume of a material. Measured in kg m⁻³. and the area perpendicular to the field. Measured in weberThe SI unit of magnetic flux. One weber is the flux through an area of 1 m² when the magnetic flux density is 1 T perpendicular to the area. (Wb). linkage through a conductor changes.. Everything else in this topic is an application of it.
The EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V). is induced whenever the flux linkageThe product of magnetic flux and the number of turns of a coil. Measured in weber-turns (Wb turns). through a circuit changes. It doesn't matter how the change happens — moving a wire, rotating a coil, switching a nearby currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). on or off. If NΦ changes, an EMF appears.
The word 'induced' means the EMF is created by the changing flux, not by a battery or cell. It's nature generating a voltageThe energyThe capacity to do work. Measured in joules (J). transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). Informal term for potential difference..
The minus sign comes from Lenz's lawThe direction of an induced EMF is such that it opposes the change in magnetic flux that produced it. This is a consequence of conservation of energyThe capacity to do work. Measured in joules (J).. (next card). It tells you the direction of the induced EMF, not its size. When calculating magnitude, you can drop the minus sign.
If the flux linkageThe product of magnetic flux and the number of turns of a coil. Measured in weber-turns (Wb turns). changes at a constant rate, the induced EMF is constant. If the rate of change varies, the EMF varies too. This is why a rotating coil produces a sinusoidal EMF.
For a coil with N turns, the total flux linkage is NΦ. A single-turn loop has flux linkage equal to Φ.

Worked Example

A coil of 200 turns has a flux linkage that decreases uniformly from 0.040 Wb turns to 0.010 Wb turns in 0.15 s. Calculate the magnitude of the induced EMF.

Show Solution

List known values

Number of turns: $N = 200$ (already accounted for in $N\Phi$)
Initial flux linkage: $N\Phi_1 = 0.040 \text{ Wb turns}$
Final flux linkage: $N\Phi_2 = 0.010 \text{ Wb turns}$
Time: $\Delta t = 0.15 \text{ s}$

Calculate the change in flux linkage

$$\Delta(N\Phi) = N\Phi_2 - N\Phi_1 = 0.010 - 0.040 = -0.030 \text{ Wb turns}$$

Apply Faraday's law

$$\varepsilon = -\frac{\Delta(N\Phi)}{\Delta t} = -\frac{-0.030}{0.15}$$

Evaluate

$$\varepsilon = \frac{0.030}{0.15} = 0.20 \text{ V}$$

Answer

$\varepsilon = 0.20$ V

Examiner Tips and Tricks

Watch the units carefully.
Flux linkage NΦ is given in Wb turns (or just Wb if $N = 1)$.
If a question gives you the flux per turn and the number of turns separately, multiply them first to get the total flux linkage before dividing by time.