Electromagnetic induction, Faraday's law and Lenz's law (QCE Physics Unit 3)

What this dot point is asking

QCAA wants you to state and apply Faraday's law in the form $\varepsilon = -N \, d\Phi / dt$, use Lenz's law to determine the direction of an induced current, work with motional EMF ($\varepsilon = B L v$) as a special case, and explain how Lenz's law is a statement of energy conservation. This dot point is the keystone of Topic 2: transformers, motors and generators all rely on it, and it dominates IA1 and IA2 in Term 2.

The answer

Magnetic flux

The magnetic flux through a flat coil of area $A$ in a uniform field $\vec{B}$ at angle $\theta$ to the area normal:

Units: weber (Wb), where 1 Wb = 1 T m$^2$.

Faraday's law

For a single conducting loop, the induced EMF around the loop equals the negative rate of change of flux:

For a coil of $N$ turns, each turn intercepts the same flux, so the EMFs add in series:

Anything that changes $\Phi$ produces an EMF:

• Changing $B$: a magnet moved toward or away from a stationary coil; a changing primary current in an adjacent coil (the basis of the transformer).
• Changing $A$: a sliding rod on rails enlarges or shrinks the circuit area (motional EMF).
• Changing $\theta$: a coil rotated in a steady field (the AC generator).

The minus sign encodes Lenz's law.

Lenz's law

The induced EMF and induced current always act in a direction that opposes the change in flux that produced them.

To find the direction of the induced current:

1. Identify the direction of the external flux through the coil and whether it is increasing or decreasing.
2. The induced current produces its own magnetic field inside the coil; this induced field points opposite to the external field if external flux is increasing, and along the external field if external flux is decreasing.
3. Use the right-hand grip rule for a current loop (fingers curl with current, thumb along the induced field) to read off the direction of the induced current.

Lenz's law is a statement of energy conservation. If the induced current reinforced the change, it would accelerate the motion or amplify the field that produced it, creating energy from nothing.

Motional EMF

A conducting rod of length $L$ moving with velocity $v$ perpendicular to a uniform field $B$ (with $L$, $v$ and $B$ mutually perpendicular) has free charges in it experiencing a magnetic force $q v B$ along the rod. Charges separate until an electric field inside the rod balances the magnetic force. The resulting EMF between the ends of the rod is:

This is exactly Faraday's law for the case where the rod is part of a circuit and changes the enclosed area: $d\Phi / dt = B \, dA / dt = B L v$.

If the rod sits in a closed circuit of resistance $R$, the induced current is $I = \varepsilon / R$, and the magnetic force on this current-carrying rod ($F = B I L$) opposes the motion. The external agent must do mechanical work equal to the electrical energy dissipated in the resistor (Lenz's law is energy conservation in disguise).

Try it: Induced EMF calculator. Enter coil turns, flux change and time interval, or rod length, field and speed.

Worked example: rotating coil (the AC generator)

A 100-turn rectangular coil of area $0.020$ m$^2$ rotates at $50$ Hz in a uniform field of $0.10$ T.

Flux through one turn: $\Phi(t) = B A \cos(\omega t)$ with $\omega = 2 \pi f = 314$ rad/s.

Induced EMF:

$\varepsilon = -N \frac{d\Phi}{dt} = N B A \omega \sin(\omega t)$.

Peak EMF:

$\varepsilon_{\max} = N B A \omega = 100 \times 0.10 \times 0.020 \times 314 = 63$ V.

This sinusoidal EMF is the output of an AC generator: the same coil rotated at a fixed frequency in a steady field, with slip rings to take the AC out to the external circuit.

How this appears in IA1 and IA2

IA1 data test. Expect a coil-in-changing-field setup with a flux-vs-time graph and questions on the EMF in each segment, or a moving-rod-on-rails diagram with motional EMF and the direction of the induced current. Markers focus on candidates who drop the $N$ factor, or who state that the induced current opposes "the field" rather than "the change in flux".

IA2 student experiment. Common IA2 designs include: a bar magnet dropped through a coil (with peak EMF measured against drop height); a coil rotated at variable frequency between magnets (peak EMF vs frequency); a sliding rod on rails through a fixed magnet (EMF vs speed). Strong reports linearise $\varepsilon$ against the variable (e.g. $\varepsilon$ vs $v$ for the sliding rod) and report a slope consistent with $B L$ within uncertainty.

Examples in context

Example 1. A Bundaberg sugar mill turbine generator spins a coil of $200$ turns, area $0.25 \text{ m}^2$, in $B = 1.2 \text{ T}$ at $50 \text{ Hz}$. Peak EMF is $\mathcal{E}_0 = N B A \omega = 200 \times 1.2 \times 0.25 \times 2\pi \times 50 = 1.88 \times 10^4 \text{ V}$. The minus sign in Faraday's law ($\mathcal{E} = -N \, d\Phi/dt$) encodes Lenz's law: the induced current opposes the change, providing the back-torque the turbine must overcome.

Example 2. A Cairns light-rail tram coasting at $14 \text{ m s}^{-1}$ across a $0.05 \text{ T}$ Earth-field component induces motional EMF along the $1.5 \text{ m}$ axle $\mathcal{E} = BvL = 1.05 \text{ V}$. While small, it influences signal-detection circuit grounding. QCAA Unit 3 EA Paper 2 sets exactly this $\mathcal{E} = BvL$ calculation, often paired with the Lenz's-law direction analysis.

Try this

Q1. State Faraday's law of electromagnetic induction. [2 marks]

• Cue. $\mathcal{E} = -N \, d\Phi/dt$.

Q2. A $50$-turn coil of area $0.04 \text{ m}^2$ in a field changing from $0.30 \text{ T}$ to $0.10 \text{ T}$ in $0.20 \text{ s}$. Calculate the induced EMF magnitude. [3 marks]

• Cue. $\Delta\Phi = A \Delta B = 0.008 \text{ Wb}$; $\mathcal{E} = N\Delta\Phi/\Delta t = 2.0 \text{ V}$.

Q3. A Bundaberg generator coil ($N = 200$, $A = 0.25 \text{ m}^2$) rotates at $50 \text{ Hz}$ in $B = 1.2 \text{ T}$. (a) Calculate the peak EMF. (b) Apply Lenz's law to identify the direction of the induced current as flux through the coil increases. (c) Discuss one IA2 design check on a model generator. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Evaluation]

• Cue. (a) $1.88 \times 10^4 \text{ V}$; (b) current opposes the increase, creating opposing field; (c) measure $\mathcal{E}$ vs $\omega$ for linearity.