SCSA WACE 2022-style practice: A coil of 200 turns and cross-sectional area 4.010^-3\ m^2 sits in a magnetic field perpendicular to its plane. The field increases uniformly from 0.10\ T to 0.55\ T in 0.030\ s. (a) Calculate the change in magnetic flux through one turn. (b) Calculate the magnitude of the average emf induced in the coil.

A 6 mark calculation rewards the flux change, the use of N turns and the rate of change. **(a) Flux change per turn.** =A\, B=(4.010^-3)(0.55-0.10)=(4.010^-3)(0.45)=1.810^-3\ Wb. **(b) Induced emf.** Faraday's law gives ||=N t: $||=200×1.810^-30.030=2000.060=12\ V.$ Markers reward =A B, multiplying by N=200, dividing by t and the final value of 12\ V.

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How does a changing magnetic flux induce an emf in a circuit?

Apply Faraday's and Lenz's laws to magnetic flux, generators and transformers

A focused answer to the WACE Year 12 Physics Unit 3 dot point on electromagnetic induction. Magnetic flux, Faraday's law, Lenz's law, induced emf, and the operation of AC generators and transformers.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

WACE wants you to define magnetic flux, use Faraday's law to calculate an induced emf, apply Lenz's law to find its direction, and explain generators and transformers in terms of changing flux. This topic closes Unit 3 by linking magnetism back to electricity: motion plus a magnetic field produces electricity.

Magnetic flux

Magnetic flux measures how much field passes through a surface,

\Phi=BA\cos\theta\quad(\text{Wb}),

where $A$ is the area of the loop and $\theta$ is the angle between the field and the normal to the loop. Flux is greatest when the field is perpendicular to the area ( $\theta=0$ ) and zero when the field lies in the plane of the loop. Crucially, an emf is induced only when flux changes, which can happen by changing $B$ , changing $A$ , or rotating the loop to change $\theta$ .

Faraday's law

The induced emf equals the rate of change of flux linkage, where $N$ is the number of turns,

\varepsilon=-N\frac{\Delta\Phi}{\Delta t}.

A larger field change, a faster change, or more turns all increase the emf. On a flux-time graph the induced emf is proportional to the gradient, so a steep change gives a large emf and a constant flux gives none.

Lenz's law and the minus sign

The negative sign is Lenz's law: the induced current flows in the direction that opposes the change in flux producing it. If the flux through a coil is increasing, the induced current creates a field opposing that increase; if it is decreasing, the induced current tries to maintain it. This is energy conservation in disguise: the opposition is why you must do work to push a magnet into a coil, and that work becomes electrical energy.

Generators

An AC generator rotates a coil in a magnetic field. As the coil turns, the angle $\theta$ between the field and the coil normal changes continuously, so the flux varies sinusoidally and the induced emf varies sinusoidally too. The emf is maximum when the coil plane is parallel to the field (flux changing fastest) and zero when the plane is perpendicular (flux momentarily steady at its peak). Slip rings connect the rotating coil to the external circuit, delivering alternating current.

Transformers

A transformer uses a changing flux to transfer energy between two coils on a shared iron core. AC in the primary creates a changing flux that the core carries to the secondary, inducing an emf there. For an ideal transformer the voltages share the turns ratio,

\frac{V_p}{V_s}=\frac{N_p}{N_s}.

Assuming no power loss, $V_pI_p=V_sI_s$ , so stepping voltage up steps current down. Transformers work only on AC because they need a continually changing flux; this is the reason mains power is distributed as high-voltage AC to cut transmission losses.

Induced emf and a transformer ratio

Coil in a changing field, then a step-down transformer

A coil of $200$ turns and area $0.015\ \text{m}^2$ sits with its plane perpendicular to a field that falls from $0.80\ \text{T}$ to $0.20\ \text{T}$ in $0.30\ \text{s}$ .

Step 1: Calculate the change in flux per turn.

Magnetic flux through one turn is $\Phi = BA$ , so a change in field $\Delta B$ produces a flux change $\Delta\Phi = A\,\Delta B$ . Here the field drops by $0.60\ \text{T}$ , so the flux decreases by:

\Delta\Phi=A\,\Delta B=(0.015)(0.80-0.20)=9.0\times10^{-3}\ \text{Wb}.

Step 2: Apply Faraday's law to find the induced emf.

Faraday's law states that the magnitude of the induced emf equals the number of turns multiplied by the rate of flux change. With $N=200$ turns and $\Delta t=0.30\ \text{s}$ :

\varepsilon=N\frac{\Delta\Phi}{\Delta t}=200\times\frac{9.0\times10^{-3}}{0.30}=6.0\ \text{V}.

Step 3: Apply the transformer turns ratio to find the secondary voltage.

A step-down transformer with $N_p=300$ primary turns and $N_s=50$ secondary turns steps the voltage down in the same ratio as the turns. The primary voltage is the $6.0\ \text{V}$ just calculated:

V_s=V_p\frac{N_s}{N_p}=6.0\times\frac{50}{300}=1.0\ \text{V}.

Stating direction clearly

When asked for direction, name the change in flux first, then say the induced current opposes it, and give the current sense (for example clockwise viewed from a stated side). Marks are lost when students quote Faraday's law but never resolve the sign with Lenz's law.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20226 marksA coil of

200

turns and cross-sectional area

4.0\times10^{-3}\ \text{m}^2

sits in a magnetic field perpendicular to its plane. The field increases uniformly from

0.10\ \text{T}

0.55\ \text{T}

0.030\ \text{s}

. (a) Calculate the change in magnetic flux through one turn. (b) Calculate the magnitude of the average emf induced in the coil.

Show worked answer →

A 6 mark calculation rewards the flux change, the use of $N$ turns and the rate of change.

(a) Flux change per turn. $\Delta\Phi=A\,\Delta B=(4.0\times10^{-3})(0.55-0.10)=(4.0\times10^{-3})(0.45)=1.8\times10^{-3}\ \text{Wb}$ .

(b) Induced emf. Faraday's law gives $|\varepsilon|=N\dfrac{\Delta\Phi}{\Delta t}$ :

|\varepsilon|=200\times\frac{1.8\times10^{-3}}{0.030}=200\times0.060=12\ \text{V}.

Markers reward $\Delta\Phi=A\Delta B$ , multiplying by $N=200$ , dividing by $\Delta t$ and the final value of $12\ \text{V}$ .

WACE 20215 marksA bar magnet is dropped north-pole-first through a vertical copper tube. Using Lenz's law, explain why the magnet falls more slowly than it would in free fall, and explain the direction of the induced current in the tube as the magnet approaches.

Show worked answer →

A 5 mark explanation needs the energy or opposition argument plus a correct current direction.

Opposing the change: As the magnet falls, the flux through each ring-like cross-section of the tube changes, inducing a current. By Lenz's law the induced current opposes the change that causes it, so it creates a magnetic field that repels the approaching magnet (the upper part of the tube acts like a north pole facing the falling north pole).
Why it slows: This repulsion exerts an upward retarding force on the magnet, so the net downward force is less than its weight and the magnet falls slower than free fall, often reaching a near-constant terminal speed.
Direction: Viewed from above as the north pole approaches, the induced current flows so as to make the near face a north pole, which is anticlockwise.
Energy: The kinetic energy not gained appears as resistive heating in the tube, consistent with conservation of energy.

Markers reward the Lenz opposition, the retarding force reducing acceleration, the anticlockwise current and the energy-conservation link.