Skip to main content
ExamExplained

Electromagnetic induction and Faraday's law (SACE Stage 2 Physics)

Syllabus dot point

“Apply Faraday's law to relate induced EMF to the rate of change of magnetic flux through a coil.”

9 min read

Quick answer

A changing magnetic flux through a coil induces an EMF (voltage). Faraday's law gives its size: ε=−NΔΦΔt\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}, where NN is the number of turns and ΔΦΔt\dfrac{\Delta\Phi}{\Delta t} is the rate of change of magnetic flux. Faster changes, more turns, or larger flux changes all give a larger induced EMF.

Jump to a section
  1. What this dot point is asking
  2. Magnetic flux
  3. Faraday's law
  4. Ways to induce an EMF
  5. Motional EMF
  6. How SACE assesses this
  7. Exam-style questions

What this dot point is asking

You need to define magnetic flux and apply Faraday's law to calculate the EMF induced when the flux through a coil changes.

EMF from a changing field

A 200-turn coil of area 0.015 m20.015\ \text{m}^2 sits in a field that drops from 0.40 T to 0 in 0.10 s

Find the average induced EMF (the coil faces the field, θ=0\theta = 0).

Initial flux: Φi=BA=(0.40)(0.015)=6.0×10−3 Wb\Phi_i = BA = (0.40)(0.015) = 6.0\times10^{-3}\ \text{Wb}.
Final flux: Φf=0\Phi_f = 0.

Change in flux: ΔΦ=0−6.0×10−3=−6.0×10−3 Wb\Delta\Phi = 0 - 6.0\times10^{-3} = -6.0\times10^{-3}\ \text{Wb}.

Induced EMF (magnitude):

∣ε∣=N∣ΔΦΔt∣=200×6.0×10−30.10=200×0.060=12 V.|\varepsilon| = N\left|\frac{\Delta\Phi}{\Delta t}\right| = 200\times\frac{6.0\times10^{-3}}{0.10} = 200\times0.060 = 12\ \text{V}.

Magnetic flux

Magnetic flux measures how much magnetic field passes through an area.

Magnetic flux

Φ=BAcos⁡θ\Phi = B A \cos\theta

Φ\Phi is the magnetic flux (weber, Wb), BB the flux density (T), AA the area (m2\text{m}^2), and θ\theta the angle between the field and the normal to the area. Flux is maximum when the field is perpendicular to the area (θ=0\theta = 0) and zero when the field lies in the plane of the loop (θ=90∘\theta = 90^\circ).

Flux changes if the field strength changes, the area changes, or the orientation of the coil relative to the field changes.

Faraday's law

The central law of induction states that the induced EMF equals the rate of change of flux linkage (flux times number of turns).

Faraday's law

ε=−NΔΦΔt\varepsilon = -N\frac{\Delta\Phi}{\Delta t}

ε\varepsilon is the induced EMF (V), NN the number of turns in the coil, and ΔΦΔt\dfrac{\Delta\Phi}{\Delta t} the rate of change of magnetic flux (Wb s−1\text{Wb s}^{-1}). The minus sign is Lenz's law (next dot point), giving the direction of the induced EMF.

So you get a bigger induced voltage by:

  • changing the flux faster (smaller Δt\Delta t),
  • using more turns (larger NN),
  • producing a larger flux change (stronger field, larger area, or bigger orientation change).

If the flux is not changing, there is no induced EMF: induction needs change.

Ways to induce an EMF

  • Moving a magnet into or out of a coil changes the flux.
  • Rotating a coil in a field changes θ\theta, hence the flux (the basis of generators).
  • Changing the current in a nearby coil changes the field and so the flux (the basis of transformers).
  • A conductor moving through a field cuts field lines, inducing a motional EMF ε=BLv\varepsilon = BLv.

These four routes cover essentially every SACE induction scenario, from a dropped magnet through a tube to a generator coil and the secondary of a transformer.

Motional EMF

A straight conductor of length LL moving at speed vv perpendicular to a field BB has an EMF induced across its ends, ε=BLv\varepsilon = BLv. This is a special case of Faraday's law: as the rod moves, the area of the circuit it forms with its rails changes, so the flux Φ=BA\Phi = BA changes at the rate BLvB L v. This is the principle behind a simple generator and the eddy-current effects seen in electromagnetic braking, and it shows that "cutting field lines" and "changing flux" are two descriptions of the same physics.

How SACE assesses this

SACE Stage 2 induction questions give a flux change (a field dropping to zero, a loop leaving a field, or a magnet passing a coil) over a stated time, and ask for the average EMF using ε=NΔΦΔt\varepsilon = N\dfrac{\Delta\Phi}{\Delta t}. The reliable method is to compute the initial and final flux as Φ=BA\Phi = BA (or BAcos⁡θBA\cos\theta when an angle is given), find the change, divide by the time, and multiply by the number of turns. The two most common slips are forgetting the factor of NN for a multi-turn coil and using the field BB in place of the flux BABA. Show the flux change as an explicit intermediate step before applying the rate, and quote the answer with units.

Common errors
Thinking a steady field induces an EMF
Only a changing flux induces an EMF. A coil sitting still in a constant field has no induced voltage.
Forgetting the number of turns
Faraday's law multiplies by NN; a 200-turn coil gives 200 times the EMF of a single loop for the same flux change.
Using field instead of flux
Flux is BAcos⁡θBA\cos\theta, not just BB. A change in area or orientation can induce an EMF even with a constant field.
Dropping the time
The EMF depends on the rate of change. The same flux change over a shorter time gives a larger EMF.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2024-style3 marks
A square conductive loop leaves a region of uniform magnetic field of magnitude 0.900 T0.900\ \text{T}. The loop has cross-sectional area 1.35×10−2 m21.35\times10^{-2}\ \text{m}^2 and leaves the field in 0.160 s0.160\ \text{s}. Calculate the average emf induced in the loop.
Show worked answer →

Faraday's law gives the magnitude of the average induced emf as ε=NΔΦΔt\varepsilon = N\dfrac{\Delta\Phi}{\Delta t}, with N=1N = 1 for a single loop.

The flux is Φ=BA\Phi = BA, falling from BABA to zero as the loop leaves:

ΔΦ=BA=(0.900)(1.35×10−2)=1.215×10−2 Wb.\Delta\Phi = BA = (0.900)(1.35\times10^{-2}) = 1.215\times10^{-2}\ \text{Wb}.

ε=ΔΦΔt=1.215×10−20.160=7.59×10−2 V.\varepsilon = \dfrac{\Delta\Phi}{\Delta t} = \dfrac{1.215\times10^{-2}}{0.160} = 7.59\times10^{-2}\ \text{V}.

1 mark for the flux change, 1 mark for applying Faraday's law, 1 mark for the answer of about 0.076 V0.076\ \text{V}.

SACE 2025-style3 marks
In a wind-turbine generator, as one magnet passes a coil the magnetic field through the coil decreases by 0.35 T0.35\ \text{T} in 0.048 s0.048\ \text{s}. The coil has 5858 loops and cross-sectional area 0.45 m20.45\ \text{m}^2. Calculate the magnitude of the emf induced in the coil.
Show worked answer →

Faraday's law: ε=NΔΦΔt\varepsilon = N\dfrac{\Delta\Phi}{\Delta t}, where the flux change comes from the changing field, ΔΦ=(ΔB)A\Delta\Phi = (\Delta B)A.

ΔΦ=(0.35)(0.45)=0.1575 Wb.\Delta\Phi = (0.35)(0.45) = 0.1575\ \text{Wb}.

ε=NΔΦΔt=58×0.15750.048=9.1350.048=190 V.\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 58\times\dfrac{0.1575}{0.048} = \dfrac{9.135}{0.048} = 190\ \text{V}.

1 mark for the flux change, 1 mark for including the 5858 turns via NN, 1 mark for the answer of about 190 V190\ \text{V} (1.9×102 V1.9\times10^2\ \text{V}).

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

ExamExplained