Electromagnetic induction: flux, Faraday and Lenz: VCE Physics Unit 3

Syllabus dot point

“Investigate and apply theoretically and practically electromagnetic induction using the concepts of magnetic flux $\Phi_B = B_\perp A$ , induced EMF $\varepsilon = -N \Delta\Phi_B / \Delta t$ (Faraday's law) and Lenz's law to determine the direction of the induced current”

11 min readUpdated 7 Oct 2026

Quick answer

Magnetic flux is $\Phi_B = B_\perp A$ , and any change in flux through a coil of $N$ turns induces an EMF $\varepsilon = -N \Delta\Phi_B / \Delta t$ (Faraday's law), with the induced current flowing in the direction that opposes the change (Lenz's law, a statement of energy conservation).

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What this dot point is asking
The answer
Examples in context
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Exam-style questions

What this dot point is asking

VCAA wants you to define magnetic flux ( $\Phi_B = B_\perp A$ ), apply Faraday's law to calculate induced EMF, and use Lenz's law to determine the direction of the induced current. You should be able to apply these ideas to a bar magnet entering a coil, a coil rotating in a field, and a loop with a changing area.

The answer

Magnetic flux

Magnetic flux through a flat loop of area $A$ in a uniform field $B$ is:

\Phi_B = B_\perp A = B A \cos\theta

where $B_\perp$ is the component of the field perpendicular to the plane of the loop and $\theta$ is the angle between the field and the normal to the loop. Units are webers (Wb), where $1$ Wb $= 1$ T m squared.

If $B$ is perpendicular to the loop, $\Phi_B = BA$ (maximum). If $B$ is parallel to the plane of the loop, $\Phi_B = 0$ .

Faraday's law (induced EMF)

When the magnetic flux through a coil changes, an EMF is induced. For a coil of $N$ turns:

\varepsilon = -N \frac{\Delta \Phi_B}{\Delta t}

The flux can change because:

The magnetic field strength $B$ changes (for example, a magnet moves toward or away from a coil).
The area of the loop changes (for example, a sliding bar on a conducting rail).
The angle $\theta$ between the field and the loop changes (for example, a coil rotating in a field, as in a generator).

The minus sign comes from Lenz's law: the induced EMF acts to oppose the change in flux. For a magnitude calculation, you can ignore the minus sign.

Lenz's law

The direction of the induced current is such that the magnetic field it produces opposes the change in flux that produced it.

Applied steps:

Identify the direction of the original magnetic flux through the loop.
Identify whether the flux is increasing or decreasing.
The induced current creates a magnetic field that opposes the change: if flux is increasing, the induced field is in the opposite direction; if flux is decreasing, the induced field is in the same direction as the original.
Apply the right-hand grip rule to find the direction of the induced current from the direction of the induced field.

Lenz's law is a consequence of conservation of energy: if the induced current reinforced the change, it would accelerate the magnet, increase the change, and produce energy from nothing.

Common configurations

Bar magnet entering a coil: As the north pole approaches, flux into the coil increases; the induced current creates a field opposing the magnet (the near face of the coil becomes a north pole), repelling it.
Bar magnet leaving a coil: Flux decreases; the induced current reverses to maintain the original flux (the near face of the coil becomes a south pole), attracting the receding magnet.
Coil rotating in a uniform field (generator): $\Phi_B = BA \cos(\omega t)$ . The EMF is $\varepsilon = NBA\omega \sin(\omega t)$ , a sinusoidal AC waveform.
Sliding bar on a rail: A bar of length $L$ moves with velocity $v$ on parallel rails through a field $B$ perpendicular to the loop. Area changes at rate $L v$ , so $\varepsilon = B L v$ .

Worked example

A square coil of 100 turns, side $0.10$ m, is rotated in a $0.40$ T field at $50$ Hz. Find the peak EMF.

Area: $A = 0.10^2 = 0.010$ m squared.

Angular frequency: $\omega = 2 \pi f = 2 \pi \times 50 = 314$ rad/s.

Peak EMF: $\varepsilon_{peak} = N B A \omega = 100 \times 0.40 \times 0.010 \times 314 = 126$ V.

Try it: Induced EMF calculator - enter turns, field, area and change in flux or rotation rate and get the induced EMF.

Common traps

Forgetting the $N$ in Faraday's law: A coil of 200 turns produces an EMF $200$ times that of a single loop.
Using the wrong area: $A$ is the area of one turn, not the total area of all turns added together. $N$ accounts for the turns separately.
Calculating flux with $B$ parallel to the loop: $\Phi_B = B A \cos\theta$ ; if $B$ lies in the plane of the loop, $\Phi_B = 0$ .
Misapplying Lenz's law: The induced current opposes the change in flux, not the flux itself. If flux is increasing into the page, the induced current creates flux out of the page. If flux is decreasing into the page, the induced current creates flux into the page (to maintain the original).
Forgetting energy conservation: Lenz's law is what makes a generator hard to turn: the induced current creates a force that opposes the motion that produces the EMF. The mechanical energy input becomes electrical energy output.

Examples in context

Example 1. Wind-turbine generator stator induction (simplified model). A wind farm such as Hornsdale uses $3.6$ MW turbines spin at $15$ rpm, driving doubly-fed induction generators through a gearbox at $1500$ rpm. Each stator coil ( $N = 200$ turns, area $0.05$ m $^2$ ) cuts through a peak rotor field $B = 1.2$ T. Peak EMF is $\varepsilon = NBA\omega = 200 \times 1.2 \times 0.05 \times 2\pi \times 25 = 1885$ V. Rotating phase-to-phase voltage is $\sim 690$ V RMS, stepped up via on-board transformer to $33$ kV for collection. Lenz's law dictates that induced current opposes the field change, providing the reaction torque the rotor's blade thrust must overcome.

Example 2. Melbourne tram regenerative-braking induction. A Yarra Trams Z-class tram braking on the Bourke Street descent converts kinetic energy to electrical energy via its traction motor acting as a generator. As the rotor (with its field) spins through the stator coils, EMF is induced at $\varepsilon = NBA\omega$ . A typical tram motor with $N = 50$ , $A = 0.04$ m $^2$ , $B = 0.8$ T at $1000$ rpm peak produces peak EMF $\varepsilon = 50 \times 0.8 \times 0.04 \times 2\pi \times 16.7 = 168$ V. The induced current opposes the rotor motion (Lenz's law), providing braking torque while delivering current back to the catenary or to onboard supercapacitors.

Try this

Q1. State Faraday's law and Lenz's law in words. [2 marks]

Cue. Faraday: induced EMF equals negative rate of change of flux ( $\varepsilon = -N \Delta\Phi/\Delta t$ ). Lenz: direction of induced current opposes the change in flux.

Q2. A circular coil of $100$ turns and area $0.02$ m $^2$ is rotated from parallel to perpendicular to a $0.5$ T field in $0.10$ s. Calculate the average induced EMF. [3 marks]

Cue. $\Delta\Phi = BA\Delta(\cos\theta) = 0.5 \times 0.02 \times 1 = 0.01$ Wb; $\varepsilon = N\Delta\Phi/\Delta t = 100 \times 0.01/0.1 = 10$ V.

Q3. Refer to the Hornsdale wind-turbine generator. (a) State the source of the rotor field. (b) Calculate peak EMF for $N = 200$ , $A = 0.05$ m $^2$ , $B = 1.2$ T at $25$ Hz. (c) Apply Lenz's law to explain the reaction torque on the rotor. [2+3+2 marks]

Cue. (a) DC excitation of rotor winding. (b) $1885$ V. (c) Induced stator current creates a field that opposes the rotor motion.

Exam-style questions

Questions in the style of VCAA 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.

Practice3 marks

A coil of 200 turns has cross-sectional area 0.040 m^2. A magnetic field perpendicular to the coil changes uniformly from 0.20 T to 0.50 T in 0.50 s. Calculate the magnitude of the average EMF induced in the coil.

Show worked answer →

Initial flux per turn: $\Phi_1 = B_1 A = 0.20 \times 0.040 = 0.008$ Wb.

Final flux per turn: $\Phi_2 = B_2 A = 0.50 \times 0.040 = 0.020$ Wb.

Change in flux per turn: $\Delta \Phi = 0.020 - 0.008 = 0.012$ Wb.

Average EMF magnitude:

$|\varepsilon| = N \frac{\Delta \Phi}{\Delta t} = 200 \times \frac{0.012}{0.50} = 4.8$ V.

Markers reward correct use of $\Phi = BA$ (with $B$ perpendicular), inclusion of $N$ , and the magnitude (sign is not needed when only magnitude is asked for).

Practice3 marks

A bar magnet is dropped north-pole-down through a vertical conducting ring. Use Lenz's law to predict the direction of the induced current in the ring as the magnet approaches the ring and as the magnet leaves below the ring.

Show worked answer →

As the magnet approaches with its north pole facing down, the downward flux through the ring increases. By Lenz's law the induced current opposes the change, so it must create a magnetic field pointing upward through the ring. Apply the right-hand grip rule: viewed from above, the induced current flows anticlockwise.

As the magnet falls below the ring with its south pole now facing up, the downward flux through the ring decreases (still pointing down but getting weaker; equivalently, the magnetic field through the ring now points up from the south pole and is increasing). By Lenz's law the induced current opposes the decrease, so it creates a magnetic field pointing downward through the ring. Viewed from above, the induced current flows clockwise.

In both stages the induced current creates a magnetic field that opposes the motion of the magnet (energy conservation): the ring repels the approaching magnet and attracts the receding one.

Markers reward identifying the change in flux, the direction of the opposing induced field, and the resulting current direction (with a clear reference frame).