Electromagnetic induction - TCE Physics (Tasmania)

Syllabus dot point

“Apply Faraday's and Lenz's laws to electromagnetic induction, generators and transformers.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min readUpdated 2 Jun 2026

Quick answer

A changing magnetic flux through a circuit induces an EMF of size $\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}$ (Faraday's law). Lenz's law, the minus sign, says the induced current opposes the change that caused it. Generators and transformers are direct applications.

What this dot point is asking

Electromagnetic induction is the reverse of the motor effect: instead of using a current to make motion, you use motion or a changing field to make a current. It is the basis of almost all electrical power generation.

Magnetic flux

Magnetic flux $\Phi$ measures how much magnetic field passes through a loop of area $A$ :

\Phi = B A \cos\theta

where $\theta$ is the angle between the field and the normal (perpendicular) to the loop. Flux is measured in webers. Flux changes if the field strength changes, the area changes, or the loop rotates relative to the field.

Magnetic flux

Magnetic flux is the product of the magnetic field component perpendicular to a surface and the area of that surface, representing the total field passing through it.

Faraday's law

Faraday found that an EMF (voltage) is induced whenever the flux through a circuit changes. The size of the induced EMF is:

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

where $N$ is the number of turns in the coil and $\dfrac{\Delta\Phi}{\Delta t}$ is the rate of change of flux. The faster the flux changes and the more turns the coil has, the larger the induced EMF. A steady, unchanging flux induces nothing; only change matters.

Motional EMF

A special case is a straight conductor of length $L$ moving at speed $v$ perpendicular to a field $B$ . The free charges in the conductor feel the magnetic force $qvB$ and pile up at the ends until an electric field balances it, giving a motional EMF:

\varepsilon = BLv

This is the same physics as Faraday's law: as the rod moves it sweeps out area, so the flux through the circuit it is part of changes at the rate $BLv$ . It explains the voltage induced across an aircraft wing moving through Earth's field and the output of a simple generator.

Lenz's law

The negative sign in Faraday's law represents Lenz's law: the induced current flows in the direction that opposes the change in flux producing it. If the flux is increasing, the induced current creates a field to oppose the increase; if it is decreasing, the induced current tries to maintain it. Lenz's law is a statement of conservation of energy. If the induced current aided the change, you would get energy for free.

Opposition costs work

Because the induced current opposes the motion that creates it, you must do work to keep a generator turning. That work is what becomes electrical energy. This is the energy-conservation heart of Lenz's law.

Generators and transformers

A generator is essentially a motor run in reverse. A coil is rotated in a magnetic field, so the flux through it changes continuously, inducing an alternating EMF. With slip rings the output is alternating current; with a split-ring commutator it can be made into pulsed DC.

A transformer changes the voltage of an AC supply. An alternating current in the primary coil produces a changing flux in an iron core, which induces an EMF in the secondary coil. For an ideal transformer:

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

A step-up transformer ( $N_s > N_p$ ) raises voltage; a step-down transformer lowers it. Energy is conserved, so for an ideal transformer $V_p I_p = V_s I_s$ . Transformers only work with AC because they need a continuously changing flux.

Worked example: induced EMF in a coil

A coil of $200$ turns has the magnetic flux through it change from $0.020\ \text{Wb}$ to $0.060\ \text{Wb}$ in $0.10\ \text{s}$ . Find the magnitude of the induced EMF.

Step 1: Calculate the rate of change of flux

Faraday's law tells us that the induced EMF depends on how quickly the flux changes, not on its absolute value. We first find the average rate of change over the given interval by dividing the change in flux by the time taken. A larger flux change over a shorter time produces a bigger EMF:

\frac{\Delta\Phi}{\Delta t} = \frac{0.060 - 0.020}{0.10} = \frac{0.040}{0.10} = 0.40\ \text{Wb s}^{-1}

Step 2: Apply Faraday's law with the number of turns

Each turn of the coil experiences the same changing flux, so the EMFs from all turns add together. This is why more turns means a larger output voltage for the same rate of flux change. Multiplying the rate of flux change by the number of turns gives the total induced EMF:

|\varepsilon| = N\frac{\Delta\Phi}{\Delta t} = (200)(0.40) = 80\ \text{V}

We take the magnitude here because the sign of the EMF (which determines the direction of the induced current via Lenz's law) depends on the geometry of the coil and the direction of the flux change, which are not specified in this problem.

Final answer: The induced EMF is $80\ \text{V}$ .

Thinking a strong field alone induces an EMF

Induction depends on a changing flux, not on the size of the field. A coil sitting still in a very strong but steady magnetic field has zero induced EMF. To induce a voltage you must change the flux by moving the magnet, rotating the coil, or varying the current that produces the field.

When tackling induction questions, identify what is causing the flux to change, apply Faraday's law for the size of the EMF, and use Lenz's law to argue the direction. For transformers, remember the turns ratio sets the voltage ratio and that real transformers lose some energy to heat in the core and windings.

Exam-style questions

Questions in the style of TASC 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.

Practice2 marks

An Airbus A350 is flying north through the Earth's magnetic field, which has a flux density of

1.90 \times 10^{-5}\ \text{T}

at a dip angle of

70^\circ

. The plane flies at

900\ \text{km h}^{-1}

with a wingspan of

64.75\ \text{m}

. Find the motional EMF generated across the wing.

Show worked answer →

A conductor moving through a field has a motional EMF of $\varepsilon = B_\perp L v$ , where $B_\perp$ is the component of the field perpendicular to both the wing and the velocity.

Convert the speed: $900\ \text{km h}^{-1} = \dfrac{900}{3.6} = 250\ \text{m s}^{-1}$ .

For a horizontally moving wing, the vertical component of the Earth's field cuts the wing: $B_\text{vert} = B\sin 70^\circ = (1.90\times10^{-5})(0.9397) = 1.785 \times 10^{-5}\ \text{T}$ .

\varepsilon = B_\text{vert} L v = (1.785\times10^{-5})(64.75)(250) = 0.289\ \text{V}.

A potential difference of about $0.29\ \text{V}$ is induced across the wingtips. Markers want the vertical field component (using the dip angle) and $\varepsilon = BLv$ .

Practice5 marks

30

-turn coil

5.00\ \text{cm}

2.00\ \text{cm}

rotates at

50\ \text{Hz}

in a magnetic field of flux density

0.2\ \text{T}

. Show that the speed of the side AB is about

3.1\ \text{m s}^{-1}

, then calculate the maximum EMF generated by the side AB.

Show worked answer →

The side AB moves in a circle of radius equal to half the $2.00\ \text{cm}$ side: $r = 0.01\ \text{m}$ . Its speed is $v = 2\pi f r$ :

v = 2\pi (50)(0.01) = 3.14\ \text{m s}^{-1},

which is about

3.1\ \text{m s}^{-1}

as required.

The maximum EMF of a single side occurs when it moves perpendicular to the field, $\varepsilon = BLv$ per turn, with $L = 0.05\ \text{m}$ :

\varepsilon_\text{turn} = (0.2)(0.05)(3.14) = 0.0314\ \text{V}.

For $30$ turns the side AB contributes $\varepsilon = (30)(0.0314) = 0.94\ \text{V}$ .

So each side AB produces a maximum EMF of about $0.94\ \text{V}$ . Markers want $v = 2\pi f r$ and $\varepsilon = NBLv$ at the peak.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Physics Level 4 (PHY415115) course document (2015 onwards) — TASC