Electromagnetic induction: HSC Physics Module 6

Syllabus dot point

“Describe and quantitatively analyse electromagnetic induction using Faraday's law (induced EMF = - N dPhi/dt) and Lenz's law, including motional EMF, eddy currents and the induction coil”

14 min readUpdated 1 Jul 2026

Quick answer

A changing magnetic flux through a circuit induces an EMF $\varepsilon = -N \, d\Phi / dt$ (Faraday's law) directed to oppose the change that produced it (Lenz's law); special cases include motional EMF $\varepsilon = BLv$ in a moving rod and eddy currents in a bulk conductor.

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

What this dot point is asking

NESA wants you to state Faraday's law in the form $\varepsilon = -N \, d\Phi / dt$ , apply Lenz's law to determine the direction of an induced current, work with motional EMF ( $\varepsilon = BLv$ ) as a special case, and explain qualitatively where eddy currents and the induction coil sit in the same framework. Faraday's law is the keystone of the rest of Module 6: transformers and motors all rely on it.

The answer

Faraday's law

For a single conducting loop linked by a magnetic flux $\Phi(t)$ , the induced EMF around the loop is:

\varepsilon = -\frac{d\Phi}{dt}

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

\boxed{\varepsilon = -N \frac{d\Phi}{dt}}

The flux through one turn is $\Phi = B A \cos \theta$ . Anything that changes $B$ , $A$ or $\theta$ produces an EMF:

Changing $B$ : a magnet moved toward or away from a stationary coil, or a changing current in a nearby circuit (the basis of the transformer).
Changing $A$ : a conducting rod sliding on rails to enlarge or shrink 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.

In practice, to find the direction of the induced current:

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

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

Motional EMF

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

\varepsilon = B L v

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

If the rod is part of a closed circuit of resistance $R$ , the induced current is $I = \varepsilon / R$ , and the magnetic force on this current-carrying rod opposes the motion (Lenz's law again).

The must-know equations

Faraday's law: $\varepsilon = -N\dfrac{d\Phi}{dt}$ , where $\Phi = BA\cos\theta$ is the flux through one turn.

Lenz's law: the induced EMF/current opposes the change in flux that produced it (the minus sign above); this is a statement of conservation of energy.

Motional EMF: $\varepsilon = BLv$ for a rod of length $L$ moving at speed $v$ perpendicular to a field $B$ , with $L$ , $v$ , $B$ mutually perpendicular.

Eddy currents

When a bulk conductor (a sheet, disc or block of metal) experiences a changing flux, induced EMFs drive circulating currents called eddy currents inside the conductor. They oppose the change in flux that produced them, so they exert a drag force on whatever is causing the flux change.

Examples:

Magnetic braking. A conducting disc spinning between the poles of a magnet has currents induced in it. These currents experience a magnetic force opposing the rotation, slowing the disc. Used in train brakes and gym equipment.
Induction cooktops. A high-frequency AC field induces eddy currents in the base of a ferromagnetic pot, dissipating energy as heat directly in the pot.
Aluminium-tube demo. A magnet dropped down an aluminium tube falls slowly because eddy currents in the tube wall oppose the change in flux as the magnet moves.

Eddy currents are useful for braking and heating, but they are unwanted losses in transformer cores and motor armatures. To reduce them, the iron in those devices is laminated (thin sheets electrically insulated from each other), which breaks the eddy-current paths.

The induction coil

An induction coil is a transformer-like device used to produce high-voltage pulses from a low-voltage DC source. It has:

A primary coil of relatively few turns wound on a soft iron core.
A secondary coil of many more turns wound on the same core.
An interrupter (a mechanical or electronic switch) that rapidly opens and closes the primary circuit.

Each time the interrupter breaks the primary current, the primary's magnetic field collapses rapidly, producing a fast $d\Phi / dt$ through the secondary. With the large turns ratio, the induced secondary EMF can be tens of kilovolts, enough to produce a spark across an air gap.

Historically used for X-ray tubes and Tesla coils. Modern car ignition coils work on the same principle: the breaker (or transistor) interrupts the primary 12 V supply many times per second, producing tens of kilovolts at the spark plugs.

Worked example: rotating coil

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

The flux through one turn is $\Phi(t) = B A \cos(\omega t)$ , where $\omega = 2 \pi f = 2 \pi \times 50 = 314$ rad/s.

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

Peak EMF:

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

This is the operating principle of the AC generator: a constant-magnitude EMF that varies sinusoidally with the rotation angle.

Try it: Induced EMF calculator for change in flux, turns and time.

Worked examples

Model 1 - flux change and Lenz's law direction. A $250$ -turn coil of area $0.012\ \text{m}^2$ has its plane perpendicular to a uniform field that drops from $0.60\ \text{T}$ to $0.35\ \text{T}$ in $0.40\ \text{s}$ . Find the magnitude of the induced EMF, and state the direction of the induced current relative to the original field.

Formula: $|\varepsilon| = N\dfrac{|\Delta\Phi|}{\Delta t}$ , with $\Delta\Phi = A\,\Delta B$ .

Substitution with units: $\Delta\Phi = 0.012\ \text{m}^2 \times (0.60 - 0.35)\ \text{T} = 3.0 \times 10^{-3}\ \text{Wb}$ .

$|\varepsilon| = 250 \times \dfrac{3.0 \times 10^{-3}\ \text{Wb}}{0.40\ \text{s}} = 1.9\ \text{V}$ (2 s.f.).

Direction: the field is decreasing, so by Lenz's law the induced current opposes the decrease - it flows so as to reinforce the original field direction inside the coil.

Marker's note: state that a decreasing flux induces a current that tries to maintain the flux (same direction as the original field), the reverse of the increasing-flux case. Confusing this is the single most common Lenz's-law error.

Model 2 - eddy-current heating on an induction cooktop. A circular region of a steel pot base of area $8.0 \times 10^{-3}\ \text{m}^2$ links a rapidly alternating flux that changes at an average rate of $|d\Phi/dt| = 0.45\ \text{Wb s}^{-1}$ per turn-equivalent loop, with an effective $N = 1$ (a single conducting loop in the pot base). Find the magnitude of the EMF induced around that loop.

Formula: $|\varepsilon| = N\dfrac{d\Phi}{dt}$ .

Substitution with units: $|\varepsilon| = 1 \times 0.45\ \text{Wb s}^{-1} = 0.45\ \text{V}$ .

Answer: $\varepsilon = 0.45\ \text{V}$ (2 s.f.) around each small conducting loop in the pot base; with many such loops in the low-resistance steel, the resulting eddy currents dissipate significant power as heat directly in the pot, which is the entire purpose of an induction cooktop.

Marker's note: the key physics point is that the pot itself, not a coil in the traditional sense, plays the role of the secondary conductor - the induction cooktop's coil supplies the rapidly changing $B$ , and Faraday's law does the rest inside the metal.

Common traps

Dropping the $N$ for multi-turn coils: A 100-turn coil with a flux change of 1 Wb produces 100 times more EMF than a single loop with the same change.
Forgetting the minus sign or misusing it: The minus sign encodes Lenz's law. Most numerical problems ask for the magnitude; a separate sentence then explains direction using Lenz's law.
Confusing flux change with flux: A large steady flux produces no EMF. Only a changing flux does.

Saying the induced current "opposes the field." It opposes the change in flux, not the field itself. If the external field is decreasing, the induced current reinforces it (it points the same way as the original field).

Mixing up motional EMF and Faraday's law as if they were two separate things. They are the same law; motional EMF is what Faraday's law gives when the changing flux is due to a changing area in a steady field.

Treating eddy-current heating as a separate phenomenon. It is Faraday's law applied to a bulk conductor, plus resistive dissipation of the induced currents.

Exam technique

"Calculate" wants a numeric answer with working shown; "explain" or "using Lenz's law" wants the direction reasoning in words (increasing flux opposed, decreasing flux reinforced), not just a number.

State the relationship before substituting: write $\varepsilon = N\,d\Phi/dt$ (or $\varepsilon = BLv$ ) before plugging in numbers, so method marks are secured even if arithmetic slips.

When reading a $\Phi$ - $t$ or a flux-linkage graph, the gradient equals $d\Phi/dt$ ; multiply by $N$ to get the EMF. Keep answers to two or three significant figures with the correct unit (volts for EMF, weber for flux).

In plain English

Picture pushing a bar magnet toward a loop of wire, like posting a letter into a metal letterbox slot. The wire does not "want" the magnetic field pushing into it, so it fights back: a current appears in the loop, all on its own, and that current makes its own little magnetic field that pushes back against the incoming magnet, like the letterbox flap springing shut against your hand. The faster you push the magnet in, or the more loops of wire there are, the harder it pushes back and the bigger the current. Pull the magnet back out, and the wire flips its trick, now trying to hold onto the field it is losing rather than resist a gain. This "always resist the change" behaviour is Lenz's law, and it is really just nature's way of making sure you cannot get electricity for free: pushing the magnet in against that resistance is what supplies the energy that becomes the current. Slide a metal rod along rails in a magnetic field instead of pushing a magnet, and exactly the same idea gives you a battery-like voltage across the rod - which is precisely how a bicycle dynamo, a hand-cranked torch and a hydro-power generator all make electricity.

Examples in context

Example 1. EMF induced in a Snowy Hydro 2.0 stator coil. Each stator coil of a Snowy 2.0 generator has $N = 18$ turns and links a peak flux of $\Phi_{\max} = 0.20 \text{ Wb}$ per turn. The flux through the coil oscillates at the electrical frequency $f = 50 \text{ Hz}$ , so $\Phi(t) = \Phi_{\max} \cos(2 \pi f t)$ . The induced EMF peaks at $\varepsilon_{\max} = N \Phi_{\max} 2 \pi f = 18 \times 0.20 \times 2 \pi \times 50 = 1131 \text{ V}$ per coil. Hundreds of coils in series produce the $20 \text{ kV}$ generator output that the step-up transformer raises to $330 \text{ kV}$ for transmission. Lenz's law ensures the induced current opposes the rotor's motion, supplying the mechanical "load" the turbine sees.

Example 2. Eddy-current braking on a Sydney Trains carriage. Sydney Metro trains use electromagnetic eddy-current brakes for high-speed retardation. A coil energised on the carriage produces $B = 0.80 \text{ T}$ across the pole face, which spans a width $L = 0.20 \text{ m}$ along the steel rail. As the train moves at $v = 30 \text{ m/s}$ , a conducting path of that width sweeping through the field develops a motional EMF $\varepsilon = BLv = 0.80 \times 0.20 \times 30 = 4.8 \text{ V}$ . Induced eddy currents in the rail dissipate the train's KE as heat. Braking force is independent of mechanical contact, eliminating brake-pad wear.

Exam-style questions

Questions in the style of NESA 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.

2023 HSC-style4 marks

A 200-turn rectangular coil of area 0.015 m^2 sits with its normal aligned with a magnetic field. The field strength changes uniformly from 0.30 T to 0.80 T over 0.20 s. Calculate the magnitude of the induced EMF and explain, using Lenz's law, the direction of the induced current.

Show worked answer →

Change in flux per turn:

$\Delta \Phi = A \, \Delta B = 0.015 \times (0.80 - 0.30) = 7.5 \times 10^{-3}$ Wb.

Magnitude of induced EMF (Faraday's law):

$|\varepsilon| = N \frac{\Delta \Phi}{\Delta t} = 200 \times \frac{7.5 \times 10^{-3}}{0.20} = 7.5$ V.

Direction by Lenz's law: the external flux through the coil is increasing in the direction of the original field, so the induced current must produce a magnetic field that opposes the increase, that is, a field pointing opposite to the external field inside the coil. Curling the right-hand fingers in the direction of the induced current with the thumb opposite to $\vec{B}$ gives the sense of the current loop.

Markers reward correct change in flux, application of Faraday's law with the factor of $N$ , the answer in volts, and a clear Lenz's law statement linking opposition to the change (not opposition to the field itself).

2020 HSC-style5 marks

A conducting rod of length 0.40 m slides at 3.0 m/s along frictionless parallel rails perpendicular to a uniform magnetic field of 0.25 T directed into the page. The rails are connected at one end through a 2.0 ohm resistor. Calculate the induced EMF, the current in the circuit, and the force needed to keep the rod moving at constant velocity.

Show worked answer →

Motional EMF:

$\varepsilon = B L v = 0.25 \times 0.40 \times 3.0 = 0.30$ V.

Induced current:

$I = \varepsilon / R = 0.30 / 2.0 = 0.15$ A.

The current in the rod sits in the external field, so there is a magnetic force on the rod. By Lenz's law it opposes the motion (otherwise energy would be created from nothing):

$F = B I L = 0.25 \times 0.15 \times 0.40 = 1.5 \times 10^{-2}$ N.

To maintain constant velocity (zero net force on the rod), an external agent must push with $1.5 \times 10^{-2}$ N in the direction of motion. The mechanical power input $P = F v = 1.5 \times 10^{-2} \times 3.0 = 4.5 \times 10^{-2}$ W equals the electrical power dissipated $P = \varepsilon I = 0.30 \times 0.15 = 4.5 \times 10^{-2}$ W, confirming energy conservation.

Markers reward the motional-EMF formula, Ohm's law step, the opposing-force direction by Lenz's law, and the energy-balance check.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marks

150

-turn coil of area

0.020\ \text{m}^2

has its plane perpendicular to a magnetic field that increases uniformly from

0.20\ \text{T}

0.55\ \text{T}

over

0.35\ \text{s}

. Calculate the magnitude of the average induced EMF.

Show worked solution →

Change in flux per turn: $\Delta\Phi = A\,\Delta B = 0.020 \times (0.55 - 0.20) = 7.0 \times 10^{-3}\ \text{Wb}$ .

Faraday's law: $|\varepsilon| = N\dfrac{\Delta\Phi}{\Delta t} = 150 \times \dfrac{7.0 \times 10^{-3}}{0.35} = 3.0\ \text{V}$ .

Marks: one for correctly computing $\Delta\Phi$ , one for applying $\varepsilon = N\,\Delta\Phi/\Delta t$ with the factor of $N$ included, one for the final answer $3.0\ \text{V}$ stated with the correct unit.

foundation2 marks

A straight conducting rod of length

0.60\ \text{m}

moves at a constant

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

perpendicular to a uniform

0.40\ \text{T}

magnetic field, with the rod, velocity and field mutually perpendicular. Calculate the motional EMF induced between the ends of the rod.

Show worked solution →

Motional EMF: $\varepsilon = BLv = 0.40 \times 0.60 \times 2.5 = 0.60\ \text{V}$ .

Marks: one for the correct formula $\varepsilon = BLv$ with values substituted, one for the answer $0.60\ \text{V}$ stated with the correct unit.

foundation2 marks

State Lenz's law and explain why it must be consistent with the law of conservation of energy.

Show worked solution →

Lenz's law. The induced EMF (and the current it drives) always acts in a direction that opposes the change in magnetic flux that produced it.

Why it must hold. If the induced current instead reinforced the change in flux, the induced magnetic effects would accelerate whatever was causing the change (for example, pulling a moving magnet in faster, or amplifying a growing field), which would extract mechanical or electrical energy from nothing and violate conservation of energy. Opposition, not reinforcement, is required so that work must always be done against the induced effect to drive the change.

Marks: one for a correct statement of Lenz's law naming "opposes the change", one for the energy-conservation reasoning (reinforcement would create energy from nothing).

core4 marks

A conducting rod

L = 0.80\ \text{m}

long sits across two horizontal frictionless rails in a uniform field

B = 0.20\ \text{T}

directed perpendicular to the plane of the rails. The rails are joined by a

5.0\ \Omega

resistor. The rod is pulled so that it travels

1.2\ \text{m}

0.60\ \text{s}

at constant velocity. Calculate **(a)** the rod's speed, **(b)** the induced EMF, **(c)** the induced current, and **(d)** the magnetic force opposing the rod's motion.

Show worked solution →

(a) $v = \dfrac{d}{t} = \dfrac{1.2}{0.60} = 2.0\ \text{m s}^{-1}$ .

(b) $\varepsilon = BLv = 0.20 \times 0.80 \times 2.0 = 0.32\ \text{V}$ .

(c) $I = \dfrac{\varepsilon}{R} = \dfrac{0.32}{5.0} = 6.4 \times 10^{-2}\ \text{A}$ .

(d) $F = BIL = 0.20 \times 6.4 \times 10^{-2} \times 0.80 = 1.0 \times 10^{-2}\ \text{N}$ , directed opposite to the rod's velocity (Lenz's law), so an external agent must supply this force to keep the rod moving at constant speed.

Marks: one for the speed, one for $\varepsilon = BLv$ correctly evaluated, one for $I = \varepsilon/R$ correctly evaluated, one for $F = BIL$ with the opposing direction stated.

core4 marks

The figure shows a graph of the magnetic flux

\Phi

linked by a

60

-turn coil against time

t

as a bar magnet is pushed steadily toward it. **(a)** Describe the relationship between

\Phi

and

t

shown. **(b)** Using the points

(0.05\ \text{s},\ 0.60\ \text{mWb})

and

(0.25\ \text{s},\ 3.00\ \text{mWb})

, calculate the gradient of the graph. **(c)** Hence calculate the magnitude of the EMF induced in the coil.

Show worked solution →

(a) The graph is a straight line with a constant positive gradient, so the flux increases at a constant rate as the magnet approaches (steady $d\Phi/dt$ ).

(b) Gradient $= \dfrac{\Delta\Phi}{\Delta t} = \dfrac{(3.00 - 0.60) \times 10^{-3}\ \text{Wb}}{(0.25 - 0.05)\ \text{s}} = \dfrac{2.40 \times 10^{-3}}{0.20} = 1.2 \times 10^{-2}\ \text{Wb s}^{-1}$ .

(c) The gradient of a $\Phi$ - $t$ graph equals $d\Phi/dt$ , so by Faraday's law $|\varepsilon| = N\dfrac{d\Phi}{dt} = 60 \times 1.2 \times 10^{-2} = 0.72\ \text{V}$ .

Marks: one for correctly describing the constant gradient (linear increase), one for a correctly evaluated gradient with units of $\text{Wb s}^{-1}$ , one for identifying the gradient as $d\Phi/dt$ , one for $|\varepsilon| = 0.72\ \text{V}$ using the factor of $N$ .

exam6 marks

Analyse how Faraday's law and Lenz's law together account for the operation and the energy behaviour of an eddy-current brake, such as a magnetic brake fitted to a train or a gym exercise bike.

Show worked solution →

Band-6 plan. (1) State Faraday's law as the origin of the induced EMFs in the bulk conductor. (2) State Lenz's law and use it to fix the direction of the eddy currents (opposing relative motion). (3) Link the induced currents to a retarding force via $F = BIL$ -type reasoning. (4) Close with the energy-conservation argument: kinetic energy of the moving conductor becomes electrical energy in the eddy currents, then heat, consistent with Lenz's law and with no energy created.
Model answer. As a conducting disc or rail moves relative to a magnet, the magnetic flux linking small closed loops within the bulk metal changes continuously. By Faraday's law, $\varepsilon = -N\,d\Phi/dt$ , this changing flux induces circulating EMFs, and because the metal is a continuous conductor these EMFs drive closed loops of current called eddy currents.

Lenz's law fixes the direction of these currents: they must act to oppose the change in flux that produced them, which here is the relative motion between the conductor and the magnet. Each eddy-current loop sits in the magnet's field and therefore experiences a magnetic force; by Lenz's law this force acts on the conductor in the direction opposing its motion, i.e. it is a retarding (braking) force, and its reaction is felt by the magnet assembly.

This is consistent with conservation of energy. The braking force does negative work on the moving conductor, so its kinetic energy decreases. That lost kinetic energy is converted into electrical energy in the eddy currents, which is immediately dissipated as heat through the conductor's resistance ( $P = I^2R$ locally). No energy is created: the induced effect always opposes the motion that generates it, so mechanical work must be done against it, and that work reappears as heat. Because there is no mechanical contact, eddy-current brakes wear no brake pads and their braking force smoothly increases with speed (faster relative motion gives a larger $d\Phi/dt$ , hence larger induced current and larger force).

Marker's note: the top band explicitly chains Faraday's law (why an EMF exists) to Lenz's law (why the eddy currents oppose the motion) to the resulting retarding force, and closes with the energy-conservation statement (KE to electrical to heat). A response that only asserts "eddy currents cause braking" without this chain caps in the middle band.

exam7 marks

Evaluate the claim that 'the induction coil and the eddy-current brake are really the same physics used for opposite purposes.' In your answer, refer to Faraday's law, Lenz's law, and the practical outcome each device is designed to achieve.

Show worked solution →

Band-6 plan. Thesis: agree, with a precise reason. (1) State the shared law (Faraday, driven by rapidly changing flux). (2) Describe the induction coil's mechanism and its designed outcome (a large voltage spike is wanted). (3) Describe the eddy-current brake's mechanism and its designed outcome (a retarding force is wanted, the induced voltage itself is incidental). (4) Weigh: same underlying law, opposite engineering goal, so the claim is largely true but needs the "opposite purposes" qualified precisely.
Model answer. The claim is substantially correct. Both devices rely on exactly the same physics: a changing magnetic flux induces an EMF according to Faraday's law, $\varepsilon = -N\,d\Phi/dt$ , with the direction of any resulting current fixed by Lenz's law so as to oppose the change that caused it. Neither device introduces any new physical principle beyond induction.

The induction coil is engineered to maximise the induced voltage. Interrupting the primary current rapidly collapses its magnetic field, producing an extremely large $d\Phi/dt$ through a many-turn secondary coil; because $\varepsilon = -N\,d\Phi/dt$ , a large $N$ and a very fast flux collapse together produce a large voltage spike, which is the desired output (used historically for spark generation and in modern ignition coils). Here the induced EMF itself is the useful product.

The eddy-current brake instead uses induction to produce a mechanical effect. Relative motion between a conductor and a magnet changes the flux linking closed loops within the bulk metal, inducing circulating eddy currents; by Lenz's law these currents create a magnetic force that opposes the relative motion, so their useful output is a retarding force, and the induced EMF and current are only a means to that end, with the electrical energy ultimately dissipated as unwanted heat.

So the claim is fair in that both devices are pure applications of Faraday's and Lenz's laws with no additional physics, but the "opposite purposes" should be stated precisely: the induction coil is optimised to deliver a large useful EMF, while the eddy-current brake is optimised to deliver a large retarding force, treating the induced EMF/current as an intermediate step rather than the goal.

Marker's note: the top band identifies the shared law explicitly (not just "both use magnets"), correctly separates what each device treats as the desired output (voltage versus force), and reaches a genuine evaluative judgement rather than simply describing both devices in parallel. Missing the "EMF is the goal" versus "force is the goal" distinction caps in the middle band.

Practise this

Sources & how we know this

Physics Stage 6 Syllabus (2017) — NESA
Physics HSC exam papers — NESA