QLDPhysicsSyllabus dot point

Topic 2: Electromagnetism

Apply Faraday's law of electromagnetic induction (induced EMF = - N dPhi/dt) and Lenz's law to determine the magnitude and direction of induced EMF, including motional EMF in a moving conductor and the induced current in a circuit

Q: 2022 QCAA-style HSC: A conducting rod of length 0.50 m slides at 4.0 m/s along frictionless parallel rails perpendicular to a uniform magnetic field of 0.30 T directed into the page. The rails are connected at one end through a 6.0 ohm resistor. (a) Calculate the motional EMF. (b) Calculate the induced current and the magnetic force on the rod. (c) Calculate the force the external agent must apply to keep the rod moving at constant velocity, and verify energy conservation.

**(a) Motional EMF.** $\varepsilon = B L v = 0.30 \times 0.50 \times 4.0 = 0.60$ V. **(b) Current and magnetic force on the rod.** $I = \varepsilon / R = 0.60 / 6.0 = 0.10$ A. $F_{\text{mag}} = B I L = 0.30 \times 0.10 \times 0.50 = 1.5 \times 10^{-2}$ N. By Lenz's law the magnetic force on the rod opposes its motion (otherwise energy would be created from nothing). **(c) External force and energy balance.** For constant velocity (zero net force), the external agent must push with $1.5 \times 10^{-2}$ N in the direction of motion. Mechanical power input: $P_{\text{mech}} = F v = 1.5 \times 10^{-2} \times 4.0 = 0.060$ W. Electrical power dissipated: $P_{\text{elec}} = \varepsilon I = 0.60 \times 0.10 = 0.060$ W. They are equal: all of the mechanical power supplied by the agent is converted to electrical power dissipated in the resistor, confirming energy conservation. Markers reward the motional-EMF formula, the Ohm's law step, the opposing-force direction by Lenz's law, and the equality of mechanical and electrical powers as evidence of energy conservation.

A focused answer to the QCE Physics Unit 3 dot point on electromagnetic induction. Faraday's law for the induced EMF in a coil, Lenz's law for the direction, the motional-EMF special case for a sliding rod, the energy-conservation argument behind the minus sign, and the standard worked examples QCAA uses in IA1 stimulus and IA2 design.

Generated by Claude OpusReviewed by Better Tuition Academy10 min answerUpdated 2026-05-19

Have a quick question? Jump to the Q&A page

What this dot point is asking

QCAA wants you to state and apply Faraday's law in the form $\varepsilon = -N \, d\Phi / dt$ , use Lenz's law to determine the direction of an induced current, work with motional EMF ( $\varepsilon = B L v$ ) as a special case, and explain how Lenz's law is a statement of energy conservation. This dot point is the keystone of Topic 2: transformers, motors and generators all rely on it, and it dominates IA1 and IA2 in Term 2.

The answer

Magnetic flux

The magnetic flux through a flat coil of area $A$ in a uniform field $\vec{B}$ at angle $\theta$ to the area normal:

\Phi = B A \cos\theta

Units: weber (Wb), where 1 Wb = 1 T m $^2$ .

Faraday's law

For a single conducting loop, the induced EMF around the loop equals the negative rate of change of flux:

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

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

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

Anything that changes $\Phi$ produces an EMF:

Changing $B$ : a magnet moved toward or away from a stationary coil; a changing primary current in an adjacent coil (the basis of the transformer).
Changing $A$ : a sliding rod on rails enlarges or shrinks 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.

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; this induced field 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 grip rule for a current loop (fingers curl with current, thumb along the induced field) to read off the direction of the induced current.

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

Motional EMF

A conducting rod of length $L$ moving with velocity $v$ perpendicular to a uniform field $B$ (with $L$ , $v$ and $B$ mutually perpendicular) has free charges in it experiencing a magnetic force $q v B$ along the rod. Charges separate 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 Faraday's law for the case where the rod is part of a circuit and changes the enclosed area: $d\Phi / dt = B \, dA / dt = B L v$ .

If the rod sits in a closed circuit of resistance $R$ , the induced current is $I = \varepsilon / R$ , and the magnetic force on this current-carrying rod ( $F = B I L$ ) opposes the motion. The external agent must do mechanical work equal to the electrical energy dissipated in the resistor (Lenz's law is energy conservation in disguise).

Try it: Induced EMF calculator. Enter coil turns, flux change and time interval, or rod length, field and speed.

Worked example: rotating coil (the AC generator)

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

Flux through one turn: $\Phi(t) = B A \cos(\omega t)$ with $\omega = 2 \pi f = 314$ rad/s.

Induced EMF:

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

Peak EMF:

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

This sinusoidal EMF is the output of an AC generator: the same coil rotated at a fixed frequency in a steady field, with slip rings to take the AC out to the external circuit.

How this appears in IA1 and IA2

IA1 data test. Expect a coil-in-changing-field setup with a flux-vs-time graph and questions on the EMF in each segment, or a moving-rod-on-rails diagram with motional EMF and the direction of the induced current. Markers focus on candidates who drop the $N$ factor, or who state that the induced current opposes "the field" rather than "the change in flux".

IA2 student experiment. Common IA2 designs include: a bar magnet dropped through a coil (with peak EMF measured against drop height); a coil rotated at variable frequency between magnets (peak EMF vs frequency); a sliding rod on rails through a fixed magnet (EMF vs speed). Strong reports linearise $\varepsilon$ against the variable (e.g. $\varepsilon$ vs $v$ for the sliding rod) and report a slope consistent with $B L$ within uncertainty.

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. For magnitude questions, work with $|\varepsilon|$ ; a separate sentence then explains direction using Lenz's law.

Confusing flux change with flux. A large steady flux produces zero 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 (its induced field points the same way as the external field).

Mixing up motional EMF and Faraday's law as separate phenomena. They are the same law; motional EMF is what Faraday's law gives when the area changes in a steady field.

Forgetting energy conservation in the moving-rod problem. Mechanical power input = electrical power dissipated. If the two do not match, recheck signs and the Lenz's-law direction of the magnetic force on the rod.

In one sentence

A changing magnetic flux through a coil induces an EMF $\varepsilon = -N \, d\Phi / dt$ (Faraday's law) directed to oppose the change that produced it (Lenz's law, which is energy conservation in disguise), with motional EMF $\varepsilon = B L v$ in a moving rod as the special case where the changing flux is due to a changing area.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2023 QCAA-style5 marksA 250-turn rectangular coil of area 0.012 m^2 sits in a uniform magnetic field with its normal parallel to the field. The field strength is reduced uniformly from 0.40 T to 0.10 T over 0.50 s. (a) Calculate the magnitude of the average induced EMF. (b) Use Lenz's law to determine the direction of the induced current as seen from the side of the coil that the original field points out of. (c) If the coil has a total resistance of 5.0 ohms, calculate the average induced current and the energy dissipated during the field change.

Show worked answer →

A 5-mark answer needs the EMF, the direction by Lenz's law, the current, and the energy.

(a) Induced EMF.

$\Delta \Phi = A \, \Delta B = 0.012 \times (0.40 - 0.10) = 3.6 \times 10^{-3}$ Wb per turn.

$|\varepsilon| = N \frac{|\Delta \Phi|}{\Delta t} = 250 \times \frac{3.6 \times 10^{-3}}{0.50} = 1.8 \text{ V}$ .

(b) Direction. The external flux through the coil is decreasing in the original direction. By Lenz's law the induced current must oppose the decrease, that is, it must produce a magnetic field inside the coil pointing in the same direction as the original field. Using the right-hand rule for a current loop, the induced current flows counter-clockwise as seen from the side that the original field points out of.

(c) Current and energy.

$I = \varepsilon / R = 1.8 / 5.0 = 0.36 \text{ A}$ .

$E = \varepsilon I \, \Delta t = 1.8 \times 0.36 \times 0.50 = 0.324 \text{ J}$ .

Equivalently $E = I^2 R \, \Delta t = 0.36^2 \times 5.0 \times 0.50 = 0.324$ J.

Markers reward the correct flux change, $N$ factored in for the EMF, a Lenz's law direction tied to "opposes the change" (not "opposes the field"), and the energy in joules.

2022 QCAA-style4 marksA conducting rod of length 0.50 m slides at 4.0 m/s along frictionless parallel rails perpendicular to a uniform magnetic field of 0.30 T directed into the page. The rails are connected at one end through a 6.0 ohm resistor. (a) Calculate the motional EMF. (b) Calculate the induced current and the magnetic force on the rod. (c) Calculate the force the external agent must apply to keep the rod moving at constant velocity, and verify energy conservation.

Show worked answer →

(a) Motional EMF.

$\varepsilon = B L v = 0.30 \times 0.50 \times 4.0 = 0.60$ V.

(b) Current and magnetic force on the rod.

$I = \varepsilon / R = 0.60 / 6.0 = 0.10$ A.
$F_{\text{mag}} = B I L = 0.30 \times 0.10 \times 0.50 = 1.5 \times 10^{-2}$ N.

By Lenz's law the magnetic force on the rod opposes its motion (otherwise energy would be created from nothing).

(c) External force and energy balance. For constant velocity (zero net force), the external agent must push with $1.5 \times 10^{-2}$ N in the direction of motion.

Mechanical power input: $P_{\text{mech}} = F v = 1.5 \times 10^{-2} \times 4.0 = 0.060$ W.
Electrical power dissipated: $P_{\text{elec}} = \varepsilon I = 0.60 \times 0.10 = 0.060$ W.

They are equal: all of the mechanical power supplied by the agent is converted to electrical power dissipated in the resistor, confirming energy conservation.

Markers reward the motional-EMF formula, the Ohm's law step, the opposing-force direction by Lenz's law, and the equality of mechanical and electrical powers as evidence of energy conservation.