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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

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 Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. How this appears in IA1 and IA2
  4. Examples in context
  5. Try this

What this dot point is asking

QCAA wants you to state and apply Faraday's law in the form ε=NdΦ/dt\varepsilon = -N \, d\Phi / dt, use Lenz's law to determine the direction of an induced current, work with motional EMF (ε=BLv\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 AA in a uniform field B\vec{B} at angle θ\theta to the area normal:

Φ=BAcosθ\Phi = B A \cos\theta

Units: weber (Wb), where 1 Wb = 1 T m2^2.

Faraday's law

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

ε=dΦdt\varepsilon = -\frac{d\Phi}{dt}

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

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

Anything that changes Φ\Phi produces an EMF:

  • Changing BB: a magnet moved toward or away from a stationary coil; a changing primary current in an adjacent coil (the basis of the transformer).
  • Changing AA: a sliding rod on rails enlarges or shrinks the circuit area (motional EMF).
  • Changing θ\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:

  1. Identify the direction of the external flux through the coil and whether it is increasing or decreasing.
  2. 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.
  3. 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 LL moving with velocity vv perpendicular to a uniform field BB (with LL, vv and BB mutually perpendicular) has free charges in it experiencing a magnetic force qvBq 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:

ε=BLv\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: dΦ/dt=BdA/dt=BLvd\Phi / dt = B \, dA / dt = B L v.

If the rod sits in a closed circuit of resistance RR, the induced current is I=ε/RI = \varepsilon / R, and the magnetic force on this current-carrying rod (F=BILF = 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 0.0200.020 m2^2 rotates at 5050 Hz in a uniform field of 0.100.10 T.

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

Induced EMF:

ε=NdΦdt=NBAωsin(ωt)\varepsilon = -N \frac{d\Phi}{dt} = N B A \omega \sin(\omega t).

Peak EMF:

εmax=NBAω=100×0.10×0.020×314=63\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 NN 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 ε\varepsilon against the variable (e.g. ε\varepsilon vs vv for the sliding rod) and report a slope consistent with BLB L within uncertainty.

Examples in context

Example 1. A Bundaberg sugar mill turbine generator spins a coil of 200200 turns, area 0.25 m20.25 \text{ m}^2, in B=1.2 TB = 1.2 \text{ T} at 50 Hz50 \text{ Hz}. Peak EMF is E0=NBAω=200×1.2×0.25×2π×50=1.88×104 V\mathcal{E}_0 = N B A \omega = 200 \times 1.2 \times 0.25 \times 2\pi \times 50 = 1.88 \times 10^4 \text{ V}. The minus sign in Faraday's law (E=NdΦ/dt\mathcal{E} = -N \, d\Phi/dt) encodes Lenz's law: the induced current opposes the change, providing the back-torque the turbine must overcome.

Example 2. A Cairns light-rail tram coasting at 14 m s114 \text{ m s}^{-1} across a 0.05 T0.05 \text{ T} Earth-field component induces motional EMF along the 1.5 m1.5 \text{ m} axle E=BvL=1.05 V\mathcal{E} = BvL = 1.05 \text{ V}. While small, it influences signal-detection circuit grounding. QCAA Unit 3 EA Paper 2 sets exactly this E=BvL\mathcal{E} = BvL calculation, often paired with the Lenz's-law direction analysis.

Try this

Q1. State Faraday's law of electromagnetic induction. [2 marks]

  • Cue. E=NdΦ/dt\mathcal{E} = -N \, d\Phi/dt.

Q2. A 5050-turn coil of area 0.04 m20.04 \text{ m}^2 in a field changing from 0.30 T0.30 \text{ T} to 0.10 T0.10 \text{ T} in 0.20 s0.20 \text{ s}. Calculate the induced EMF magnitude. [3 marks]

  • Cue. ΔΦ=AΔB=0.008 Wb\Delta\Phi = A \Delta B = 0.008 \text{ Wb}; E=NΔΦ/Δt=2.0 V\mathcal{E} = N\Delta\Phi/\Delta t = 2.0 \text{ V}.

Q3. A Bundaberg generator coil (N=200N = 200, A=0.25 m2A = 0.25 \text{ m}^2) rotates at 50 Hz50 \text{ Hz} in B=1.2 TB = 1.2 \text{ T}. (a) Calculate the peak EMF. (b) Apply Lenz's law to identify the direction of the induced current as flux through the coil increases. (c) Discuss one IA2 design check on a model generator. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Evaluation]

  • Cue. (a) 1.88×104 V1.88 \times 10^4 \text{ V}; (b) current opposes the increase, creating opposing field; (c) measure E\mathcal{E} vs ω\omega for linearity.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

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.

ΔΦ=AΔB=0.012×(0.400.10)=3.6×103\Delta \Phi = A \, \Delta B = 0.012 \times (0.40 - 0.10) = 3.6 \times 10^{-3} Wb per turn.

ε=NΔΦΔt=250×3.6×1030.50=1.8 V|\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=ε/R=1.8/5.0=0.36 AI = \varepsilon / R = 1.8 / 5.0 = 0.36 \text{ A}.

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

Equivalently E=I2RΔt=0.362×5.0×0.50=0.324E = I^2 R \, \Delta t = 0.36^2 \times 5.0 \times 0.50 = 0.324 J.

Markers reward the correct flux change, NN 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.

ε=BLv=0.30×0.50×4.0=0.60\varepsilon = B L v = 0.30 \times 0.50 \times 4.0 = 0.60 V.

(b) Current and magnetic force on the rod.

I=ε/R=0.60/6.0=0.10I = \varepsilon / R = 0.60 / 6.0 = 0.10 A.
Fmag=BIL=0.30×0.10×0.50=1.5×102F_{\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 1.5×1021.5 \times 10^{-2} N in the direction of motion.

Mechanical power input: Pmech=Fv=1.5×102×4.0=0.060P_{\text{mech}} = F v = 1.5 \times 10^{-2} \times 4.0 = 0.060 W.
Electrical power dissipated: Pelec=εI=0.60×0.10=0.060P_{\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.

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