OCR A A-level Physics · H556 · 6.3.3

Electromagnetic induction: Faraday’s law, Lenz’s law and transformers

Explain how changing magnetic flux generates an e.m.f., read generator graphs and investigate search coils and transformers.

8 OCR learning outcomes8 worked examples3 interactive activities49 marks of practice

The induced magnetic effect opposes the change in linked flux. The responses shown require a closed conducting coil.

Start with a change-versus-size check

A coil has a large, constant magnetic flux linkage. What induced e.m.f. does Faraday’s law predict?

Your prediction

Separate the size of flux linkage from its rate of change.

By the end, you should be able to

Distinguish B, Φ and NΦ, and identify the angle to the coil normal.
Use Faraday’s law for magnitudes, means and graph gradients, and Lenz’s law for direction.
Relate rotating-coil orientation to flux-linkage and e.m.f. waveforms.
Plan search-coil and transformer measurements, with clear controls and uncertainty evaluation.
Use ideal transformer voltage/current ratios and separate them from real losses.

Start with magnetic field patterns and direction rules if those are unfamiliar. Charge and current supplies the circuit conventions.

Magnetic flux and flux linkage: Φ = BA cos θ

Magnetic flux measures the magnetic field passing through an area. For a flat coil in a uniform field, use the component of B perpendicular to its plane:

Φ = BA cos θ
Flux linkage λ = NΦ = NBA cos θ

B is magnetic flux density in tesla, A is area in m², and θ is the angle between B and the normal to the coil. The normal is perpendicular to the plane. Φ is the flux through one turn; NΦ is the linkage for N turns each linking the same flux. If different turns link different fluxes, add their contributions.

Edge-on coil with its plane, perpendicular normal and magnetic field arrow; theta is between field and normal, not field and plane — The coil’s normal fixes the cosine angle. A chosen normal also fixes the sign of flux.

Three orientations in a uniform field
θ to normal	Coil plane relative to B	Φ
0°	Perpendicular	+BA, greatest positive flux
90°	Parallel	0
180°	Perpendicular, normal reversed	−BA, greatest negative flux

The unit of flux is the weber, Wb: 1 Wb = 1 T m² = 1 V s. A change of linkage of 1 Wb in 1 s gives mean e.m.f. magnitude 1 V. Turns are a count, so linkage may be written Wb or Wb turns. “Turns” is a useful reminder that N is already included.

Worked example 1 · flux through a tilted coil

A 180-turn coil has area 24.0 cm² in a 0.280 T uniform field. Its normal is at 35.0° to B.

A = 24.0 × 10⁻⁴ = 0.00240 m². Φ = 0.280 × 0.00240 × cos 35.0° = 5.50 × 10⁻⁴ Wb. NΦ = 180Φ = 0.0991 Wb turns.

If 35° were instead measured from the plane, the angle to the normal would be 55°. Convert area using 1 cm² = 10⁻⁴ m²; do not use the conversion for a length.

Faraday’s law and Lenz’s law

Faraday’s law: induced e.m.f. magnitude equals the rate of change of magnetic flux linkage. Linkage can change by changing B, the linked area, the orientation, or the number of linked turns. Merely having a field present is insufficient.

Mean ε = −Δ(NΦ)/Δt
Instantaneous ε = −d(NΦ)/dt

The first equation uses a finite interval. The second describes the local gradient, the limiting rate at an instant. An e.m.f. can exist across an open coil; a continuous induced current needs a closed conducting path.

Lenz’s law: any induced current produces a magnetic effect that opposes the change in flux causing it. The minus sign expresses this direction convention. Define a positive coil normal and a corresponding positive circuit direction before interpreting a numerical sign; swapping voltmeter leads reverses the displayed sign.

Find direction in three steps

Identify the original flux direction and whether it is increasing or decreasing.
Choose an induced field that opposes that increase or decrease. If the original flux decreases, the induced field tries to maintain it.
Use the right-hand grip rule to find conventional current. Viewed from one coil face, anticlockwise current makes that face north; clockwise makes it south.

Closed coil: face viewed from the magnet
Magnet motion	Near coil face	Current viewed from magnet
North pole approaches	North, repels	Anticlockwise
North pole retreats	South, attracts	Clockwise
South pole approaches	South, repels	Clockwise
South pole retreats	North, attracts	Anticlockwise
Magnet stationary, flux steady	No induced pole from this process	No induced current

Why Lenz’s law fits energy conservation

When moving a magnet induces current, the magnetic reaction resists the imposed motion. Work supplied by the mover transfers energy to the circuit, such as heating a resistor. A current that assisted the flux change would allow increasing energy without the required input. Faraday gives the e.m.f.; Lenz explains the opposing response.

Worked example 2 · mean e.m.f. during a turn

A 120-turn coil, A = 0.00360 m² and B = 0.250 T, turns its normal from 0° to 60° in 0.0800 s.

Initial linkage = NBA = 0.108 Wb turns. Final linkage = NBA cos 60° = 0.0540 Wb turns. Δ(NΦ) = 0.0540 − 0.108 = −0.0540 Wb turns, so mean ε = −(−0.0540)/0.0800 = +0.675 V in the chosen reference direction.

The magnitude is 0.675 V. N is already included in the linkage change, so multiplying by 120 again would be wrong. This is a mean over the interval, rather than the instantaneous value at either end.

Quick retrieval check: does an induced field always oppose the original field?

No. It opposes the change. If the original flux is decreasing, the induced field acts in the same direction to oppose that decrease.

Flux–time graphs: e.m.f. is the negative gradient

Read the vertical axis before choosing a formula.

Vertical axis	Signed e.m.f.
Flux linkage NΦ	−gradient
Flux through one turn Φ	−N × gradient, for fixed N
Perpendicular field component B_⊥	−NA × gradient, for fixed N and A

For a curved graph, draw a tangent at the requested instant and use a large triangle on that tangent. A line between two separate points gives an average rate. Convert axis prefixes: 1 mT ms⁻¹ = 1 T s⁻¹, but 1 mWb s⁻¹ = 10⁻³ V for a linkage gradient.

Worked example 3 · a rise, a pause and a fall

In the original graph below, linkage rises from 0 to +0.300 Wb turns over 0.060 s, stays constant to 0.100 s, then falls to −0.100 Wb turns at 0.140 s.

0–0.060 s: ε = −0.300/0.060 = −5.00 V.
0.060–0.100 s: gradient is zero, so ε = 0 V.
0.100–0.140 s: ε = −(−0.100 − 0.300)/0.040 = +10.0 V.

The final segment gives twice the magnitude because its linkage changes twice as fast. The ideal straight-line corners do not specify a finite instantaneous e.m.f. exactly at the join; real changes are smoothed over a finite time.

Aligned original graphs: linkage rises 0 to 0.300 Wb turns in 60 milliseconds, stays flat to 100 milliseconds, falls to minus 0.100 by 140 milliseconds; emf is minus 5, zero, then plus 10 volts — Linkage is continuous; its gradient changes between intervals. The lower graph uses the same time axis and sign reference.

Worked example 4 · a tangent to a field-component graph

At one instant dB_⊥/dt = −8.00 T s⁻¹. A coil has 90 turns and area 18.0 cm².

ε = −NA dB_⊥/dt = −90 × 0.00180 × (−8.00) = +1.296 V ≈ +1.30 V. The field-component gradient alone is not a voltage; multiply by NA.

Flux-change calculator: predict before changing a setting

This calculation assumes the stated initial and final fluxes are through each turn. It reports the mean e.m.f. in the Faraday-law reference direction.

Turns NInterval / msInitial flux / mWbFinal flux / mWb

Δ(NΦ) = −0.128 Wb turns; mean ε = +3.200 V.

Predict the effect of doubling N from 160 to 320.
Reset, then halve the interval from 40 to 20 ms. Explain why this has the same effect on magnitude.
Reset, then choose equal initial and final fluxes. Finally make the final flux greater than the initial flux: what changes in the sign?

Check the calculator tasks

Doubling N doubles the linkage change and mean e.m.f. to +6.400 V. Halving time doubles the rate and also gives +6.400 V. Equal endpoint fluxes give zero mean e.m.f.; this does not prove that the instantaneous e.m.f. was zero throughout an unspecified interval. An increasing flux gives negative mean e.m.f. in the fixed reference convention.

Simple a.c. generator: flux linkage and e.m.f.

A coil rotates in a magnetic field. Its normal changes angle to B, changing flux linkage. Two slip rings, each connected to one coil end, rotate with the coil. Stationary brushes maintain electrical contact with the external circuit. The alternating e.m.f. changes sign every half turn. A split-ring commutator belongs to a different output arrangement.

For uniform rotation in a uniform field, with θ = ωt and the normal initially parallel to B:

NΦ = NBA cos(ωt)
ε = NBAω sin(ωt)
ω = 2πf; ε_max = 2πfNBA

These equations follow from Faraday’s law; the OCR core requirement is to explain the simple a.c. generator and use flux-change reasoning. The sinusoidal form makes its graph relationships explicit.

One rotation in the chosen phase reference
Normal angle	Flux linkage	E.m.f.
0°	+NBA maximum	0
90°	0, falling fastest	+ε_max
180°	−NBA minimum	0
270°	0, rising fastest	−ε_max
360°	+NBA maximum	0

The graphs are a quarter cycle apart. At maximum or minimum linkage, its gradient is zero. At a linkage zero crossing, the gradient magnitude is greatest for this sinusoidal motion. The e.m.f. depends on how quickly linkage changes, rather than on its size.

Worked example 5 · generator amplitude and period

N = 150, A = 0.00750 m², B = 0.320 T and rotation frequency f = 20.0 Hz.

Peak linkage = NBA = 0.360 Wb turns. T = 1/f = 0.0500 s. ω = 2πf = 125.7 rad s⁻¹. Peak ε = NBAω = 45.2 V. At θ = 90°, linkage is zero and ε = +45.2 V.

Double f alone: peak ε doubles and period halves, while peak linkage stays 0.360 Wb turns. Halve B and double f together: peak ε stays 45.2 V, period halves and peak linkage halves.

Aligned sinusoidal graphs for a 20 hertz generator: linkage cosine amplitude 0.360 Wb turns and emf sine amplitude 45.2 V, quarter cycle apart — Both graphs have period 50.0 ms. At 12.5 ms, linkage passes through zero and e.m.f. reaches its positive peak.

Oscilloscope readings

Period = horizontal divisions per cycle × time base. Frequency = 1/period. Peak voltage = displacement from the zero-voltage line × volts per division; peak-to-peak voltage is twice the peak for a centred sine wave. If comparing rms readings with a sinusoidal peak, use V_rms = V_peak/√2. This last conversion is supporting measurement knowledge, rather than a separate 6.3.3 outcome.

Induction lab: magnet motion and rotating-coil graphs

Open the induction explorer

Magnet + coil

Choose North and Towards. Predict Repel before running the motion.
Choose Away and predict Attract. Explain the change in coil-face polarity using linked flux.
Try Stationary and No force. A field can be present without an induced e.m.f.
Compare slow and fast motion and 100, 200 and 400 turns. Explain the relative e.m.f. with Faraday’s law. The magnet activity reports relative strength, rather than calibrated voltages.

Rotating coil

Use Reset the activity, then Rotating coil. Baseline peak linkage is 0.400 Wb turns and peak e.m.f. about 12.6 V.
Move Angle of coil normal to B, θ to 0°, 90°, 180° and 270°. Compare the signs and zeros with the table above.
Choose Field, B and change the factor from ×1 to ×2. Predict the new peak e.m.f. of about 25.1 V.
Choose Frequency, f, then double its factor. Explain why peak e.m.f. doubles while peak linkage is unchanged. Each new variable starts at factor ×1.

Check the lab conclusions

Approach gives repulsion; retreat gives attraction; steady flux gives no induced response. Doubling B, N or rotation frequency doubles the ideal generator’s peak e.m.f. Doubling frequency also halves period. Doubling B or N changes peak linkage; doubling frequency alone does not.

Search-coil practical: investigate magnetic flux

A search coil is a small coil with known N and area A connected to a voltage-measuring instrument. It responds to changing linked flux. A stationary coil in a steady field produces no sustained signal, so it cannot directly display a static B using the induced-voltage method.

Method 1: move or flip a coil in a static field

Place a small search coil in an approximately uniform field, with its normal along the field. Connect it to a high-input-impedance oscilloscope or suitable voltage recorder.
Record the zero-signal baseline. Withdraw the coil fully into a negligible-field region, or flip it through 180° while keeping it in the same field.
Capture the entire signed voltage pulse with sufficient sample rate. Subtract the baseline offset and find its area by integration or numerical sums εΔt.
Use ∫εdt = −Δ(NΦ). Withdrawal from parallel alignment to negligible field changes linkage by magnitude NAB; a 180° flip changes it by magnitude 2NAB.
Repeat the motion and compare inferred B. Changing movement time changes the pulse shape and peak, but the ideal full-pulse area is fixed by the endpoints.

Worked example 6 · infer B from a flip pulse

A 200-turn search coil of area 3.00 cm² is flipped 180° in a uniform static field. The magnitude of its complete signed voltage–time area is 0.00600 V s.

|Δ(NΦ)| = 2NAB, so B = 0.00600/(2 × 200 × 3.00 × 10⁻⁴) = 0.0500 T. Dividing by NA instead would give twice the correct field because a flip goes from +NAB to −NAB.

Method 2: a fixed coil in an alternating field

A low-voltage a.c. source drives a long solenoid. Put the search coil in its central uniform region, keeping position and orientation fixed, and record its voltage trace. An oscilloscope measures amplitude and period; a suitable a.c. voltmeter can measure rms voltage when its frequency response is appropriate.

A low-voltage AC supply drives a solenoid; a separate small search coil inside its uniform central region connects to an oscilloscope, with its normal along the solenoid field — The search coil senses changing flux magnetically. Its leads form a separate measuring circuit; the solenoid supply does not conduct current into them.

If B(t) = B₀ sin(2πft), a fixed search coil gives peak e.m.f. ε₀ = 2πfNAB₀|cos θ|, where θ is measured from its normal. At fixed N, A, f and B₀, vary θ from 0° to 90° and plot ε₀ against cos θ. If an angle is measured from the plane, the equivalent factor is sine. Draw the normal rather than guessing the trig function.

Controls and evaluation

Keep supply frequency and field amplitude fixed during an orientation sweep; recheck them rather than relying on a supply dial.
Use a small coil in a uniform region. A large coil in a nonuniform field measures linked flux over its area, rather than B at a single point.
Align the normal carefully and use a guide for a 180° flip. Account for area uncertainty, including squared length effects for a circular coil.
Choose suitable volts/division and time base, avoid clipping, check probe settings and remove baseline offset before integrating.
Repeat measurements to assess random spread; repeats do not correct a wrong area, angle or calibration.

Quick retrieval check: will a faster complete flip always give a larger inferred B?

No. It may give a larger peak voltage, but the full signed pulse area for the same endpoints remains 2NAB. Inferring B from peak alone without accounting for timing would be invalid.

Transformers: changing flux links two insulated coils

A simple transformer has primary and secondary coils on a shared laminated iron core. Alternating primary current produces changing core flux. This flux links the secondary turns and induces an alternating voltage. The coils are electrically insulated; the core guides magnetic flux, rather than carrying current from one circuit to the other.

Separate insulated primary and secondary windings on a closed laminated iron core, with primary connected to AC and secondary to a resistive load — Energy transfers through the electromagnetic coupling. The ideal model links the same changing core flux through each turn.

V_s/V_p = N_s/N_p = I_p/I_s (ideal transformer)
V_pI_p = V_sI_s (ideal, resistive-load model)

The changing flux per turn is shared, so Faraday’s law makes induced voltage proportional to N. Use the same voltage measure on both sides, such as rms. The ideal current ratio follows from power conservation; a step-up transformer increases voltage and reduces current for the same transferred power. Frequency stays the same.

Worked example 7 · a step-down transformer and its load

An ideal transformer has N_p = 600, N_s = 150 and V_p = 12.0 V rms. A 9.00 Ω resistor is connected to the secondary.

V_s = 12.0 × 150/600 = 3.00 V rms. I_s = V_s/R = 0.333 A rms. Output power = V_s²/R = 1.00 W, so I_p = 1.00/12.0 = 0.0833 A rms.

The secondary current is four times the primary current here, because its voltage is one quarter as large. Label primary and secondary before setting a ratio.

Why steady d.c. gives no sustained secondary voltage

Switching the primary on or off changes its current and the linked core flux, so the secondary can show brief voltage pulses. Once a d.c. current has settled and flux is steady, Faraday’s law gives zero sustained secondary e.m.f. A.c. keeps changing the flux. A core can have nonzero flux while its induced secondary voltage is zero.

Real transformer losses

Windings have resistance and heat by I²R; suitable low-resistance conductors reduce this loss.
Changing core flux induces eddy currents. Thin insulated laminations restrict their loops and reduce heating.
Repeated magnetisation transfers energy to the core as hysteresis loss. A suitable soft magnetic core reduces it.
Leakage flux fails to link both coils fully. A well-assembled closed core and suitable winding arrangement improve coupling.

η = P_out/P_in; P_loss = P_in − P_out

Worked example 8 · real input power and current

A transformer supplies 6.00 V rms and 0.800 A rms to a resistive load. Its efficiency is 85.0%; primary voltage is 24.0 V rms. Assume primary voltage and current are in phase for this simple power calculation.

P_out = 6.00 × 0.800 = 4.80 W. P_in = 4.80/0.850 = 5.65 W. I_p = P_in/24.0 = 0.235 A. Loss power = 5.647 − 4.80 = 0.847 W.

Do not apply ideal equality of input and output power here. For real a.c. measurements, rms V × rms I gives real power only when voltage and current are in phase; a suitable wattmeter measures real input power directly.

Application: efficient electricity transmission

At fixed transferred power, P = VI gives a smaller current at higher voltage. Resistive line loss I²R then falls. Step-up transformers allow high transmission voltages; step-down transformers provide suitable use voltages. Transformers transfer energy, rather than creating extra power.

Transformer practical: test voltage ratio and investigate efficiency

Use suitable insulated coils, a laminated iron core and an isolated low-voltage a.c. supply. Choose turn counts and loads within the apparatus ratings; high turns ratios can produce unexpectedly high secondary voltages.
With supply off, assemble a closed core and connect a.c. voltmeters across the primary and secondary. Leave the secondary unloaded for a simple voltage-ratio investigation.
Keep N_p, supply frequency, measured V_p and core assembly fixed. Change N_s with coil taps or known coils, switching the supply off before rewiring.
Use several values and repeats. Measure V_p at each setting and keep peak/rms conventions consistent.
Plot V_s against N_s. The ideal gradient is V_p/N_p. Alternatively plot V_s/V_p against N_s/N_p, expecting a gradient of 1.
Compare the fitted gradient and uncertainty with the ideal prediction. Consider meter resolution, repeated spread, supply variation, core joints and loading. Do not force an intercept through zero without checking the data.

Original voltage-ratio data: Np = 600, Vp = 6.00 V rms
Ns / turns	Measured Vs / V rms	Ideal Vs / V rms
100	0.98	1.00
200	1.97	2.00
300	2.96	3.00
400	3.96	4.00
500	4.96	5.00

A best fit has gradient about 0.00995 V per turn, compared with the ideal 6.00/600 = 0.0100 V per turn. If each voltage has uncertainty ±0.03 V, a gradient difference alone does not establish disagreement; use a justified gradient uncertainty as well as uncertainty in V_p.

For a loaded efficiency investigation, connect a rated resistor and measure output voltage/current. Measure real input power with suitable equipment, or explicitly state the unity-power-factor approximation if using rms V and I. Compare η at different loads and record heating. Efficiency is undefined as P_out/P_in when both are zero; real unloaded transformers may still consume input power.

Transformer lab: ratios, traces and d.c. transients

Open the transformer lab. The separate lab offers A Level mode, winding presets, loads and twin oscilloscope traces.

Select A Level. With the baseline a.c. settings (12 V rms, 600 primary turns, 300 secondary turns, 24 Ω load), predict V_s = 6 V and I_s = 0.25 A. At 90% efficiency, output power is 1.50 W and input power about 1.67 W.
Set Efficiency, η to 100% and compare ideal input/output powers. Then reduce it to 90%: input current rises to supply the loss while the model keeps its turns-ratio voltage unchanged.
Open Oscilloscope settings and choose Use one linked scale. Compare actual amplitudes. Auto-fitting each trace can make unequal voltages look similar in height.
Use the Step-up, Step-down and 1 : 1 presets. Predict the ratio before reading it. The output frequency is the input frequency.
Switch to DC. Observe the transient and settled output. Use the supply control to test switch-off. Explain each observation through the changing core flux.

Model assumption: this lab keeps the ideal turns-ratio voltage while representing losses through input power/current. It omits real voltage regulation, core saturation and unloaded magnetising losses. Use the measured-data practical above when judging a physical transformer.

Check the transformer tasks

At baseline, Vs = 6 V, Is = 0.25 A and Pout = 1.50 W. At 90% efficiency Pin = 1.667 W and Ip = 0.139 A; at 100%, Pin = 1.50 W and Ip = 0.125 A. Steady d.c. gives no sustained secondary output. Linked oscilloscope scales permit a direct height comparison.

Common electromagnetic-induction exam mistakes

Flux versus linkage: multiply a per-turn flux by N once. Do not add N to a gradient already labelled NΦ.
Plane versus normal: mark the perpendicular normal before selecting cosine or sine.
Graph height versus gradient: large constant linkage gives zero induced e.m.f.
Mean versus instantaneous: an endpoint calculation is an average; an instantaneous value needs the local gradient.
Direction without change: state whether the original flux increases or decreases, then identify the opposing magnetic effect.
Faster generator, same amplitude: higher rotation frequency increases the rate of change and peak e.m.f.; also shorten the period.
Iron core as a conducting connection: the insulated coils couple magnetically; secondary charge is not supplied through the core.
Ideal current ratio for a lossy transformer: use efficiency and power when ideal assumptions do not hold.
Unit conversion: cm² to m² uses 10⁻⁴, and milliseconds to seconds uses 10⁻³.

Retrieve six ideas without looking back

Which angle belongs in BA cos θ?
What does the minus sign in Faraday’s law describe?
What is the e.m.f. when NΦ is constant?
At which generator orientations are e.m.f. magnitudes greatest?
Why is the linkage change 2NAB for a full flip?
How do voltage and current change in an ideal step-up transformer?

Answers: angle to the normal; the opposing response/sign convention; zero; normal at 90° or 270° to B; +NAB changes to −NAB; voltage rises and current falls for the same transferred power.

Original OCR A induction exam practice · 49 marks

Attempt these questions before opening the answers. Show equations, SI conversions and intermediate working. The questions and point schemes are original PhysicsUK practice, informed by assessment themes rather than copied past-paper wording.

1. Flux and flux linkage [3 marks]

A 200-turn coil has area 18.0 cm² in a uniform 0.240 T field. The field makes 40.0° with the coil normal. Calculate the flux through one turn and the flux linkage. State the SI unit of flux.