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Rotating-coil generators

Key idea: A coil rotating steadily in a uniform magnetic field has sinusoidal flux linkage and induced e.m.f. Faster rotation increases both the e.m.f.'s frequency and its peak.

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  • GCE A-Level H2 Physics 2027
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A generator transfers mechanical energy to electrical energy by rotating a coil in a magnetic field. As a coil rotates, the flux through it rises and falls, so Faraday’s law induces an e.m.f. that changes direction every half-turn: an alternating e.m.f.

The e.m.f. of a rotating coil

Take a coil of N turns, each of area A, rotating at a steady angular speed ω in a uniform field B. If its normal is parallel to the field at t = 0, the angle between the normal and the field at time t is ω t, so the flux linkage is

NΦ = NBA cos ω t

By Faraday’s law, the induced e.m.f. is the negative rate of change of this:

E = -d(NΦ)/dt = NBAω sin ω t

The e.m.f. is sinusoidal, with peak value E₀ = NBAω. More turns, a stronger field, a larger coil or faster rotation all give a larger peak. Its frequency is the rotation frequency f = ω/2π: one cycle per turn.

When the plane of the coil is at right angles to the field, the flux linkage is largest, but at that instant it is not changing, so the e.m.f. is zero. When the plane is parallel to the field, the flux linkage is zero but changing fastest, so the e.m.f. is largest. The two curves are a quarter of a cycle apart.

Linkage and e.m.f. during one rotation

Normalised linkage follows cos(2πt/T) and normalised e.m.f. follows sin(2πt/T). E.m.f. is zero at linkage extrema and largest in magnitude when linkage crosses zero.

Scroll across the figure to read all labels.

Normalised linkage follows cos(2πt/T) and normalised e.m.f. follows sin(2πt/T). E.m.f. is zero at linkage extrema and largest in magnitude when linkage crosses zero.Normalised linkage follows cos(2πt/T) and normalised e.m.f. follows sin(2πt/T). E.m.f. is zero at linkage extrema and largest in magnitude when linkage crosses zero.
Calculated for steady rotation in uniform B with the coil normal parallel to B at t = 0. Each curve is divided by its own positive peak: λ₀ = NBA, ε₀ = NBAω. Horizontal position is t/T; vertical values are dimensionless, not quantities in the same physical units.
Open full-size graph
What to noticeLinkage height and its gradient reach maxima a quarter-cycle apart. Faster rotation increases ε₀ but leaves λ₀ unchanged at fixed N, B and A.
View figure data
Figure data: Linkage and e.m.f. during one rotation
Fraction of one rotation, t/T (1)Flux linkage, λ/λ₀E.m.f., ε/ε₀
0.0000001.000e+000.000e+00
0.0083339.986e-015.234e-02
0.0166679.945e-011.045e-01
0.0250009.877e-011.564e-01
0.0333339.781e-012.079e-01
0.0416679.659e-012.588e-01
0.0500009.511e-013.090e-01
0.0583339.336e-013.584e-01
0.0666679.135e-014.067e-01
0.0750008.910e-014.540e-01
0.0833338.660e-015.000e-01
0.0916678.387e-015.446e-01
0.1000008.090e-015.878e-01
0.1083337.771e-016.293e-01
0.1166677.431e-016.691e-01
0.1250007.071e-017.071e-01
0.1333336.691e-017.431e-01
0.1416676.293e-017.771e-01
0.1500005.878e-018.090e-01
0.1583335.446e-018.387e-01
0.1666675.000e-018.660e-01
0.1750004.540e-018.910e-01
0.1833334.067e-019.135e-01
0.1916673.584e-019.336e-01
0.2000003.090e-019.511e-01
0.2083332.588e-019.659e-01
0.2166672.079e-019.781e-01
0.2250001.564e-019.877e-01
0.2333331.045e-019.945e-01
0.2416675.234e-029.986e-01
0.2500006.123e-171.000e+00
0.258333-5.234e-029.986e-01
0.266667-1.045e-019.945e-01
0.275000-1.564e-019.877e-01
0.283333-2.079e-019.781e-01
0.291667-2.588e-019.659e-01
0.300000-3.090e-019.511e-01
0.308333-3.584e-019.336e-01
0.316667-4.067e-019.135e-01
0.325000-4.540e-018.910e-01
0.333333-5.000e-018.660e-01
0.341667-5.446e-018.387e-01
0.350000-5.878e-018.090e-01
0.358333-6.293e-017.771e-01
0.366667-6.691e-017.431e-01
0.375000-7.071e-017.071e-01
0.383333-7.431e-016.691e-01
0.391667-7.771e-016.293e-01
0.400000-8.090e-015.878e-01
0.408333-8.387e-015.446e-01
0.416667-8.660e-015.000e-01
0.425000-8.910e-014.540e-01
0.433333-9.135e-014.067e-01
0.441667-9.336e-013.584e-01
0.450000-9.511e-013.090e-01
0.458333-9.659e-012.588e-01
0.466667-9.781e-012.079e-01
0.475000-9.877e-011.564e-01
0.483333-9.945e-011.045e-01
0.491667-9.986e-015.234e-02
0.500000-1.000e+001.225e-16
0.508333-9.986e-01-5.234e-02
0.516667-9.945e-01-1.045e-01
0.525000-9.877e-01-1.564e-01
0.533333-9.781e-01-2.079e-01
0.541667-9.659e-01-2.588e-01
0.550000-9.511e-01-3.090e-01
0.558333-9.336e-01-3.584e-01
0.566667-9.135e-01-4.067e-01
0.575000-8.910e-01-4.540e-01
0.583333-8.660e-01-5.000e-01
0.591667-8.387e-01-5.446e-01
0.600000-8.090e-01-5.878e-01
0.608333-7.771e-01-6.293e-01
0.616667-7.431e-01-6.691e-01
0.625000-7.071e-01-7.071e-01
0.633333-6.691e-01-7.431e-01
0.641667-6.293e-01-7.771e-01
0.650000-5.878e-01-8.090e-01
0.658333-5.446e-01-8.387e-01
0.666667-5.000e-01-8.660e-01
0.675000-4.540e-01-8.910e-01
0.683333-4.067e-01-9.135e-01
0.691667-3.584e-01-9.336e-01
0.700000-3.090e-01-9.511e-01
0.708333-2.588e-01-9.659e-01
0.716667-2.079e-01-9.781e-01
0.725000-1.564e-01-9.877e-01
0.733333-1.045e-01-9.945e-01
0.741667-5.234e-02-9.986e-01
0.750000-1.837e-16-1.000e+00
0.7583335.234e-02-9.986e-01
0.7666671.045e-01-9.945e-01
0.7750001.564e-01-9.877e-01
0.7833332.079e-01-9.781e-01
0.7916672.588e-01-9.659e-01
0.8000003.090e-01-9.511e-01
0.8083333.584e-01-9.336e-01
0.8166674.067e-01-9.135e-01
0.8250004.540e-01-8.910e-01
0.8333335.000e-01-8.660e-01
0.8416675.446e-01-8.387e-01
0.8500005.878e-01-8.090e-01
0.8583336.293e-01-7.771e-01
0.8666676.691e-01-7.431e-01
0.8750007.071e-01-7.071e-01
0.8833337.431e-01-6.691e-01
0.8916677.771e-01-6.293e-01
0.9000008.090e-01-5.878e-01
0.9083338.387e-01-5.446e-01
0.9166678.660e-01-5.000e-01
0.9250008.910e-01-4.540e-01
0.9333339.135e-01-4.067e-01
0.9416679.336e-01-3.584e-01
0.9500009.511e-01-3.090e-01
0.9583339.659e-01-2.588e-01
0.9666679.781e-01-2.079e-01
0.9750009.877e-01-1.564e-01
0.9833339.945e-01-1.045e-01
0.9916679.986e-01-5.234e-02
1.0000001.000e+00-2.449e-16

The graph divides each quantity by its own peak, so both fit on the same axis; equal heights do not mean equal values.

Check your understanding 1 Before changing the speed

In the model below, keep N, B and A fixed. Predict the factors by which the peak flux linkage and peak e.m.f. change when the rotation frequency rises from 5 to 10 Hz. Enter new peak divided by old peak.

Show answer

NBA is unchanged, so its factor is 1. Since E₀ = NBA(2π f), doubling f doubles the peak e.m.f.: its factor is 2.

Test the frequency change

Use rotating-coil mode. Set the frequency to 5 Hz, note both peaks, then set it to 10 Hz without changing the turns, field or area. Compare the peaks and the time between successive cycles.

t = 0.00 s

A 100-turn coil of area 100 square centimetres rotates at 5 hertz in a 0.2 tesla field; the peak e.m.f. is 6.28 volts. Flux linkage 200 milliwebers, e.m.f. 0 volts.

T
cm²
Hz
More settings
Flux linkage, NΦ
200 mWb
e.m.f., ε
0 V
Peak e.m.f., NBAω
6.28 V
Flux linkage against time
Induced e.m.f. against time

Try this

0 of 1 done

  1. Stop the coil where the e.m.f. is largest: pause it or drag it round. (not done yet)

Check your understanding 2 Read the two traces

Describe the changes in the two peaks and the period when the frequency doubles. Explain why a larger e.m.f. peak can occur without a larger flux-linkage peak. You can use the equations if you cannot run the model.

Show answer

The peak flux linkage stays fixed and the peak e.m.f. doubles. The period halves from 0.20 to 0.10 s. The same change in flux linkage now happens in half the time, so its maximum rate of change doubles. Check that your explanation connects the e.m.f. to the gradient of the linkage trace, rather than its height.

Worked example 1

Peak e.m.f. of a generator coil

Problem

A coil of 200 turns, each of area 1.0 × 10⁻³ m², rotates at 25 revolutions per second in a uniform magnetic field of 0.30 T. Find the peak flux linkage and the peak e.m.f.

Show full solution

  1. Peak flux linkage

    Reason

    The flux linkage is largest when the plane of the coil is at right angles to the field.

    Working

    Answer

    NBA = (200)(0.30)(1.0 × 10⁻³) = 0.060 Wb

  2. Angular speed

    Method

    Convert the frequency to angular speed.

    Reason

    E₀ = NBAω needs ω in rad s⁻¹.

    Working

    Answer

    ω = 2π f = 2π(25) = 157 rad s⁻¹

  3. Peak e.m.f.

    Working

    Answer

    E₀ = NBAω = (0.060)(157) = 9.4 V

Guided practice 2

Twice as fast

About 2 min

Problem

The same coil now rotates at 50 revolutions per second. Find the new peak flux linkage and peak e.m.f.

Find both peaks

Unit: Wb
Show the steps Optional working
Unit: Wb
Unit: V

Hints

Hint 1

which contains ω

Which of the expressions NBA and NBAω contains the quantity that changes?

Show full solution

  1. Both peaks

    Working

    Answer

    The peak flux linkage is still 0.060 Wb. The peak e.m.f. doubles to (0.060)(2π × 50) = 19 V, because the flux linkage changes twice as fast.

Spot the mistake 3

Most flux, most e.m.f.?

About 2 min

Learner claim

A student says the e.m.f. of a generator coil is largest when the plane of the coil is at right angles to the field, because then the most field lines pass through it. What is the e.m.f. at that instant?

Try this before viewing the solution

E.m.f. when the plane is at right angles to the field
Show solution step by step

  1. Rate, not amount

    Method

    Look at how the flux linkage is changing at that instant.

    Reason

    At its largest value, the flux linkage stops rising and starts falling, so for an instant it is not changing.

    Working

    Answer

    The rate of change of flux linkage is zero.

  2. E.m.f.

    Working

    Answer

    The e.m.f. is zero. It is largest a quarter-turn later, when the plane is parallel to the field.

Slip rings and energy

Each end of the coil is connected to a slip ring that turns with it. Carbon brushes press on the rings and connect them to the external circuit, so the alternating e.m.f. reaches the circuit without the wires twisting.

When a load is connected, the induced current in the coil feels forces in the magnetic field. By Lenz’s law, they produce a torque that opposes the rotation. To keep the coil turning steadily, the engine or turbine driving it must do work against this torque, and that work is the electrical energy delivered. The more current drawn, the harder the generator is to turn. With the output circuit open there is an e.m.f. but no load current, so no torque due to that current. A real generator still needs work to overcome friction and heating losses.

Exam-style question 1

Half the turns, twice the frequency

5 marks

Examination question

A generator coil is rewound with half the number of turns and driven at twice the rotation frequency. The area and the field are unchanged.

(a) State and explain the effect on the peak flux linkage and on the peak e.m.f. [2 marks]

(b) State the effect on the frequency of the output. [1 mark]

(c) A resistive load is connected. Explain why the driver must now do more work to keep the coil turning at the same speed. [2 marks]

Write your answer before viewing the mark scheme

Show solution step by step

  1. (a) Peaks

    2 marks

    Method

    Apply the changes to NBA and NBAω.

    Reason

    Only the peak e.m.f. depends on ω.

    Working

    Answer

    The peak flux linkage NBA halves. The peak e.m.f. NBAω is unchanged, because Nω is unchanged.

  2. (b) Frequency

    1 mark

    Reason

    The e.m.f. completes one cycle per revolution.

    Working

    Answer

    The output frequency doubles.

  3. (c) Work

    2 marks

    Working

    Answer

    The induced current in the coil produces a torque that opposes the rotation (Lenz’s law). The driver must do work against this torque, and this work supplies the electrical energy delivered to the load.

Check your understanding 3 Find the rotation rate

A coil rotates steadily in a uniform field. Its peak flux linkage is 0.040 Wb and its peak e.m.f. is 12 V. At t = 0 the linkage is at its positive maximum. Find the rotation frequency and the time of the first positive e.m.f. maximum.

Unit: Hz
Show the steps Optional working
Unit: Hz
Unit: s
Show answer

ω = E₀/(NBA) = 12/0.040 = 300 rad s⁻¹, so f = 300/(2π) = 47.7 Hz.

The e.m.f. is positive as the linkage falls from its maximum. Its first maximum is at ω t = π/2, giving t = π/(2 × 300) = 5.24 × 10⁻³ s. This is T/4, not T/2.

Common mistakes

  • Putting the e.m.f. peak where the flux linkage peaks; they are a quarter of a cycle apart.
  • Using the frequency f instead of the angular speed ω = 2π f in E₀ = NBAω.
  • Thinking faster rotation increases the peak flux linkage; it increases only the rate of change.
  • Thinking a generator produces energy; it transfers energy from whatever turns it.

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Syllabus and review details

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