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Electromagnetic inductionEdexcel A-Level Physics: Revision notes

Section 1

Electromagnetic induction

Electromagnetic induction is the production of an e.m.f. in a conductor when the magnetic flux linkage through it changes. If the conductor forms a complete circuit, the e.m.f. drives an induced current.

The flux linkage can change in two ways:

  • Relative movement. A magnet moves into or out of a coil (or a wire cuts across field lines). The flux through the coil changes.
  • A changing current in a linked coil. The current in a primary coil changes, so the field it produces changes, and the flux linkage of a nearby secondary coil changes. This happens when a switch is closed or opened, or with an alternating current.

If the flux linkage is steady, no e.m.f. is induced, even if the field is strong.

Key termselectromagnetic induction
Common mistake

Saying an e.m.f. is induced because there is a magnetic field. It is induced because the flux linkage changes; a stationary magnet inside a coil induces nothing.

Section 2

Factors affecting the induced e.m.f.

Moving magnet and coil. The induced e.m.f. is larger if:

  • the magnet moves faster (flux linkage changes more quickly)
  • the magnet is stronger (larger flux density)
  • the coil has more turns (larger flux linkage)
  • the coil has a larger area

Changing current in a linked coil. The e.m.f. induced in the secondary is larger if:

  • the current in the primary changes faster (for example, a higher-frequency a.c. or a quicker switch)
  • the current change is larger
  • the secondary has more turns
  • a soft iron core guides the flux through the secondary so that more of it is linked
Key termsrate of change of flux linkage

Section 3

Faraday's law

Faraday's law: the magnitude of the induced e.m.f. is equal to the rate of change of flux linkage.

E=−d(NΦ)dt\mathcal{E} = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

For a constant rate of change this is E=−Δ(NΦ)Δt\mathcal{E} = -\dfrac{\Delta(N\Phi)}{\Delta t}. The minus sign shows that the e.m.f. opposes the change producing it (Lenz's law).

Worked example. A 400-turn coil of area 2.5×10−32.5\times10^{-3} m² is perpendicular to a field that falls uniformly from 0.80 T to 0.20 T in 0.30 s.

Δ(NΦ)=400×2.5×10−3×0.60=0.60\Delta(N\Phi) = 400 \times 2.5\times10^{-3} \times 0.60 = 0.60 Wb (turns)

∣E∣=0.60/0.30=2.0|\mathcal{E}| = 0.60 / 0.30 = 2.0 V.

Key termsFaraday's law
Exam tip

Work out the change in flux linkage first (final minus initial, with N included), then divide by the time. Forgetting N is the most common lost mark.

Section 4

Lenz's law and conservation of energy

Lenz's law: the direction of an induced e.m.f. (and any current it drives) is always such as to oppose the change that produces it.

Example: when the north pole of a magnet approaches a coil, the induced current makes the near end of the coil a north pole, which repels the magnet. Withdrawing the magnet makes the near end a south pole, which attracts it back.

This is a consequence of conservation of energy. Work must be done against the opposing force to move the magnet, and that work becomes electrical energy. If the induced current helped the change, the magnet would accelerate while also supplying electrical energy, which would create energy from nothing.

Key termsLenz's law

Must know

  • E.m.f. is induced only when the flux linkage changes
  • Faster change, stronger field and more turns give a larger e.m.f.
  • E=−d(NΦ)/dt\mathcal{E} = -\mathrm{d}(N\Phi)/\mathrm{d}t, with the minus sign from Lenz's law
  • Lenz's law: the induced effect opposes the change
  • Lenz's law is conservation of energy
  • For a coil: ∣E∣=Δ(NΦ)/Δt|\mathcal{E}| = \Delta(N\Phi)/\Delta t

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Electromagnetic induction

  1. A bar magnet is pushed, north pole first, into one end of a long solenoid that is connected to a sensitive centre-zero voltmeter.
    Explain, using the principle of conservation of energy, why the induced current must produce a magnetic field that opposes the movement of the magnet.2 marks
  2. A flat coil of 400 turns and cross-sectional area 2.5 × 10⁻³ m² is placed with its plane perpendicular to a uniform magnetic field. The flux density falls uniformly from 0.80 T to 0.20 T in 0.30 s.
    The same fall in flux density, from 0.80 T to 0.20 T, now takes 0.60 s instead of 0.30 s. Calculate the new average induced e.m.f. and explain your answer.2 marks
  3. A flat rectangular coil of 50 turns and area 6.0 × 10⁻³ m² is held with its plane perpendicular to a uniform magnetic field of flux density 0.40 T. It is rotated through 90° about an axis lying in its plane, in a time of 0.050 s, so that its plane finishes parallel to the field lines.
    Calculate the initial flux linkage of the coil and the average e.m.f. induced in it during the rotation.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).