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NotesPhysics HLTopic 4.4Applications of electromagnetic induction
Back to Physics HL Topics
4.4.26 min read

Applications of electromagnetic induction (Physics HL)

IB Physics · Unit 4

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Contents

  1. 1The AC generator
  2. 2The emf equation
  3. 3Turning faster: bigger, more frequent peaks
  4. 4Coil angle, flux and emf
  5. 5In the exam
The big idea: A generator is a motor run backwards: you spin a coil in a magnetic field and it makes a current.
Everyday example: Squeeze a wind-up torch and the bulb glows with no battery inside — you are spinning a coil to make electricity.

From spinning coil to AC

  • A motor uses current to spin a coil; a generator does the reverse.
  • As the coil turns, the magnetic flux through it changes smoothly.
  • By Faraday's law this induces an emf (a driving voltage).
  • The flux rises and falls as the coil rotates, so the emf comes out as a sine wave.
  • This is alternating current (AC).
Why a sine wave?: When the coil is flat to the field the flux is largest but momentarily unchanging, so the emf is zero. When the coil is edge-on to the field the flux is changing fastest, so the emf is at its peak. One full turn gives one full cycle of the sine.

The output is described by the rotational frequency of the coil: ω is the angular frequency (how fast it spins, in rad s⁻¹). The faster you spin it, the bigger and the more frequent the peaks.

Flux and emf as the coil turns

A coil turning in a uniform field with its flux and emf against time, and the emf when it turns twice as fast
The emf peaks where the flux is zero; twice as fast: twice the peak, half the period.

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For a coil of N turns and area A rotating with angular frequency ω in a magnetic field B, the induced emf follows a sine curve.

Peak emf: Its highest value — the peak emf ε₀ — is reached each time the coil is edge-on to the field.
Not a data-booklet line: it comes from Faraday's law with Φ = BA cos ωt. The peak ε₀ = BANω is reached when the coil is edge-on to the field.
instantaneous induced emf (V)
peak (maximum) emf (V)
magnetic flux density (T)
area of the coil (m²)
number of turns on the coil
angular frequency of rotation (rad s⁻¹)
Four ways to get a bigger emf: ε₀ = BANω, so the peak voltage grows if you increase any of:

- B — a stronger magnet - A — a larger coil - N — more turns of wire - ω — spin it faster
IB-style questionDetermine[2 marks]

A flat coil of 200 turns, each of area 0.015 m², spins at an angular frequency of 50 rad s⁻¹ in a uniform magnetic field of flux density 0.40 T. Determine the peak emf of the generator.

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Two things change together: Spin the coil faster and the flux changes more quickly, so the emf peaks are taller; each turn also takes less time, so the peaks come more often.

Peak emf

  • ε₀ = BANω
  • ω × 2 ⇒ peak × 2

Period

  • T = 2π/ω
  • ω × 2 ⇒ period ÷ 2
Changing the speed on a graph: Multiply the peak by the speed factor and divide the period by it. Changing B, A or N changes only the peak, not the period.

Worked example — a slower generator

A generator gives a peak emf of 12 V with a period of 20 ms. Its rotation speed is changed to 1.5 times as fast. Find the new peak emf and period.

Solution

  1. Peak ∝ ω:
  2. Period ∝ 1/ω:

Final answer

Peak 18 V, period 13 ms.

Measure the angle θ between the field and the normal to the coil. The flux and the emf are always a quarter turn out of step:

θ (normal to B)Flux Φ = BA cos θemf
0°: coil faces the fieldlargest, BAzero
90°: coil edge-onzerolargest, BANω
180°: faces it againlargest, −BAzero
When is the emf biggest?: When the plane of the coil is parallel to the field (normal at 90°): the flux is zero there but changing fastest.
Where the energy comes from: When a current is drawn, the induced current opposes the rotation (Lenz), so the generator is harder to turn. The work done turning it becomes the electrical energy delivered.

Worked example — average emf over a quarter turn

A 500-turn coil of area 0.020 m² faces a 0.10 T field. It turns to edge-on in 5.0 ms. Find the average emf.

Solution

  1. Flux change per turn:
  2. Faraday's law:

Final answer

Average emf 200 V.

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Where it shows up: The rotating coil is HL only (D.4). It comes up in Paper 1A and Paper 2.
PaperWhat they ask
Paper 1A'Which change increases the peak emf?', 'what are the new peak and period?', 'at which angle is the emf largest?'
Paper 2Explain how a rotating coil induces an emf; draw the new emf–time graph after a change of speed; determine an average emf over part of a turn.
Three easy marks: (1) The emf is largest when the flux is zero (coil edge-on). (2) Speed × k: peak × k, period ÷ k. (3) Average emf = NΔΦ/Δt, with Φ = BA cos θ at the start and the end.
IB-style questionExplain[3 marks]

A rectangular coil rotates at a steady rate in a uniform magnetic field. Explain how an emf is induced in it.

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A coil rotates in a uniform magnetic field.

the angle between the normal to the coil and the field at which the instantaneous emf is zero.
[1 mark]

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