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c059741
NotesPhysics HLTopic 4.4Applications of electromagnetic induction
Back to Physics HL Topics
4.4.23 min read

Applications of electromagnetic induction (Physics HL)

IB Physics • Unit 4

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Contents

  • The AC generator
  • The emf equation
  • rms values of AC
  • Transformers
  • In the exam
The big idea: Squeeze a wind-up torch and the bulb glows with no battery inside — you are spinning a coil to make electricity. A generator is a motor run backwards: instead of using current to spin a coil, you spin a coil in a magnetic field and it makes a current.

As the coil turns, the magnetic flux through it changes smoothly, so by Faraday's law it induces an emf (a driving voltage). Because the flux rises and falls as the coil rotates, 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.

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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. Its highest value — the peak emf ε₀ — is reached each time the coil is edge-on to the field.

Given in the data booklet — the AC generator emf, with peak value ε₀ = BANω.
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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Why we need rms: An AC voltage is constantly changing — its average over a cycle is zero, which is useless for describing how much power it delivers. So we quote the root-mean-square (rms) value instead.

The rms value is the steady DC value that would deliver the same average power (the same heating effect in a resistor). When the mains is called "230 V", that 230 V is the rms voltage.
Given in the data booklet — rms is the peak divided by √2 (for a sine wave).
rms current (A) / rms voltage (V)
peak current (A) / peak voltage (V)
≈ 1.41 (the sine-wave factor)
IB-style questionDetermine[2 marks]

A mains supply has a peak voltage of V₀ = 325 V. Determine the rms voltage of the supply.

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What a transformer does: A transformer changes an AC voltage up or down. Two coils share an iron core: AC in the primary (Np turns) makes a changing flux that induces an emf in the secondary (Ns turns).

The voltages are in the same ratio as the turns. More turns on the secondary ⇒ step-up (higher voltage); fewer turns ⇒ step-down (lower voltage).
Given in the data booklet — the ideal transformer equation (note the current ratio is inverted).
primary / secondary voltage (V)
number of turns on primary / secondary
primary / secondary current (A)
An ideal transformer conserves power: An ideal (100% efficient) transformer wastes no energy, so the power in equals the power out: εp Ip = εs Is.

That is why the current ratio is upside down: if a step-down transformer halves the voltage, it doubles the current — and vice versa.
IB-style questionDetermine[2 marks]

A step-down transformer has Np = 1000 turns on the primary and Ns = 50 turns on the secondary. The primary voltage is Vp = 230 V. Determine the secondary voltage.

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Where it shows up: AC generators, rms and transformers are HL only (D.4):

- Paper 1A — a one-step 'find the rms value', 'is this step-up or step-down?', or 'what does an ideal transformer conserve?'. - Paper 2 — determine a peak emf with ε₀ = BANω, or use the transformer ratio (and power conservation) to find a secondary voltage or current.
Three easy marks: (1) rms = peak ÷ √2 (and peak = rms × √2 going the other way). (2) Voltage ratio = turns ratio; current ratio is inverted. (3) "Ideal transformer" ⇒ power in = power out (εp Ip = εs Is).
IB-style questionDetermine[2 marks]

An ideal transformer in a phone charger steps 230 V (rms) down to 5.0 V (rms). The output current to the phone is 2.0 A. Determine the current drawn from the mains by the primary coil.

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IB Exam Questions on Applications of electromagnetic induction

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How Applications of electromagnetic induction Appears in IB Exams

Examiners use specific command terms when asking about this topic. Here's what to expect:

Define

Give the precise meaning of key terms related to Applications of electromagnetic induction.

AO1
Describe

Give a detailed account of processes or features in Applications of electromagnetic induction.

AO2
Explain

Give reasons WHY — cause and effect within Applications of electromagnetic induction.

AO3
Evaluate

Weigh strengths AND limitations of approaches in Applications of electromagnetic induction.

AO3
Discuss

Present arguments FOR and AGAINST with a balanced conclusion.

AO3

See the full IB Command Terms guide →

Related Physics HL Topics

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4.1.1Newton's law of gravitation and field strength
4.1.2Kepler's laws and orbital motion
4.1.3Circular orbits and satellites
4.1.4Gravitational potential energy and escape speed
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