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7 steps to solving electromagnetic induction problems

Key idea: A fast, exam-friendly workflow for electromagnetic induction problems using NΦ, Faraday’s law, Lenz’s law and direction rules (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use magnetic flux and flux-linkage relationships.
  • Apply Faraday's and Lenz's laws to induced e.m.f. and direction.
  • Explain simple applications of electromagnetic induction, including motional e.m.f. and eddy currents.
  • Explain simple iron-core transformer operation and apply ideal transformer ratios.

1. Definitions (Must Know)

A. A 7-step induction workflow

This page is a problem-solving workflow for induction questions.

2. Key Ideas (What Earns Marks)

  1. Write flux linkage: NΦ = NBA cos θ
  2. Identify what is changing (B, A, θ, or N).
  3. Find the change in flux linkage, Δ(NΦ).
  4. Find the time taken, Δ t.
  5. Use Faraday’s law for magnitude: |ε| = |Δ(NΦ)/(Δ t)|
  6. Use Lenz’s law for direction: induced effect opposes the change producing it.
  7. If a conductor is moving in a field, use Fleming’s right-hand rule to confirm direction.
Scope note

This workflow is fully syllabus-consistent, but it’s an exam technique page rather than a new content concept.

3. Detailed Explanations

A. Why steps 1–5 work

They are just Faraday’s law written in a calculation-friendly form:

ε = -d(NΦ)/dt ≈ -Δ(NΦ)/(Δ t)

4. Common Mistakes

  • Using Φ = BA when the field is not perpendicular (missing cos θ).
  • Forgetting that direction needs a separate reasoning step (Lenz’s law / right-hand rule).
  • Mixing units (cm² vs m², ms vs s).

5. Exam Tips

  • In direction questions, talk about “flux increasing/decreasing” explicitly before applying Lenz’s law.
  • If asked for the induced current, you must know whether the circuit is closed.

6. Worked Examples

Modelled example 1

Quick magnitude question

Core

Problem

Flux linkage changes from 0.080 to 0.020 Wb turn in 0.015 s. Find the induced e.m.f. magnitude.
Study the worked solution

  1. Identify the changing quantity

    Method

    The flux linkage decreases.

    Reason

    The question gives initial and final NΦ values directly, so no separate NBA cos θ calculation is needed.

    Working

    Δ(NΦ) = 0.020-0.080 = -0.060 Wb turn

  2. Apply Faraday’s law for magnitude

    Method

    Divide the magnitude of the linkage change by the elapsed time.

    Reason

    The question asks for e.m.f. magnitude; direction is a separate Lenz’s-law step.

    Working

    |ε| = |Δ(NΦ)/(Δ t)| = |(-0.060)/0.015|

  3. Evaluate

    Method

    |ε| = 4.0 V.

    Reason

    A negative linkage change does not make a magnitude negative.

    Working

    |ε| = 4.0 V

Guided practice 2

Lenz’s-law direction workflow

About 4 min

Problem

Viewed from the front, magnetic flux directed into the page through a closed coil is increasing. State the direction of the induced magnetic field and the induced conventional current.

Try this before viewing the solution

Induced field and current

Hints

Hint 1: oppose the change
The induced field must oppose an increase of flux into the page.
Hint 2: convert field to current
Curl the fingers of your right hand in the current direction so the thumb points along the induced field.
View solution step by step

  1. Name the change

    Method

    Flux into the page is increasing.

    Reason

    Lenz’s law responds to the change, not simply to the existing field direction.

    Working

    increasing: into page

  2. Oppose the change

    Method

    The induced field points out of the page.

    Reason

    An out-of-page field opposes the increase of into-page flux.

    Working

    induced field: out of page

  3. Find current direction

    Method

    The induced conventional current is anticlockwise as viewed from the front.

    Reason

    The coil right-hand grip rule gives an out-of-page field for anticlockwise current.

    Working

    out-of-page field ↔ anticlockwise current

Challenge 3

Moving-conductor workflow transfer

Minimal support

Independent transfer

A straight conductor of effective length 0.30 m moves at 8.0 m s⁻¹ through a uniform 0.50 T field. Find the motional e.m.f. when its velocity is (i) perpendicular to the field and (ii) at 30° to the field.

Try this before viewing the solution

Hints

Hint 1: identify the effective motion
Use ε = Bℓ v sin θ; only the velocity component perpendicular to the field cuts flux.
View solution step by step

  1. Use perpendicular motion

    Method

    ε = 1.2 V.

    Reason

    At 90°, the full speed cuts magnetic field lines.

    Working

    ε = Bℓ v = (0.50)(0.30)(8.0) = 1.2 V

  2. Adapt the geometry

    Method

    At 30°, use the perpendicular speed component v sin 30°.

    Reason

    The component parallel to the field does not contribute to motional e.m.f.

    Working

    ε = Bℓ v sin 30°

  3. Evaluate the changed case

    Method

    ε = 0.60 V.

    Reason

    The perpendicular speed component is half the original speed.

    Working

    ε = (0.50)(0.30)(8.0)(0.5) = 0.60 V

7. Mind Stretchers

Mind stretcher 1: Open circuit vs closed circuitExtension

If a coil is open-circuited, can there still be an induced e.m.f.? Can there be an induced current?

Show Answer

An induced e.m.f. can exist, but no induced current flows because the circuit is open.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027