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E.m.f. Induced in a Moving Conductor

Key idea: Derive and use ε = Blv for a straight conductor moving in a uniform magnetic field, including direction via Fleming’s right-hand rule (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain simple applications of electromagnetic induction, including motional e.m.f. and eddy currents.

1. Definitions (Must Know)

A. Motional e.m.f.

When a straight conductor of length l moves through a magnetic field, charges in the conductor experience a magnetic force. This separates charge and produces an induced potential difference (an e.m.f.).

For the common perpendicular case:

ε = Blv

where B is flux density (T), l is conductor length in the field (m), and v is speed (m s⁻¹).

2. Key Ideas (What Earns Marks)

  • Magnetic force on a charge: F_B = qvB (for motion perpendicular to B).
  • Charge separation sets up an electric field E that grows until qE balances qvB: E = vB
  • Potential difference across the ends: ε = Δ V = El = Blv
  • Direction of induced e.m.f.: Fleming’s right-hand rule.
Geometry assumptions

The result ε = Blv assumes v ⟂ B and the conductor is perpendicular to v. If the geometry changes, you must include the appropriate sine factor.

3. Detailed Explanations

A. Derivation of ε = Blv (perpendicular case)

Motional e.m.f. in a moving conductorA vertical conducting rod moves right through a magnetic field into the page. Positive and negative charges separate along the rod and labelled arrows show velocity, magnetic field and charge force.B into page (×)+−velocity vforce on positive chargerod length ℓε = Bℓv
Scroll diagram horizontally to read all labels.
For perpendicular B, rod length ℓ and velocity v, magnetic forces separate charge until the open-circuit p.d. is ε = Bℓv.

Take a conductor moving with speed v perpendicular to a uniform field B.

Free electrons in the conductor experience magnetic force magnitude: F_B = qvB

This drives charge separation until an electric field E forms such that: F_E = qE = F_B = qvB ⇒ E = vB

The induced potential difference across length l is: ε = Δ V = El = (vB)l = Blv

B. Direction

Use Fleming’s right-hand rule: thumb for conductor motion, first finger for magnetic field and second finger for conventional current. For an open conductor, the second finger gives the direction positive charge is driven and therefore the high-potential end.

4. Common Mistakes

  • Using ε = Blv when the motion is not perpendicular to the field.
  • Mixing up Fleming’s left-hand rule (force) with right-hand rule (induction).
  • Using the full conductor length instead of the length inside the field region.

5. Exam Tips

  • If asked for magnitude only, use magnitudes and state direction in one sentence after.
  • If asked to “show that”, write the balance step qE = qvB explicitly.

6. Worked Examples

Modelled example 1

Find the induced e.m.f.

Core

Problem

A 0.20 m conductor moves at 4.0 m s⁻¹ perpendicular to a uniform 0.50 T field and to its own length. Find the induced e.m.f.
Study the worked solution

  1. Confirm the geometry

    Method

    Use the perpendicular result ε = Blv.

    Reason

    The conductor, velocity and field have the mutually perpendicular geometry required by the simple expression.

    Working

    ε = Blv

  2. Evaluate

    Method

    ε = 0.40 V.

    Reason

    The stated 0.20 m is the active length within the field.

    Working

    ε = (0.50)(0.20)(4.0) = 0.40 V

  3. Connect the mechanism

    Method

    Charge separation grows until the internal electric force balances magnetic force.

    Reason

    qE = qvB gives E = vB, then ε = El = Blv.

    Working

    High-potential end follows Fleming’s right-hand rule.

Guided practice 2

Find speed needed for a target e.m.f.

About 4 min

Problem

A perpendicular setup must generate 2.0 V using B = 0.80 T and active length 0.10 m. Find the required speed.

Try this before viewing the solution

Unit: m s⁻¹

Hints

Hint 1: isolate speed
Divide the target e.m.f. by both B and l.
View solution step by step

  1. Rearrange

    Method

    v = ε/(Bl).

    Reason

    All three geometry factors are perpendicular and fixed.

    Working

    v = ε/Bl

  2. Evaluate

    Method

    v = 25 m s⁻¹.

    Reason

    Substitute the target e.m.f., field and active length.

    Working

    v = 2.0/(0.80)(0.10) = 25 m s⁻¹

Common misconception 3

Find the magnetic flux density

Find and correct the mistake

Learner claim

A perpendicular 0.15 m conductor moving at 6.0 m s⁻¹ produces 0.54 V. A learner writes B = ε lv. Diagnose and calculate.

Try this before viewing the solution

Unit: T

View solution step by step

  1. Repair the rearrangement

    Method

    B = ε/(lv).

    Reason

    Length and speed multiply B in the original equation.

    Working

    B = ε/lv

  2. Evaluate

    Method

    B = 0.60 T.

    Reason

    Divide 0.54 V by (0.15)(6.0).

    Working

    B = 0.54/(0.15)(6.0) = 0.60 T

Examiner practice 4

Find the effective length inside the field

3 marks

Examination question

A conductor moves perpendicularly through B = 0.40 T at 3.0 m s⁻¹ and produces 0.12 V. Find its active length and explain “active”. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Rearrange

    1 mark

    Method

    l = ε/(Bv).

    Reason

    The simple perpendicular geometry supports ε = Blv.

    Working

    l = ε/Bv

  2. Evaluate

    1 mark

    Method

    l = 0.10 m.

    Reason

    Divide the induced e.m.f. by field times speed.

    Working

    l = 0.12/(0.40)(3.0) = 0.10 m

  3. Interpret active length

    1 mark

    Method

    It is the conductor length inside the field that cuts flux.

    Reason

    Segments outside the field make no contribution to this motional e.m.f.

    Working

    Active length = 0.10 m.

Challenge 5

Motion at an angle

Minimal support

Independent transfer

A 0.20 m conductor moves at 5.0 m s⁻¹ in a 0.30 T field, with velocity 40° to vector B. Find the e.m.f. and explain why the parallel velocity component contributes nothing.

Try this before viewing the solution

Hints

Hint 1: use the flux-cutting component
Replace v by v sin 40°.
View solution step by step

  1. Resolve velocity

    Method

    v_⊥ = v sin 40°.

    Reason

    Only motion across field lines produces magnetic separation along the conductor.

    Working

    v_⊥ = 5.0 sin 40°

  2. Calculate e.m.f.

    Method

    ε = 0.193 V ≈ 0.19 V.

    Reason

    Use the perpendicular velocity in Blv_⊥.

    Working

    ε = (0.30)(0.20)(5.0) sin 40° = 0.193 V

  3. Explain the parallel component

    Method

    Motion parallel to vector B gives no magnetic force.

    Reason

    q vector v × vector B = 0 for parallel vectors.

    Working

    Only v_⊥ contributes.

7. Mind Stretchers

Mind stretcher 1: Link to Faraday’s lawExtension

Explain briefly how ε = Blv is consistent with Faraday’s law.

Show Answer

Moving the conductor changes the area of the circuit in the magnetic field, so the flux linkage changes with time.

Faraday’s law says an induced e.m.f. occurs when d(NΦ)/dt ≠ 0. In this geometry, that rate of change produces the result ε = Blv.

Mind stretcher 2: Energy transferExtension

A conductor moves through a magnetic field and drives current in an external resistor. Where does the electrical energy in the resistor come from?

Show Answer

It comes from the mechanical work done to keep the conductor moving through the field.

The induced current produces a magnetic effect that opposes the motion (Lenz’s law), so an external force must do work. That work is converted into electrical energy (and then dissipated as heat in the circuit).

8. Optional (Enrichment)

A. “Cutting flux” wording

Some questions say a conductor “cuts” magnetic field lines. That is just a geometric way to describe a changing flux linkage as the conductor moves.

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