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Electromagnetic Induction

Emf induced in a moving conductor

  • 3 things to know
  • A2 Unit 4 Section A

Overview

Recall questions on Emf induced in a moving conductor.

What you need to know

Answer each one before you reveal it.

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Key term

Why is an electromotive force induced in a straight conductor moved at right angles to a uniform magnetic field?

Numbers

Calculate the electromotive force induced in a rod 0.40 m long moving at 6.0 m s⁻¹ at right angles to a field of 0.25 T.

Key term

An emf of 0.60 V is induced in a rod 0.40 m long moving at right angles to a field of 0.25 T. Work back to find its speed

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why is an electromotive force induced in a straight conductor moved at right angles to a uniform magnetic field?

    Model answer

    The conductor sweeps out area and so cuts through field lines, which changes the flux linked with the circuit and induces an electromotive force.

  2. 2

    Give the equation for the electromotive force induced in a straight conductor and define every symbol.

    Model answer

    E = Blv, where E is the induced emf in volts, B the flux density in tesla, l the length of the conductor in the field in metres and v its speed at right angles to the field in m s⁻¹.

  3. 3

    Derive E = Blv from Faraday's law.

    Model answer

    In a time t the conductor sweeps out an area lvt, so the flux cut is Blvt, and dividing by t gives an induced emf of Blv.

  4. 4

    What happens to the induced electromotive force if the conductor moves parallel to the field?

    Model answer

    It is zero, because no field lines are cut and the flux linked with the circuit does not change.

  5. 5

    How is the direction of the induced current in a moving conductor found?

    Model answer

    With Fleming's right-hand rule, taking the thumb along the motion and the first finger along the field.

  6. 6

    Calculate the electromotive force induced in a rod 0.40 m long moving at 6.0 m s⁻¹ at right angles to a field of 0.25 T.

    Model answer

    E = Blv = 0.25 × 0.40 × 6.0, which is 0.60 V.

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