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Magnetic Fields

Magnetic fields due to a straight wire and a solenoid

  • 3 things to know
  • Component 3

Overview

Recall questions on the shape and direction of the field around a straight wire and inside a solenoid, the right-hand grip rule for each, B = μ₀I/2πa and B = μ₀nI with every symbol, the permeability of free space, and worked field calculations.

What you need to know

Answer each one before you reveal it.

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Key term

State the equation for the flux density inside a long solenoid, defining each symbol and its unit.

Key term

Why does the flux density inside a long solenoid not depend on its radius?

Key term

Why is the field weaker near the ends of a solenoid than at its centre?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Describe the shape of the magnetic field around a long straight current-carrying wire.

    Model answer

    The field lines are concentric circles centred on the wire, lying in planes at right angles to it. They are closest together near the wire, where the field is strongest.

  2. 2

    How is the direction of the field around a straight wire found?

    Model answer

    The right-hand grip rule is used, with the right thumb pointing along the conventional current. The curled fingers then give the sense in which the field lines run around the wire.

  3. 3

    State the equation for the flux density near a long straight wire, defining each symbol and its unit.

    Model answer

    The equation is B = μ₀I/(2πa). B is the flux density in T, I the current in A, a the perpendicular distance from the wire in m and μ₀ the permeability of free space, 4π × 10⁻⁷ H m⁻¹.

  4. 4

    How does the field around a long straight wire vary with distance from it?

    Model answer

    The flux density is inversely proportional to the distance from the wire. Doubling the distance halves the flux density.

  5. 5

    Calculate the flux density 20 mm from a long straight wire carrying 5.0 A.

    Model answer

    B = μ₀I/(2πa) = (4π × 10⁻⁷ × 5.0)/(2π × 0.020) = 5.0 × 10⁻⁵ T. Multiplying 5.0 × 10⁻⁵ T by 2π × 0.020 m and dividing by 4π × 10⁻⁷ H m⁻¹ returns a current of 5.0 A.

  6. 6

    Describe the shape of the magnetic field of a long current-carrying solenoid.

    Model answer

    Inside the solenoid the field is uniform and parallel to the axis, with evenly spaced straight lines. Outside it looks like the field of a bar magnet, spreading from one end round to the other.

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