Home · A-Level 9702 · Magnetic fields · Force on a current-carrying conductor · Practice
A2 · Practice questions · Force on a current-carrying conductor

The push that defines the tesla.

Six original Cambridge-style questions on F = BIL sin θ, the definition of magnetic flux density and the tesla, Fleming's left-hand rule, and the current balance.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Sine of the angle between current and field.

01
Calculation
[2 marks]

A straight wire has 8.0 cm of its length in a uniform magnetic field of flux density 0.12 T and carries a current of 6.5 A. Calculate the force on the wire when the current is at right angles to the field, and when it makes an angle of 40° with the field.

  • F = BIL = 0.12 × 6.5 × 0.080 = 6.2 × 10⁻² N. ✓
  • F = BIL sin θ = 0.0624 × sin 40° = 4.0 × 10⁻² N. ✓
02
Analysis
[3 marks]

Define magnetic flux density, and hence define the tesla.

  • Magnetic flux density is the force per unit current per unit length acting on a conductor placed in the field, B = F / IL. ✓
  • The definition applies to a straight conductor placed at right angles to the field (so that sin θ = 1). ✓
  • One tesla is the flux density that produces a force of 1 N on each 1 m of a conductor carrying 1 A at right angles to the field: 1 T = 1 N A⁻¹ m⁻¹. ✓
03
Calculation
[3 marks]

A U-shaped magnet stands on a top-pan balance. A stiff horizontal wire, held in a clamp that is not touching the balance, passes between the poles at right angles to the field; the length of wire in the field is 5.0 cm. When a current of 2.4 A is passed through the wire the balance reading increases by 0.27 g. Calculate the flux density between the poles, and explain why the reading increases even though the wire is not resting on the magnet.

  • The extra force on the balance is F = mg = 0.27 × 10⁻³ × 9.81 = 2.65 × 10⁻³ N. ✓
  • B = F / IL = 2.65 × 10⁻³ / (2.4 × 0.050) = 2.2 × 10⁻² T (22 mT). ✓
  • The field exerts an upward force on the wire, so by Newton's third law the wire exerts an equal downward force on the magnet, which the balance registers as extra weight. ✓
04
Calculation
[3 marks]

A straight conductor of length 0.15 m carrying a current of 3.2 A is turned slowly in a uniform magnetic field. The largest force it experiences is 0.096 N. Calculate the flux density of the field, and the angle between the current and the field when the force is 0.048 N.

  • The maximum force occurs at 90°, so B = Fmax / IL = 0.096 / (3.2 × 0.15) = 0.20 T. ✓
  • 0.048 = 0.096 sin θ, so sin θ = 0.50. ✓
  • θ = 30° (or 150°) between the current and the field. ✓
05
Analysis
[2 marks]

A horizontal wire carries a current towards the east through a region where the magnetic field is directed vertically downwards. State the direction of the magnetic force on the wire, naming the rule used, and state what happens to the force when the current is reversed.

  • By Fleming's left-hand rule (first finger down along the field, second finger east along the current) the thumb points north, so the force on the wire is horizontal, towards the north. ✓
  • Reversing the current reverses the force, which then acts towards the south with the same magnitude. ✓
06
Calculation
[4 marks]

A straight horizontal wire of mass per unit length 0.012 kg m⁻¹ lies at right angles to a uniform horizontal magnetic field of flux density 0.45 T directed towards the north. The current in the wire is adjusted until the magnetic force exactly supports the weight of the wire.

(a) Calculate the current required. (b) State the direction in which the current must flow. (c) The wire is now turned in the horizontal plane so that it makes an angle of 60° with the field. Calculate the current now required to support the wire.

  • (a) Per metre of wire the weight is 0.012 × 9.81 = 0.118 N, and this must equal BI × 1 m. ✓
  • I = 0.118 / 0.45 = 0.26 A. ✓
  • (b) For an upward force with the field towards the north, the left-hand rule requires the current to flow towards the east. ✓
  • (c) Now F = BIL sin 60°, so I = 0.118 / (0.45 × sin 60°) = 0.30 A. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.