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9702 · 20.2

Force on a current-carrying conductor — practice questions

Practice and worked examples for 9702 Force on a current-carrying conductor. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A straight wire of length 0.25 m carrying a current of 3.0 A is placed in a uniform magnetic field of flux density 0.50 T. The wire is oriented at an angle of 60° to the magnetic field lines. Calculate the force acting on the wire.

Show solution outline
  1. Identify the given values: B = 0.50 T, I = 3.0 A, L = 0.25 m, θ = 60°.
  2. Recall the formula: F = BILsinθ.
  3. Substitute the values: F = (0.50 T) × (3.0 A) × (0.25 m) × sin(60°).
  4. Calculate sin(60°) ≈ 0.866.
  5. F = 0.50 × 3.0 × 0.25 × 0.866 ≈ 0.32475 N.
  6. State the answer with appropriate significant figures: F ≈ 0.32 N.

Worked example 2

An electron is accelerated to a speed of 3.0 x 10^6 m/s and enters a region of uniform magnetic field of flux density 5.0 mT. The electron's velocity is perpendicular to the magnetic field. Calculate the magnitude of the force on the electron. (Charge of an electron, e = 1.60 x 10^-19 C)

Show solution outline
  1. Identify the given values: B = 5.0 mT = 5.0 x 10^-3 T, Q = e = 1.60 x 10^-19 C, v = 3.0 x 10^6 m/s, θ = 90°.
  2. Recall the formula for the force on a charged particle: F = BQvsinθ.
  3. Since the velocity is perpendicular to the field, θ = 90° and sin(90°) = 1. The formula simplifies to F = BQv.
  4. Substitute the values into the simplified formula: F = (5.0 x 10^-3 T) × (1.60 x 10^-19 C) × (3.0 x 10^6 m/s).
  5. Calculate the product: F = (5.0 × 1.60 × 3.0) × 10^(-3 - 19 + 6) N.
  6. F = 24 × 10^-16 N.
  7. State the answer in standard form with appropriate significant figures: F = 2.4 x 10^-15 N.
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