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JAMB Physics 2005 Question 6

Question 6

Magnetism

A proton moving with a speed of 1.0 × 106ms−1 through a magnetic field of 1.0 T experiences a magnetic force of magnitude 8.0 × 10−14 N. The angle between the proton's velocity and the field is [Proton charge = 1.6 × 10−19C]

  1. A.45°.
  2. B.30°.
  3. C.60°.
  4. D.90°.
Show the answer and explanation

Answer: B (30°.)

Explanation

F = qvB sin θ: sin θ = 8.0 × 10−14 ÷ (1.6 × 10−19 × 106 × 1.0) = 0.5, so θ = 30°.