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JAMB Physics 2012 Question 41

Question 41

Magnetism

A particle carrying a charge of 1.0 × 10−8C enters a magnetic field at 3.0 × 102 ms−1 at right angles to the field. If the force on this particle is 1.8 × 10−8 N, what is the magnitude of the field?

  1. A.6.0 × 10−1T
  2. B.6.0 × 10−2T
  3. C.6.0 × 10−3T
  4. D.6.0 × 10−4T.
Show the answer and explanation

Answer: C (6.0 × 10−3T)

Explanation

F = Bqv, so B = F/(qv) = 1.8 × 10−8 ÷ (1.0 × 10−8 × 3.0 × 102) = 6.0 × 10−3 T.