A particle of charge 5C moves perpendicularly to a magnetic Held of magnitude C.01 T….

Physics WAEC 1993

A particle of charge 5C moves perpendicularly to a magnetic Held of magnitude C.01 T. If the velocity of the charge is 1.5ms’2, calculate the magnitude of the force exerted on the particle

  • 0.050 N
  • 0.075N checkmark
  • 0.0300N
  • 3.300N
  • 0.000N

The correct answer is: B

Explanation

F = eV = 0.01 x 5 x 1.5 = 0.075N
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