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 t...

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

Answer Details

The force experienced by a charged particle moving in a magnetic field is given by the formula F = Bqv, where B is the magnetic field strength, q is the charge of the particle, and v is the velocity of the particle. In this case, the charge on the particle is 5C, the magnetic field strength is 0.01T, and the velocity of the particle is 1.5ms^-1. Thus, substituting the given values into the formula, we get: F = (0.01T)(5C)(1.5ms^-1) = 0.075N Therefore, the magnitude of the force exerted on the particle is 0.075N. So, the correct option is (b) 0.075N.