Answer to Question #312730 in Physics for cols

Question

A particle of charge 4.25 x 10-⁹C moves at 3.0 x 10 ⁴ m/s at 60° up the x-y plane through a uniform magnetic field of

5.0 x 10-³T directed along the +z - axis. Find the magnetic force on a particle.

Answer 1

The magnetic force on a particle is given by

$$F=qvB\sin\theta\\ ={\rm 4.25*10^{-9}\:C*3.0*10^4\: m/s}\\ *5.0*\rm 10^{-3}\: T*\sin30^\circ=3.4*10^{-7}\: N$$