A particle with a charge of 70 C and a speed of 59 ms enters

A particle with a charge of +7.0 ?C and a speed of 59 m/s enters a uniform magnetic field whose magnitude is 0.41 T. For each of the cases in the drawing, find the magnitude of the magnetic force on the particle.

Solution

F = q B v sin 30 = 7.0 * 10^-6 * 0.41 * 59 * 0.5

= 8.46 * 10^-5 N

F = q B v sin 90 = 7.0 * 10^-6 * 0.41 * 59 * 1

= 1.69 * 10^-4 N

F = q B v sin 150 = 7.0 * 10^-6 * 0.41 * 59 * 0.5

= 8.46 * 10^-5 N