Question

A proton travels with a speed of 4.90 106 m/s at an angle of 56° with the direction of a magnetic field of magnitude 0.250 T in the positive x-direction. What is the magnitude of the magnetic force on the proton?

Answer 1

F = 1510.21 x 10^-16N

Step-by-step explanation:

F = Bqv Sin (angle)

F = 0.250 x 1.6 x 10^-19 x 4.9 x 10^6 x sin(56)

F = 1510.21 x 10^-16N