Question

A proton is launched with a speed of 3.20 * 10⁶ m/s perpendicular to a uniform magnetic field of 0.150 T in the positive z direction. (a) What is the magnitude of the magnetic force experienced by the proton?

Answer 1

`7.68 * {10}^( - 13) \: N`

Step-by-step explanation:

Given:

q = 1.6×10^-19

v = 3.20×10⁶ m/s

B = 0.150 T

θ = 90°

F = qvB sinθ

`= 1.6 * {10}^( - 19) * 3.2 * {10}^(6) * 0.15 * sin {90}^(o)`

`= 7.68 * {10}^( - 13) \: N`