Question

Find the magnitude of the magnetic force on a proton moving at 2.5×105 m/s perpendicular to a 0.30 T magnetic field.

Answer 1

Answer:

F = 1.2×10^-14 N

Explanation:

Given

charge of proton = q = 1.6×10^-19 C

magnetic field = B = 0.3 T

velocity = V = 2.5 ×10^5 m/s

magnetic force = F = ?

F = qvB

F = ( 1.6×10^-19 ) × ( 2.5×10^5 ) × ( 0.3 )

F = 1.2×10^-14 N