Question

An +9.7 C charge moving at 0.75 m/s makes an angle of 45∘ with a uniform, 1.5 T magnetic field. What is the magnitude of the magnetic force F that the charge experiences?

Answer 1

F = 7.72 N

Step-by-step explanation:

The magnetic force on the charge can be given by the following formula:

`F = qvB Sin\theta`

where,

F = magnetic force = ?

q = magnitude of charge = 9.7 C

v = speed of charge = 0.75 m/s

B = magnetic field = 1.5 T

θ = angle = 45°

Therefore,

`F = (9.7\ C)(0.75\ m/s)(1.5\ T)Sin45^(o)`

F = 7.72 N