A proton moving at 4.80 10⁶ m/s through a magnetic field of 1.52 T experiences a magnetic force of magnitude 7.95 10⁻¹³ N. What is the angle between the proton's velocity and the field?

Angle between the proton's velocity and the field is 42.86°

Explanation:

We know that

Force = Charge x velocity x Magnetic field x sin of angle between velocity and field.

F = qvB sinθ

Here we have

q = 1.602 x 10⁻¹⁹ C

v = 4.80 x 10⁶ m/s

B = 1.52 T

F = 7.95 x 10⁻¹³ N

Substituting

F = qvB sinθ

7.95 x 10⁻¹³ = 1.602 x 10⁻¹⁹ x 4.80 x 10⁶ x 1.52 sinθ

θ = 42.86°

Angle between the proton's velocity and the field is 42.86°