A proton moving at 8.00 106 m/s through a magnetic field of magnitude 1.72 T experiences a magnetic force of magnitude 7.20 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest. Enter only positive values between 0 and 360.)

19.1 deg

Explanation:

v = speed of the proton = 8 x 10⁶ m/s

B = magnitude of the magnetic field = 1.72 T

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N

θ = Angle between proton's velocity and magnetic field

magnitude of magnetic force on the proton is given as

F = q v B Sinθ

7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ

Sinθ = 0.327

θ = 19.1 deg