Answer to Question #340191 in Electricity and Magnetism for Ruby

Question

A proton moving horizontally at 4.5 x 10 ⁴ m/s enters a uniform magnetic field of 0.075 T directed upward 45⁰ with the horizontal. Find the magnitude of the magnetic force acting on the proton.

Answer 1

$F=qvB\sin 45°=3.8\cdot 10^{-16}~N.$