A proton moves with a velocity of 2.0 × 10^6 m/s perpendicular to a uniform magnetic field of 0.50 T. What is the magnitude of the magnetic force acting on the proton?

Core Concept

The magnetic force on a moving charge is given by F = qvB sinθ, where q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and magnetic field vectors. For a proton, q = 1.6 × 10^-19 C. Since the motion is perpendicular to the field, θ = 90° and sinθ = 1. Therefore, F = (1.6 × 10^-19 C)(2.0 × 10^6 m/s)(0.50 T) = 1.6 × 10^-13 N.