A proton is projected with a velocity of 7.0 km/s into a magnetic field of 0.60 T perpendicular to the motion of the proton. What is the magnitude of the magnetic force that acts on the proton? (e = 1.60 × 10^-19C) A) 0 N B) 3.4 × 10^-16 N C) 4.2 × 10^-16 N D) 13 × 10^-16 N E) 6.7 × 10^-16 N

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle is given by F = qvB sin(θ). Here, q = 1.60 × 10^-19 C, v = 7.0 × 10^3 m/s, B = 0.60 T, and θ = 90° (since the field is perpendicular), so F = (1.60 × 10^-19 C)(7.0 × 10^3 m/s)(0.60 T) = 6.72 × 10^-16 N, which is approximately 6.7 × 10^-16 N.

A Proton Is Projected with a Velocity of 7