A proton moving with a velocity of 4.0 × 10⁴ m/s enters a magnetic field of 0.20 T. If the angle between the velocity of the proton and the direction of the magnetic field is 60°, what is the magnitude of the magnetic force on the proton? (e = 1.60 × 10^-19C) A) 1.8 × 10^-15 N B) 0.60 × 10^-15 N C) 1.1 × 10^-15 N D) 2.2 × 10^-15 N E)3)3 × 10^-15 N

Question

Quizplus · Accepted Answer

The formula for the magnetic force on a charged particle moving in a magnetic field is F = qvBsinθ, where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Plugging in the given values, we get F = (1.60 × 10^-19 C)(4.0 × 10^14 m/s)(0.20 T)sin60° = 1.1 × 10^-15 N. Therefore, the correct choice is C.

A Proton Moving with a Velocity of 4