A proton travels at a speed of 5.0 × 10⁷ m/s through a 1.0-T magnetic field.What is the magnitude of the magnetic force on the proton if the angle between the proton's velocity and the magnetic field vector is 30°? (e = 1.60 × 10^-19C) A)2.0 × 10^-14 N B)4.0 × 10^-14 N C)2.0 × 10^-12 N D)4.0 × 10^-12 N

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle moving in a magnetic field is given by:

F = qvBsinθ

Where:
q = charge of the particle = charge of a proton = 1.60 × 10^-19 C
v = velocity of the particle = 5.0 × 10^7 m/s
B = magnetic field strength = 1.0 T
θ = angle between the velocity and magnetic field vectors = 30°

Plugging in the values:

F = (1.60 × 10^-19 C)(5.0 × 10^7 m/s)(1.0 T)sin(30°)
F = 4.0 × 10^-12 N

So, the magnitude of the magnetic force on the proton is 4.0 × 10^-12 N, which corresponds to choice D.

A Proton Travels at a Speed of 5