A proton moving eastward with a velocity of 5.0 × 10³ m/s enters a magnetic field of 0.20 T pointing northward. What is the magnitude and direction of the force that acts on the proton? A) 0 N B) 1.6 × 10^-16 N upwards C) 1.6 × 10^-16 N downwards D) 1.1 × 10^-16 N eastwards E) 4.4 × 10^-16 N westwards

Question

Quizplus · Accepted Answer

The force on a charged particle moving in a magnetic field is given by the equation:
F = qvB sin(theta)
where F is the force, q is the charge on the particle, v is the velocity of the particle, B is the magnetic field, and theta is the angle between the velocity and magnetic field vectors.

In this case, the proton has a positive charge and is moving eastward, which is perpendicular to the northward magnetic field. Therefore, the angle between the velocity and magnetic field vectors is 90 degrees and sin(theta) = 1. Plugging in the given values, we get:
F = (1.6 x 10^-19 C)(5.0 x 10^13 m/s)(0.20 T)(1) = 1.6 x 10^-16 N

Since the force is perpendicular to both the velocity and the magnetic field, it acts upwards in the y-direction. Therefore, the answer is B: 1.6 x 10^-16 N upwards.

A Proton Moving Eastward with a Velocity of 5