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 force on the proton? 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 force on a charged particle moving with a velocity v in a magnetic field B can be calculated using the formula F = qvBsinθ, where q is the charge of the particle and θ is the angle between v and B.

In this case, the charge on the proton is the elementary charge e, which is 1.60 × 10^-19 C. Substituting the given values, we get:

F = evBsinθ = (1.60 × 10^-19 C) × (4.0 × 10^14 m/s) × (0.20 T) × sin 60°

F = 1.12 × 10^-4 N

Rounding off to the appropriate number of significant figures, we get:

F = 1.1 × 10^-15 N

Therefore, the best choice is C.

A Proton Moving with a Velocity of 4