A proton moving eastward with a velocity of 5.0 km/s enters a magnetic field of 0.20 T pointing northward. What are the magnitude and direction of the force that the magnetic field exerts on the proton? (e = 1.60 × 10^-19C) 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

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Quizplus · Accepted Answer

The magnetic force on a charged particle is given by F = qvB sin(θ). Here, θ = 90° since velocity and magnetic field are perpendicular, so sin(θ) = 1. F = (1.60 × 10^-19 C)(5.0 × 10^3 m/s)(0.20 T) = 1.6 × 10^-16 N. Using the right-hand rule, the force direction is upwards.

A Proton Moving Eastward with a Velocity of 5