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

Question

Quizplus · Accepted Answer

The force exerted by a magnetic field on a moving charge is given by the equation F = qvBsinθ, where q is the charge (1.60 × 10^-19 C for a proton), v is the velocity (5.0 × 10^3 m/s), B is the magnetic field strength (0.20 T), and θ is the angle between the velocity and the magnetic field direction (90 degrees since the velocity is eastward and the field is northward). The sine of 90 degrees is 1, so F = (1.60 × 10^-19 C)(5.0 × 10^3 m/s)(0.20 T) = 1.6 × 10^-16 N. The direction of the force is given by the right-hand rule, which indicates the force is upwards for a positively charged particle like a proton.

A Proton Moving Eastward with a Velocity of 5