A proton enters a uniform magnetic field directed into the page. The proton initially travels horizontally to the right and follows a circular path. Explain why the speed of the proton remains constant throughout its circular motion in the magnetic field, and identify the direction of the magnetic force acting on the proton at the instant it enters the field.

The magnetic force on a moving charge is
F = q(v×B). For a proton (q>0) moving to the right (v to the right) in a field directed into the page (B into page), v×B points upward. Thus the force is always perpendicular to the velocity, so it can change direction but not the magnitude of v. Because the force is perpendicular to the velocity, the work done by the magnetic force is zero (W = F·dr = 0). With no work done, the kinetic energy of the proton does not change, so its speed remains constant throughout the circular motion.

At the instant it enters the field the magnetic force is directed upward (towards the top of the page).