Question

An electron moving horizontally to the right with speed v enters a region where a uniform magnetic field exists pointing into the page with magnitude B. As a result the electron executes uniform circular motion with radius r. What is the total work done by the magnetic field during a half cycle of the electron’s motion in the field?

Answer 1

W = 0

Step-by-step explanation:

When an electron moves perpendicular to a uniform B-field. If the field is in a vacuum, the magnetic field B is the dominant factor determining the motion. Since the magnetic force F is perpendicular to the direction of travel, an electron follows a curved path in a magnetic field. The electron continues to follow this curved path until it forms a complete circle. Another way to look at this is that the magnetic force F is always perpendicular to velocity v, so that it does no work on the charged particle. The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected but not the speed.

W = F*d*Cos ∅ = F*d*Cos 90° = 0 where d is the displacement.

The pic shown can help to understand the explanation.