A loop of current-carrying wire has a magnetic dipole moment of 5.0 *10^-4 A.m².If the dipole moment makes an angle of 57° with a magnetic field of 0.35 T, what is its potential energy? A)-9.5 x 10^-5 J B)-1.5 x 10^-4 J C)-1.8 x 10^-4 J D)+1.5 x 10^-4 J E)+9.5 x 10^-⁵ J

Question

Quizplus · Accepted Answer

The potential energy of a magnetic dipole in a magnetic field is given by U = -m * B * cos(theta), where m is the magnetic dipole moment, B is the magnetic field strength, and theta is the angle between the dipole moment and the magnetic field. In this case, m = 5.0 * 10^-4 A*m^2, B = 0.35 T, and theta = 57°. Plugging these values into the equation, we get U = -5.0 * 10^-4 A*m^2 * 0.35 T * cos(57°) = -9.5 * 10^-5 J.

A Loop of Current-Carrying Wire Has a Magnetic Dipole Moment