A rigid circular loop has a radius of 0.20 m and is in the xy-plane. A clockwise current I is carried by the loop, as shown. The magnitude of the magnetic moment of the loop is 0.75 A ∙ m². A uniform external magnetic field, B = 0.20 T in the positive x-direction, is present. An external torque changes the orientation of the loop from one of lowest potential energy to one of highest potential energy. The work done by this external torque is closest to

Question

Quizplus · Accepted Answer

The potential energy of a magnetic dipole in an external magnetic field is given by U = -m · B, where m is the magnetic moment and B is the external magnetic field. Initially, the loop is in the xy-plane with its magnetic moment antiparallel to the external field, so U = -0.75 A · m² · 0.20 T = -0.15 J. Finally, the loop is aligned with the external field, so U = 0.75 A · m² · 0.20 T = 0.15 J. The difference in potential energy is U(final) - U(initial) = 0.15 J - (-0.15 J) = 0.30 J. Therefore, the work done by the external torque is W = U(final) - U(initial) = 0.30 J, which is closest to choice B.

A Rigid Circular Loop Has a Radius of 0