A loop of current-carrying wire has a magnetic dipole moment of 5.0 * 10^-4 A.m².The moment initially is aligned with a 0.50-T magnetic field.To rotate the loop so its dipole moment is perpendicular to the field and hold it in that orientation, you must do work of: A)0 J B)2.5 * 10^-4 J C)-2.5* 10^-4 J D)1.0* 10^-3 J E)-1.0 *10^-3 J

Question

Quizplus · Accepted Answer

The work done to rotate a magnetic dipole in a magnetic field is given by W = μB(1 - cosθ). Initially, θ = 0° (aligned), and finally, θ = 90° (perpendicular), so W = μB. Here, μ = 5.0 x 10^-4 A·m² and B = 0.50 T, thus W = 5.0 x 10^-4 * 0.50 = 2.5 x 10^-4 J.

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