A bar magnet is suspended in a uniform magnetic field of field strength 0.36 T . If the magnetic moment of the bar magnet is 4 Am ^2 then the work done in rotating the magnet from its most stable position to its most unstable position is:

Step-by-step solution

The potential energy U of a magnetic dipole with magnetic moment M in a uniform magnetic field B is given by U = - M B = -MB , where is the angle between M and B . The most stable position corresponds to = 0^ , where U_ initial = -MB (0^ ) = -MB . The most unstable position corresponds to = 180^ , where U_ final = -MB (180^ ) = MB . The work done W in rotating the magnet is the change in potential energy: W = U_ final - U_ initial = MB - (-MB) = 2MB . Given M = 4 Am ^2 and B = 0.36 T , we have W = 2 4 0.36 = 8 0.36