Q.The work done in rotating a magnet with magnetic dipole moment m, through $90^\circ$ from its magnetic meridian, is n times the work done to rotate it through $60^\circ$. Find the value of n.

Whenever the torque on a magnetic dipole produces an angular displacement, work is done, and this work is stored as magnetic potential energy in the dipole's new orientation -- directly analogous to the potential energy of an electric dipole in an electric field. Integrating the torque $\tau=mB\sin\theta$ from a reference angle gives $U_m=-mB\cos\theta$, where $\theta$ is the angle between the dipole moment $\vec m$ and the field $\vec B$.

Three special orientations follow directly: at $\theta=0^\circ$ ($\vec m$ parallel to $\vec B$), $U_m=-mB$, the minimum possible value -- the most STABLE orientation; at $\theta=180^\circ$ ($\vec m$ antiparallel to $\vec B$), $U_m=+mB$, the maximum value -- the most UNSTABLE orientation; and at $\theta=90^\circ$ (perpendicular), $U_m=0$. The WORK DONE in rotating the dipole from its stable position through some angle $\theta$ is the resulting rise in potential energy, $W(\theta)=U(\theta)-U(0)=mB(1-\cos\theta)$ -- a form that appears repeatedly in numerical problems asking for the work needed to rotate a magnet through a given angle, or from one named position to another (such as from the most stable to the most unstable orientation, where $W=2mB$).