A thin non-conducting ring of radius $R$ has a total charge $Q$ distributed uniformly over its circumference. The ring is rotated about its axis with a constant angular velocity $\omega$. A small magnetic dipole of magnetic moment $\vec{m}$ is placed at the center of the ring such that $\vec{m}$ is perpendicular to the plane of the ring. The torque experienced by the dipole is:

Solution

1. The rotating ring acts as a circular current loop with current $I=\frac{Q\omega }{2\pi }$.
2. The magnetic field at the center of a current loop is directed along the axis of the loop.
3. The magnetic dipole moment $\vec{m}$ is given to be perpendicular to the plane of the ring, meaning it is parallel to the axis.
4. The torque on a magnetic dipole is $\vec{\tau }=\vec{m}\times \vec{B}$.
5. Since $\vec{m}$ and $\vec{B}$ are parallel, the cross product is zero. Hence the answer is (A).