Question

A charge Q is spread uniformly over an insulated ring of radius ‘R’. If the ring is rotated with angular velocity $‘{\omega }^{\prime }$ with respect to the normal axis, the magnetic moment of the ring is

• A. $\frac{1}{2}Q\omega R^2$
• B. $\frac{Q\omega R^2}{4}$
• C. $\frac{1}{3}Q\omega R^2$
• D. $Zero$

Answer 1

The correct option is A $\frac{1}{2}Q\omega R^2$

Magnetic moment = Angular momentum $\times \frac{q}{2m}$
$=Iw\times \frac{q}{2m}=m{R}^{2}w\times \frac{q}{2m}=\frac{Qw{R}^2}{2}$