Question

A charge $q$ is spread uniformly over an isolated ring of radius $R$. The ring is rotated about its natural axis with an angular velocity $\omega$. Magnetic dipole moment of the ring is

• A. $\frac{q \omega R^{2}}{2}$
• B. $\frac{q \omega R}{2}$
• C. $q \omega R^{2}$
• D. $\frac{q \omega}{2 R}$

Answer 1

Answer: (A)

Given, charge $=q$
Radius of isolated ring $=R$
Angular velocity of ring $=\omega$
Magnetic dipole moment of the ring,
$M=n i A, n=1, M=i A$
The current in the ring, $i=q \times f$
where, $f=$ frequency of charge
$\omega=2 \pi f$
$f=\frac{\omega}{2 \pi}$
$ \Rightarrow i=q \cdot \frac{\omega}{2 \pi}$
Area, $A=\pi R^{2}$
$\Rightarrow M=q \cdot \frac{\omega}{2 \pi} \cdot \pi R^{2}=\frac{1}{2} q \omega R^{2}$