A charge $q$ is spread uniformly over an isolated ring of radius $R$. The ring is rotated about its natural…
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
$\frac{q \omega R^2}{2}$
$\frac{q \omega R}{2}$
$q \omega R^2$
$\frac{q \omega}{2R}$
Solution
Magnetic dipole moment
$\mathrm{M}=n i \mathrm{~A}, n=1, \mathrm{M}=\mathrm{iA}$
The current in the ring, $i=q / t=q \times f$ Where, $f=$ frequency of charge
$\begin{aligned}
& \mathrm{i}=q \frac{\omega}{2 \pi}\left(\because \omega=2 \pi f \Rightarrow f=\frac{\omega}{2 \pi}\right) \\
& \therefore \quad \mathrm{M}=q \cdot \frac{\omega}{2 \pi} \cdot \pi \mathrm{R}^2=\frac{1}{2} q \omega R^2
\end{aligned}$