Question

A ring of radius $R$ is charged uniformly with charge $+Q$. It can be rotated with angular velocity $\omega$ about (i) its axis or (ii) its diameter. Choose the correct options.

• A. In case-(i) the magnetic field at any instant is in direction of $\omega$ and in case-(ii) the magnetic field is opposite to $\omega$.
• B. The magnetic moment in case-(i) is more than magnetic moment in case-(ii).
• C. The magnitude of magnetic field at centre of the ring in case-(i) is more than that in case-(ii)
• D. In case-(i) as well as case-(ii) the magnetic field on axis of rotation is in direction of $\omega$.

Answer 1

Answer: (B), (C), (D)

The correct options are
B The magnetic moment in case-(i) is more than magnetic moment in case-(ii).
C The magnitude of magnetic field at centre of the ring in case-(i) is more than that in case-(ii)
D In case-(i) as well as case-(ii) the magnetic field on axis of rotation is in direction of $\omega$.

$i_1=\frac{Q}{\frac{2\pi }{\omega }}=\frac{Q\omega }{2\pi }$
$\frac{M_1}{L_1}=\frac{Q}{2m}⨀$
$\Rightarrow M_1=\frac{Q}{2m}\left(m{R}^2\omega \right)$
$=\frac{Q\omega R^2}{2}$
At $O$,
$B_1=\frac{{\mu }_0{i}_1}{2R}$
$=\frac{{\mu }_{0}Q\omega }{4\pi R}⨀$
$d{i}_2=\frac{2dq\omega }{2\pi }=\frac{dq\omega }{\pi }$
$\frac{M_2}{L_2}=\frac{Q}{2m}$
$\Rightarrow M_2=\frac{Q}{2m}\times \left(\frac{m{R}^2}{2}\right)\omega =\frac{Q\omega R^2}{4}\uparrow$
At $O$,
$d{B}_2=\frac{{\mu }_0}{4\pi }\times \frac{2\pi d{i}_2{r}^2}{(r^2+x^2{)}^{3/2}}$
$=\frac{{\mu }_0}{2\pi }\times \frac{d{M}_2}{R^3}(d{M}_2=\pi r^{2}d{i}_2)$
$B_2=\frac{{\mu }_0}{2\pi R^3}\times \int d{M}_2$
$=\frac{{\mu }_0{M}_2}{2\pi R^3}\uparrow$
$=\frac{{\mu }_0}{2\pi R^3}\times \frac{Q\omega R^2}{4}$
$=\frac{{\mu }_{0}Q\omega }{8\pi R}$