Question

A non-conducting disc of mass $M$ and radius $R$ has a uniform surface charge density $\sigma$ and rotates with an angular velocity $\omega$ about its axis. The magnetic dipole moment and angular momentum are related as

• A. $\overrightarrow{\mu }=\frac{\pi \sigma R^2}{2M}\overrightarrow{L}$
• B. $\overrightarrow{\mu }=\frac{\pi \sigma R^2}{M}\overrightarrow{L}$
• C. $\overrightarrow{\mu }=-\frac{\pi \sigma R^2}{2M}\overrightarrow{L}$
• D. $\overrightarrow{\mu }=-\frac{\pi \sigma R^2}{M}\overrightarrow{L}$

Answer 1

The correct option is A $\overrightarrow{\mu }=\frac{\pi \sigma R^2}{2M}\overrightarrow{L}$

Let us consider an elemental ring of radius $r$ and thickness $dr$.


The charge on the element is,

$dq=\sigma dA=\sigma \left(2\pi rdr\right)$

The magnetic moment of the ring,

$d\mu =\left(dI\right)A=\left(dI\right)\pi r^2$

The current in the elemental ring

$dI=\left(dq\right)f=\left(\sigma dA\right)\frac{\omega }{2\pi }=\left(\sigma 2\pi rdr\right)\frac{\omega }{2\pi }=\sigma \omega rdr$

The magnetic moment of elemental ring,

$d\mu =\left(\sigma \omega rdr\right)\pi r^2=\pi \sigma \omega r^{3}dr$

Now, total magnetic moment will be

$\mu =\int d\mu ={\int }_0^R\pi \sigma \omega r^{3}dr=\frac{1}{4}\pi \sigma \omega R^4$

The magnetic moment vector $\overrightarrow{\mu }$ is parallel to $\overrightarrow{\omega }$ if $\sigma$ is positive.

$\overrightarrow{\mu }=\frac{1}{4}\pi \sigma R^4\overrightarrow{\omega }.........\left(1\right)$

The angular momentum of the disc is,

$\overrightarrow{L}=\left(\frac{1}{2}M{R}^2\right)\overrightarrow{\omega }........\left(2\right)$

From equations $\left(1\right)\&\left(2\right)$, we get

$\therefore \overrightarrow{\mu }=\left(\frac{\pi \sigma R^2}{2M}\right)\overrightarrow{L}$

Hence, option $\left(a\right)$ is the correct answer.