Question

A non-conducting disc of radius $R$, having uniformly distributed positive charge $Q$ on the surface, is rotating about its axis with uniform angular velocity $\omega$. The magnetic moment of the disc is-

• A. $\frac{Q\omega R^2}{2}$
• B. $\frac{Q\omega R^2}{3}$
• C. $\frac{Q\omega R^2}{4}$
• D. $\frac{Q\omega R^2}{5}$

Answer 1

The correct option is C $\frac{Q\omega R^2}{4}$

As given the disc of radius $R$ having uniformly distributed charge $Q$ is rotating about its axis with uniform angular velocity $\omega .$


Let us consider an elementary ring of radius $x$ and thickness $dx$.

The magnetic moment due to this elemental ring is,

$dM=di\times dA.......\left(1\right)$

So, area of elementary strip, $dA=2\pi xdx$

Charge distributed on the elemental $\left(dQ\right)$ ring is,

$dQ=\frac{Q}{\pi R^2}\times 2\pi xdx=\frac{2Qxdx}{R^2}$

As we know, equivalent current due to rotation of charges on the strip is,

$di=\frac{dQ}{\left(\frac{2\pi }{\omega }\right)}=\frac{2Qxdx}{R^2}\frac{\omega }{2\pi }$

Putting this value in (1) we get,

$dM=\frac{2Qxdx}{R^2}\frac{\omega }{2\pi }\times \pi x^2=\frac{\omega Q}{R^2}{x}^{3}dx$

On integrating both sides with proper limit we get,

${\int }_0^{M}dM=\frac{\omega Q}{R^2}{\int }_0^R{x}^{3}dx$

$M=\frac{\omega Q}{R^2}{\left[\frac{x^4}{4}\right]}_0^R=\frac{\omega Q}{R^2}\left(\frac{R^4}{4}\right)$

$\therefore M=\frac{\omega Q{R}^2}{4}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

Why this question? To understand the concept of magnetic moment due to distribution of charges.