Question

A charge $q$ is uniformly distributed over the volume of a uniform sphere of mass $m$ and radius $R$, which rotates with an angular velocity $\omega$ about the axis passing through its centre. The ratio of the magnetic momentum and the angular momentum will be

• A. $\frac{q}{m}$
• B. $\frac{q}{2m}$
• C. $\frac{2q}{m}$
• D. $\frac{2q}{5m}$

Answer 1

The correct option is B $\frac{q}{2m}$

Given,

A sphere of radius $R$ having charge $q$ and mass $m$

Consider a volume element $dV$.

From the figure, $dV=r^{2}Sin\theta d\varphi d\theta dr$

Charge present in this $dV$ element is given by

$\frac{q}{\left(\frac{4}{3}\pi R^3\right)}dV$

When the sphere rotates with angular velocity $\omega$, the charge also revolves around the axis of rotation which, in turn, causes a current to flow.

The current is given by:

$i=\frac{Q}{T}=\frac{q}{\left(\frac{4}{3}\pi R^3\right)}dV\times \frac{\omega }{2\pi }$

It's magnetic moment will be

$dM=iA=(\frac{q}{\left(\frac{4}{3}\pi R^3\right)}dV\times \frac{\omega }{2\pi })\times \left(\pi r^{2}Si{n}^2\theta \right)$

Total Magnetic moment is given by:

$M={\int }_0^R{\int }_0^{\pi }{\int }_0^{2\pi }(\frac{q}{\left(\frac{4}{3}\pi R^3\right)}\times \frac{\omega }{2\pi })\times \left(\pi r^{2}Si{n}^2\theta \right)r^{2}Sin\theta drd\theta d\varphi$

$M={\int }_0^R{\int }_0^{\pi }{\int }_0^{2\pi }\left(\frac{3q\omega }{8\pi R^3}\right)r^{4}drSi{n}^3\theta d\theta d\varphi$

$M={\int }_0^R{\int }_0^{\pi }\left(\frac{2\pi \times 3q\omega }{8\pi R^3}\right)r^{4}drSi{n}^3\theta d\theta$

$M={\int }_0^R(\frac{4}{3}\times \frac{2\pi \times 3q\omega }{8\pi R^3})r^{4}dr$

$M=\frac{4}{3}\times \frac{2\pi \times 3q\omega }{8\pi R^3}\times \frac{R^5}{5}$

$M=\frac{q{R}^2\omega }{5}$

Angular momentum of sphere is given by

$L=I\omega =\frac{2}{5}M{R}^2\omega$

Then,

$\frac{M}{L}=\frac{\left(\frac{q{R}^2\omega }{5}\right)}{\left(\frac{2}{5}M{R}^2\omega \right)}$

$\Rightarrow \frac{M}{L}=\frac{q}{2M}$

Hence, the correct answer is OPTION B.