Question

A non conducting solid sphere of radius $r=0.5\text{m}$ and mass $m=2.5\text{kg}$ has a charge of $Q=10\text{C}$ distributed uniformly over its volume. If the sphere is rotated about a diameter with an angular speed of $\omega =25\text{rad/s}$. Find the magnetic moment (in ${\text{Am}}^2$) and angular momentum (in ${\text{kgm}}^2/\text{s}$) of the solid sphere.

• A. $12.5;12.5$
• B. $12.5;6.25$
• C. $12.5;12.25$
• D. $6.25;12.5$

Answer 1

The correct option is B $12.5;6.25$

Let us consider an infinitesimal spherical shell of radius $x$ in the sphere as shown in the figure,



Let $dq$ be the charge present in the shell.

As the solid sphere as charge $Q$ over its volume.

$dq=\rho dV$

$\Rightarrow dq=\rho \times \left(2\pi xdx\right)\left(x\cos \theta d\theta \right)$

When the sphere rotates with angular speed $\omega$. Let $di$ be the current in the disc.

$di=dq\times \frac{\omega }{2\pi }$

$\therefore di=\frac{\rho \times \left(2\pi x^2\cos \theta dxd\theta \right)\omega }{2\pi }$

From this, Magnetic moment of the disc $d\mu =\left(di\right)\times$ (Area of elemental disc)

$\Rightarrow d\mu =\frac{\rho \times \left(2\pi x^2\cos \theta dxd\theta \right)\omega }{2\pi }\times \pi x^2{\cos }^2\theta$

Integrating on both sides we get,

$\Rightarrow \mu =\pi \rho \omega \underset{0}{\overset{r}{\int }}{x}^{4}dx\times \underset{\frac{-\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\cos }^3\theta d\theta$

$\Rightarrow \mu =\pi \rho \omega \times \frac{r^5}{5}\times (\sin \theta -\frac{{\sin }^3\theta }{3}){|}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}$

$\Rightarrow \mu =\frac{3Q\omega }{4r^3}\times \frac{r^5}{5}\times \frac{4}{3}$

$\Rightarrow \mu =\frac{Q\omega r^2}{5}$

Substituting the data given data we get,

$\mu =\frac{10\times 25\times (0.5{)}^2}{5}\Rightarrow \mu =12.5{\text{Am}}^2$

Since, $\mu =\frac{Q}{2m}L$

$\Rightarrow L=\frac{12.5\times 2\times 2.5}{10}=6.25\text{kg}{\text{m}}^2/\text{s}$

Hence, option (b) is the correct answer.