Question

Find the ratio of the magnetic moment to the mechanical moment.

• A. $\frac{p_m}{M}=\frac{q}{2m}$
• B. $\frac{p_m}{M}=\frac{q}{m}$
• C. $\frac{p_m}{M}=\frac{2q}{m}$
• D. $\frac{p_m}{M}=\frac{q}{3m}$

Answer 1

The correct option is A $\frac{p_m}{M}=\frac{q}{2m}$

Here, a charge $q$ is uniformly distributed over the volume of a uniform ball of mass $m$ and radius $R$ which rotates with an angular velocity $\omega$ about the axis passing through its center.
The volume of the ball is
$V=\frac{4}{3}\pi R^3$
The charge $dq$ contained by the volume $dV$ of the ball is
$dq=\frac{q}{\frac{4}{3}\pi R^3}dV$............................................(1)
Also, current $dI$ constituted by the same element is
$dI=\frac{dq}{dt}$
$dI=\frac{3q}{4\pi R^3}dVdt$
$dI=\frac{3q}{4\pi R^3}dV\frac{\omega }{2\pi }$............$($since, $\omega =2\pi n$ and $dt=\frac{1}{n})$
The magnetic moment along the axis of rotation in a circular arc is
$P_m={\int }_0^{R}dIdS$
$P_m={\int }_0^R\frac{3q}{4\pi R^3}dV\frac{\omega }{2\pi }.\pi r^2{\sin }^2\theta dr$
$P_m={\int }_0^R\frac{3q}{4\pi R^3}dV\frac{\omega }{2}.r^2{\sin }^2\theta dr$
Now, integrating for whole volume of ball we get
$P_m={\int }_0^{\frac{\pi }{2}}{\int }_0^R\frac{3q}{4\pi R^3}\left(2\pi r^2\sin \theta d\theta \right)\frac{\omega }{2}.r^2{\sin }^2\theta dr$
$P_m=\frac{3\omega q}{4R^3}{\int }_0^{\frac{\pi }{2}}{\int }_0^R{r}^4.{\sin }^3\theta d\theta dr$
$P_m=\left(\frac{3\omega q}{4R^3}\right)\left(\frac{4R^5}{5}\right)\left(\frac{1}{3}\right)$
$P_m=\frac{q{R}^2\omega }{5}$
The mechanical moment is,
$M=\frac{2}{5}m{R}^2\omega$
Hence, the ratio of the magnetic moment to the mechanical moment is
$\frac{P_m}{M}=\frac{\frac{1}{5}{R}^2\omega q}{\frac{2}{5}m{R}^2\omega }$
$\frac{P_m}{M}=\frac{q}{2m}$