A conducting solid sphere of radius $R$ and mass $M$ carries a charge $Q$. The sphere is rotating about an axis passing through its center with a uniform angular speed $\omega$. The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as $\alpha \frac{Q}{2 M}$. The value of $\alpha$ is _______ .

Solution

$\begin{aligned} & d M=d I A \\ & A=\pi r^2=\pi(R \sin \theta)^2 \\ & d I=\frac{d a}{T}=\frac{\sigma(2 \pi r)(R d \theta) \omega}{2 \pi} \\ & d I=\frac{\sigma 2 \pi R^2 \omega \sin \theta d \theta}{2 \pi}\end{aligned}$ $\mathrm{dI}=\sigma \mathrm{R}^2 \omega \sin \theta \mathrm{~d} \theta$ Magnetic dipole moment : $M=\int d M=\int_0^\pi \sigma R^2 \omega \pi R^2 \sin ^3 \theta d \theta$ $M=\sigma R^4 \omega \pi \int_0^\pi \sin ^3 \theta d \theta \quad\left(\because \int_0^\pi \sin ^3 \theta d \theta=\frac{4}{3}\right)$ $\mathrm{M}=\left(\frac{\mathrm{Q}}{4 \pi \mathrm{R}^2}\right) \mathrm{R}^4 \omega \pi\left(\frac{4}{3}\right)$ Magnetic dipole moment $\mathrm{M}=\frac{\mathrm{QR}^2 \omega}{3}$ Angular momentum $\mathrm{L}=\left(\frac{2}{5} \mathrm{MR}^2\right) \omega$ $\frac{\mathrm{M}}{\mathrm{~L}}=\frac{\mathrm{QR}^2 \omega}{3 \times \frac{2}{5} \mathrm{MR}^2 \omega}=\frac{\mathrm{Q}}{2 \mathrm{M}}\left(\frac{5}{3}\right)$ $\alpha=\frac{5}{3}=1.67$

Hence, the correct answer is 1.66.