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 _______ .

\(\begin{aligned} & d M=d I A \\ & A=\pi r^2=\pi(R \sin \theta)^2 \\ & d I=\frac{d a}{T}=\frac{\sigma(2 \pi r)(R d \theta) \omega}{2 \pi} \\ & d I=\frac{\sigma 2 \pi R^2 \omega \sin \theta d \theta}{2 \pi}\end{aligned}\)
\(\mathrm{dI}=\sigma \mathrm{R}^2 \omega \sin \theta \mathrm{~d} \theta\)
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
\(\mathrm{L}=\left(\frac{2}{5} \mathrm{MR}^2\right) \omega\)
\(\frac{\mathrm{M}}{\mathrm{~L}}=\frac{\mathrm{QR}^2 \omega}{3 \times \frac{2}{5} \mathrm{MR}^2 \omega}=\frac{\mathrm{Q}}{2 \mathrm{M}}\left(\frac{5}{3}\right)\)
\(\alpha=\frac{5}{3}=1.67\)