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 $$\alpha \frac{Q}{2M}$$. The value of $$\alpha$$ is ______.

Let the surface charge of the conducting sphere be uniformly distributed.
All the charge therefore resides on the outer surface.

Surface charge density:
$$\sigma = \frac{Q}{4\pi R^{2}}$$

Take the rotation axis as the z-axis. Consider a thin surface strip between polar angles $$\theta$$ and $$\theta + d\theta$$.

• Radius of the circular ring in this strip: $$r = R\sin\theta$$.
• Width of the strip: $$R\,d\theta$$.
• Area of the strip: $$dA = 2\pi R^{2}\sin\theta\,d\theta$$.
• Charge on the strip: $$dq = \sigma\,dA = \frac{Q}{4\pi R^{2}}\;(2\pi R^{2}\sin\theta\,d\theta) = \frac{Q}{2}\sin\theta\,d\theta.$$

The ring rotates with the sphere at angular speed $$\omega$$. Current in the ring:

$$dI = \frac{dq}{T} = \frac{dq\,\omega}{2\pi}.$$

Magnetic moment of a current loop: $$d\mu = dI \times (\text{area of ring})$$, and the loop area is $$\pi r^{2} = \pi R^{2}\sin^{2}\theta$$.

Therefore $$d\mu = \frac{dq\,\omega}{2\pi}\;(\pi R^{2}\sin^{2}\theta) = \frac{\omega R^{2}}{2}\;dq\,\sin^{2}\theta = \frac{\omega R^{2}}{2}\; \bigl(\frac{Q}{2}\sin\theta\,d\theta\bigr)\sin^{2}\theta = \frac{Q\omega R^{2}}{4}\sin^{3}\theta\,d\theta.$$

Total magnetic dipole moment:

$$\mu = \int_{0}^{\pi} d\mu = \frac{Q\omega R^{2}}{4}\int_{0}^{\pi} \sin^{3}\theta\,d\theta.$$

Using $$\int_{0}^{\pi}\sin^{3}\theta\,d\theta = \frac{4}{3}$$, we obtain

$$\mu = \frac{Q\omega R^{2}}{4}\left(\frac{4}{3}\right) = \frac{Q\omega R^{2}}{3}.$$

Moment of inertia of a solid sphere about a diameter:
$$I = \frac{2}{5}MR^{2}.$$

Angular momentum about the same axis:
$$L = I\omega = \frac{2}{5}MR^{2}\omega.$$

Ratio of magnitudes:

$$\frac{\mu}{L} = \frac{\dfrac{Q\omega R^{2}}{3}} {\dfrac{2}{5}MR^{2}\omega} = \frac{Q}{3}\;\frac{5}{2M} = \frac{5Q}{6M}.$$

The question writes this ratio as $$\alpha \dfrac{Q}{2M}$$, so

$$\alpha\,\frac{Q}{2M} = \frac{5Q}{6M} \;\;\Longrightarrow\;\; \alpha = \frac{5}{3} \approx 1.666.$$

Hence the required value lies in the range 1.65 - 1.67.