Question 9

Step 1: For a conducting solid sphere, the charge $Q$ resides entirely on the surface. The magnetic dipole moment $\mu$ of a rotating spherical shell of charge is given by: $\mu = \frac{1}{3} Q R^2 \omega$

Step 2: The angular momentum $L$ of a solid sphere rotating about its diameter is given by: $L = I \omega = \frac{2}{5} M R^2 \omega$

Step 3: Calculate the ratio of the magnetic dipole moment to the angular momentum: $\frac{\mu}{L} = \frac{\frac{1}{3} Q R^2 \omega}{\frac{2}{5} M R^2 \omega} = \frac{5}{6} \frac{Q}{M}$

Step 4: Equate this to the given expression $\alpha \frac{Q}{2M}$: $\alpha \frac{Q}{2M} = \frac{5}{3} \cdot \frac{Q}{2M}$ $\alpha = \frac{5}{3} \approx 1.666...$ Rounding off to two decimal places, we get $\alpha = 1.67$.