Q: If two charged particle each of charge q mass m are connected to the ends of a rigid massless rod and is rotated about an axis passing through the centre and perpendicular to length. Then find the ratio of magnetic moment to angular momentum.

Sol: $\large M = n i A$

$\large M = 2 \times \frac{q \omega}{2 \pi} (\pi (\frac{l}{2})^2)$

$\large M = \frac{q \omega l^2}{4}$

$\large L = 2 (m r^2 \omega) = 2 (m \frac{l^2}{4} \omega)$

$\large L = \frac{m l^2 \omega}{2} $

Hence , $\large \frac{M}{L} = \frac{q}{2 m} $