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Q.
A particle of charge $q$ and mass $m$ moves in a circular orbit of radius $r$ with angular
speed $\omega$. The ratio of the magnitude of its magnetic moment to that of its angular
momentum depends on
Solution:
The effective current $i=\frac{q\, \omega}{2 \pi}$ and $A=\pi r^{2}$.
Magnetic moment $M=i A=\frac{1}{2} q \, \omega r^{2}$
Angular moment $L=I\, \omega=m r^{2} \, \omega$
$ \Rightarrow \frac{M}{L}=\frac{q}{2 m}$