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(C) The angular momentum $L$ of a particle of mass $m$ moving in a circular orbit of radius $r$ with angular speed $\omega$ is given by $L = I\omega =

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$q$ and $m$

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(C) The angular momentum $L$ of a particle of mass $m$ moving in a circular orbit of radius $r$ with angular speed $\omega$ is given by $L = I\omega = (mr^2)\omega$.The magnetic moment $M$ associated with the circulating charge is given by $M = IA$,where $I$ is the equivalent current and $A$ is the area of the orbit.The current $I = \frac{q}{T} = \frac{q}{2\pi/\omega} = \frac{q\omega}{2\pi}$.The area $A = \pi r^2$.Thus,$M = \left(\frac{q\omega}{2\pi}\right)(\pi r^2) = \frac{1}{2}q\omega r^2$.The ratio of the magnetic moment to the angular momentum is $\frac{M}{L} = \frac{\frac{1}{2}q\omega r^2}{mr^2\omega} = \frac{q}{2m}$.Therefore,the ratio depends only on the charge '$q$' and the mass '$m$' of the particle.

$A$ 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

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