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(A) The current $i$ produced by the rotating charge $q$ is given by $i = \frac{q}{T} = \frac{q}{2\pi / \omega} = \frac{q\omega}{2\pi}$.The magnetic mo

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$\frac{q}{2m}$

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(A) The current $i$ produced by the rotating charge $q$ is given by $i = \frac{q}{T} = \frac{q}{2\pi / \omega} = \frac{q\omega}{2\pi}$.The magnetic moment $M$ of the system is $M = iA$,where $A = \pi R^2$ is the area of the circle traced by the particles.$M = \left( \frac{q\omega}{2\pi} \right) (\pi R^2) = \frac{q\omega R^2}{2}$.The angular momentum $L$ of the system about the centre is $L = I\omega$,where $I$ is the moment of inertia.For two particles of mass $m$ at distance $R$ from the axis,$I = mR^2 + mR^2 = 2mR^2$.Thus,$L = (2mR^2)\omega = 2mR^2\omega$.The ratio of the magnitudes is $\frac{M}{L} = \frac{q\omega R^2 / 2}{2mR^2\omega} = \frac{q}{4m}$.Wait,re-evaluating the current: Since there are two charges,the total current $i = \frac{2q}{T} = \frac{2q\omega}{2\pi} = \frac{q\omega}{\pi}$.Then $M = iA = (\frac{q\omega}{\pi})(\pi R^2) = q\omega R^2$.Then $\frac{M}{L} = \frac{q\omega R^2}{2mR^2\omega} = \frac{q}{2m}$.

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