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(A) The magnetic moment '$\mu$' of a particle moving in a circular orbit is given by $\mu = I A$, where '$I$' is the current and '$A$' is the area of

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$q$ and $m$

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(A) The magnetic moment '$\mu$' of a particle moving in a circular orbit is given by $\mu = I A$, where '$I$' is the current and '$A$' is the area of the orbit.For a particle of charge '$q$' moving with angular speed '$\omega$',the current '$I$' is $I = \frac{q}{T} = \frac{q\omega}{2\pi}$.The area of the orbit is $A = \pi r^2$.Thus, $\mu = \left(\frac{q\omega}{2\pi}\right)(\pi r^2) = \frac{q\omega r^2}{2}$.The angular momentum '$L$' of the particle is $L = mvr = m(\omega r)r = m\omega r^2$.The ratio of the magnitude of magnetic moment to angular momentum is $\frac{\mu}{L} = \frac{q\omega r^2 / 2}{m\omega r^2} = \frac{q}{2m}$.This 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 upon

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