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Question

(C) The magnetic moment $M$ of a particle of charge $q$ moving in a circular orbit of radius $r$ with speed $v$ is given by $M = I A = (\frac{q}{T}) (

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$\frac{q}{2 m}$

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(C) The magnetic moment $M$ of a particle of charge $q$ moving in a circular orbit of radius $r$ with speed $v$ is given by $M = I A = (\frac{q}{T}) (\pi r^2) = (\frac{q \omega}{2 \pi}) (\pi r^2) = \frac{1}{2} q \omega r^2$.The angular momentum $L$ of the particle is given by $L = mvr = m(\omega r)r = m \omega r^2$.The ratio of the magnitude of magnetic moment to angular momentum is $\frac{M}{L} = \frac{\frac{1}{2} q \omega r^2}{m \omega r^2} = \frac{q}{2m}$.This ratio is known as the gyromagnetic ratio.

$A$ 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 is

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