A charge 'q' is spread uniformly over an isol | vedclass

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(C) The equivalent current '$I$' produced by the rotating ring is given by the charge per unit time: $I = \frac{q}{T} = \frac{q \omega}{2 \pi}$ where

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$\frac{q \omega R^2}{2}$

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(C) The equivalent current '$I$' produced by the rotating ring is given by the charge per unit time: $I = \frac{q}{T} = \frac{q \omega}{2 \pi}$ where '$T = \frac{2 \pi}{\omega}$' is the time period of rotation. The magnetic dipole moment '$M$' of a current loop is given by the product of current '$I$' and the area '$A$' of the loop: $M = I 	imes A$ Since the area of the ring is '$A = \pi R^2$',we have: $M = \left( \frac{q \omega}{2 \pi} ight) 	imes (\pi R^2)$ $M = \frac{q \omega R^2}{2}$

$A$ charge '$q$' is spread uniformly over an isolated ring of radius '$R$'. The ring is rotated about its natural axis with angular speed $\omega$. The magnetic dipole moment of the ring is

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