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(B) The magnetic moment $M$ of a current loop is given by $M = iA$,where $i$ is the current and $A$ is the area of the loop.For a ring of radius $R$,t

Vedclass · Accepted Answer

$\frac{1}{2}Q\omega R^2$

Vedclass · Answer

(B) The magnetic moment $M$ of a current loop is given by $M = iA$,where $i$ is the current and $A$ is the area of the loop.For a ring of radius $R$,the area $A = \pi R^2$.The current $i$ produced by the rotating charge $Q$ is $i = \frac{Q}{T}$,where $T$ is the time period of one revolution.Since the angular speed is $\omega$,the time period $T = \frac{2\pi}{\omega}$.Thus,$i = \frac{Q}{(2\pi / \omega)} = \frac{Q\omega}{2\pi}$.Substituting these into the formula for magnetic moment:$M = \left( \frac{Q\omega}{2\pi} ight) 	imes (\pi R^2) = \frac{1}{2}Q\omega R^2$.

$A$ ring of radius $R$,made of an insulating material,carries a charge $Q$ uniformly distributed on it. If the ring rotates about the axis passing through its centre and normal to the plane of the ring with a constant angular speed $\omega$,then the magnitude of the magnetic moment of the ring is:

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