A disc of radius r carrying a positive charge | vedclass

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(D) The magnetic moment $M$ of a rotating charged disc is given by $M = \frac{q}{2m} L$,where $L$ is the angular momentum of the disc.For a disc of ma

Vedclass · Accepted Answer

$\frac{q\omega r^2 B \sin 	heta}{4}$,clockwise

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(D) The magnetic moment $M$ of a rotating charged disc is given by $M = \frac{q}{2m} L$,where $L$ is the angular momentum of the disc.For a disc of mass $m$ and radius $r$,the moment of inertia is $I = \frac{1}{2}mr^2$. Thus,the angular momentum is $L = I\omega = \frac{1}{2}mr^2\omega$.Substituting $L$ into the expression for $M$,we get $M = \frac{q}{2m} \left( \frac{1}{2}mr^2\omega ight) = \frac{q\omega r^2}{4}$.The torque $	au$ acting on the magnetic dipole in a magnetic field $B$ is given by $\vec{	au} = \vec{M} 	imes \vec{B}$.The magnitude of the torque is $	au = MB \sin 	heta = \frac{q\omega r^2}{4} B \sin 	heta$.Using the right-hand rule for the direction of the magnetic moment and the cross product,the torque is directed clockwise.

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