Question

A charge $q$ is spread uniformly over an insulated loop of radius $r$. If it is rotated with an angular velocity $\omega$ with respect to normal axis then the magnetic moment of the loop is

• A. $\frac{1}{2}q\omega r^2$
• B. $\frac{4}{3}q\omega r^2$
• C. $\frac{3}{2}q\omega r^2$
• D. $q\omega r^2$

Answer 1

The correct option is A $\frac{1}{2}q\omega r^2$

Let us take an element at an angle $\theta$ subtending an angle $d\theta$

The charge $dq$ the element has can be written as

$dq=\frac{q}{2\pi }d\theta$

We know that $i=\frac{dq}{dt}=\frac{q}{2\pi }\times \frac{d\theta }{dt}$

The time $dt$ can be written as

$dt=\frac{d\theta }{w}$

Hence $i=\frac{qw}{2\pi }$

magnetic moment = $iA$

Hence Manetic moment $M=\frac{qw}{2\pi }\times \pi r^2=\frac{qw{r}^2}{2}$