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

Answer 1

Answer: (D)

Consider an element of the loop AB, subtending angle $d\theta$ at the centre of loop $O$,

So, charge present $d\theta$ is given by
$q^{\prime }=\frac{q}{2\pi }\cdot d\theta$
Then, current $i=\frac{d{q}^{\prime }}{dt}=\frac{d}{dt}\frac{q}{2\pi }\cdot d\theta =\frac{q}{2\pi }\cdot \frac{d\theta }{dt}$
$i=\frac{q}{2\pi }\cdot \omega$
$[$ Since, $\omega =\frac{d\theta }{dt}]$
Therefore, magnetic moment of current loop
$M=iA=i\times \pi r^2=\frac{q}{2\pi }\cdot \omega \pi r^2$
$\Rightarrow M=\frac{1}{2}q\omega r^2$