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
$q'=\frac{q}{2 \pi} \cdot d \theta$
Then, current $i=\frac{d q^{\prime}}{d t}=\frac{d}{d t} \frac{q}{2 \pi} \cdot d \theta=\frac{q}{2 \pi} \cdot \frac{d \theta}{d t}$
$i=\frac{q}{2 \pi} \cdot \omega$
$\left[\right.$ Since, $\left.\omega=\frac{d \theta}{d t}\right]$
Therefore, magnetic moment of current loop
$M=i A=i \times \pi r^{2}=\frac{q}{2 \pi} \cdot \omega \pi r^{2}$
$\Rightarrow M=\frac{1}{2} q \omega r^{2}$