Question

A rod has a total charge $Q$, uniformly distributed along its length $L$. If the rod rotates with angular velocity $\omega$ about its end, compute its magnetic moment.

• A. $\frac{Q\omega L^2}{2}$
• B. $\frac{Q\omega L^2}{6}$
• C. $\frac{Q\omega L^2}{3}$
• D. $\frac{Q\omega L^2}{4}$

Answer 1

The correct option is B $\frac{Q\omega L^2}{6}$


Take differential elements $dl$ on the rod at a distance $l$ from the pivoted end as shown in diagram.

Assuming charge per unit length of the rod is $\lambda$.

$\therefore \lambda =\frac{Q}{L}$

So, the charge on a differential element of length $dl$ will be

$dq=\lambda dl$

The current $dI$ due to rotation of this charge $dq$ is given by

$dI=\frac{dq}{\left(2\pi /\omega \right)}=\frac{\omega }{2\pi }dq=\frac{\omega }{2\pi }\lambda dl$

Thus, the magnetic moment of this differential current loop will be

$d\mu =\left(dI\right)\left(\pi l^2\right)=\left(\frac{\omega }{2\pi }\lambda dl\right)\pi l^2=\frac{\omega \lambda }{2}{l}^{2}dl$

To find total magnetic moment, we integrate

$\mu =\frac{\omega \lambda }{2}{\int }_0^L{l}^{2}dl=\frac{\omega \lambda L^3}{6}$

Substituting $\lambda =\frac{Q}{L}$, we obtain

$\mu =\frac{Q\omega L^2}{6}$

Hence, option $\left(b\right)$ is the correct answer.