Question

A rod of length $l$ having uniformly distributed charge $Q$ is rotated about one end with constant frequency $f$. Its magnetic moment is

• A. $\pi fQ{l}^2$
• B. $\frac{\pi fQ{l}^2}{3}$
• C. $\frac{2\pi fQ{l}^2}{3}$
• D. $2\pi fQ{l}^2$

Answer 1

The correct option is B $\frac{\pi fQ{l}^2}{3}$


Let the linear charge density of the rod is $\lambda$.

So, charge on $dx$ will be

$dq=\lambda dx$

The magnetic moment of length $dx$ is given by

$dM=\left(dI\right)A$

$dM=(\frac{dq}{T})\pi x^2=\left(f\lambda dx\right)\pi x^2(\because f=1/T)$

$dM=\pi f\lambda \left(x^{2}dx\right)$

$\therefore$ The dipole moment of the rod will be

$M=\int dM={\int }_0^l\pi f\lambda \left(x^{2}dx\right)$

$M=\pi f\lambda {\left[\frac{x^3}{3}\right]}_0^l=\pi f\lambda \times \frac{l^3}{3}$

$\therefore M=\frac{\pi Qf{l}^2}{3}(\because \lambda =Q/l)$

Hence, option $\left(B\right)$ is correct.