Question

An insulating thin rod of length $l$ has a linear charge density $\rho \left(x\right)={\rho }_0\frac{x}{l}$ on it. The rod is rotated about an axis passing through the origin $(x=0)$ and perpendicular to the rod. If the rod makes $n$ rotations per second, then the time averaged magnetic moment of the rod is:

• A. $\frac{\pi }{4}n\rho l^3$
• B. $n\rho l^3$
• C. $\frac{\pi }{3}n\rho l^3$
• D. $\pi n\rho l^3$

Answer 1

The correct option is A $\frac{\pi }{4}n\rho l^3$

Given:
Rod length $=l$
linear charge density $\rho \left(x\right)={\rho }_0\frac{x}{l}$
Rotation of rod $=n\text{per second}$


We know that,
Magnetic moment, $M=NIA$
$\because dQ=\rho dx$

$\therefore dI=\frac{dQ}{2\pi }.\omega$


$dM=dI\times A$

$dM=\frac{\omega }{2\pi }.\frac{{\rho }_0}{l}.x\pi x^{2}dx$

$\Rightarrow M=\frac{{\rho }_0}{l}n\pi {\int }_0^l{x}^{3}dx$

$\Rightarrow M=\frac{\pi }{4}n\rho l^3$

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