Question

An insulating rod of length $l$ carries a charge $q$ distributed uniformly on it. The rod is pivoted at its mid-point and is rotated at a frequency $f$ about a fixed axis perpendicular to the rod and passing through the pivot. The magnetic moment of the rod system is

• A. $\frac{1}{12}\pi qf{l}^2$
• B. $\pi qf{l}^2$
• C. $\frac{1}{6}\pi qf{l}^2$
• D. $\frac{1}{3}\pi qf{l}^2$

Answer 1

The correct option is A $\frac{1}{12}\pi qf{l}^2$

Magnetic moment ($m$) of a loop having current $I$ in it is given by $I\times A$ (where $A$ is the area of the loop)
Let $dq$ be the small charge present on the rod at a distance $x$ on it.


$I=dq\times f$
$dm=I(\pi x^2$)
$⟹dm=dq\times f\times \pi x^2$
$⟹dm=\frac{q}{m}dx\times f\times \pi x^2$

On integration we get
$\int dm=\int \frac{q}{l}dx\times f\times \pi x^2$
$M=\frac{q}{l}f\pi {\int }_{\frac{-l}{2}}^{\frac{l}{2}}{x}^{2}dx$
$M=\frac{q}{l}\times f\times \pi [\frac{x^3}{3}{]}_{\frac{-l}{2}}^{\frac{l}{2}}$
$⟹M=\frac{q}{l}\times f\times \pi [\frac{l^3}{24}+\frac{l^3}{24}$]

$\therefore M=q\times f\times \pi \left[\frac{l^2}{12}\right]$