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 $\frac{1}{12}\pi qf{l}^2$

Consider a small element $dx$ at a distance $x$ from the pivot.

Charge on element $dq=\frac{q}{l}dx$

The rod sweeps a circle of radius $\frac{l}{2}$

Charge $dq$ rotating at frequency $f$ will give rise to current $di=dq\times f=\frac{q}{l}fdx$

Magnetic moment of this element $dx$ is $d\mu =di\times A=\frac{q}{l}fdx\times \pi x^2$

where $A$ is the area of the circle of radius $x$.

$\therefore \mu ={\int }_0^{l}d\mu ={\int }_{-l/2}^{l/2}\frac{q}{l}f\times \pi x^{2}dx=\frac{\pi qf}{l}(\frac{{\left(\frac{l}{2}\right)}^3}{3}-\frac{{\left(\frac{-l}{2}\right)}^3}{3})=\frac{\pi qf{l}^2}{12}$