Question

An insulating rod of length l carries a charge q distributed uniformly on it. The rod is pivoted at an end and is rotated at a frequency f about a fixed perpendicular axis. The magnetic moment of the system is

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

Answer 1

Answer: (D)

$\frac{1}{3}\pi qf{l}^2$

Consider a small element $dl$ at a distance $l$ from the pivot.

Charge on element = $dq=\frac{q}{l}dl$
The rod sweeps a circle of radius $l$

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

Magnetic moment of this element $dl$ is
$dM=diA=\frac{q}{l}fdl\left(\pi l^2\right)$
$M=\int dM=f\frac{q}{l}\pi {\int }_0^l{l}^{2}dl=f\frac{q}{l}\pi \frac{l^3}{3}=f\left(\frac{q}{l}\right)\pi \frac{l^3}{3}$
$M=\frac{fq{l}^2\pi }{3}$