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

From given,
Let us consider a small part of rod as shown
So, the magnetic moment of that small part is
$dm=IA=\left(dqf\right)\pi x^2=\left(\frac{q}{l}dx\right)f\pi x^2$
On integrating
$\begin{array}{cc}\int dm& ={\int }_{-l/2}^{l/2}\frac{q}{l}dxf\pi x^2\\ & =q/lf\pi \int x^{2}dx\end{array}$
$=\frac{q}{l}f\pi [\frac{{\left(l/2\right)}^3}{3}-\frac{{(-l/2)}^3}{3}]=\frac{q\pi f{l}^2}{12}$