Question

An insulating rod of length $l$ carries a charge $q$ uniformly distributed on it. The rod is pivoted at one of its ends, and it is rotated at a frequency $f$ about a fixed perpendicular axis in the horizontal plane. The magnetic moment of the rod is

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

Answer 1

The correct option is D $\frac{\pi qf{l}^2}{3}$


Let us suppose a small element of the rod of length $dx$ at a distance $x$ from the axis of rotation.

Total charge on length $l=q$
$\therefore$ charge on the element $dx$ is given by,

$dq=\frac{q}{l}dx$

$\therefore$ magnetic moment for this element is,

$dm=\left(di\right)\pi x^2$

$\int dm=\int \left(\frac{dq}{T}\right)\pi x^2$

$m=\int \left(\frac{qdx}{l\times \frac{2\pi }{\omega }}\right)\pi x^2$

$\Rightarrow m=\frac{\omega q\pi }{2\pi l}{\int }_{x=0}^{x=l}{x}^{2}dx$

$=\frac{\omega q}{2l}.{\left[\frac{x^3}{3}\right]}_{x=0}^{x=l}$

$\Rightarrow m=\frac{l^2\omega q}{6}=\frac{l^2\times 2\pi fq}{6}$

$m=\frac{\pi fq{l}^2}{3}$

Hence, option $\left(D\right)$ is the correct answer.