Question

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

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

Answer 1

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


Considering a small element of length $dx$ at a distance $x$ from the centre of rod.


The element will be rotating in a circular path of radius $x$

$A=\pi x^2$

The charge on this element is $dq=\frac{q}{l}\times dx$

Equivalent current due to one complete rotation of this charge is,

$i=\frac{dq}{T}=dq.f$

Magnetic moment due to this small element will be,

$d\mu =\left[\right(\frac{q}{l}\left)dx\right]f.\pi x^2$

Thus net magnetic moment due to the rod,

$\mu =\int d\mu =\frac{\pi qf}{l}\underset{\frac{-l}{2}}{\overset{\frac{l}{2}}{\int }}{x}^{2}dx$

Taking the limits from $x=\frac{-l}{2}$ to $x=\frac{+l}{2}$

$\mu =\frac{\pi qf}{l}[\frac{x^3}{3}]{|}_{\frac{-l}{2}}^{\frac{l}{2}}$

$\mu =\frac{\pi qf}{3l}[\frac{l^3}{8}-(-\frac{l^3}{8})]$

$\mu =\frac{\pi qf}{3l}\times \frac{2l^3}{8}$

$\mu =\frac{\pi qf{l}^2}{12}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (c) is the correct answer.