question_answer

A non-conducting V-shaped rod of length \[2\ell \] has linear charge density \[\lambda \] as shown in figure. The rod is rotated about point \[O\] with an angular velocity \[\omega \] in the horizontal plane. The magnetic moment of system will be

A)  \[\frac{\lambda \omega {{R}^{3}}}{6}\]               

B)  \[\frac{\lambda \omega {{\ell }^{3}}}{3}\]

C)  zero                                     

D)  \[\frac{\lambda {{\omega }^{2}}{{\ell }^{2}}}{3}\]

Correct Answer: C

Solution :

The charge on small element \[dq=\lambda \,dx\] Current, \[di=\frac{dq}{T}=\frac{(d\ell )\omega }{2\pi }\] Magnetic moment, \[\mu =\int_{{}}^{{}}{(di)A=\int\limits_{0}^{\ell }{\frac{\omega \cdot dq}{2\pi }\times \pi \,{{x}^{2}}}}\] \[=\frac{\omega \ell }{32}\left( \frac{{{\ell }^{3}}}{3} \right)=\frac{\omega \lambda {{\ell }^{3}}}{6}\] Similarly, same magnetic moment is there due to another rod in opposite direction, hence net becomes zero.