question_answer

An insulating thin rad of length l has a linear charge density \[\rho (x)\,\,=\,\,{{\rho }_{0}}\,\frac{x}{l}\]on it. The rod is rotated about an axis passing through the origin \[(x\,\,=\,\,0)\]and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is- [JEE Main 10-Jan-2019 Morning]

A) \[\frac{\pi }{3}\,n\,\,\rho {{l}^{3}}\]

B) \[\frac{\pi }{4}\,n\,\,\rho {{l}^{3}}\]

C) \[n\,\,\rho {{l}^{3}}\]

D) \[\pi n\,\,\rho {{l}^{3}}\]

Correct Answer: B

Solution :

[b]

Charge of element \[=dq=\rho dx=\frac{{{\rho }_{0}}x}{l}dx\] Magnetic moment of this charge element

\[=\,\,\frac{dqvr}{2}\]

\[dM=\,\frac{{{\rho }_{0}}xdx}{l}\,\times \omega x\,\times \,\frac{x}{2}\]

\[dM=\,\frac{{{\rho }_{0}}\omega {{x}^{3}}dx}{2l}\,\]

\[M=\int\limits_{0}^{\ell }{\frac{{{\rho }_{0}}\omega {{x}^{3}}dx}{2l}}\,=\,\frac{{{\rho }_{0}}\omega }{2l}\,\,\int\limits_{0}^{\ell }{{{x}^{3}}dx}\]

\[=\frac{\rho \omega }{2l}\,\,\frac{{{l}^{4}}}{4}\,=\,\frac{{{\rho }_{0}}\omega {{l}^{3}}}{8}\,\,(Put\,\omega =\,2\pi n)\]

\[=\,\,\frac{2\pi n{{\rho }_{0}}\omega {{l}^{3}}}{8}\,=\,\frac{\pi }{4}\,n\,\rho \,{{l}^{3}}\]