Question 260 of 276

An insulating thin rod of length $l$ has a linear charge density $\rho \left(x\right)={\rho }_{0}\dfrac{x}{l}$ on it. The rod is rotated about an axis passing through the origin $\left(x=0\right)$ and perpendicular to the rod. If the rod makes $n$ rotations per second, then the time averaged magnetic moment of the rod is:

A$\dfrac{\pi }{4}n{\rho }_{0}{l}^{3}$B$n{\rho }_{0}{l}^{3}$C$\pi n{\rho }_{0}{l}^{3}$D$\dfrac{\pi }{3}n{\rho }_{0}{l}^{3}$

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Answer: (A) $\dfrac{\pi }{4}n{\rho }_{0}{l}^{3}$

$\omega =2\pi nrad{s}^{-1}$


$dQ=\rho dx$

$=\dfrac{{\rho }_{0}}{l}xdx$

Current due to elementary charge $dQ$

$dI=\dfrac{dQ\omega }{2\pi }$

Magnetic dipole moment

$dM=dI\times A$

$\int dM=\int _{0}^{l}\dfrac{\omega }{2\pi }\times \dfrac{{\rho }_{0}}{l}\times xdx\pi {x}^{2}$

$M=\dfrac{{\pi n\rho }_{0}{l}^{3}}{4}$