Question

An insulating thin rod of length $l$ as a $x$ linear charge density $p\left(x\right)={\rho }_0\frac{x}{l}$ 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

• A. $\frac{\pi }{4}n\rho l^3$
• B. $n\rho l^3$
• C. $\pi n\rho l^3$
• D. $\frac{\pi }{3}n\rho l^3$

Answer 1

The correct option is A $\frac{\pi }{4}n\rho l^3$

$\because M=NIA$

$dq=\lambda dx$ and $A=\pi x^2$

$\int dm=\int \left(x\right)=\frac{{\rho }_{0}x}{l}dx.\pi x^2$

$M=\frac{n{\rho }_0\pi }{l}\cdot {\int }_{0}l{x}^3\cdot dx=\frac{n{\rho }_0\pi }{l}\cdot \left[\frac{L^4}{4}\right]$

$M=\frac{n{\rho }_0\pi l^3}{4}$ or $\frac{\pi }{4}n\rho l^3$.