JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Magnetic Effects of Current

Each charge element on the spinning rod is a tiny current loop. An element $dx$ at distance $x$ holds charge $dq={\rho }_0\frac{x}{\ell }dx$, and rotating $n$ times a second it is equivalent to a current $di=dqn$. The loop it traces has area $\pi x^2$, so its moment is $dM=di\pi x^2=\pi n{\rho }_0\frac{x^3}{\ell }dx$. Integrating from axis to tip, $M=\frac{\pi n{\rho }_0}{\ell }{\int }_0^{\ell }{x}^{3}dx=\frac{\pi n{\rho }_0}{\ell }\cdot \frac{{\ell }^4}{4}=\frac{\pi }{4}n{\rho }_0{\ell }^3$. The rising charge density weights the outer parts of the rod, which also enclose the most area, so the moment grows as the cube of the length.