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(C) Consider a small element of length $dx$ at a distance $x$ from the origin. The charge on this element is $dq = \rho(x) dx = \rho_0 \frac{x}{l} dx$

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$\frac{\pi}{4} n \rho_0 l^3$

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(C) Consider a small element of length $dx$ at a distance $x$ from the origin. The charge on this element is $dq = \rho(x) dx = \rho_0 \frac{x}{l} dx$.As the rod rotates with frequency $n$ (rotations per second),the angular velocity is $\omega = 2\pi n$. The current $di$ produced by this rotating charge element is $di = \frac{dq}{T} = dq \cdot n = \rho_0 \frac{x}{l} dx \cdot n$.The magnetic moment $dM$ of this circular current loop of radius $x$ is $dM = di \cdot A = di \cdot (\pi x^2)$.Substituting $di$: $dM = (\rho_0 \frac{x}{l} dx \cdot n) \cdot \pi x^2 = \frac{\pi n \rho_0}{l} x^3 dx$.Integrating from $x = 0$ to $x = l$: $M = \int_0^l \frac{\pi n \rho_0}{l} x^3 dx = \frac{\pi n \rho_0}{l} [\frac{x^4}{4}]_0^l = \frac{\pi n \rho_0}{l} \cdot \frac{l^4}{4} = \frac{\pi}{4} n \rho_0 l^3$.

An insulating thin rod 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:

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