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(D) Let the linear charge density be $\lambda = \frac{q}{l}$.Consider a small element of length $dx$ at a distance $x$ from the pivot.The charge on th

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$\frac{1}{3} \pi qfl^2$

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(D) Let the linear charge density be $\lambda = \frac{q}{l}$.Consider a small element of length $dx$ at a distance $x$ from the pivot.The charge on this element is $dq = \lambda dx = \frac{q}{l} dx$.As the rod rotates with frequency $f$,this charge element moves in a circle of radius $x$ with time period $T = \frac{1}{f}$.The equivalent current due to this rotating charge is $dI = \frac{dq}{T} = dq \cdot f = \frac{q}{l} f dx$.The magnetic moment $dm$ of this current loop is $dm = dI \cdot A$,where $A = \pi x^2$ is the area of the circle.$dm = (\frac{q}{l} f dx) (\pi x^2) = \frac{\pi qf}{l} x^2 dx$.Integrating from $x = 0$ to $x = l$ to find the total magnetic moment $M$:$M = \int_0^l \frac{\pi qf}{l} x^2 dx = \frac{\pi qf}{l} [\frac{x^3}{3}]_0^l = \frac{\pi qf}{l} \cdot \frac{l^3}{3} = \frac{1}{3} \pi qfl^2$.

An insulating rod of length $l$ carries a charge $q$ distributed uniformly on it. The rod is pivoted at an end and is rotated at a frequency $f$ about a fixed perpendicular axis. The magnetic moment of the system is

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