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(B) Consider a small element of length $dx$ at a distance $x$ from the axis of rotation. The charge on this element is $dq = \frac{Q}{l} dx$.Since the

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$\frac{\pi f Q l^2}{3}$

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(B) Consider a small element of length $dx$ at a distance $x$ from the axis of rotation. The charge on this element is $dq = \frac{Q}{l} dx$.Since the rod rotates with frequency $f$,the current $di$ associated with this rotating charge element is $di = dq \cdot f = \frac{Q}{l} f dx$.The magnetic moment $d\mu$ of this circular loop of radius $x$ is given by $d\mu = (di) \cdot A$,where $A = \pi x^2$ is the area of the loop.Substituting the values,we get $d\mu = \left( \frac{Q f}{l} dx \right) (\pi x^2) = \frac{\pi f Q}{l} x^2 dx$.To find the total magnetic moment $\mu$,we integrate $d\mu$ from $x = 0$ to $x = l$:$\mu = \int_0^l \frac{\pi f Q}{l} x^2 dx = \frac{\pi f Q}{l} \left[ \frac{x^3}{3} \right]_0^l = \frac{\pi f Q}{l} \cdot \frac{l^3}{3} = \frac{1}{3} \pi f Q l^2$.

$A$ rod of length $l$ having uniformly distributed charge $Q$ is rotated about one end with constant frequency $f$. Its magnetic moment is:

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