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

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$\frac{q\omega l^2}{6}$

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(A) Consider a small element of length $dx$ at a distance $x$ from the axis of rotation.The charge on this element is $dq = \frac{q}{l} dx$.The time period of rotation is $T = \frac{2\pi}{\omega}$.The current $di$ due to this rotating element is $di = \frac{dq}{T} = \frac{(q/l) dx}{2\pi/\omega} = \frac{q\omega}{2\pi l} dx$.The magnetic moment $dM$ of this small current loop is $dM = (di) \cdot A = (di) \cdot (\pi x^2) = \left( \frac{q\omega}{2\pi l} dx \right) (\pi x^2) = \frac{q\omega}{2l} x^2 dx$.Integrating from $x = 0$ to $x = l$ to find the total magnetic moment $M$:$M = \int_{0}^{l} \frac{q\omega}{2l} x^2 dx = \frac{q\omega}{2l} \left[ \frac{x^3}{3} \right]_{0}^{l} = \frac{q\omega}{2l} \cdot \frac{l^3}{3} = \frac{q\omega l^2}{6}$.

In the given diagram,a rod is rotating with angular velocity $\omega$. The mass of this rod is $m$,the charge is $q$,and the length is $l$. Find the magnetic moment of this rod.

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