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An insulating rod of length $ \ell$ carries a charge $q$ uniformly distributed on it. The rod is pivoted at its centre and is rotated with a frequency $F$ about a fixed axis perpendicular to the rod and passing through the pivot. The magnetic moment of the rod system is

$\begin {array} {1 1} (a)\;Zero & \quad (b)\; \pi qF\ell^2 \\ (c)\;\pi qF\ell^2/3 & \quad (d)\;\pi qF\ell^2/6 \end {array}$


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Consider a small element at a distance $x$ of small width $dx.$
Magnetic moment of the small element is
$ dm = \large\frac{\bigg( \large\frac{q}{l} dx \bigg) \omega}{2 \pi}$ $ \pi x^2$
Integrating from $ \large\frac{–l}{2} \: to\: \large\frac{\ell}{2}$ give the final answer
Ans : (d)


answered Feb 21, 2014 by thanvigandhi_1
edited Sep 19, 2014 by thagee.vedartham

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