A rod of length $l$ having uniformly distributed charge $Q$ is rotated about one end with constant frequency $'f'.$ Its magnetic moment
Diffcult
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Charge on the differential element $d x, d q=\frac{Q}{\ell} . d x$
equivalent current $di$ $=\mathrm{f} \mathrm{dq}$
$\therefore $ magnetic moment of this element ${\rm{d}}\mu = \left( {{\rm{di}}} \right){\rm{NA}}$
$(\mathrm{N}=1)$
$=\left(\pi \mathrm{x}^{2}\right) \mathrm{f} \frac{\mathrm{Q}}{\ell} \mathrm{d} \mathrm{x}$
$ \Rightarrow {\rm{m}} = \int_0^\mu {\rm{d}} \mu = \frac{{\pi {\rm{fQ}}}}{\ell }\int_0^\ell {{{\rm{x}}^2}} ;\mu = \frac{1}{3}\pi {\rm{fQ}}{\ell ^2} \ldots \ldots $
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