In the given diagram a rod is rotating with angular velocity $\omega $. Mass of this rod is $m$ charge $q$ and length $l$ then find out magnetic moment of this rod
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$ \frac{M}{L} =\frac{q}{2 m} $
$ M =\frac{q L}{2 m}=\frac{q I \omega}{2 m} $
$=\frac{q \omega}{2 m} \times \frac{m \ell^{2}}{3}\left(I=\frac{m \ell^{2}}{3}\right)$
$=\frac{q \omega \ell^{2}}{6} $
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