Question 261 of 276

A charge $q$ is spread uniformly over an insulated loop of radius $r$. If it is rotated with an angular velocity $\omega$ with respect to normal axis then magnetic moment of the loop is:

A$\dfrac{3}{2}q\omega {r}^{2}$B$\dfrac{1}{2}q\omega {r}^{2}$C$q\omega {r}^{2}$D$\dfrac{4}{3}q\omega {r}^{2}$

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Answer: (B) $\dfrac{1}{2}q\omega {r}^{2}$

$\dfrac{M}{L}=\dfrac{q}{2m}$$\Rightarrow M=\dfrac{q}{2m}\times m{r}^{2}\omega$$\Rightarrow M=\dfrac{q\omega {r}^{2}}{2}$.