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(C) The magnetic moment $\mu$ of a current loop is given by $\mu = iA$,where $i$ is the current and $A$ is the area of the loop.For an electron revolv

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$\frac{e}{2 m} L$

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(C) The magnetic moment $\mu$ of a current loop is given by $\mu = iA$,where $i$ is the current and $A$ is the area of the loop.For an electron revolving in a circular orbit of radius $r$ with period $T$,the equivalent current is $i = \frac{e}{T}$.The area of the orbit is $A = \pi r^2$.Thus,$\mu = \left(\frac{e}{T}\right) \pi r^2$.The period $T$ is related to the orbital velocity $v$ by $T = \frac{2 \pi r}{v}$,so $\frac{1}{T} = \frac{v}{2 \pi r}$.Substituting this into the expression for $\mu$,we get $\mu = e \left(\frac{v}{2 \pi r}\right) \pi r^2 = \frac{evr}{2}$.The angular momentum of the electron is $L = mvr$,which implies $vr = \frac{L}{m}$.Substituting $vr$ into the expression for $\mu$,we obtain $\mu = \frac{e}{2} \left(\frac{L}{m}\right) = \frac{e}{2m} L$.

An electron moving around the nucleus with an angular momentum $L$ has a magnetic moment ( $e=$ charge on electron,$m=$ mass of electron )

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