The magnetic moment of an electron (e) revolv | vedclass

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(B) The magnetic moment $\vec{\mu}$ of a current loop is given by $\vec{\mu} = I \vec{A}$.For an electron of charge $-e$ revolving in an orbit of radi

Vedclass · Accepted Answer

$\vec{\mu}_{L} = -\frac{e \vec{L}}{2m}$

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(B) The magnetic moment $\vec{\mu}$ of a current loop is given by $\vec{\mu} = I \vec{A}$.For an electron of charge $-e$ revolving in an orbit of radius $r$ with speed $v$,the equivalent current is $I = \frac{-e}{T} = \frac{-ev}{2\pi r}$.The area of the orbit is $A = \pi r^2$.Thus,$\vec{\mu} = I \vec{A} = \left( \frac{-ev}{2\pi r} \right) (\pi r^2) = \frac{-evr}{2}$.The orbital angular momentum is $\vec{L} = mvr$.Substituting $vr = \frac{L}{m}$,we get $\vec{\mu} = -\frac{e}{2m} \vec{L}$.

The magnetic moment of an electron $(e)$ revolving in an orbit around a nucleus with an orbital angular momentum $\vec{L}$ is given by:

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