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Updated on : 2022-09-05

Solution

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$M=lA$

Where $l$ =current and $A$ =area of cross-section

if $r$ is the radius of orbit then

$M=(te )(πr_{2})$

=$v2πr e πr_{2}=2evr $

Sice $mvr$ = angular momentum $(L)$

So, $M=(2m_{e}2 )L$ (in vector form), since magnetic dipole moment of the electron and angular momentum are in opposite direction.

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