Question

Obtain an expression for the orbital magnetic moment of an electron rotating about the nucleus in an atom.

Solution
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Consider an electron revolving around a nucleus in an atom. Since the electron is a negatively charged particle, so atom will act as a current-carrying loop and its magnetic dipole moment is given by
$$ M=lA $$
Where $$ l$$ =current and $$ A $$ =area of cross-section
if $$ r $$ is the radius of orbit then
$$ M=\left( \frac { e }{ t } \right) \left( \pi r^{ 2 } \right) $$
=$$ \frac { e }{ \frac { 2\pi r }{ v } } \pi r^2=\frac{evr}{2} $$
Sice $$ mvr$$ = angular momentum $$ (L) $$
So, $$ M=\left( \frac { 2 }{ 2m_{ e } } \right) L $$ (in vector form), since magnetic dipole moment of the electron and angular momentum are in opposite direction.

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