Question
As the electron revolves in the second Bohr orbit in the hydrogen atom, the corresponding current is $($about$) 1.3 \times 10^{-4} A$. If the area of the orbit is $($about$) 1.4 \times 10^{-19} m ^2$, what is the $($approximate$)$ equivalent magnetic moment?

Answer

$M=I A=\left(1.3 \times 10^{-4}\right)\left(1.4 \times 10^{-19}\right)$
$=1.82 \times 10^{-23} A \cdot m ^2 \text { is the (approximate) equivalent magnetic moment. }$

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