As the electron revolves in the second Bohr orbit in the hydrogen… - Vidyadip

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?

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Answer

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

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