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(B) The magnetic moment $\mu$ of a current loop is given by the product of current $i$ and the area $A$ of the loop.$1$. The current $i$ produced by a

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$\pi n e r^2$

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(B) The magnetic moment $\mu$ of a current loop is given by the product of current $i$ and the area $A$ of the loop.$1$. The current $i$ produced by an electron revolving in an orbit is given by $i = q \cdot f$,where $f$ is the frequency of revolution.$2$. Given that the electron makes $n$ revolutions per second,the frequency $f = n$. Thus,$i = e \cdot n$.$3$. The area $A$ of the circular orbit of radius $r$ is $A = \pi r^2$.$4$. The magnetic moment $\mu = i \cdot A = (e n)(\pi r^2) = \pi n e r^2$.Therefore,the correct option is $B$.

In a hydrogen atom,an electron of mass $m$ and charge $e$ revolves in an orbit of radius $r$ making $n$ revolutions per second. If the mass of hydrogen nucleus is $M$,the magnetic moment associated with the orbital motion of the electron is

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