In the Bohr model, an electron moves in circular orbit around the...

The magnetic moment $M$ of a circular current loop is given by $M = IA$.

The current $I$ due to the orbiting electron is $I = \dfrac{e}{T} = \dfrac{ev}{2\pi r}$.

The area of the circular loop is $A = \pi r^2$.

Substituting $I and A$ into the formula for magnetic moment:

$M = \left(\dfrac{ev}{2\pi r}\right)(\pi r^2) = \dfrac{evr}{2}$

Multiplying and dividing by the mass of the electron $m$:

$M = \dfrac{e}{2m}(mvr) = \dfrac{e}{2m}L$

According to Bohr's quantization rule, the angular momentum $L$ of the electron in the $n^{th}$ state is $L = \dfrac{nh}{2\pi}$.

Substituting $L$ into the expression for $M$:

$M = \left(\dfrac{e}{2m}\right)\dfrac{nh}{2\pi}$

Answer: \left(\dfrac{e}{2m}\right)\dfrac{nh}{2\pi}