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

[!TLDR] Treating the orbiting electron as a tiny current loop of area $\pi r^2$ carrying current $I=ev/2\pi r$ gives $m_{orb}=IA=\dfrac12evr=\dfr …

Model the electron as moving with constant speed v in a circular orbit of radius r about the nucleus (Fig. 11.4). If it completes one revolution (distance $2\pi r$) in time T, its speed is $v=2\pi r/T$, so the associated current is $I=\dfrac{e}{T}=\dfrac{ev}{2\pi r}$. Treating the orbit as a current loop of area $A=\pi r^2$, the orbital magnetic moment is $m_{orb}=IA=\dfrac{ev}{2\pi r}\times\pi r^2=\dfrac12evr$. Since the orbital angular momentum of the electron is $L=m_evr$ (with $m_e$ the electron's mass), eliminating $vr=L/m_e$ gives $m_{orb}=\dfrac{e}{2m_e}L$ -- the orbital magnetic moment is proportional to the orbital angular momentum, with the (universal, electron-specific) constant of proportionality $\dfrac{e} …

Model the orbiting electron as a current loop ($I=ev/2\pi r$, area $\pi r^2$), take $m_{orb}=IA$, then su …

Dropping the factor of $\dfrac12$ (it comes from $A=\pi r^2$ combined with $I\propto1/r$, so the r's don't fully cancel the way they might be assumed to); or forgetting that $\vec m_ …