Q.An electron in an atom is revolving round the nucleus in a circular orbit of radius $5.3\times10^{-11}$ m, with a speed of $2\times10^6$ m s$^{-1}$. Find the resultant orbital magnetic moment and the orbital angular momentum of the electron. (charge on electron $e=1.6\times10^{-19}$ C, mass of electron $m_e=9.1\times10^{-31}$ kg.)

An electron revolving around the nucleus in a circular orbit of radius r at speed v is, physically, a tiny current loop, with associated current $I=\dfrac{ev}{2\pi r}$ and orbital magnetic moment $m_{orb}=IA=\dfrac12evr$. Since the electron's orbital angular momentum is $L=m_evr$, this can be rewritten as $m_{orb}=\dfrac{e}{2m_e}L$ -- proportional to the angular momentum, with the electron's gyromagnetic ratio $e/(2m_e)$ as the constant of proportionality. Because the electron carries negative charge, $\vec m_{orb}$ points OPPOSITE to $\vec L$.

Apart from this orbital contribution, an electron also has an intrinsic spin angular momentum, entirely apart from any orbital motion, which similarly gives rise to a spin magnetic dipole moment (again pointing opposite to the spin angular momentum, for the same charge-sign reason). The total magnetic moment of an atom is the vector sum of the orbital and spin contributions of every one of its electrons. …