Magnetic moment and angular momentum of electron I recently got to know about something really interesting.
These are as follows:
1: The magnetic moment of an electron is, $\cfrac{ev}{2πr}$, where $e$ is the charge of the electron, $v$ is its velocity, and $r$ is the radius of the orbit it revolves.
2: The direction of the magnetic moment of the electron is anti-parallel to the direction of angular momentum.
3: The ratio $\cfrac ML$, where M is the magnetic moment, and L is the angular momentum, is constant $\cfrac e{2m}$.
Are these facts somewhat or in some way related to the spin quantum number?
 A: What you described is called the gyromagnetic ratio
between magnetic moment ($M$) and orbital angular momentum ($L$)
of the electron
$$\gamma_l=\frac{M}{L}=\frac{e}{2m}$$
where $e$ and $m$ are charge and mass of the electron.
However this relation holds only for the orbital angular momentum.
The corresponding relation for the spin angular momentum ($S$)
experimentally turned out to be
$$\gamma_s=\frac{M}{S}=\frac{e}{m}$$
Notably, this ratio is not the same, but double of the $\gamma_l$ above.
A: Yes and no. While nothing you listed is relativistic, L and M will take on only discrete values based on quantum effects in the bohr model. The orbits are integer numbers of wavelengths of the electron’s debroglie matter-wave. In that sense, yes quantum.
But the formulae considered are all classical Maxwell stuff. The field for the dipole comes from the classical (non-relativistic) Bio-savart Law for a point charge:
$$\vec{B} = \frac {\mu_0 q \vec{v} \times \vec{r}}{4 \pi |r|^2}, \text{ } q=e<0$$
(The following article is easy and quick to pan through, see Moving Charges Create Magnetic Fields: https://www.school-for-champions.com/science/magnetic_field_moving_charges.htm )
This is how the magnetic field is created. $\vec{v} \times \vec{r}$ by the right-hand rule points a direction we know define as up, making $\vec{B}$ down due to $q<0$.
The field maximum is inside the electron loop on the plane of orbit. A closed-current loop is commonly referred to as a magnetic dipole with $IA$ its magnetic dipole moment $\vec{M}$, direction down as of $\vec{B}$, where $A$ is enclosed area, and current $I=\tfrac{dq}{dt}=fe$ , frequency times charge (the electron’s charge times how many times the electron passes per unit time). $\implies M = feA$:
$$M = feA = (\tfrac{v}{2 \pi r})e(\pi r^2) = \tfrac{1}{2} evr$$
$$\vec{L}= I \vec{\omega} = (m|r|^2) (\vec{v} \times  \tfrac{\vec{r}}{|r|^2})=m\vec{v} \times \vec{r}$$
Where $\vec{L}$ points up by $v \times r$ as noted in the question.
$$\implies \frac{M}{L} = \frac{e}{2m}<0$$
