Magnetic field created by the electron spin I would like to know if there is some expression for the magnetic field generated by a electron spin.
As far as I know, the spin provides a magnetic moment to the electron and a magnetic dipoles momentum $\vec m$ produces a field:
$H = {1\over4\pi}  \left({3\vec r(\vec m\cdot\vec r)\over r^5} - {\vec{m}\over r^5}\right)$ 
However, the spin can be measured in each at any direction so, is the previous expression valid for the electron spin?
 A: Separation of up down spin electrons yields a 50 / 50 split. If an additional gradient magnetic field at 90 degrees to all up spin electrons is applied then the up / down spin ratio returns to a 50 / 50 split ? Does the electron have spin or is the a magnetic gradient imposing spin on the electron? How could all spin up electrons end up being 50 / 50 spin up and spin down?
A: The magnetization of for example an iron magnet is almost completely due to the spin moment of the electrons. I have not checked OP's expression, but the field outside a magnet should be the integral over all those electrons. 
At close range, the dipole-dipole interaction between electrons is usually negligible in comparison with the exchange interaction (which is an electrostatic term due to the Pauli principle). But it is there. It also plays a role in the muon-electron interaction. An electron has a dipole magnetic field.
See also Lubos Motl's answer to a previous query and his blog post.
A: Rewriting my comment in answer form:
The issue is that spin is an intrinsic property of a particle, there isn't a classical analogue to this property as it is purely a quantum mechanical phenomenon. So the electron itself isn't to be viewed as a spinning sphere of charge and as such doesn't create a magnetic field because of it.
