How exactly does an electrons spin cause it to have a magnetic moment? So I understand you can rearrange the equation for a single electron orbiting in a circle $(\mu=evr/2)$ and substitute in the associated velocity for angular momentum $(v=L/mr)$ where angular momentum is the electrons spin angular momentum. But what does make sense to me is how to electrons spin is actually causing it to behave like a current, since the charge isn't moving how I would traditionally imagine charges in a current moving. Also if the spin angular momentum is causing the magnetic dipole does that mean the magnetic dipole is also present in free electrons and in fact has nothing to do with the classical model of the electron orbiting a nucleus?
 A: Yes, the magnetic moment due to spin has indeed nothing to do with the classical idea of a charge orbiting - or indeed moving - in any classical fashion. In fact, the accurate prediction of the anomalous magnetic moment of the electron by quantum electrodynamics is one of the hallmark experimental confirmations of QED. And even the non-anomalous g-factor of 2 is already famously different from anything you could get from simple classical charge geometries.
Quantum mechanics, even less so than classical mechanics, does not really deal in human-readable "causes". When we write down the traditional Lagrangian for a classical electromagnetic field coupled to a current as involving the term $A_\mu j^\mu$, and then postulate that the current is given by a Dirac spinor quantum field as $j^\mu \propto \bar{\psi}\gamma^5\psi$, then the theory produces a value for the expectation value of the interaction energy in the case where the particle remains undisturbed (i.e. with no change in momentum) that is proportional to $\vec \sigma \cdot \vec B$, where $\vec \sigma$ is the spin operator and $\vec B$ an external magnetic field. That's it, and since this computation is so essential to QED, it or at least an outline of it can be found in many textbooks on the subject.
