# Why does spin, and not total angular momentum, couple to the magnetic field in a Stern-Gerlach experiment?

I understand that the dynamical principle behind a Stern-Gerlach experiment is the coupling of a magnetic dipole moment to a (non-homogeneous) magnetic field. I also understand that when an electric charge is spinning (more generally: when an electric charge has angular momentum), it also has a magnetic dipole moment, proportional to the angular momentum.

My question is: doesn't the "orbital" angular momentum $$\mathbf{L} = \mathbf{r} \times \mathbf{p}$$ (with $$\mathbf{r}$$ and $$\mathbf{p}$$ given relative to the laboratory's frame) also have its own magnetic moment? In that case, when the particle is going through the magnetic field, under what conditions may I describe the dynamics (the interaction hamiltonian) via $$\mathbf{S} \cdot \mathbf{B}$$ instead of $$\mathbf{J} \cdot \mathbf{B}$$, with $$\mathbf{J} = \mathbf{L} + \mathbf{S}$$?

Simple answer is that the Stern-Gerlach experiment does couple to the total angular momentum. The original experiment was done with silver atoms, which have an electron in their outermost orbital with $$L=0$$ so that there would be no interaction with the magnetic field other than that of the spin.
Edit: The electron configuration for silver is $$[\text{Kr}]4d^{10}5s^1$$ by the way, the $$5s$$ electron is the one I'm talking about.