I believe it is in Rabi's paper "On the Process of Space Quantization" published in 1935 if you can't find the 1931 paper you might enjoy that.
I am going to try and paraphrase the derivation he did in that although I am not sure how much it relates to your question though so apologies if you already know this already. Do check the paper Rabi has lots of pretty pictures and also goes into a lot more depth than I am able to do here.
When an atom has no external field acting and spin $l$, its angular momentum can be expressed as $$F_1=\left(l+\frac{1}{2}\right)$$ $$F_2=\left(l-\frac{1}{2}\right)$$ with units $\hbar$. According to Rabi, the positive moments in such a nucleus will have higher energy $F_1=(l+\frac{1}{2})$ and the negative moments will have lower energy $F_2=(l-\frac{1}{2})$.
Note that the energy difference between states $\Delta E=h\Delta f$ is due to the interactions between protons and electrons in the atom
Now to your question of what happens when an external field is applied. Rabi states without any proof that magnetic field splits into $2(l+\frac{1}{2})$ levels when the field is applied.
If we imagine there is some state that is neither affected by the external magnetic field nor the nuclear fields then we have the The energy shif of the levels with respect to this state is given by $$E_M=-\frac{\Delta E }{2(l+\frac{1}{2})}$$. For for a positive moment we have $$E_M=\frac{\Delta E }{2(l+\frac{1}{2})}$$.
For each state the effective moment is given by $$\frac{\partial{E_M}}{\partial{H}}$$. It goes on quite a bit longer than this, but see how that makes sense so far.