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Assume $eB>0$; reversing the sign only exchanges the spin labels. Since $\mathbf{B}\cdot \mathbf{\sigma }=B{\sigma }_3$, the Pauli term shifts the spinless energies by $\pm \hslash {\omega }_c/2$. ThusAfter collecting equal energies, the spin splitting of Landau levels isThe level $\ell =0$ contains only $(n,sigma)=(0,-1)$ and has degeneracy $D$. Every level $\ell \ge 1$ contains $(\ell ,-1)$ and $(\ell -1,+1)$ and has degeneracy $2D$.

For $N$ noninteracting electrons, the Pauli exclusion principle fills these states from the bottom. If $0\le N\le D$, every electron fits in the zero-energy level andFor $N>D$, writeThen the first $k$ positive levels are full and $r$ states in level $k+1$ are occupied. The ground-state energy isEquivalently, onThe graph is continuous and piecewise linear: it is flat up to $N=D$, then has slopes $\hslash {\omega }_c,2\hslash {\omega }_c,3\hslash {\omega }_c,\dots$, with kinks at