 Solution

For an electron of charge $-e$, magnetic minimal coupling gives $H=(\mathbf{p}+e\mathbf{A}{)}^2/(2m)$. In Landau gauge, $\mathbf{A}=(0,Bx,0)$, and fixing the conserved $p_y$ turns the Hamiltonian into a harmonic oscillator centered at $x=-p_y/(eB)$. The Landau levels are thereforeIn a rectangle of dimensions $L_x,L_y$, consecutive momenta differ by $2\pi \hslash /L_y$, so consecutive oscillator centers differ by $2\pi \hslash /(eB{L}_y)$. Centers lying in the sample supply $eBA/(2\pi \hslash )$ orbitals per level. Neglecting the spin-field interaction leaves two degenerate spin states, rather than removing spin from Pauli counting. Thus each level holds $D=eBA/(\pi \hslash )$ electrons, to leading order for a large sample.

In the specified field range, $N\le D\le 2N$. At zero temperature fill the lowest level with $D$ electrons and the next with $2N-D$. This ground-state filling of two spin-degenerate Landau levels givesAt the lower endpoint the first two levels are exactly full; at the upper endpoint only the lowest level is full. Both giveBetween them the total energy is a concave quadratic and is larger than this common endpoint value. Filling reorganizes as the magnetic degeneracy increases; ignoring spin splitting still requires retaining the twofold spin degeneracy in this calculation.