 Solution

For weak fields in a linear isotropic material, magnetization is $\mathbf{M}={\chi }_m\mathbf{H}$, with magnetic susceptibility ${\chi }_m$. Paramagnetism has ${\chi }_m>0$, while diamagnetism has ${\chi }_m<0$. In either case, $\mathbf{B}={\mu }_0(\mathbf{H}+\mathbf{M})={\mu }_0(1+{\chi }_m)\mathbf{H}$. These weak responses generally disappear when the applied field is removed, unlike hysteretic ferromagnetism.

A paramagnet can contain permanent magnetic moments from unpaired Electrons or atomic angular momentum. Alignment lowers the energy $-\mathbf{\mu }\cdot \mathbf{B}$, while thermal disorder opposes it. For independent classical moments of magnitude $\mu$, their weak-field average projection is ${\mu }^{2}B/(3k_{B}T)$, soThis is Curie's law; it assumes weak magnetization and no important interactions between moments. With quantum angular momentum $J$, replace ${\mu }^2$ by $(g{\mu }_B{)}^{2}J(J+1)$. Strong-field magnetization saturates instead of growing linearly without limit. Conduction Electrons also exhibit Pauli paramagnetism: only occupations near the Fermi surface readily redistribute between spin states, giving a nearly temperature-independent susceptibility when $T\ll T_F$.

Diamagnetism is an induced orbital response opposing the applied magnetic field. For bound Electrons, a weak field changes orbital motion and induces an opposing magnetic moment. For an isotropic collection of atoms, the elementary orbital estimate iswhere the sum runs over the Electrons of one atom. It is usually small and weakly temperature dependent. Delocalized Electrons also have orbital Landau diamagnetism; for an ideal three-dimensional free Electron gas with a quadratic isotropic dispersion and electron spin with $g=2$, ${\chi }_L=-{\chi }_P/3$. Both spin and orbital contributions must be included to determine the total response. Diamagnetism is not a consequence of permanent dipoles choosing to align against the field; that orientation would increase their energy.