 Solution

Take two localized spin orientations with moments $\pm {\mu }_B$ along the applied magnetic field. This assumes an electron-like spin $g$ factor of two. Their energies are $\mp {\mu }_{B}B$, so the one-moment partition function is $Z_1=2\cosh x$, with $x={\mu }_{B}B/(k_{B}T)$. The thermal mean moment is $k_{B}T{\partial }_B\ln Z_1={\mu }_B\tanh x$, and the spin-one-half paramagnetic magnetization isFor $|x|\ll 1$, $M\simeq n{\mu }_B^{2}B/(k_{B}T)$. With the same SI convention $\chi =\partial M/\partial H$ and $B\simeq {\mu }_{0}H$,This Curie law is a zero-field or weak-field susceptibility. At arbitrary field the differential response is $(C/T){\mathrm{sech}}^{2}x$, and $M$ saturates at $n{\mu }_B$; extending the linear law to unlimited field would contradict that finite saturation magnetization.