 Solution

A magnetic field favors alignment of the magnetic moments of a paramagnet, reducing the number of substantially occupied magnetic configurations at fixed temperature. Its magnetic entropy therefore decreases as the field magnitude increases, explaining the negative sign of $(\partial S/\partial B{)}_T=-\alpha B/T^2$.

Choose the magnetic Helmholtz free energy convention $dF=-SdT-mdB$. Equality of its mixed partial derivatives gives the magnetic Maxwell relationThe magnetic moment inferred from an entropy derivative, integrated at fixed $B$, isThe given entropy derivative alone cannot determine the temperature-independent function $m_0(B)$. With the usual independent-paramagnet condition $m\to 0$ as $T\to \infty$ at fixed $B$, $m_0(B)=0$, giving Curie's law, $m=\alpha B/T$. The sketch uses that additional condition. The inverse-$T$ expression applies in the weak-field regime; its divergence cannot describe the actual low-temperature saturation of a finite set of magnetic moments.

For reversible quasistatic thermal isolation, $dS=0$. With positive field-dependent heat capacity $C_B=T(\partial S/\partial T{)}_B$,Reducing a positive $B$ therefore lowers $T$: this is adiabatic demagnetization. Equivalently, reducing $B^2$ cools for either field sign. Reduced alignment would raise magnetic entropy at fixed $T$; cooling compensates so total entropy remains constant. Insulation without reversibility would allow additional entropy production and need not follow this adiabat exactly.