= Solution
Differentiation of the restricted free energy gives the <magnetization>
$$
\begin{aligned}
M_{\mathrm{restricted}}
&=-\frac1V\left(\frac{\partial A_{\mathrm{mag}}}{\partial B}\right)_{T,V}\\
&=\boxed{\frac{N\mu}{V}\tanh\frac{\mu B}{k_{\mathrm B}T}}.
\end{aligned}
$$
In the weak-field limit $|\mu B|\ll k_{\mathrm B}T$, $\tanh x\simeq x$, so
$$
\boxed{M_{\mathrm{restricted}}\simeq\frac{N\mu^2B}{Vk_{\mathrm B}T}
=\frac{4N\mu_{\mathrm B}^2B}{Vk_{\mathrm B}T}.}
$$
\b[The printed coefficient is inconsistent with the stated moment $\mu=2\mu_{\mathrm B}$.] The requested expression without the factor four would follow from $\mu=\mu_{\mathrm B}$ in this restricted model. This discrepancy is present in the original PDF, rather than introduced by its transcription.
For completeness, differentiating the full thermal free energy gives
$$
\boxed{M_{\mathrm{full}}=\frac{N\mu}{V}
\frac{e^{-\beta\Delta}\sinh x}{1+e^{-\beta\Delta}\cosh x}.}
$$
Its weak-field form is
$$
\boxed{M_{\mathrm{full}}\simeq\frac{N\mu^2B}{Vk_{\mathrm B}T}
\frac1{1+e^{\theta/T}}
=\frac{4N\mu_{\mathrm B}^2B}{Vk_{\mathrm B}T(1+e^{\theta/T})}.}
$$
At $k_{\mathrm B}T\gg\Delta$ this approaches $2N\mu_{\mathrm B}^2B/(Vk_{\mathrm B}T)$, so restoring the two nonmagnetic ground states does not recover the printed coefficient either. The inverse-temperature <Curie law> form is valid, but the population factor and moment squared must be retained consistently.
Back to article page
