= Solution
\b[Component chemical potentials.] Define the per-spin <quantum concentration> $n_Q=\lambda_T^{-3}=(mk_BT/(2\pi\hbar^2))^{3/2}$. For a classical component with one-particle <potential energy> $U_\sigma(\mathbf r)$, the local <Helmholtz free energy> density is $k_BTn_\sigma[\ln(n_\sigma/n_Q)-1]+n_\sigma U_\sigma$. Its derivative with respect to <number density> gives the full local <chemical potential>, including the magnetic <energy>. With the up state defined as the lower level,
$$
\boxed{\mu_\uparrow=k_BT\ln\frac{n_\uparrow}{n_Q}-\mu_BB,
\qquad
\mu_\downarrow=k_BT\ln\frac{n_\downarrow}{n_Q}+\mu_BB.}
$$
The <Bohr magneton> $\mu_B$ is a <magnetic moment>; $\mu_BB$ is an <energy>. The factor two for the two spin states is not already included in $n_Q$.
\b[Total equilibrium chemical potential.] If spin conversion is allowed, a particle can be added in either spin state at the same equilibrium free-<energy> cost. Thus
$$
\boxed{\mu=\mu_\uparrow=\mu_\downarrow,}
$$
not their sum or zero. The total <particle number>, or a reservoir, fixes this common value; the supplied information does not specify a numerical <chemical potential>. A useful local expression is $\mu=k_BT\ln[n/(2n_Q\cosh x)]$, with $x=\mu_BB/(k_BT)$, whose density dependence is found below.
\b[Why it is uniform and common.] Moving a small number $dN$ of one species from $\mathbf r_1$ to $\mathbf r_2$ changes the <free energy> by $[\mu_\sigma(\mathbf r_2)-\mu_\sigma(\mathbf r_1)]dN$. If this difference were nonzero, a transfer could lower the <free energy>. Equilibrium under diffusion therefore requires each full <chemical potential> to be spatially constant. A spin flip transferring $dN$ from down to up similarly changes <free energy> by $(\mu_\uparrow-\mu_\downarrow)dN$, proving equality when spin relaxation is permitted. This is <chemical equilibrium of a classical spin gas>. Mere thermal contact at fixed separate spin populations would not enforce the latter equality; full spin equilibrium is the necessary assumption. A spatially varying <magnetic field> is balanced by a spatially varying density, not by a varying full <chemical potential>.
\b[Magnetization.] Invert the component equations at the common $\mu$:
$$
n_\uparrow(\mathbf r)=n_Qe^{\beta\mu}e^{x(\mathbf r)},\qquad
n_\downarrow(\mathbf r)=n_Qe^{\beta\mu}e^{-x(\mathbf r)}.
$$
The two levels have moments $+\mu_B$ and $-\mu_B$ along the field. The <magnetization>, <magnetic moment> per unit <volume>, is therefore
$$
\boxed{M(\mathbf r)=\mu_B(n_\uparrow-n_\downarrow)
=2\mu_Bn_Qe^{\beta\mu}\sinh x(\mathbf r).}
$$
Its SI units are $\mathrm{A\,m^{-1}}$. This expression neglects interactions and magnetic feedback of the gas on the specified external field.
\b[High-temperature response.] For $|x|\ll1$, $\sinh x\simeq x$ and $\cosh x\simeq1$, yielding
$$
\boxed{M(\mathbf r)\simeq
\frac{2\mu_B^2 n_Qe^{\beta\mu}}{k_BT}B(\mathbf r)
\simeq\frac{n(\mathbf r)\mu_B^2}{k_BT}B(\mathbf r).}
$$
At fixed <number density>, this is the <Curie law>: the susceptibility with respect to $B$ is $n\mu_B^2/(k_BT)$. The conventional dimensionless susceptibility with respect to $H$ is approximately $\chi_v=\mu_0n\mu_B^2/(k_BT)$ when $B\simeq\mu_0H$. At fixed $\mu$, however, $n_Qe^{\beta\mu}$ changes with <temperature>, so $M$ itself need not scale simply as $1/T$. The moment per particle has the Curie behaviour; the density constraint must be stated.
\b[Total density.] Adding the two populations gives
$$
\boxed{n(\mathbf r)=2n_Qe^{\beta\mu}\cosh x(\mathbf r),\qquad
\frac{M(\mathbf r)}{n(\mathbf r)}=\mu_B\tanh x(\mathbf r).}
$$
For a finite container and fixed $N$, normalization gives $e^{\beta\mu}=N/[2n_Q\int\cosh x(\mathbf r)\,d^3r]$. In a weak field the leading density is uniform, with changes only of order $x^2$. Particles preferentially occupy regions where the magnitude of the field lowers one of the spin energies, even though the common <chemical potential> remains uniform.
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