= Solution
Below $T_N$, a small perpendicular field produces canting of the two ordered sublattices. In the isotropic two-sublattice mean-field model, the induced component is $B_\perp/\lambda$: the reduced ordered moment at higher <temperature> tilts more easily and the two effects cancel. Thus $\chi_\perp$ remains constant until the transition.
A field parallel to the maintained ordering axis instead increases the magnitude of one sublattice <magnetization> and decreases the other. At zero <temperature> saturated localized <spins> have essentially no such amplitude response in this model, so $\chi_\parallel\to0$. Thermal population changes make it easier to redistribute the longitudinal moments as <temperature> rises, giving the growing longitudinal susceptibility that meets $\chi_\perp$ at $T_N$. These directions presume the domain or a weak anisotropy fixes the ordered axis; in a perfectly isotropic freely rotating sample the Néel vector can reorient.
Above $T_N$ there is no chosen magnetic axis, so the response is isotropic. Antiferromagnetic correlations oppose uniform <magnetization>, giving the mean-field <Curie–Weiss law>[Curie–Weiss] form
$$
\boxed{\chi(T)=\frac{C}{T+T_N},\qquad
C=\frac{\mu_0(g\mu_B)^2S(S+1)}{3k_Ba^3},\quad
k_BT_N=2J_1S(S+1).}
$$
At $T_N$ it equals $C/(2T_N)=\mu_0/\lambda$, consistent with the transverse branch; at large <temperature> it decreases as $1/T$. A negative Weiss <temperature> characterizes this uniform antiferromagnetic response, rather than a ferromagnetic divergence at positive $T_N$. The illustrative mean-field curves in <past-exam-of-the-condensed-matter-physics-major-option-course-of-the-university-of-oxford/2018/3/a/solution> make these limits visible.
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