= Solution
Choose the positive spontaneous branch for $H_z=0$. Its <magnetization> starts at $M_s(0)$, decreases with <temperature>, and reaches zero continuously at $T_c$. It remains zero for $T_c\leq T<2T_c$. There is an equally valid negative branch below $T_c$; domain averaging can instead give zero net moment.
A fixed small positive <magnetic field intensity> selects the positive branch. At zero <temperature> it still saturates at $M_s(0)$; at intermediate <temperature> it increases the <magnetization> relative to the zero-field branch. It rounds the zero-field transition and leaves a positive <magnetization> above $T_c$, tending in the linear-response region to $C H_z/(T-\Theta)$. The <magnetization> therefore does not vanish at $T_c$ in the nonzero-field curve.
For a concrete labelled sketch, the figure uses the illustrative <mean-field theory> equation
$$
m=\tanh\!\left(\frac{m+h}{t}\right),\qquad
m=\frac{M_z}{M_s(0)},\quad t=\frac{T}{T_c},\quad h=\frac{\mu\mu_0H_z}{k_BT_c}.
$$
It shows $h=0$ and a fixed small $h=0.01$. The curve is an illustrative equilibrium equation of state, not a determination of the precise critical exponents of GdN. At $h>0$ no symmetry-breaking transition remains along this temperature sweep.
\Image[/past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2018-gadolinium-magnetization.png]
{title=Spontaneous and field-selected <magnetization> in the two-direction model}
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