= Solution
For a uniformly magnetized bilayer that relaxes to one of its easy directions after a saturating in-plane field, the in-plane remanent component is
$$
\boxed{M_{r,\parallel}=M_s\cos\theta(t_{\rm Co}).}
$$
It is therefore approximately zero when the easy axis is perpendicular at the thin-Co end, and increases towards $M_s$ as the easy axis rotates into the plane. The graph should show an increasing crossover between about $1$ and $4\,\mathrm{nm}$, but \b[the source does not specify its exact shape]. A straight line is not a consequence of the two endpoint orientations.
The trend reflects competition between interface-driven perpendicular <magnetic anisotropy> and volume or demagnetizing contributions favouring in-plane magnetization. An illustrative effective anisotropy is $K_{\rm eff}=K_v+K_s^{\rm eff}/t_{\rm Co}-\mu_0M_s^2/2$, with sign convention chosen so positive $K_{\rm eff}$ favours the normal. A continuous rotation can require higher-order anisotropy terms; the simple sign change alone need not produce a smooth canting interval.
The relation assumes a well-defined remanent direction and no complete cancellation by domains. A vibrating-sample magnetometer measures total moment: if the result is reported without dividing by magnetic volume, the moment per area additionally scales as $t_{\rm Co}M_s\cos\theta$.
\Image[/past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-2-magnetic-memory.png]
{title=Illustrative remanence crossover and independent two-layer magnetic switching}
{description=The remanence interpolation is a chosen schematic, not fitted data. The hysteresis loop assumes equal magnetic moments, distinct coercivities and independent abrupt reversal.}
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