= Solution
For a linear induced response, reversing $I$ reverses $H_0$ and the sample <magnetization>, while also reversing the field gradient. Their product in $F_z=\mu_0V\chi_{\rm eff}H_0H'_0$ therefore keeps the same sign and magnitude:
$$
\boxed{\mathbf F(-I)=\mathbf F(I).}
$$
This is also explicit in the $I^2$ dependence in part (d), for both the shape-consistent and printed alternatives. The conclusion assumes no independent permanent moment, remanence or <hysteresis>; those would require a different response model.
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