 Solution

In the small-body approximation, write ${\chi }_{\mathrm{eff}}=\chi /(1+N_d\chi )$, with $\chi =\mu -1$. The induced magnetic dipole moment is $\mathbf{m}=VM\hat{\mathbf{z}}=V{\chi }_{\mathrm{eff}}{H}_0\hat{\mathbf{z}}$. Its effective interaction potential energy at fixed source current isThe factor $1/2$ accounts for inducing the moment: differentiating $-m(z)B_0(z)$ as if the moment were permanent would double the force. Equivalently, the instantaneous dipole force is $F_z=md{B}_0/dz$. Thus the force on a small linearly magnetizable body isFrom part (b), $d{H}_0/dz=-3a^{2}Iz/[2(a^2+z^2{)}^{5/2}]$. HenceFor the ideal slender rod, ${\chi }_{\mathrm{eff}}\simeq \mu -1$. If the source's printed magnetisation is imposed instead, its corresponding conditional answer isFor $\mu >1$ and $z>0$, both expressions point toward the loop, where the applied magnetic field is stronger; for $z<0$ the direction reverses toward the loop plane. A material with diamagnetism, ${\chi }_{\mathrm{eff}}<0$, reverses this attraction. The force is in newtons, since a magnetic dipole moment times a field gradient has units $\mathrm{A}{\mathrm{m}}^2\mathrm{T}{\mathrm{m}}^{-1}$. Both formulas assume linear response, negligible saturation and approximately uniform applied field across the sample.