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For constant scalar response $\mathbf{m}={\alpha }_B{\mathbf{B}}_0$ and a body small compared with the applied field's spatial variation scale, the gradient of the induced magnetic dipole energy gives $\mathbf{F}={\alpha }_B\nabla B_0^2/2$. Along an axis this is $F_z=md{B}_0/dz$. For an ellipsoidal body, put ${\chi }_{\mathrm{eff}}=\chi /(1+N_d\chi )$ to obtain $F_z={\mu }_{0}V{\chi }_{\mathrm{eff}}{H}_0{H}_0^{\prime }$. Positive effective magnetic susceptibility attracts the body towards stronger field; a negative response repels it. Reversing the source current leaves the force unchanged because both the induced moment and the field gradient reverse. Shape and orientation enter through the demagnetizing factor, not merely through volume.

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.