Table 3.

Exact solutions for the force and torque between coaxial magnets

Axial magnetization

Diametric magnetization

$${\mathbf{f}}_{\parallel }^{1\to 2}=-{\mu }_0{\overline{\mathrm{R}}}_1{\overline{\mathrm{R}}}_2{\eta }_f({\mathrm{R}}_2,{\mathrm{L}}_1,{\mathrm{L}}_2,\mathrm{d})({\mathbf{M}}_{\parallel 1}·{\mathbf{M}}_{\parallel 2}){\widehat{\mathbf{e}}}^{1\to 2}$$ (16)

$${\displaystyle {\mathbf{f}}_{\perp }^{1\to 2}=\frac{{\mu }_0}{2}{\overline{\mathrm{R}}}_1{\overline{\mathrm{R}}}_2{\eta }_f({\mathrm{R}}_2,{\mathrm{L}}_1,{\mathrm{L}}_2,\mathrm{d})({\mathbf{M}}_{\perp 1}·{\mathbf{M}}_{\perp 2}){\widehat{\mathbf{e}}}^{1\to 2}}$$ (17) $${\displaystyle {\mathbf{t}}_{\perp }^{1\to 2}=\frac{{\mu }_0}{6}{\overline{\mathrm{R}}}_1^2{\overline{\mathrm{R}}}_2{\zeta }_t({\mathrm{R}}_2,{\mathrm{L}}_1,{\mathrm{L}}_2,\mathrm{d})({\mathbf{M}}_{\perp 1}\times {\mathbf{M}}_{\perp 2})}$$ (18)