Table 5.

Implementation of the exact solutions for magnetic force and torque

Group	Auxiliary expressions
FT1	$${\eta }_f({\mathrm{R}}_2,{\mathrm{L}}_1,{\mathrm{L}}_2,\mathrm{d}):=\eta ({\mathrm{R}}_2,\mathrm{d}+{\mathrm{L}}_1+{\mathrm{L}}_2)+\eta ({\mathrm{R}}_2,\mathrm{d}-{\mathrm{L}}_1-{\mathrm{L}}_2)-\eta ({\mathrm{R}}_2,\mathrm{d}-{\mathrm{L}}_1+{\mathrm{L}}_2)-\eta ({\mathrm{R}}_2,\mathrm{d}+{\mathrm{L}}_1-{\mathrm{L}}_2)$$ $${\zeta }_t({\mathrm{R}}_2,{\mathrm{L}}_1,{\mathrm{L}}_2,\mathrm{d}):=\zeta ({\mathrm{R}}_2,\mathrm{d}+{\mathrm{L}}_1+{\mathrm{L}}_2)+\zeta ({\mathrm{R}}_2,\mathrm{d}-{\mathrm{L}}_1-{\mathrm{L}}_2)-\zeta ({\mathrm{R}}_2,\mathrm{d}-{\mathrm{L}}_1+{\mathrm{L}}_2)-\zeta ({\mathrm{R}}_2,\mathrm{d}+{\mathrm{L}}_1-{\mathrm{L}}_2)$$
FT2	$$\left\{\begin{array}{ccc}{\displaystyle \eta ({\mathrm{R}}_2,x):=x\left(f_6-f_7\right)/\ell ,}\hfill & \zeta ({\mathrm{R}}_2,x):=\left((2(1+{\mathrm{R}}_2^2)-x^2)f_6+f_8\right)/\ell \hfill & \mathrm{for}({\mathrm{R}}_2,x)\ne (1,0)\hfill \\ \eta ({\mathrm{R}}_2,x):=0,\hfill & \zeta ({\mathrm{R}}_2,x):=4\hfill & \mathrm{for}({\mathrm{R}}_2,x)=(1,0)\hfill \end{array}\right.$$ $$f_6({\mathrm{R}}_2,x):=k\mathcal{C}\left(k_c,1,0,-1\right),f_7({\mathrm{R}}_2,x):=\mathcal{C}\left(k_c,1-\xi k,\xi ,\xi -k\right)$$ $$f_8({\mathrm{R}}_2,x):=\mathcal{C}\left(k_c,1-\xi k,4{\mathrm{R}}_2+3\xi x^2,4{\mathrm{R}}_2+3\xi x^2-k(4\xi {\mathrm{R}}_2+3x^2)\right),\xi :=\min (1,{\mathrm{R}}_2)/\max (1,{\mathrm{R}}_2)$$ $${\displaystyle k({\mathrm{R}}_2,x):={\mathrm{R}}_2/{\ell }^2,k_c({\mathrm{R}}_2,x):=+\sqrt{1-k^2},\ell ({\mathrm{R}}_2,x):=\left(\sqrt{{(1+{\mathrm{R}}_2)}^2+x^2}+\sqrt{{(1-{\mathrm{R}}_2)}^2+x^2}\right)/2}$$