The force exerted by the conductor AB on the loop CDFG shown in figure is

Force exerted by AB on sides CD and FG is zero Force by AB on GC
$\begin{array}{l}{\mathrm{F}}_1=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_1{\mathrm{i}}_2\mathrm{l}}{2\mathrm{\pi r}}=\frac{4\mathrm{\pi }\times {10}^{-7}\times 20\times 10\times 20}{2\mathrm{\pi }\times 1}\\ =8\times {10}^{-4}\mathrm{N} (\mathrm{attraction})\end{array}$
Force by AB on DF
${\mathrm{F}}_2=\frac{4\mathrm{\pi }\times {10}^{-7}20\times 10\times 20}{2\mathrm{\pi }\times 10}=0.8\times {10}^{-4}\mathrm{N}$
Net force $={\mathrm{F}}_1-{\mathrm{F}}_2=8\times {10}^{-4}-0.8\times {10}^{-4}$
$=7.2\times {10}^{-4}\mathrm{N}$ (attraction)