The resultant force on the current loop due to a long current carrying conductor will be

Force on and are equal but opposite so their net will be zero.
Force between two parallel conductors carrying currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\)
\(\mathrm{F}=\frac{\mu_{0}}{2 \pi} \frac{\mathrm{I}_{1} \mathrm{I}_{2} l}{\mathrm{r}}\)
where \(r=\) distance between two parallel
\(\mathrm{F}_{\mathrm{PS}} =\frac{10^{-7} \times 2 \times 20 \times 20 \times 15 \times 10^{-2}}{2 \times 10^{-2}} \)
\(=6 \times 10^{-4} \mathrm{~N} \)
\(\mathrm{~F}_{\mathrm{QR}} =\frac{10^{-7} \times 2 \times 20 \times 20 \times 15 \times 10^{-2}}{12 \times 10^{-2}} \)
\( =1 \times 10^{-4} \mathrm{~N} \)
\(\mathrm{~F}_{\text {net }} =\mathrm{F}_{\mathrm{PS}}-\mathrm{F}_{\mathrm{QR}} \)
\(=6 \times 10^{-4}-1 \times 10^{-4}=5 \times 10^{-4} \mathrm{~N}\)