What is the net force on the square coil?

Force due to straight wire on the current-carrying coil is, $F=F_{AB}+F_{BC}+F_{CD}+F_{AD}$

$F_{BC}\text{ and} F_{AD}$are equal and opposite therefore will cancel each other effect

$F_{AB}=\frac{{\mu }_0\left(2 \mathrm{A}\right)\left(1 \mathrm{A}\right)\left(15\times {10}^{-2} \mathrm{m}\right)}{2\mathrm{\pi }\left(2\times {10}^{-2} \mathrm{m}\right)}$

$F_{CD}=-\frac{{\mu }_0\left(2 \mathrm{A}\right)\left(1 \mathrm{A}\right)\left(15\times {10}^{-2} \mathrm{m}\right)}{2\mathrm{\pi }\left(0.12 \mathrm{m}\right)}$

Net force on the square coil $ABCD$is, $$\begin{gathered} F=\frac{{\mu }_0\left(2 \mathrm{A}\right)\left(1 \mathrm{A}\right)\left(15\times {10}^{-2} \mathrm{m}\right)}{2\mathrm{\pi }\left(2\times {10}^{-2} \mathrm{m}\right)}-\frac{{\mu }_0\left(2 \mathrm{A}\right)\left(1 \mathrm{A}\right)\left(15\times {10}^{-2} \mathrm{m}\right)}{2\mathrm{\pi }\left(0.12 \mathrm{m}\right)} \\ F=25\times {10}^{-7} \mathrm{N} \end{gathered}$$