$$
\begin{aligned}
I_{1} &=20 \mathrm{~A}, I_{2}=20 \mathrm{~A} \\
r_{1} &=2 \mathrm{~cm} \\
r_{2} &=(10+2) \mathrm{cm}=12 \mathrm{~cm}
\end{aligned}
$$



The given situation is shown in the figure.
The forces on arm $S R$ and $P Q$ are equal in magnitude and opposite in direction and also having same line of action, hence they cancel to each other.
$\therefore$ Resultant force on the current loop PQRS,
$$
\begin{aligned}
F &=F_{P S}-F_{Q R} \\
&=\frac{\mu_{0}}{2 \pi} \cdot \frac{I_{1} I_{2} \times l}{r_{1}}-\frac{\mu_{0}}{2 \pi} \cdot \frac{I_{1} I_{2} l}{r_{2}} \\
&=\frac{\mu_{0}}{2 \pi} \cdot I_{1} I_{2} l\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right] \\
=2 \times 10^{-7} \times 20 \times 20 \times 15 \times 10^{-2}\left[\frac{1}{2 \times 10^{-2}}-\frac{1}{12 \times 10^{-2}}\right] \\
=5 \times 10^{-4} \mathrm{~N}
\end{aligned}
$$