As we know that, magnetic field due to a long wire at a distance $x$ from it is given by $B=\frac{\mu_{0} I}{2 \pi x},$ where, $I$ is the current flowing through the wire $\beta \propto I$
Magnetic flux associated with the square loop,
$$
\phi \propto \beta \propto I
$$
Now, if the current increase, then $\phi$ also increases. Direction of long wire will be $\otimes$.
This means, magnetic field due to induced current will be opposite to the existing magnetic field, i.e. according to Lenz's law, The induced current in the loop will be in the anti-clockwise direction. Now,

since, wires attract each other, if current flowing through them is in same direction and repel each other, if currents are in opposite direction.
$\therefore$ Part $C D$ of the loop will experience a force of repulsion, whereas part $A B$ will experience attraction. Parts $B C$ and $A D$ will not experience any force. Thus, the overall force will be a force of repulsion because $A B$ is closer to straight. The force between two current carrying conductors is inversely proportional to the distance between them
$\because$
$$
\begin{array}{l}
F \propto \frac{1}{r} \\
r_{1} < r_{2}
\end{array}
$$
So,
$$
F_{C D}>F_{A B}
$$
$$
F_{\text {net }}=F_{C D}-F_{A B}
$$
Hence, the loop will moves away from the wire.