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A square loop, carrying a steady current $I$, is placed in a horizontal plane near a long straight conductor carrying a steady current $I_1$ at a distance $d$ from the conductor as shown in figure. The loop will experience

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Verified Answer
The correct answer is:
a net attractive force towards the conductor

$$
\begin{aligned}
& \mathbf{F}_2=-\mathbf{F}_4 \\
& \mathbf{F}_1=\frac{\mu_0 I_1 I l}{2 \pi d} \\
& \mathbf{F}_2=\frac{\mu_0 I_1 I l}{2 \pi(d+l)} \\
& \mathbf{F}_1>\mathbf{F}_3 \\
& F_{\text {net }}=F_1-F_3
\end{aligned}
$$
So, wire atrract loop.
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