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Step-by-Step Solution
Step 1: Understand the Scenario
A square loop ABCD carrying current $i$ is placed coplanar and near a long straight conductor XY carrying current $I$. We want to determine the net force on the loop due to the magnetic interaction between the currents in the square loop and the conductor.
Step 2: Identify Relevant Formula
The magnetic force $F$ on a segment of length $l$ of a wire carrying current $i_1$, placed at a distance $r$ from another long straight wire carrying current $i_2$, can be found using the expression:
$F = \frac{\mu_0}{4\pi}\,\frac{2\,i_1\,i_2\,l}{r}.
$
This force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.
Step 3: Calculate Forces on Individual Sides
Side AB
Let the distance of side AB from the conductor XY be $\frac{L}{2}$. The length of AB is $L$. Using the formula above, the force on AB is:
$F_1 = \frac{\mu_0}{4\pi}\,\frac{2\,I\,i\,L}{\left(\frac{L}{2}\right)} \;=\; \frac{\mu_0\,I\,i}{\pi}.
$
This force acts towards the long straight conductor (assuming currents are such that the force is attractive).
Side CD
Let the distance of side CD from the conductor XY be $\frac{3L}{2}$ (since square ABCD is of side $L$). Again, using the force formula:
$F_2 = \frac{\mu_0}{4\pi}\,\frac{2\,I\,i\,L}{3\,\left(\frac{L}{2}\right)}
\;=\; \frac{\mu_0\,I\,i}{3\pi}.
$
This force acts away from the conductor (the direction depends on relative orientation of the currents).
Step 4: Find the Net Force
The net force on the square loop is the difference between these two forces, taking into account their directions:
$F_{\text{net}} \;=\; F_1 \;-\; F_2
=\; \frac{\mu_0\,I\,i}{\pi} \;-\; \frac{\mu_0\,I\,i}{3\pi}
=\; \left(1 - \frac{1}{3}\right)\,\frac{\mu_0\,I\,i}{\pi}
=\; \frac{2}{3}\,\frac{\mu_0\,I\,i}{\pi}.
$
Simplifying,
$F_{\text{net}} \;=\; \frac{2\,\mu_0\,I\,i}{3\,\pi}.
$
Step 5: Final Answer
The net force on the square loop ABCD is
$ \displaystyle \frac{2\,\mu_0\,I\,i}{3\,\pi}. $
