A square loop ABCD of side L carrying a current \(I_1\) is placed at a distance \((L/3)\) from a conductor coplaner with a straight conductor XY carrying current \(I_2\) as shown in figure. The net force on the loop will be (\(μ_0\) = magnetic permeability)
Show Hint
Parallel currents attract and opposite currents repel; the near side wins.
Step 1: Understanding the Figure:
The wire XY carries \(I_2\) upward. In the loop, side AB (nearest the wire) carries \(I_1\) upward, so its current is parallel to \(I_2\) and is attracted toward the wire. Side CD (far side) carries \(I_1\) downward, antiparallel to \(I_2\), so it is repelled away from the wire. The forces on the sides BC and DA are equal and opposite and cancel.
Step 2: Force on AB (distance \(\frac L3\)):
\[ F_{AB} = \frac{\mu_0I_1I_2}{2\pi\,(L/3)}\cdot L = \frac{3\mu_0I_1I_2}{2\pi} \]
directed toward the wire.
Step 3: Force on CD (distance \(\frac L3 + L = \frac{4L}{3}\)):
\[ F_{CD} = \frac{\mu_0I_1I_2}{2\pi\,(4L/3)}\cdot L = \frac{3\mu_0I_1I_2}{8\pi} \]
directed away from the wire.
Step 4: Net force:
The two act in opposite directions, so
\[ F = \frac{3\mu_0I_1I_2}{2\pi} - \frac{3\mu_0I_1I_2}{8\pi} = \frac{12 - 3}{8\pi}\mu_0I_1I_2 = \frac{9\mu_0I_1I_2}{8\pi} \]
This is directed toward the wire.
Step 5: Why the other options are wrong.
\(\frac{3\mu_0I_1I_2}{4\pi}\) and \(\frac{3\mu_0I_1I_2}{8\pi}\) are the sum or only the far-side force. \(\frac{\mu_0I_1I_2}{3\pi}\) does not match either side.
Final Answer:
The net force is \(\frac{9\mu_0I_1I_2}{8\pi}\), option (C).
\[ \boxed{\frac{9\mu_0I_1I_2}{8\pi}} \]
