Question

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)

• A. \(\frac{μ_0I_1I_2}{3π}\)
• B. \(\frac{3μ_0I_1I_2}{8π}\)
• C. \(\frac{9μ_0I_1I_2}{8π}\)
• D. \(\frac{3μ_0I_1I_2}{4π}\)

Hint:

Parallel currents attract and opposite currents repel; the near side wins.

Answer 1

The Correct Option is C

Step 1: Use field at each side:
Field of the wire at distance $r$ is $B = \frac{\mu_0I_2}{2\pi r}$. The force on a side of length $L$ is $F = I_1LB$.

Step 2: Near side:
$r = \frac L3$: $B_1 = \frac{3\mu_0I_2}{2\pi L}$, $F_1 = I_1L B_1 = \frac{3\mu_0I_1I_2}{2\pi}$ (attraction).

Step 3: Far side:
$r = \frac{4L}{3}$: $B_2 = \frac{3\mu_0I_2}{8\pi L}$, $F_2 = \frac{3\mu_0I_1I_2}{8\pi}$ (repulsion).

Step 4: Subtract:
$F_1 - F_2 = \frac{12 - 3}{8\pi}\mu_0I_1I_2 = \frac{9\mu_0I_1I_2}{8\pi}$.

Final Answer:
The net force is 9 mu0 I1 I2 / (8 pi), option (C). \[ \boxed{\frac{9\mu_0I_1I_2}{8\pi}} \]