If a square loop ABCD carrying a current i is placed near and coplanar with a long straight conductor XY carry

Question

If a square loop ABCD carrying a current i is placed near and coplanar with a long straight conductor XY carrying a current I, what will be the net force on the loop?

Accepted Answer

1. **Force on Parallel Sides:** The net force on the loop is the resultant of the forces acting on the sides parallel to the long straight conductor. The forces on the sides perpendicular to the conductor cancel each other out due to symmetry.
2. **Formula:** The force between two parallel current-carrying conductors of length $L$ separated by distance $d$ is given by $F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$ .
3. **Application:** 
   - Let the side length of the square loop be $L$. Based on the standard NEET problem configuration (NEET 2016), the distance of the near side (AB) from the conductor is usually taken as $L/2$.
   - **Force on Near Side (Attractive):** $F_1 = \frac{\mu_0 I i L}{2\pi (L/2)} = \frac{\mu_0 I i}{\pi}$. This force acts towards the conductor (attractive) because currents are likely parallel (or determines direction).
   - **Force on Far Side (Repulsive):** The distance to the far side (CD) is $L/2 + L = 3L/2$. The force is $F_2 = \frac{\mu_0 I i L}{2\pi (3L/2)} = \frac{\mu_0 I i}{3\pi}$. This force acts away from the conductor.
4. **Net Force:** Since $F_1 > F_2$ and they act in opposite directions:
   $$F_{net} = F_1 - F_2 = \frac{\mu_0 I i}{\pi} - \frac{\mu_0 I i}{3\pi}$$
   $$F_{net} = \frac{\mu_0 I i}{\pi} \left( 1 - \frac{1}{3} \right) = \frac{\mu_0 I i}{\pi} \left( \frac{2}{3} \right)$$
   $$F_{net} = \frac{2\mu_0 I i}{3\pi}$$
   Note: The length $L$ cancels out in the final expression, making the force dependent only on the currents and constants.

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