A square loop ABCD carrying a current $i$ is placed near and coplanar with a long straight conductor XY carryi

Question

A square loop ABCD carrying a current $i$ is placed near and coplanar with a long straight conductor XY carrying a current $I$. The net force on the loop will be:

Accepted Answer

1. **Identify the Magnetic Field:** The long straight conductor XY produces a non-uniform magnetic field $B$ at a distance $r$ given by $B = \frac{\mu_0 I}{2\pi r}$ . 
2. **Force on Perpendicular Sides:** For the sides of the square loop perpendicular to the wire XY, the magnetic forces are equal in magnitude but opposite in direction, meaning they cancel each other out . 
3. **Force on Parallel Sides:** Let the side of the square be $L$. In the standard NEET 2016 configuration, the distance from the wire XY to the nearest side (AB) is $r_1 = L/2$. The distance to the farther side (CD) is $r_2 = L + L/2 = 3L/2$. 
4. **Calculate Component Forces:** The force $F$ on a wire of length $L$ in a magnetic field is $F = iLB$. 
   - Force on AB: $F_{AB} = i L \left( \frac{\mu_0 I}{2\pi (L/2)} \right) = \frac{\mu_0 I i}{\pi}$. 
   - Force on CD: $F_{CD} = i L \left( \frac{\mu_0 I}{2\pi (3L/2)} \right) = \frac{\mu_0 I i}{3\pi}$. 
5. **Net Force Calculation:** Since the currents in side AB and side CD flow in opposite directions relative to the field orientation, the forces are in opposite directions (one attractive, one repulsive). The net force is: 
   $$F_{net} = F_{AB} - F_{CD} = \frac{\mu_0 I i}{\pi} - \frac{\mu_0 I i}{3\pi} = \frac{2\mu_0 I i}{3\pi}$$.

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