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NEET PHYSICSHard

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

A

μ0Ii2π\frac{\mu_0 I i}{2\pi}

B

2μ0IiL3π\frac{2\mu_0 I i L}{3\pi}

C

μ0IiL2π\frac{\mu_0 I i L}{2\pi}

D

2μ0Ii3π\frac{2\mu_0 I i}{3\pi}

Step-by-Step Solution

  1. Identify the Magnetic Field: The long straight conductor XY produces a non-uniform magnetic field BB at a distance rr given by B=μ0I2πrB = \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 LL. In the standard NEET 2016 configuration, the distance from the wire XY to the nearest side (AB) is r1=L/2r_1 = L/2. The distance to the farther side (CD) is r2=L+L/2=3L/2r_2 = L + L/2 = 3L/2.
  4. Calculate Component Forces: The force FF on a wire of length LL in a magnetic field is F=iLBF = iLB.
  • Force on AB: FAB=iL(μ0I2π(L/2))=μ0IiπF_{AB} = i L \left( \frac{\mu_0 I}{2\pi (L/2)} \right) = \frac{\mu_0 I i}{\pi}.
  • Force on CD: FCD=iL(μ0I2π(3L/2))=μ0Ii3πF_{CD} = i L \left( \frac{\mu_0 I}{2\pi (3L/2)} \right) = \frac{\mu_0 I i}{3\pi}.
  1. 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: Fnet=FAB−FCD=μ0Iiπ−μ0Ii3π=2μ0Ii3πF_{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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