Q1.Two infinitely long parallel conducting wires A and B carry currents I and 2I, respectively, in the same direction. The wire A has uniform mass per unit length λ and lies on an insulated floor. The wire B is kept fixed at a height h above the floor. The minimum magnitude of h so that the wire A does not rise from the floor is: [g is the acceleration due to gravity and μ₀ is the permeability of free space.]
- Aμ₀I²/(2πλg)
- Bμ₀I²/(πλg)Answer
- C2μ₀I²/(πλg)
- D4μ₀I²/(πλg)
Parallel currents in the same direction attract, so wire B pulls A upward with a force per unit length μ₀(I)(2I)/(2πh) = μ₀I²/(πh). Wire A stays on the floor only while that does not exceed its weight per unit length λg, and the limiting case μ₀I²/(πh) = λg gives the minimum height h = μ₀I²/(πλg).
Source: NEET 2026 re-exam (21 June), Code 50, Q15