Two infinitely long parallel conducting wires A and B carry currents I and 2I respectively in the same direction. Wire A has uniform mass per unit length λ and lies on an insulated floor. Wire B is kept fixed at a height h above the floor. The minimum magnitude of h so that wire A does not rise from the floor is: [g = acceleration due to gravity, µ₀ = permeability of free space]

Solution & Explanation

Principle: two parallel wires carrying currents in the same direction attract; the force per unit length between them is f = µ₀·I₁·I₂/(2π·d), where d is their separation. Here wire A (current I) lies on the floor and wire B (current 2I) is fixed a height h above it, currents in the same direction, so B pulls A upward. Upward force per unit length on A: f = µ₀·(I)·(2I)/(2π·h) = µ₀·I²/(π·h). Wire A rests on the floor with weight per unit length λ·g pressing it down. A just begins to lift off when the magnetic pull equals the weight: µ₀·I²/(π·h) = λ·g. Solve for h: h = µ₀·I²/(π·λ·g). This matches option B. Why 'minimum h': the upward force grows as h shrinks (f ∝ 1/h). At this h the pull exactly balances gravity; any smaller h gives a larger pull and lifts A. So this h is the smallest height at which A still stays on the floor.