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…

Question

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.)

Accepted Answer

(3) (μ₀I²)/(πλ g). Given: wire A on the floor carrying I, wire B fixed at height h carrying 2I in the same direction; A has mass per unit length λ. Idea: parallel currents in the same direction attract, so B pulls A upwards; A stays down while its weight is at least as large as that pull. Attractive force per unit length: F/L=(μ₀(I)(2I))/(2π h)=(μ₀I²)/(π h). Weight per unit length: λ g. A does not rise while (μ₀I²)/(π h)≤λ g. Rearranging: h≥(μ₀I²)/(πλ g). So the minimum height is (μ₀I²)/(πλ g) — bring B any closer and A lifts off.

Step-by-step solution

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