Two infinitely long parallel conducting wires A and B carry currents I and 2 I, respectively, in the same direction. The wire A has uniform mass per unit length 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 ₀ is the permeability of free space.]

Step-by-step solution

The magnetic force per unit length between two parallel wires carrying currents I₁ and I₂ separated by a distance h is given by: f = ₀ I₁ I₂ 2 h Given I₁ = I and I₂ = 2I , the upward attractive force per unit length on wire A due to wire B is: f = ₀ (I)(2I) 2 h = ₀ I^2 h The downward gravitational force per unit length on wire A is its weight per unit length: W = g For wire A not to rise from the floor, the upward magnetic force must be less than or equal to the downward gravitational force: f W ₀ I^2 h g h ₀ I^2 g