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 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.]
C. _0 I ^2 g
The magnetic force per unit length between two parallel wires carrying currents I_1 and I_2 separated by a distance h is given by: f = _0 I_1 I_2 2 h Given I_1 = I and I_2 = 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 Therefore, the minimum magnitude of h is _0 I^2 g . Answer: _0 I ^2 g
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