A long straight wire carrying current of `25A` rests on a table as shown in figure. Another wire PQ of length 1m, mass `2*5g` carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

Explanation

Force per unit length on a current carrying wire PQ due to long straight current carrying wire, carrying same current at a distance from it is given by:
`F/l = mu_(0)/(4 pi) (2I.I)/h`
` rArr " " F = mu_(0)/(4 pi) (2I^(2)l)/h`
Here, I = current carried by both wires
Since, the wires are carrying currents in opposite directions the force on wire PQ will be repulsive in nature. Hence, magnetic force will move the wire up but gravity is acting downward. Let us assume that wire is allowed to be moved slowly without any acceleration so at a height when magnetic force gets balanced by weight of wire then it will remain in equilibrium at that point.
`mg = F`
` rArr " " mg = mu_(0)/(4 pi) (2I^(2)l)/h rArr h = mu_(0)/(4 pi) (2I^(2)l)/(mg) `
Here, `m = 2.5 g = 2.5 xx 10^(-3) kg`
` I = 25 A, h = ? `
`g = 9.8 m//s^(2) , l = 1 m`
`rArr" " h = (4pi xx 10^(-7) xx 2 xx (25)^(2) xx 1)/(4 pi xx 2.5 xx 10^(-3) xx 9.8) `
` rArr " " h = 5.1 xx 10^(-3) m = 5.1 mm`