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

The magnetic field produced by long straight wire carrying current of 25A rests on a table on small wire `B=(mu_(0)I)/(2pih)`
the magnetic force on small conductor is
`F=Bilsintheta=Bil`
Force applied on PQ balance the weight of small current carrying wire.
`F=mg=(mu_(0)I^(2)l)/(2pih)`
`h=(mu_(0)I^(2)l)/(2pimg)=(4pixx10^(-7)xx25xx25xx1)/(2pixx2.5xx10^(-3)xx9.8)=51xx10^(-4)`
`h=0.51cm`