Two long straight parallel current carrying conductors each of length `l` and current `i` are placed at a distance `r_(0)`. Show that the total work done by an external agent in slowly reducing their distance of separation to `(r_(0))/(2)` is `(mu_(0))/(2pi)i^(2)ln(2)`

Explanation

The force acting on the conductor 2 is F=ilB
` = il (mu_0 i)/(2pir) = (mu_0i^2 r)/(2pi r) `
his force does a work dW in displacing the conductor 2 by a distance dr
`dW = vecF.vecdr = (mu_0i^2 r)/(2pi r) (-dr) ( therefore theta = 180^@)`
Then, the total work done is
` W = int dW = - (mu_0i^2l)/(2pi) int_(r_0)^( (r_0)/(r_0) ) (dr)/(r ) = (mu_0i^2 l)/(2pi) ln 2`