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_(0)i^(2)l)/(2pir)`
This force does a work `dW` in displacing the conductor 2 by a distance `dr`
`dW=vec(F).vec(dr)=(mu_(0)i^(2)l)/(2pir)(-dr)( :.theta=180^(@))`
Then, the total work done is
`W=intdW=-(mu_(0)i^(2)l)/(2pi)underset(r_(0))overset((r_(0))/(2))int(dr)/(r)=(mu_(0)i^(2)l)/(2pi)ln2`