The arrangement is as shown below

Find the potential energy of the loop.

Explanation

Magnetic moment of a small element of the loop.
`dM=i_(2)Ldx`
The direction of the magnetic moment is perpendicular to the plane of paper pointing inwards.
`dU=-dvecMvecB=-dMB`, where B is the magnetic field at the position of this element.
`i.e., B=(mu_(0))/(4pi)(2i_(1))/(a+x)`
`therefore dU=-(mu_(0))/(4pi)2i_(1)i_(2)1((dx)/(a+x))`
`therefore U=-(mu_(0))/(4pi)2i_(1)i_(2)lunderset(0)overset(b)int(dx)/(a+x)`
`=(mu_(0))/(4pi)2i_(1)i_(2)llog_(e)((a+b)/(a))`