Dear Purvesh, 

  In principle, given the simple geometry that you have, you can calculate it straightforwardly. In the square loop you can apply Ampere's law for obtaining the magnetic field. This an usual exercice of magnetostatics, you have a full solution in

   Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall

                     Problem 5.37.

And once you have the magnetic field B, you can calculate the Lorentz force taking into account all the details of the distances and where the magnetic fields are coupled in fact, i.e.

             F= n v x B

being n the density of charge, v their velocity and B the magnetic indution field. Or directly on the currents

          F= I1 d x B

where d here is the length of the wire considered that the magnetic field B works on it. Notice that the current I2 really is not necessary to use due to the action reaction equal forces.

     Sorry I have forgotten to say clearly the B is the sum of the magnetic field created by the loop and the wire with I2 current.

Let's consider the forces acting on a square loop. The field generated by this loop and acting on it will produce no net force, no matter that parallel edges will repel each other. This leaves us with field generated by a straight wire and its action on a square.  It is easy to see that the forces acting on squares' edges perpendicular to the wire will perfectly cancel out (but will produce non-vanishing torque!). We need to find the field at the distances d-a/2 and d+a/2 from the wire; generated by current I2.  Its value, B(r),  may be found using Ampere's law: 2\pi r B(r) = \mu_0 I2.  Finally, |F| = d ( B(r-a/2) - B(r+a/2) ).  The net force vector will be perpendicular to the wire, pointing either to it or in opposite direction - depending on directions of both currents.