Three distinct current carrying wires intersect a finite rectangular plane ABCD. The current in left wire and the loop is I1. The direction of current in left most wire and right most loop is downwards as shown in figure. The current I2 through middle wire is adjusted so that the path integral of the total magnetic field along the perimeter of the rectangle is zero, that is, ∮ABCDAB→⋅dl→=0. Then the current I2 is :