A wire loop in the shape of a square of side a carries a current
I. If an infinitely long straight wire carrying the same current passes through the center perpendicular to the plane of the loop, what is the net force on the loop?

Consider a square wire loop of side a carrying current I, lying in a horizontal plane. An infinitely long straight wire, also carrying current I, passes perpendicularly through the center of the loop (i.e., along the vertical axis). The long straight wire produces a cylindrically symmetric magnetic field around itself. This field circles the wire and lies in planes perpendicular to the wire, decreasing in strength with distance from the wire, following: Now consider the four sides of the loop: ·    The loop is symmetric about the central axis (the position of the straight wire), so for every element of current on one side of the loop, there is a corresponding element on the opposite side. ·     The magnetic field direction and magnitude at points on opposite sides are such that the Lorentz forces on these elements are equal in magnitude but opposite in direction. The magnetic force on a current-carrying conductor in a magnetic field is given by: This force depends on the orientation of the wire segment and the magnetic field at that point. What Happens on Each Side: ·   Two sides parallel to the x-axis: The magnetic field varies across their lengths, but because the wire passes through the center, the two sides are mirror images. The force on one side is exactly cancelled by the force on the opposite side. ·    Two sides parallel to the y-axis: Same argument applies; the forces due to the field on these sides are also equal and opposite. Since the loop is symmetrical about the wire, and the current in the loop is the same on all sides, the net force on the loop is zero.