A rigid square loop of side ‘ a ’ and carrying current I 2 is lying on a horizontal surface near a long current I 1 carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be:

Step-by-step solution

Force on segment 1-2 and 3-4 will get cancel out. Force on segment 4-1, F 1 = μ 0 I 1 I 2 2 π a × a toward Right Force on segment 2-3: F 2 = μ 0 I 1 I 2 2 π ( 2 a ) × a   toward left Net force on loop: F n e t = F 1 - F 2 toward Right = μ 0 I 1 I 2 2 π - μ 0 I 1 I 2 4 π = μ 0 I 1 I 2 4 π toward Right or repulsive