The resultant force (in μ N ) on the current loop P Q R S due to a long current-carrying conductor (current = 20 A ) will be 20 A ​

Step-by-step solution

Force on S R and P Q are equal but opposite so their net will be zero. Force between two parallel conductors carrying currents I 1   a n d     I 2 F = μ 0 2 π I 1 I 2 l r Where r = distance between two parallel conductors F P S = 10 - 7 × 2 × 20 × 20 × 15 × 10 - 2 2 × 10 - 2               = 6 × 10 - 4   N F Q R = 10 - 7 × 2 × 20 × 20 × 15 × 10 - 2 12 × 10 - 2