Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are l 1 = + 10 A , l 2 = - 13 A , l 3 = + 10 A , l 4 = + 7 A , l 5 = - 12 A and l 6 = + 18 A . The magnetic induction at a perpendicular distance of 10 cm from the cable is μ 0 = 4 π × 10 - 7 W b A - m

Step-by-step solution

Net current due to all wires, i n e t = i 1 + i 2 + i 3 + i 4 + i 5 + i 6 i n e t = 10 + 13 + 10 + 7 - 12 + 18 = 20 A We know, magnetic field due to an infinitely long straight conductor at a perpendicular distance r from it is given by B = μ 0 i 2 π r = μ 0 i n e t 2 π r where, i = current in wire and r = perpendicular distance. = 4 π × 10 - 7 × 20 2 π × 10 × 10 - 2 = 4 × 10 - 5 B = 40 μ T