50 question solution Q50. The force exerted by the conductor AB on the loop CDFG shown in figure is 1) 8 × 10–4 N attraction 2) 7.2 × 10–4 N attraction 3) 7.2 × 10–4 N repulsion 4) 4 × 10–4 N repulsion Share with your friends Share 6 Vijay Shankar Yadav answered this Dear Student, There would not be any force on CD and GF part Conductor AB will exert force on CG towards AB and on FD away from AB so Fnet=iGC×μ0iAB2π×1×10-2×lGC-iDF×μ0iAB2π×10×10-2×lDFFnet=10×2×10-7×2010-2×20×10-2-10×2×10-7×2010×10-2×20×10-2Fnet=7.2×10-4N attraction (towards AB) Regards 9 View Full Answer Mohamed Mafaz answered this good question. -5