A square loop PQRS carrying a current of 6.0A is placed near a long wire carrying 10A as shown in figure (35-E10). (a) Show that the magnetic force acting on the part PQ is equal and opposite to that on the part RS. (b) Find the magnetic force on the square loop.
I2 = 6A
I1 = 10A
FPQ
= 8.4 × 10–4N (Towards right)
= 4 × 10–4 + 36 × 10–5 = 7.6 × 10–4N
Net force towards down
= (8.4 + 7.6) × 10–4 = 16 × 10–4 N
