Question

the cross section of five long parallel wires, separated by a distance d = 10 cm and have lengths of 21.0 m. Each wire carries a current of 5.6 A. Currents 1, 3, and 5 are out of the page, while currents 2 and 4 are into the page. Each wire experiences a magnetic force due to the other wires. What is the magnitude and direction of the net magnetic force on wire 1?

Answer 1

As we know wires which have current in same direction attract each other and the wires which have current Repel each other.

So wires 3 and 5 attract wire 1 and wires 2 and 4 Repel wire 1 so

Force on wire 1 due to all wires

F = - F12 + F13 - F14 + F15

= - μoI2L/2πd + μoI2L/2π(2d) - μoI2L/2π(3d) +μoI2L/2π(4d) = μoI2L/2πd(-1 +1/2 - 1/3 + 1/4)

2×10-7×(5.6)2×21/0.1(-0.58)

= - 7.68×10-4 T

Direction = away from the wire becuare Repulsive force is more then attractive force.