Question

# the cross section of five long parallel wires, separated by a distance d = 10 cm...

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?

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.

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