In the figure five long parallel wires in the xy plane are s

In the figure, five long parallel wires in the xy plane are separated by distance d = 6.36 cm, have lengths of 10.0 m, and carry identical currents of 3.96 A out of the page. Each wire experiences a magnetic force due to the other wires. What is the magnitude of the net magnetic force on (a) wire 1, (b) wire 2, (c) wire 3, (d) wire 4, (e) wire 5?

Solution

Here ,

as the magnetic force between two wires is given as

F = u0 * I1 * I2/(2pi * d)

a) for the net force on wire 1

F1 = u0 * I^2/(2pi) * (1/d + 1/(2d) + 1/(3d) + 1/(4d))

F1 = 4pi *10^-7 * 3.96^2/(2pi * 0.0636) * (1/1 + 1/2 + 1/3 + 1/4)

F1 = 0.0001027 N

the force on wire 1 is 0.0001027 N

b)

for the wire 2

F2 = u0 * I^2/(2pi) * (-1/d + 1/(2d) + 1/(3d) + 1/(d))

F2 = 4pi *10^-7 * 3.96^2/(2pi * 0.0636) * (1/2 + 1/3 + 1/4)

F2 = 0.0000534 N

the force on wire 2 is 0.0000534 N

c)
for the force on wire 3

F3 = u0 * I^2/(2pi) * (1/d + 1/(2d) - 1/(d) - 1/(2d))

F3 = 0 N

the net force on wire 3 is 0 N

d)

for the force on wire 4

F4 = u0 * I^2/(2pi) * (-1/d + 1/(2d) + 1/(3d) + 1/(d))

F4 = 4pi *10^-7 * 3.96^2/(2pi * 0.0636) * (1/2 + 1/3 + 1/4)

F4 = 0.0000534 N

the force on wire 4 is 0.0000534 N

e)

for the force on wire 5

F5 = u0 * I^2/(2pi) * (1/d + 1/(2d) + 1/(3d) + 1/(4d))

F5 = 4pi *10^-7 * 3.96^2/(2pi * 0.0636) * (1/1 + 1/2 + 1/3 + 1/4)

F5 = 0.0001027 N

the force on wire 5 is 0.0001027 N