Two long parallel wires are separated by 429 cm and carry cu

Two long parallel wires are separated by 4.29 cm and carry currents 1.05 A and 4.61 A. Find the magnitude of the magnetic force that acts on a 3.67-m length of either wire.

Solution

Magnetic force per unit length between two wires:

F/L = u*I1*I2/(2*pi*r)

where

u = permeability of free space = 1.26*10^-6

I1 = 1.05 A

I2 = 4.61 A

r = 4.29 cm = 0.0429 m

So, F/L = 1.26*10^-6*(1.05)*(4.61)/(2*pi*0.0429)

So, F/L = 2.26*10^-5 N/m

So, force on 3.67 m length of either wire:

F = 2.26*10^-5*3.67 = 8.29*10^-9 N <--------answer