Question

"two long parallel wires that are 0.30 m apart carry currents of 5.0 a and 8.0 a in the opposite direction. find the magnitude of the force per unit length that each wire exerts on the other wire and indicate if the force is attractive or repulsive. (μ0 = 4π × 10-7 t • m/a) "

Answer 1

The magnetic field B generated by wire 1 at distance r (where wire 2 is located) is

`B= (\mu _0 I_1)/(2 \pi r)`
where I1 is the current in wire 1. So, the force exerted on a segment
`\Delta L` of wire 2 is

`F=I_2 \Delta L B`
By susbstituting B inside the expression of F, we find the formula for the force per unit of length between the two wires:

`(F)/(\Delta L)=\mu_0 (I_1 I_2)/(2 \pi r)`
where r is the distance between the two wires.
By susbstituting the numbers of the problem, we get

`(F)/(\Delta L)=(4\pi \cdot 10^(-7) Tm/A ) ((5.0 A)(8.0 A))/(2 \pi (0.30 m))=2.7 \cdot 10^(-5)N`