Question

A 0.474 m long wire carrying 6.39 A ofcurrent is parallel to a second wirecarrying 3.88 A of current in the samedirection. If the magnetic forcebetween the wires is 5.72 x 10-5 N,how far apart are they?[?] m

Answer 1

We will have the following:

First, we recall that:

`F=(I_1I_2L\mu_0)/(2d\pi)`

Then, we are given:

L= 0.474 m

I1 = 6.39A

I2 = 3.88A

F = 5.72*10^-5 N

So:

`\begin{gathered} 5.72\ast10^(-5)=((4\pi\ast10^(-7))(6.39)(3.88)(0.474))/(2\pi(d))\Rightarrow d=((4\pi\ast10^(-7))(6.39)(3.88)(0.474))/(2\pi(5.72\ast10^(-5))) \\ \\ \Rightarrow d=0.041090082797...\Rightarrow d\approx0.04 \end{gathered}`

So, the distance between the wires is approximately 0.04 meters.