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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

Respuesta :

We will have the following:

First, we recall that:

[tex]F=\frac{I_1I_2L\mu_0}{2d\pi}[/tex]

Then, we are given:

L= 0.474 m

I1 = 6.39A

I2 = 3.88A

F = 5.72*10^-5 N

So:

[tex]\begin{gathered} 5.72\ast10^{-5}=\frac{(4\pi\ast10^{-7})(6.39)(3.88)(0.474)}{2\pi(d)}\Rightarrow d=\frac{(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}[/tex]

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

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