EMI - induced current on two parallel wire

[jewhitmo]

I would like to quantify the EMI on a circuit and I am confused on how to do that. I have two parallel wires that are laying on top of each other. I have tried to calculate the flux linkage and came up with:

$$\Lambda$$ = [l*$$\mu$$_o*I/(2pi)]*ln((d-a)/a)

where a=radius of each wire, d=distance between the center of the wires, l=length of the wires, I=current on wire 1, and $$\mu$$_o*=permeability of free space.

I don't understand how this would induce a current on wire 2 though.

I'd really appreciate any help with this issue.

Thanks,
J

 

[adriank]

A constant current will not induce a current in another wire; induced current is proportional to the change in the original current.

 

[jewhitmo]

Thank you for your reply. You are right. I'm sorry I forgot to indicate that the problem is occurring because inrush current. So, there would be a change in current. Do you know how to calculate the change in current on wire 2 due to a change in current on wire 1?

 

[Defennder]

It depends on how exactly the current is changing. You don't necessarily need a change in current; a change in magnetic flux through some pre-defined surface or where magnetic flux lines are being cut suffices as well.

 

[jewhitmo]

If I could measure exactly how the current is changing on wire 1 do you know how I would be able to calculate the change in current on wire 2? Does the flux linkage equation apply here? If so, how does it translate into a change in current on wire 2?

Thanks!

 

[Defennder]

You have to use Faraday's law here:
$$Emf = -\frac{d\Phi_B}{dt}$$.
So this gives you the induced emf on wire 2. To get the induced current, divide emf by resistance of wire 2. That's all I can say if you don't provide any more information on the setup.

 

[berkeman]

You can have both inductive and capacitive crosstalk. The capacitive crosstalk is just coupling from wire-to-wire. Inductive coupling requires tso loops to couple the energy. So don't ignore the return paths of the wires in your inductive crosstalk calculation.

 

[jewhitmo]

thanks for the responses.