# Two long wires: induced electric field

1. Jun 6, 2008

### Pacopag

1. The problem statement, all variables and given/known data
Please see the attached figure. We have two parallel long wires carrying a time varying current I(t), but running in opposite directions. We wish to know the induced electric field between the wires, but only in the plane containing the two wires.
Sorry that I don't know how to do vectors in latex. E, B, dl and da should all be vectors.

2. Relevant equations
Ampere's law
$$\int {\bf B}. d{\bf l} = \mu I$$ where the integral is taken over a closed Amperian loop and I is the current inclosed by the loop.

$$\int {\bf E}. d{\bf l} = -{d\over{dt}}\int {\bf B}. d{\bf a}$$
where the left-hand integral is taken over a closed loop and the right hand integral is taken over the area defined by that closed loop.

3. The attempt at a solution
First of all, I set up the coordinate system as shown in the figure (I hope you can see it), with the y-axis along one of the wires, and the other wire located at x=d.
Now, for the wire that is on the y-axis, the magnetic field is found quite easily from Ampere's law (this is pretty standard). I get,
$$B_1 = {{\mu I}\over{2\pi x}}$$.
Now, for the other wire at x=d, we can just shift the previous expression over by d, so the total magnetic field is
$$B = {{\mu I}\over{2\pi}}\left[{1\over x}+{1\over{x+d}}\right]$$.
Now we can turn to Faraday's law. Choosing the contour to be the dotted rectangle in the figure, I get for the right-hand side of Faraday's law (noting that B and da are in the same direction)
$$-{d\over{dt}}\int {\bf B}. d{\bf a} = -{\mu l\over{2\pi}}{{dI}\over{dt}}\ln\left[{{(d-x)(2d-x)}\over{x(x+d)}}\right]$$
Now, I don't know if this is right so far. But the bigger problem is that I don't really know how to handle the left-hand side of Faraday's law, particulary because I can't intuitively determine which way the electric field is pointing along the contour (i.e. the dotted one in the figure). I'm pretty sure that only the vertical sides will contribute, and symmetry tells me that both vertical sides will give the same contribution. So I'm guessing the left-hand side would give
$$\int {\bf E}. d{\bf l}=\pm 2E(x)$$,
but I don't know which sign to use.

I hope this is clear enough to understand, and that someone can help me to make sense of this. Thanks.

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Last edited: Jun 6, 2008
2. Jun 7, 2008

### G01

Hmm I see one mistake while glancing all of this over. To describe the field component in between the wires that is created by the second wire you have:

$$\frac{\mu I}{2\pi}\frac{1}{x+d}$$

I believe this should be:

$$\frac{\mu I}{2\pi}\frac{1}{x-d}$$

Also for the direction:

HINT: Right hand rule.

3. Jun 7, 2008

### Pacopag

That is what I had first, but this gives the right-side of Faradays law to be zero, which tells me that either the electric field is zero, or the contributions from the vertical parts of the rectangular contour cancel eachother. The latter case would definitely be the right one, but even so, I can't determine the electric field from this.

4. Jun 7, 2008

### Pacopag

Ok. I know that the magnetic field between the wires should be pointing in the same direction. So if I make it
$$-\frac{\mu I}{2\pi}\frac{1}{x-d}$$
Then I get the right sign between the wires. The right hand side of Faraday is
$${{\mu l}\over{2\pi}}{{dI}\over{dt}}\ln\left[{{x^2}\over{(x-d)^2}}\right]$$
This looks like a nicer answer: arg of ln is dimensionless, symmetric about d/2 (but is it correct?).
Then
$$E(x) = {{\mu l}\over{4\pi}}{{dI}\over{dt}}\ln\left[{{x^2}\over{(x-d)^2}}\right]$$

As for the sign of E, I'm still unclear because I don't know how I is behaving in time (i.e. I don't know it's sign). But, provided there are no other mistakes, at least this should be correct up to a minus sign.

Last edited: Jun 7, 2008
5. Jun 7, 2008

### G01

This looks good to me. As for the sign, you can just find the sign for t=0, your initial condition, since you don't know how the current is evolving with time.

6. Jun 7, 2008

### Pacopag

Excellent. Thank you very much G01.

7. Jun 7, 2008

### G01

Anytime. Good Job!