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## Homework Statement

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.

## Homework Equations

Ampere's law

[tex]\int {\bf B}. d{\bf l} = \mu I[/tex] where the integral is taken over a closed Amperian loop and I is the current inclosed by the loop.

Faraday's law

[tex]\int {\bf E}. d{\bf l} = -{d\over{dt}}\int {\bf B}. d{\bf a}[/tex]

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.

## 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,

[tex]B_1 = {{\mu I}\over{2\pi x}}[/tex].

Now, for the other wire at x=d, we can just shift the previous expression over by d, so the total magnetic field is

[tex] B = {{\mu I}\over{2\pi}}\left[{1\over x}+{1\over{x+d}}\right][/tex].

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)

[tex] -{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][/tex]

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

[tex]\int {\bf E}. d{\bf l}=\pm 2E(x)[/tex],

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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