homework-and-exercisesinduction
I've been asked to calculate the inductance per unit length for two wires or radius $a$ separated by $2d$ where $2d>>a$.
Starting from $\int_{s} B.dS = LI$ Im not sure what surface to take?
For each wire the field at a distance r away is given by $\int_{l} B.dl = \mu_0 I$ and by superposition $B$ in the first integral is their sum. And so $$LI= \int_{s} B_1.dS + \int_{s} B_2.dS $$ where $B_1$ and $B_2$ are the contributions from the two wires.
A few hints:
The field at the middle wire is "indeterminate" since there is a singularity due to the current in the middle wire. If you sketch the field as a function of $x$ you would get something like this:
(this is the plot of $\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}$ courtesy of Wolfram Alpha)
The zeros in the field are easily seen as occurring roughly at ±0.6 a - consistent with your calculation.
The magnetic field lines will look something like this:
where the 'x' marks the point where the field is null.
Now for the oscillation, you need to consider the B field that is experienced by the wire in the middle. As you already surmised, the sum of the fields of the two wires on either side is zero - but you need to ask yourself what is the field when you displace a small distance in x.
One way to do that is to take the derivative of the sum of the two fields - or you can just write down the expression for the field due to the wires at ±a at a position $dx$. This would be of the form
$$B \propto \frac{1}{a+dx}-\frac{1}{a-dx}$$
Doing a Taylor expansion (and figuring out the constant of proportionality) you will see that the field is indeed proportional to the displacement in x. This means that the central wire will experience a force that is proportional to the displacement. Convince yourself that the force is restoring - that is, points back to the middle - and you will have all you need to calculate the frequency.
You can then repeat the same procedure for the vertical field; but eyeballing it, I have the feeling that there is no change of direction of the field and therefore no oscillation. But I'll leave it up to you to take it from here - that's the policy on this site for homework-like questions.
Feel free to ask more questions if you don't make progress with these hints.
When dealing with infinitely long line charges (basically a cylindrical geometry) calculating the potential relative to infinity becomes a problem. You have to establish a reference (ground/earth) at a finite location. So, your result of an infinite potential difference is not incorrect, although it is confusing the first time you see it.
This site provides a description of why this happens. You are effectively grounding one end of your distribution, then stacking an infinite amount of charge down the line.
Best Answer
Answer if anyone is interested. In the end the areas outside the inner edges of the wire cancelled by symmetry and so the surface i was looking for was the area enclosed.