genxium said:
Hi Jolb, thank you for the reply.
Regarding the notations, sorry for the inconvenience, however
1. I quoted them from the book elements of engineering electromagnetics (6th edition) by Nannapaneni Narayana Rao so as to avoid stupid mistakes while using my own notations.
2. It's nothing about American or European or Asian style notation :) "Contour abcd" comprises of "paths ab, bc, cd, da"; Electric/magnetic fields directions should have no relationship with the integral paths whilst the paths are also directional.
Maybe it's an engineering notation, then. With my physics background, my instinct would be to give a separate variable to the lengths of the segments (perhaps w and l) so that I can more easily keep track of units--for example, an expression like lw
2 can be easily seen to have dimensions of length
3, whereas I might confuse myself with something like ab
2. But aside from convenience, it really doesn't matter, and I was able to figure out what you're saying.
You are right that the magnetic fields do not depend on the orientation of the contour used to calculate the fields, but the contour must have some orientation for the calculation, and it helps to show on the diagram which orientation you used when comparing with the algebra.
I got confused by your diagram because I thought the red arrows were orientation arrows (I assumed they were orientation arrows since drawing the magnetic field would imply a particular sign for J, which the problem hasn't specified.) The corrected diagram is more clear.
Now back to the topic of J_s, I think it's an easy way to treat J_s as a "confined to the interface" current density. However if the current is only proportional to ab \; or \; da, the metric of J_s will be changed from "A/m^2" to "A/m".
I'm fine with it but the book does not mention any change of metrics in this chapter, and I'm also concerned about that whether the current density "near" the interface should be treated in this way.
Well, on a basic level (magnetostatics), you don't need any fancy math like "change of metrics" (do you mean "measure" instead of "metric"?). I think you might be confusing yourself with thinking about fancy math instead of the basic vector calculus that's all over EM. (Maybe I'm totally misinterpreting you--do you mean "unit" instead of "metric"? In physics and math, the word "metric" typically has something to do with geometry.)
Let me sketch the main ideas here, and please forgive me for using my own notation. (Also, let me work this out "in vacuum"--so I'll work with B rather than H, since we're not too concerned with the properties of the media. It is trivial to do the generalization to linear media.) First of all, let me write down the theorem that I call "Green's Theorem."
\oint _C \vec{B}\cdot\vec{dl}=\iint_{C'}\left (\nabla\times\vec{B} \right )\cdot\vec{da}
Hopefully that makes sense to you. The symbol C denotes the closed contour (like the one in your diagram), while C' denotes a surface bounded by the contour C.
From Maxwell's equations we have:
\nabla\times\vec{B}=\vec{J}
Where we have set the permeability of free space μ to be equal to 1. Plugging this into the above equation is what you were using here:
<br />
\oint _C \vec{B}\cdot\vec{dl}=\iint_{C'}\vec{J}\cdot\vec{da}
Now the problem you're having is evaluating the integral on the right hand side. Instead of worrying about measures or "metrics", the point is that the quantity on the right is just the total flux of current through the surface C'. You often see this in physics books written as:
\iint_{C'}\vec{J}\cdot\vec{da}=I_{enclosed}
Since J is a surface current density, you'll want to think of it in the following way: First, we can treat the interface of the two media as a 2-dimensional sheet of conductor, infinite in extent (along both axes).***[See bottom] If we wanted to measure the magnitude of current flowing in this infinite 2D conductor, we would pick some line of length L in the conductor and measure how much current crosses that line. This would be equal to J*L. From this you can see that J would have units of current/length.
If you think of it this way you'll find I
enclosed=(J amperes/meter)(ab meters enclosed).
I think what you're attempting to do is to evaluate the integral \iint_{C'}\vec{J}\cdot\vec{da} to find I
enclosed, instead of just realizing it is very trivial. That sort of integral would require some sort of measure theory/distribution theory (you might stick a dirac delta in there to show all the current is isolated on the interface, and then do the integral rigorously), but why bother? Keep in mind the example of using basically the same reasoning to calculate the magnetic field at a distance r from a current-carrying wire (1D conductor).
*** -- Note: you are right that in real-world scenarios, the current is not confined exactly to the 2-dimensional interface of media. There are fantastically intricate things going on at the molecular scale, and there is indeed a thickness to the "surface current." But your problem deals with the basic theoretical aspects, so we are not concerned with those practicalities; plus the diagram does indicate that all the J is on the surface. In that case, the easiest way to get an answer is to do what I recommended and treat the interface as a 2D conductor.
