How do Amperian Loops work with two infinite slabs of current?

[hotcommodity]

Hello,

This isn't a homework question, just a curious question. I've been working on problems involving Ampere's Law and slabs of current density J.

I've drawn a diagram to illustrate my question:
http://www.freeimagehosting.net/t748n

Here, two infinite slabs of current are placed one on top of the other, in the xz-plane. The magnitudes and directions of Jin and Jout are such that the magnetic field outside the slabs is zero (i.e. B = 0 for y>d and y<-d). Also, I assume that the current densities vary with distance inside the slab (because if they didn't, I could simply take my entire Amperian loop inside of each slab centered on the middle of the slab). Therefore, if the current densities are functions of distance y, I have to hang my Amperian loop outside of the slabs where the magnetic field is zero.

I'm trying to use Ampere's Law to find the magnetic field inside each slab (i.e. 0<y<d and -d<y<0). The textbook I've been working with (among other sources) draw the Amperian loops similar to loops 1 and 2 in the diagram. The red part of each loop is the part I'm solving for. Each of the other 3 segments either cancel by symmetry, or the field is zero along that path.

So here's where I'm confused: I don't understand why loops 1 and 2 are correct (or are they?). Loop 1 doesn't seem to account for the field produced by the bottom slab of current. It would make more sense (to me) if loops 3 and 4 were the correct loops. As loop 3 accounts for the field produced by the bottom slab, and loop 4 accounts for the field produced by the top slab. Can anyone explain this?

Thanks in advance!

 

[frogjg2003]

Both work, it's just a lot easier to use 1 and 2 over 3 and 4 because 1 only depends on Jin. 2, 3, and 4 all depend on both Jin and Jout, so having something that only depends on one is a lot easier. The beauty of Ampere's Law is that the current outside has no effect on the integrals. If all the slabs are infinite in all but the y direction, then there is no edge effects to take account of and the field only varies with y.

In fact, loop 2 is just an extension of loop 1 into the next region. Loops 3 and 4 are different loops.

 

[hotcommodity]

Hi frogjg2003, thanks for the reply.

So you're saying loops 1 and 3 will give me the same field?

 

[frogjg2003]

If you know how J varies with y over all y, then anyone loop (say 1 or 3) will give you the correct field in that region because you can just calculate the enclosed current. You can choose either of the two loops (2 or 4) that deal with the other region to calculate the field in that region.
In short, yes, they'll give the same field.

 

[hotcommodity]

Excellent! Thanks.

That question has been frustrating me for days. Time to go back and re-work some problems.

Thanks frogjg2003!