# Surface current density problem

1. Sep 9, 2012

### ppoonamk

1. The problem statement, all variables and given/known data

A static surface current density Js(x,y) is confined to a narrow strip in the xy-plane In this
static problem ∇ ⋅Js = 0. Show that the line-integral of Js along any cross-section of the strip will yield the same value for the total current I. (The direction of dl in these 2D line-integrals is perpendicular to the line segment; these are not ordinary line-integrals but rather surface integrals in which the third dimension z has shrunk to zero.) Show that I =∫ Jsxdy =∫ Jsydx, where the integrals are over the width of the strip at any desired cross-section.

2. Relevant equations

Gauss’s theorem: ∫∫ (∇ ⋅Js)dxdy =∫ Js ⋅dl

3. The attempt at a solution

I don not understand this line The direction of dl in these 2D line-integrals is perpendicular to the line segment; these are not ordinary line-integrals but rather surface integrals in which the third dimension z has shrunk to zero.

The RHS of Gauss’s theorem = I
LHS:
Integrating from x1 to x2 where x2-x1 =Δx, dl is perpendicular to the line segment.
∫ Js ⋅dl= JsyΔx
similarly Integrating from y1 to y2 where y2-y1 =Δy, dl is perpendicular to the line segment.
∫ Js ⋅dl= JsxΔy

Is this right? How do I prove the 2nd part of the question?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 11, 2012

### susskind_leon

You need to be careful with your dl.
∫∫ (∇ ⋅Js)dxdy =∫ Js ⋅dl
The way you wrote Gauss' theorem down (in 2d) suggests that dl is tangential to the circumference of the cross section. Given that on the LHS you have ∇ ⋅Js = 0 tells you that I=0, which is indeed true if you define I as the current that goes AROUND your cross-section.
However, what you are actually looking for is ∫ Js ⋅ n dA, n being perpendicular to the surface.
So this is actually a tricky question for which you need to think a little bit about the divergence theorem in 2d and 3d and so on.
Also, instead of Gauss' theorem, you could use Stokes' theorem. But you need to think about stuff very carefully.