Surface current density problem

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SUMMARY

The discussion centers on the surface current density problem in the xy-plane, specifically demonstrating that the line-integral of the static surface current density Js along any cross-section yields the same total current I. The relationship I = ∫ Jsx dy = ∫ Jsy dx is established, where the integrals are evaluated over the strip's width. The application of Gauss's theorem and the need for careful consideration of the direction of the differential line element dl are emphasized, particularly in relation to the divergence theorem and Stokes' theorem.

PREREQUISITES
  • Understanding of surface current density and its representation in physics.
  • Familiarity with Gauss's theorem and its application in two dimensions.
  • Knowledge of Stokes' theorem and its implications in vector calculus.
  • Basic concepts of line and surface integrals in vector fields.
NEXT STEPS
  • Study the application of Gauss's theorem in two-dimensional scenarios.
  • Explore Stokes' theorem and its relationship to surface integrals.
  • Investigate the divergence theorem in both two and three dimensions.
  • Practice problems involving line and surface integrals in electromagnetic contexts.
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism and vector calculus, as well as anyone involved in advanced problem-solving related to current density and integral theorems.

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Homework Statement



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.


Homework Equations



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

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?
 
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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.
 

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