Finding Current Flux Through Closed Surface

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SUMMARY

The discussion focuses on calculating the total current flux through a closed surface defined by a cubic volume of 1m³, with a current density given by J=xy²(ĥx+ĥy+ĥz) A/m². The flow direction of the current is determined as J/|J| = (ĥx+ĥy+ĥz)/√3, and the magnitude of the current density is |J| = xy²√3 A. The approach involves using symmetry to simplify the calculation of the flux through the six surfaces of the cube, with particular attention to the integrals for surfaces where x or y equals zero, resulting in zero current flux.

PREREQUISITES
  • Understanding of vector calculus, specifically surface integrals.
  • Familiarity with current density concepts in electromagnetism.
  • Knowledge of Cartesian coordinates and their application in three-dimensional space.
  • Basic principles of symmetry in physical problems.
NEXT STEPS
  • Study the application of the Divergence Theorem in calculating flux through closed surfaces.
  • Learn about electric flux and its relationship to current density in electromagnetic theory.
  • Explore the concept of surface integrals in vector calculus.
  • Investigate the implications of symmetry in solving physical problems involving vector fields.
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Students and professionals in physics, particularly those studying electromagnetism, as well as engineers working with electrical current and flux calculations.

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Homework Statement



Let J=xy^2(\hat{x}+\hat{y}+\hat{z}) A/m^2 denote the electrical current density filed -i.e, current flux per unit area - in a region of space represented in Cartesian coordinates. A current density of J=xy^2(\hat{x}+\hat{y}+\hat{z}) A/m^2 implies the flow of electrical current in the direction J/abs(J)= (\hat{x}+\hat{y}+\hat{z})/\sqrt{3} with a magnitude of abs(J) = xy^2\sqrt{3} Amperes (A) per unit area.

Calculate the total current flux J*Ds through a closed surface S enclosing a cubic volume V = 1m^3 with vertices at (x,y,z,) = (0,0,0) and (1,1,1) m.


Homework Equations





The Attempt at a Solution



I believe that I can use a fair amount of symmetry for this problem because this flux should be the sum of six surface integrals. I have done a fair amount of things with electric flux in the past..and I feel that this should be very similar however my mind seems to believe drawing a blank is the best solution. I tried to start with the integral of abs(J)*(\hat{z})dxdy=total current enclosed/eo. Is that kind of the right idea? Or should I be trying to set it up differently.
 
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Postpone the symmetry idea and start with one of the six surface integrals. Direction of J is a constant (unit vector) and so is ##d\vec S##. Only thing that varies is the magnitude. Write out the integral.
 
∫xy^2(\hat{z})dxdy

I kind of had this before. I then was thinking along the lines well okay for when I did electric flux if I had a electric field E and I needed to find the charge enclosed and I was given a charge density I would start:

E*2*π*r^2=ρ*π*r^2/εo
then things would cancel and I would get E=ρ/(2*εo)

However, I am unsure how to progress to that point for the current flux. I'm sure its some simple thing I'm not putting together
 
OK, you picked the one where symmetry does come in useful: z=0. Check with the opposite one (z=1) and bingo, two down, four to go.

Note that two of those four have magnitude of J = 0 ( the ones with x or y equal to zero ).

Leaves some work to do, still, but you'll manage !
 

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