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Finding current flux

  1. Aug 31, 2014 #1
    1. The problem statement, all variables and given/known data

    Let J=xy^2([itex]\hat{x}[/itex]+[itex]\hat{y}[/itex]+[itex]\hat{z}[/itex]) 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([itex]\hat{x}[/itex]+[itex]\hat{y}[/itex]+[itex]\hat{z}[/itex]) A/m^2 implies the flow of electrical current in the direction J/abs(J)= ([itex]\hat{x}[/itex]+[itex]\hat{y}[/itex]+[itex]\hat{z}[/itex])/[itex]\sqrt{3}[/itex] with a magnitude of abs(J) = xy^2[itex]\sqrt{3}[/itex] 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.


    2. Relevant equations



    3. 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)*([itex]\hat{z}[/itex])dxdy=total current enclosed/eo. Is that kind of the right idea? Or should I be trying to set it up differently.
     
    Last edited: Aug 31, 2014
  2. jcsd
  3. Aug 31, 2014 #2

    BvU

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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.
     
  4. Aug 31, 2014 #3
    ∫xy^2([itex]\hat{z}[/itex])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
     
  5. Aug 31, 2014 #4

    BvU

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    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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