# Finding current flux

1. Aug 31, 2014

### Gee Wiz

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

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.

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)*($\hat{z}$)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. Aug 31, 2014

### BvU

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.

3. Aug 31, 2014

### Gee Wiz

∫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

4. Aug 31, 2014

### BvU

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 !