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Consider a wire carrying a current $I$, I need to find the current density distribution in the wire of a cylinder shape. Let the density function be $j(x,y)$, in the circle $D:x^2+y^2<r^2$.

We have the simple relation that \begin{align} \iint_{D}j(x,y)dxdy=I \end{align}

Considering the magnetic field generated by the current, the density has a trend getting in the center because we know that when two wires enjoying the same direction of current parrallel to each other, they will have ampere force pointing to each other. However, the Coulomb force keep them apart. That is to say, the ampere force will equal the Coulomb force when they have the distribution. Thus the following relation is established: \begin{aligned} \forall(x,y)\in D,\iint_{D}\frac{\mu_0 j(u,v)}{2\pi \rho}e_{\rho}dudv=\iint_{D}-\frac{ j(u,v)}{2\pi\epsilon_0 \rho w^2}e_{\rho}dudv \end{aligned} this relation is set up, where $e_\rho$ is the unit vector pointing from $(x,y)$ to $(u,v)$, $w$ is the speed, $\rho=\sqrt{(u-x)^2+(v-y)^2}$ is the distance, So \begin{aligned} \forall(x,y)\in D,\iint_{D}\frac{j(u,v)}{\rho}e_{\rho} dudv=0 \end{aligned} But I have no idea how to evaluate $j(x,y)$, thanks for your attention!

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I don't think your physical analysis is right. The equation isn't dimensionally correct, since $\mu_0$ doesn't have the same units as $1/(\epsilon_0 w)$. Also, there is no Coulomb repulsion, because the wire is electrically neutral everywhere. One suggestion would be to exploit the cylindrical symmetry by writing $j$ simply as a function of the radial coordinate $r$. I think the result is trivial for DC, since the electric field is constant due to the boundary conditions and symmetry, and $j=\sigma E$. –  Ben Crowell Nov 22 '12 at 15:49
    
@BenCrowell But the ampere force cannot be canceled out by the symmetry electric force, and that's why I think the analysis is needed. Thanks you for pointing out my mistake, I'll try to improve it –  Golbez Nov 22 '12 at 15:53
    
I see you've edited your equation to put a $w^{-2}$ on the right-hand side rather than a $w^{-1}$. Although this fixes the units, it doesn't make the equation physically correct. –  Ben Crowell Nov 22 '12 at 16:12
    
@BenCrowell Why is it incorrect? Thanks for your reply. –  Golbez Nov 22 '12 at 16:20
    
It's incorrect because it seems to be based on assuming an equilibrium between radial magnetic forces and radial electrical forces, but the radial electrical forces don't exist, and there is no reason to assume such an equilibrium. You also simply haven't given any detailed justification for it. –  Ben Crowell Nov 22 '12 at 17:06

1 Answer 1

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If the wire is long, then the solution has symmetry with respect to translation along the axis, as you've correctly assumed. The electric field lines are therefore parallel to the axis, and there are no radial electrostatic forces. The electric field is $E=\Delta V/L$, where $\Delta V$ is the voltage difference and $L$ is the length of the wire, so the electric field is constant across the radial cross-section of the wire. (Gauss's law also tells us that the net charge density $\rho$ is zero everywhere.) The current density is therefore $\sigma E=\sigma \Delta V/L$, where $\sigma$ is the conductivity.

The magnetic force you have in mind would be observable as a radial mechanical compression of the wire, but it doesn't affect $j$. The current density is not a physical object and doesn't feel forces.

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Emm, I mean the magnetic field generated by current will give Lorentz force to the electrons, thus shrinking affecting the current density. Is this statement wrong? Thank you. –  Golbez Nov 22 '12 at 16:23
    
@Golbez: Yes, your statement is wrong. The Lorentz force is canceled by microscopic forces, not by any macroscopically observable radial electric field. The radial component of the electric field is zero, as explained in my answer. The microscopic-level forces are partly embodied in the relation $j=\sigma E$, which is not consistent with the type of model you seem to have in mind. You seem to be modeling this as an electron gas in an evacuated cylindrical chamber. In that model, you can't have solutions with the longitudinal symmetry you assumed; the electrons would accelerate. –  Ben Crowell Nov 22 '12 at 17:16
    
Thanks for your reply. –  Golbez Nov 23 '12 at 6:31
    
@BenCrowell if, instead of in a wire, one created a beam of electrons moving from point a to b, such as in a linear accelerator, the moving electrons would indeed be attracted. So, why not within a wire? –  abalter Apr 25 at 19:48

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