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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.

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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    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.2012-11-22

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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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    @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?2014-04-25