# Current Density in conductor in DC Mode

I was trying to calculate the current density in a conductor subject to DC voltage but I come to an absurdity. I took a circular copper conductor.

Situation scheme

I have considered the Drude Model : $$j = E\sigma$$.

We have a current so we have a magnetic field in the conductor : $$B(r) = \frac{\mu_0 }{r} \int_{0}^{r}\sigma E(r)r .dr$$ It seems to be the equivalent of $$B = \frac{\mu_0 I}{2\pi r}$$ for inside the conductor. I have assumed that the electric field was pretty uniform in the conductor so we have : $$B(r) = \mu_0 \sigma E r$$ Following the definition of $$j = ne v$$ we have $$v=\frac{j}{ne}$$ (ie. $$v=\frac{\sigma E}{ne}$$).

So now if we applies Lorentz Law to the flow of electron we have that magnetic lorentz isn't insignifiant because : $$v \wedge B = \frac{\sigma^2 E^2}{ne}\mu_0 r$$ So we would have a force that concentrate electron in the conductor but we always says that in DC we have an equal distribution in the conductor. So where is my mistake or my confusion ?

The Lorentz force $$\rho({\bf v}\times {\bf B})$$ is cancelled by a radial $${\bf E}$$ field that occurs because of the Coulomb force between the conduction elecrons and the positive ions enforces charge neutrality. The charge density does change by a tiny amount to set up this field. It's a kind of radial Hall effect.