A confusion about the direction of the current density I'm contradicting my premise again and again and I have no idea what I'm doing wrong. Say there is a given current density in a charged infinite cylinder, $J$ in the $\hat{z}$ direction. It follows charges are moving in the $\hat{z}$ direction. If we look at Ohm microscopic law, we have:
$\vec{J}=\sigma \frac{\vec{F}}{q}=\sigma \frac{q(\vec{E}+\vec{v}\times \vec{B})}{q}=\sigma(\vec{E}+\vec{v}\times \vec{B})$. We assumed earlier $\vec{v}=v\hat{z}$ and the magnetic field is in the tangential direction using the right hand rule, meaning $\vec{B}=B\hat{\theta}$. Putting it in, we get:
$\vec{J} = \sigma(\vec{E}+v\hat{z}\times B\hat{\theta})=\sigma(E\hat{r}-vB\hat{r})$
Meaning $J$ is in the $\hat{r}$ directions, in contradiction to my assumption.  Where did I go wrong?
 A: If the current is moving in the $\hat z$ direction and the magnetic field is in the $\hat \theta$ direction, then the moving charges are going to feel a force in the $\hat r$ direction; this is a non-negotiable consequence of the Lorentz force law.
The only way to sustain $\mathbf J \propto \hat z$ is if there is a compensating radial electric field.  Qualitatively, this radial field is due to a negative induced charge density within the wire, corresponding to a buildup of positive charge on the wire's surface.  Because the charges moving in the wire are moving extremely slowly, the magnitude of the magnetic force on them (and therefore the corresponding radial electric field) is extremely small.

We can be more quantitative.  Let $\mathbf J=J(r)\hat z$ be the current density. Since $\mathbf J = -en_e \mathbf v_e$, where $e$ is the elementary charge and $n_e$ is the free electron density, we can determine the equation of motion for $\mathbf J$:
$$\frac{d\mathbf J}{dt} = -en_e \frac{d\mathbf v_e}{dt} = -en_e\left( -\frac{ e}{m_e}(\mathbf E + \mathbf v_e \times \mathbf B) - \frac{\mathbf v_e}{\tau}\right)= \frac{n_ee^2}{m_e}\left(\mathbf E + \mathbf v_e \times \mathbf B\right)+ \frac{n_e e}{\tau} \mathbf v_e$$
where $\tau$ is meant to model the time between electron collision events.  The term $-\mathbf v_e/\tau$ models the frictional loss due to Joule heating; this approach is called the Drude model.
In the steady state, $\frac{d\mathbf J}{dt} =0$.  Since $\mathbf v_e \propto \hat z$ and by symmetry we know that $\mathbf B \propto \hat \theta$, we have that
$$\frac{n_ee^2}{m_e}\left(\mathbf E + \frac{JB}{n_e e}\hat r\right) - \frac{J}{\tau} \hat z = 0$$
$$ \implies  \mathbf E  = -\frac{JB}{n_e e} \hat r + \frac{m_e J}{n_e e^2 \tau}\hat z \qquad (1)$$
The quantity $\sigma_0 \equiv \frac{n_e e^2\tau}{m_e}$ is called the Drude conductivity.  The electric and magnetic fields must also satisfy the steady-state Maxwell equations, which I will express in cylindrical coordinates with cylindrical symmetry ($\partial_\theta ,\partial_z \rightarrow 0$):
$$\begin{align}\nabla \cdot \mathbf E = \frac{\partial E_r}{\partial r} + \frac{E_r}{r} = \frac{e(n_0-n_e)}{\epsilon_0} \qquad &(2)\\
\nabla \times \mathbf E = -\frac{\partial E_z}{\partial r} \hat \theta = 0 \qquad &(3) \\
\nabla \times \mathbf B =  \left(\frac{\partial B}{\partial r}+\frac{B}{r}\right)\hat z = \mu_0 \mathbf J \qquad &(4)\end{align}$$
where $n_0$ is the equilibrium density of electrons and nuclei in the material.  Note that the $\nabla \cdot \mathbf B = 0$ is trivially satisfied under the assumptions we've made so far.
Equation (3) tells us that $\frac{\partial}{\partial r}\left(\frac{J}{n_e}\right)=0$, so $\frac{\partial E_r}{\partial r} = -\frac{J}{n_e e} \frac{\partial B}{\partial r}$.  Equation (2) becomes
$$-\frac{J}{n_ee} \left(\frac{\partial B}{\partial r} + \frac{B}{r}\right) =  -\frac{\mu_0J^2}{n_ee} =\frac{e(n_0-n_e)}{\epsilon_0} $$
Letting $n_e = n_0(1+\hat n)$, we find that
$$ \hat n \simeq \frac{\epsilon_0 \mu_0 J^2}{n_0^2 e^2} = \left(\frac{J}{n_0ec}\right)^2 = \left(\frac{v_e}{c}\right)^2\qquad (5)$$
Recalling that the drift velocity in a normal wire is on the order of cm/s, this is of order $10^{-20}$ and therefore completely negligible.  It follows that $J\sim n_e$ is essentially constant, and from (4),
$$\frac{1}{r}\frac{\partial( rB)}{\partial r} = \mu_0 J$$
$$\implies B = \frac{\mu_0 J r}{2}$$
where we have imposed the demand that $B$ be well-defined at $r=0$.  We finally have
$$\mathbf E = -\frac{\mu_0 J^2 r}{2n_e e}\hat r + \frac{J}{\sigma_0}\hat z$$
$$\mathbf B = \frac{\mu_0 J r}{2} \hat \theta$$
$$\mathbf J = J \hat z = \text{const}$$
with corrections of order $\left(\frac{v_e}{c}\right)^2$.
A: The relation between $\bf J$ and the field is
$${\bf J} = \sigma {\bf E}$$
So the moving electrons produce a magnetic field, but only ${\bf E}$ appears in the definition of ${\bf J}$
