Electromagnetic field of a spinning cylinder Let us consider an infinite cylinder of axis $(Oz)$ and radius $R$ spinning at a constant radial velocity $\omega$. We assume that this cylinder is made of a metal that is assumed to be a conductor (and that we know every characteristics about it). 
At time $t=0$ se set a magnetic field $\mathbf{B}_0=B_0\mathbf{e_z}$ where $B$ is constant and uniform.

Describe what will happen after the transitory regime. 

Well, this questions seems tough for me. It seems like the conductor is spining in a magnetic field so an induced $\mathbf j$ will be created and this new source will generate an electromagnetic field $(\mathbf E,\mathbf B)$. 
In the conductor, I will assume that $\nabla \cdot \mathbf{E}=\nabla \cdot \mathbf{B}=0$, $\mathbf{j}=\sigma \mathbf{E}$ where $\sigma$ is the surfacic charge generated by the rotation, $\nabla \wedge \mathbf{E}=-\partial _t \mathbf{B}$ and $\nabla \wedge \mathbf{B}=\mu_0 \mathbf{j}=\mu_0\sigma \mathbf{E}$. 

How can I sing the generated $(\mathbf E,\mathbf B)$ field? I don't knwo $\sigma$!

 A: We can think of the conductor as a lattice of non-motile positive charges and motile negative charges. In that case, the spinning conductor makes the charges move in uniform circular motion. For simplicity let's take the $\omega$ to be positive and thus have anticlockwise spinning. Now that imparts a tangential velocity to each charge. Let's call that $v$. When the magnetic field is switched on, a force acts on the charge according to the Lorentz force $q v \times B$ which is towards the centre for negative charges and outward for positive charges (of course the positive charges don't move only negative charges do). This implies that a transitory $j$ is set up away from the centre but after the transition a potential gradient develops due to the shift in the charge distribution. Consider standard cylindrical coordinates $(r, \theta, z)$. Clearly, the problem has cylindrical symmetry and so will the fields, potentials and charge distributions. Then we have $q E(r) = q v(r) \times B$. We have $v = \omega \times \mathbf{r}$ and $|v| = \omega r$ and hence $E(r) = \omega r B_0$.
