# 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 $$\sigma$$!

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 $$E(r) = \omega r B_0$$.