# Magnetic field inside a cylinder which is rotating in a non-constant angular velocity

 P: 4 Thanks for the reply. But it did not address my question, I would like to know why in this problem there is a certainty that the Electric field is not changing with time ? id est, look at Ampere's Law after Maxwell correction: $\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t}\$ and of course, the integral form of this equation: $\oint_{\partial S} \mathbf{B} \cdot \mathrm{d}\mathbf{l} = \mu_0 I_S + \mu_0 \varepsilon_0 \frac {\partial \Phi_{E,S}}{\partial t}$ I could, with rather ease, derive the manual's solution if I knew that $\frac{\partial \mathbf{E}} {\partial t} \$ is zero. Any notions about why E is constant in time? Thanks in advance
 P: 15 Magnetic field inside a cylinder which is rotating in a non-constant angular velocity The answer is simple. (I will call sigma -> rho) if dE/dt (partial derivative) = 0 ==> (Gauss law) d$\rho$/dt = 0 ==> (math) $\rho$ is constant in time. (stationary current) And you can see in the problem data, that they didn't say anything about the function $\rho$ .
 P: 4 Again, thanks for the reply. The $\nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}$ formula applies only to E filed that are not circular. [ I mean in order to derive the total electric field inside the cylinder you will have to find the E in the theta direction as well] According to Faraday's Law: $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$ Which means that a changing in time magnetic field induces a circular E field. How can I infer that the circular E field is not changing by time? [ prior to calculating the magnetic field - because then i just use faraday law to see that]