Curl of electric field is not zero in the case of a steady current in a loop Say we got conducting circular loop connected to a battery . The electric field inside the loop  obeys equation $\vec{J}=\sigma \vec{E}$.
Since the current flows in a circumferential way around the loop the electric field will be circumferential as well which implies that the curl of the electric field will be non zero.
Which is a contradiction ,what's wrong in the above reasoning.
Thank you
 A: You can easily see that for a stationnary circuit the curl of the electric field should be $0$ (as you mentionned, this comes from the Maxwell-Faraday equation).
However, this is not contradictory with the fact that $\overrightarrow{E}$ seems to be circumferential. $\overrightarrow{J}$ is indeed circumferential but $\overrightarrow{E}$ is not. That's because $\overrightarrow{J} = \sigma \overrightarrow{E}$ only holds in a ohmic conductor, and you must have a generator in your circuit (otherwise $\overrightarrow{J} = \overrightarrow{E} = \overrightarrow{0}$), and the electric field in your generator is actually opposite to the current, in such a way that $\int_C \overrightarrow{E} \cdot \overrightarrow{dl} = 0$ (so no contradiction with $\overrightarrow{\nabla} \times \overrightarrow{E} = \overrightarrow{0}$).
A: Note that the curl of the electric field is not necessary to be zero in all cases. It's only valid for electrostatics.
Maxwell's equation here is valid through out the electrodynamics which given by :
$$\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}$$
$$\nabla\times\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$$
$$\nabla\cdot\mathbf{B}=0$$
$$\nabla\times\mathbf{B}=\mu_0\left(\mathbf{J}+\epsilon_0\frac{\partial \mathbf{E}}{\partial t}\right)$$
