The condition for a field $\vec{E}$ to be conservative is: $\nabla \times \vec{E}=\vec{0}$. In electrostatics, $\nabla \times \vec{E}=\vec{0}$ is followed strictly, but Faraday's law says that: $$\nabla \times \vec{E}=-\frac{\partial}{\partial t}\vec{B}$$ Does this mean when $\frac{\partial}{\partial t}\vec{B}\neq\vec{0}$, the electric field is not conservative?
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