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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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    $\begingroup$ yes, the definition of a conservative vector field V is $\nabla \times \textbf{V} = 0$ therefore if $\nabla \times \textbf{V} \neq 0$ the field is non conservative $\endgroup$ – tomph Jan 9 '17 at 11:02
  • $\begingroup$ Related: physics.stackexchange.com/q/31672/2451 $\endgroup$ – Qmechanic Jan 9 '17 at 11:40
  • $\begingroup$ Yes and you have answered your question urself! $\endgroup$ – M.H.Muhamadi Jan 9 '17 at 11:51
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Yes, the electric field produced by a varying magnetic field not only isn't, but cannot be conservative, by exactly the reasoning you have mentioned. This is crucial to the operation of electric generators: they establish just such a non-conservative field in the wires attached to them through the principle of Faraday's law, and this allows the charges circulating through the circuit (usually electrons) to do sustained work in powering electrical devices. The energy to create the field and do this work, of course, comes from the mechanical input to the generator.

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Actually, the electrostatic field is conservative and the field which is produced by the time variation of magnetic field is "Electric field and not an Electrostatic field" and it is a non- conservative field.

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