Closed loop via integration

Question

if an integral around a closed loop is 0, then why is curl of the electric field not 0? We know that the Work along a loop is 0, but the electric field also does work. Therefore, again shouldn't curl of electric field be 0?

Answer 1

I'm guessing you know that \begin{align} \text{curl}(\boldsymbol{E}) = - \dfrac{\partial \boldsymbol{B}}{\partial t} \end{align} So, when there are no time-varying magnetic fields in place, the curl vanishes. Under such circumstances, the integral of $\boldsymbol{E}$ along a closed loop is $0$. In such situations, the work done by the electric field (when traversed along a closed loop) is also $0$. In such situations, we call the electric field conservative.

However, this is a special case of a general phenomenon. Sometimes, we can have time-varying magnetic fields, which means the curl doesn't vanish, and in such situations, you can have \begin{align} \oint_{\gamma} \boldsymbol{E} \cdot d\boldsymbol{l} \neq 0 \end{align} ($\gamma$ is a closed loop in space) Often what happens in introductory E&M courses is that to get you familiar with the vector calculus, you often work only in the "electrostatic and magentostatic" regimes, where the fields don't vary with time. This is why you may have been told that "the integral of E-field along a closed loop is always $0$". However, it is extremely important to remember that this is only valid in the static case.