Answer
See proof
Work Step by Step
Starting from the left side of the law: \[ \oint_C \mathbf{E} \cdot d\mathbf{r} = -\iint_S \frac{\partial \mathbf{B}}{\partial t} \cdot \mathbf{n} \, dS \] We have: \[ \oint_C \mathbf{E} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{E}) \cdot \mathbf{n} \, dS \] where we have used Stokes' theorem for the line integral. Hence, \[ \iint_S (\nabla \times \mathbf{E}) \cdot \mathbf{n} \, dS + \iint_S \frac{\partial \mathbf{B}}{\partial t} \cdot \mathbf{n} \, dS = 0 \Rightarrow \iint_S \left((\nabla \times \mathbf{E}) + \frac{\partial \mathbf{B}}{\partial t}\right) \cdot \mathbf{n} \, dS = 0 \] This implies: \[ (\nabla \times \mathbf{E}) + \frac{\partial \mathbf{B}}{\partial t} = 0 \] So, a changing magnetic field \(\mathbf{B}\) in time gives rise to an electric field \(\mathbf{E}\) circulating around it.
