I was beginning to learn about the vector and scalar potential formulations of classical E & M where you choose a $\phi$ and an $\vec{A}$ such that: \begin{equation} \begin{aligned} &\vec{B} \equiv \vec{\nabla} \times \vec{A} \\ &\vec{E} \equiv-\nabla \phi-\frac{\partial \vec{A}}{\partial t}. \end{aligned} \end{equation} I was curious what would happen if $\phi$ was such that it was simply equal to the emf from Faraday's law, \begin{equation} \phi=-\frac{\partial}{\partial t} \oint \vec{B} \cdot d \vec{A}, \end{equation} So then, \begin{equation} \begin{aligned} \frac{\partial \phi}{\partial t} &=-\frac{\partial^{2}}{\partial t^{2}} \oint \vec{B} \cdot d \vec{S} \\ &=-\frac{\partial^{2}}{\partial t^{2}} \oint \left(\vec{\nabla} \times \vec{A}\right) \cdot d \vec{S}. \end{aligned} \end{equation} Then Stoke's theorem says \begin{equation} \begin{aligned} & \frac{\partial \phi}{\partial t}=-\frac{\partial^{2}}{\partial t^{2}} \int \vec{A} \cdot d \vec{r} \\ \Rightarrow & \vec{\nabla} \frac{\partial \phi}{\partial t}=-\vec{\nabla}\left(\frac{\partial^{2}}{\partial t^{2}} \int\vec{A} \cdot d \vec{r}\right) \\ \Rightarrow &-\frac{\partial}{\partial t}(\nabla \phi)=\frac{\partial^{2} \vec{A}}{\partial t^{2}} \\ \Rightarrow &-\nabla \phi=\frac{\partial A}{\partial t} \end{aligned} \end{equation} Which means: \begin{equation} \vec{E}=-\nabla \phi-\frac{\partial \vec{A}}{\partial t}=0 \end{equation} So in the special case where the scalar potential is equal to the emf, the electric field vanishes.
Is this derivation correct? And if so, what are the implications of this result? I feel that there is some significance here but I'm not sure what it is.