What is the physical meaning of Faraday's law in terms of scalar potential $\phi$ and vector potential $\vec{A}$?

Question

What is the physical meaning of Faraday's law in terms of scalar potential $\phi$ and vector potential $\vec{A}$?

Answer 1

1. The source-free Maxwell equations $$\vec{\nabla} \cdot \vec{B} ~=~ 0 \qquad ``\text{no magnetic monopole"}$$ $$ \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} ~=~ \vec{0}\qquad ``\text{Faraday's law"}$$ in terms of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$ are identically satisfied/tautologies, cf. e.g. this related Phys.SE post.

2. Conversely, the source-free Maxwell equations imply the local existence of $A^{\mu}$, cf. e.g. this Phys.SE post.