The Faraday-Maxwell law says that
$$ \nabla \times \vec{E} = – \frac{\partial \vec{B}}{\partial t}$$
Which is always explained as a time-varying magnetic field producing an electric field. But can we go the other way and say that a rotational electric field produces a time-varying magnetic field?
Electromagnetism – Can Rotational Electric Field Produce Time Varying Magnetic Field?
electromagnetism
Best Answer
An equation simply says that two quantities have to be the same. It's not strictly accurate to use a word like "produces" which implies a logical directionality; the left hand side and right hand side are on equal footing, logically speaking.
In practice, often one side is "given" and the other side is "something to solve for." If you are solving for a vector field $\vec{V}$ as a function of space, then given its divergence and curl in terms of known functions, and assuming that the vector field dies off asymptotically fast enough, you can solve for $\vec{V}$ using the Helmholtz decomposition. Applying this logic to Faraday's law, we can solve for an unknown electric field using the known "source terms", which include $\partial \vec{B}/\partial t$ (as well as the charge density). This leads to the language that $\partial \vec{B}/\partial t$ "producing" $\vec{E}$. Additionally, experimentally, it is possible to produce a time varying magnetic field; the fact that it is relatively easy to control the magnetic field experimentally in some situations also makes it natural to think of the equation in this direction (at least sometimes).
You are right, that if you thought of $\vec{E}$ as a known quantity as a function of time and space and wanted to solve for the unknown magnetic field, you could use Faraday's law to integrate $\nabla \times \vec{E}$ with respect to time and recover $\vec{B}$. This situation is not very common in practice, but there is nothing logically wrong with it.
In a more general situation, we don't know $\vec{E}$, $\vec{B}$, or the charges and currents as a function of time. If we know their value and time derivatives at some initial time, we can use Maxwell's equations (plus additional equations telling us how the charges and currents evolve) to solve for all of the fields, charges, and currents in terms of the initial conditions. Then, it's not really correct to say that either $\nabla \times \vec{E}$ produces $\vec{B}$, or that $\partial\vec{B}$ produces $\vec{E}$, but rather that the evolution of all the quantities is mixed together in a complicated way.