I thought I might join this discussion. One related thread is this Insights article by
@rude man https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/
that received very mixed reviews.
There is one criticism in there of how can it be possible to use an ## E_s ## in the case of an inductor and have ## E_s=-E_m ## in the conductor when ## \nabla \times E_s=0 ##, while ## \nabla \times E_m=-\dot{B} ## is non-zero. I just figured out the other day how this is possible : For one loop around the inductor ring we have ## \oint \vec{E}_m \cdot ds=-\dot{\Phi} ##, while for ## E_s ## we are in the adjacent wire in the coil when we go once around, so that ## \oint \vec{E_s} \cdot ds=\dot{\Phi} ##, rather than the zero that Stokes theorem would insist upon, with ## \nabla \times E_s=0 ##. Meanwhile, with the ## E_s ## we get to amplify the ## E_m ##, so that if we have a coil of ## N=1000 ##, we get a reading of ## \int E_s \cdot ds ## that is one thousand times the reading we would get from ## \oint E_m \cdot ds ##.
This splitting of ##E_{total} ## into ##E_s ## and ## E_m ## has one other difficulty that I recognize: If you have a changing ## B ## outside the loop of the circuit, we do have ## \nabla \times E_m=0 ## for the area inside the circuit, so there is zero EMF. However, if we compute ## E_m ## from symmetry by drawing a ring around the ## B ## with the ring passing through the circuit, we get a non-zero ## E_m ##, and thereby must necessarily introduce an opposing ## E_s ## or we could have an ## E_m ## across a resistor in that circuit, and Ohm's law would be violated. This makes it far easier to work with EMF's in this case, and not introduce the ## E_m ## and ## E_s ##.
My overall opinion is that the ## E_s ## and ## E_m ## does have a good amount of merit at times, while at other times we are better off just working with EMF's. One other previous related thread on Physics Forums is the following, where in post 192
@cnh1995 has a solution whose methodology I detailed in post 193. His solution uses an electrostatic potential (he called it ##V_{ab} ## but it isn't a voltage so it might be better designated as ## U_{ab} ##), and he gets an answer much easier than I got by solving the Kirchhoff loops. (See post 152). Here is the link:
https://www.physicsforums.com/threads/faradays-law-circular-loop-with-a-triangle.926206/page-6