There is a point written in my book (S.L. Arora- Simplified Physics Part-1)-
Electric Potential is meaningful only for electric fields produced by stationary charges. It has no meaning for electric fields set up by magnetic induction.
What are they trying to convey by saying that it has no meaning for electric fields set up magnetic induction. I tried googling it and understood it a bit, but there is still no clarity.
Electric fields can be generated by a stationary electric charge or by a changing magnetic field in accordance with the two equations: $$\vec {\mathbf{E}} = \frac{k_eq}{r^2}.\hat{\mathbf{r}} \; ; \quad \oint \vec{\mathbf{E}}.d\vec{\mathbf{s}} = \frac{-d{\Phi_B}}{dt}$$
The (induced) electric field described by the second equation, (also called the General form of Faraday's law) is a $\mathbf{non-conservative \ field}$.
The scalar electric potential is defined as: $$V \equiv \frac{U_E}{q}$$ and as $U_E$ (potential energy of the charge-field system) can only be defined for a system where a conservative force acts in between the field and charge, the field given by the second equation, that is, the field generated by the changing magnetic field cannot have a potential attributed to it.
Hope this helps.
Potential functions are derived in Conservative fields, Conservative fields are fields in which the line integral is path independent. meaning if I were to go from point a to point b, the amount of work I would have to do against the field in order to move something from a to b would be independent of the path that I take. this is crucial in defining a potential function at a point as without path independence, the whole concept is completely redundant as different work would be needed to go from the same 2 points depending on the path I take, and hence the use of potential difference between 2 points to find the amount of work done on an object, wouldn't have a fixed answer
Mathematically, You can prove that for a Conservative field, it can be written as $\nabla \phi $where $\phi$ is a scalar function,$ \nabla $acting here as the gradient operator. From the fundamental theorem of line integrals the line integral in this form would have an answer$ \phi(b) -\phi(a)$ where a and b are positions in space, which has nothing to do with the path inbetween.
Electric fields created via induction require that $ \nabla × E$ be non zero, using stokes theorem we can see that this means the E field generated is not path independent and the concept of a potential at a point breaks down due to reasons mentioned above
