What is the relationship between the electric field induced by a uniform and infinitely wide magnetic field? I have been contemplating on how to solve for an $e$-field produced by a changing $b$-field, and when simplifying the differential equation, I ran into a problem where the integral form of Maxwell-Ampere's law does not describe physical reality - but instead gives us a mathematical model to describe what happens to a circuit loop under a changing $b$-field.  The problem is, what if I want to actually understand the real $e$-field at every point due to a changing $b$-field? I don't have an interest in knowing the sum total of the $e$-field along an arbitrary loop.
$\nabla\times E = - \frac {\partial B} {\partial t}$
Wouldn't the curl of $E$ of an infinitely wide and uniform $b$-field be zero?
If so, then how would a loop inside the $b$-field have an induced voltage?
Basically, I've been beaten over the head with the integral forms of Maxwell's equations, which only serve some real world practical problems, but cannot fully explain what is actually occurring the reality.
Am I wrong in believing this?
Edit 1:

Suppose we have a dynamic b-field that points in the x direction.  It only varies in the z direction.
$\nabla\times E = - \frac {\partial B} {\partial t}$
$\frac {\partial E_y} {\partial z} + \frac {\partial E_z} {\partial y} = - \frac {\partial B_x} {\partial t} $
$\partial E_y\partial y + \partial E_z \partial z = - \frac {\partial B_x} {\partial t} \partial z \partial y $
$E_z = 0 \leftarrow$ intuition tells me this, I don't see how it could have a value.  Maybe this is incorrect.
$\partial E_y\partial y = - \frac {\partial B_x} {\partial t} \partial z \partial y $
$\partial E_y = - \frac {\partial B_x} {\partial t} \partial z $
in finite form
$\Delta E_y = - \frac {\partial B_x} {\partial t} \Delta z$
We can see how knowing $\Delta E$ can be beneficial for a circuit loop. After all, the driver in a circuit is potential difference or the $\Delta V$ - which is $\Delta E \cdot L$ but this doesn't help us much outside of circuits.  How would I know the absolute value of E, rather than its relative value?
I know this question deviates a little from my original one, but I think it's strongly related.  Pointing me to any resources will be helpful.
 A: You cannot solve this equation to obtain a unique  E field IT IS IMPOSSIBLE.  you also need to specify the divergence of the E field  as
$$\nabla × (E +\nabla V) = -\partial B / \partial t $$
I can use any value for the scalar function V and still satisfy faradays equation.
as
$ \nabla × (\nabla V) = 0 $.
given I have a source current that produces a changing magnetic field, what is the induced E field?
the system of equations I would have to solve would also need to specify the divergence of E ,
Well in an electrically neutral wire, the net charge density is zero ,thus we can say
$$\nabla \cdot E = 0$$
using this fact We can solve for the E field, by writing maxwells equations in the POTENTIAL FORMULATION ( search this up)  using the value that Div E = 0
I'm not going to talk about specifically how to solve the equations, Gauge freedom, or how to solve the time varying poisson equation. However it starts by  assuming that
(1) $$\nabla × A = B$$
Why can I assume this? Well this follows from
(2) $$\nabla \cdot B = 0 $$
Taking  the divergence of both sides of equation 1, yields  equation 2, however in this form it is actually easier to solve since there is a freedom on A that I can add the gradient of a scalar function onto this, and taking the curl of this, Still gives me B, ie there are infinitely many A's that give me B back In return. Exploiting this freedom makes it easier to solve.
When substituting equation (1) into Gauss law, faradays law, amperes law,  We obtain the solution that
$E= -\partial A/\partial t$
$\nabla^{2}A - (1/c^{2}) (\partial^{2}A/ \partial^{2}t) = -\mu J$
The solution to this equation is called " retarded potential " for A
This is a solution to maxwells equations provided we are only interested in the INDUCED  electric field , and not one that encorporates a specific field due to a charge density
