Trying to find direction of force on a stationary charge in a changing magnetic field using faraday's law Suppose I have a stationary charge in an area where the magnetic field is constantly changing. Now from my understanding of Faraday's law, the direction of the force it experiences seems to be ambiguous. If I pick a circular loop and the charge is on the point P of the loop, then the rate of change of flux would tell me that the electric field points tangentially to the point P. But if instead of that particular circular loop, I had a loop with a  different orientation going through the same point P, then the direction of the electric field would again be tangential to the loop at P but in a different direction. So I can have all sorts of (real this time) wire loops in the area and the charges in the wires would feel a force in different directions depending on how I shape the loop. But then in what direction would a stationary charge feel the force?
 A: TL;DR: It depends upon the details of the magnetic field you have going on, and the source of the force would be the electric field induced by the changing magnetic field. Also, unless there's something exceptionally clever someone can come up with, thinking about loops will not help you here.
Let me first recall what a changing magnetic field really means. Not in terms of loops, but in terms of Maxwell's equations, which at the end of the day are the equations from which all else should follow. In particular, Faraday's law is written
$$
\nabla\times\vec E = -\frac{\partial \vec B}{\partial t}.
$$
This is a differential equation which, in conjunction with the other three Maxwell equations completely determine what the electric and magnetic fields are, given a set of charges/currents and boundary conditions. The important thing I want to note about the differential form of this equation is that it's just a differential equation: there is no ambiguity to be found here involving "what" loop we should choose or anything of the sort.
So, if we already knew, for some reason, what the magnetic field was supposed to be everywhere and how it was changing, there would be no ambiguity in solving Maxwell's equations for the electric field that goes with it. Though this calculation may be very difficult in practice and would certainly depend upon the details of whatever magnetic field you have going on. Once the electric and magnetic field are both known, then it's a matter of putting those into the Lorentz force law
$$
\vec F = q(\vec E + \vec v\times\vec B).
$$
So at no point here is there any ambiguity...it's just that what the answer will end up being is not obvious at any stage. But since the answer will really genuinely depend upon the details of the magnetic field, maybe that shouldn't be a surprise.
With that in mind, let's think a little bit about what we can say in terms of loops and why those will probably not give us much of a computational advantage here.
The connection to the integral form of the equation, which is what you're referring to whenever you talk about loops, comes from integrating both sides of the above equation over a surface $S$ and then applying Stokes' theorem to the LHS to write find the integral of the electric field over the boundary curve $C$:
$$
\int_C\vec E\cdot d\vec\ell = -\frac{d}{dt}\int_S \vec B\cdot d\vec A.
$$
The left hand side of this is what we would normally call the induced EMF and on the right we have what we would call the derivative of the magnetic flux, usually written in the form
$$
\varepsilon = -\frac{d\Phi_B}{dt}.
$$
The important point here is that this equation now holds for any surface $S$ we like, which is often where the confusion enters. It is worth noting though that the fact that this equation holds for any surface ensures that this statement is actually equivalent to the differential form, but the any is actually very important to this.
This is actually the same problem we encounter when first learning how to use Gauss' law to find an electric field, but in my experience it's usually easier to spot the limitations of the integral form of Gauss' law than it is to spot the same for Faraday's law. So let's take a moment to review the limitations in Gauss' law.
The integral and differential statements of this equation, next to each other, are
$$
\nabla\cdot\vec E = \rho/\epsilon_0,\ \ \ \ \int_S\vec E\cdot d\vec A = \frac{1}{\epsilon_0}\int_V\rho dV,\ \ \ \ \Phi_E = \frac{Q_{in}}{\epsilon_0}.
$$
Here the first form is, of course, differential, and the second form is found by integrating the differential form over a volume $V$ with bounding surface $S$, which we usually call the "Gaussian" surface.
Gauss' law, of course, works for any Gaussian surface we might care to think about. But anyone who has taking any course covering Gauss' law will know that when you have a spherically symmetric charge distribution (point charge), you should choose your surface to be a sphere. If you're looking at an (infinite) line of charge, you should use a Gaussian cylinder. If you have an infinite plane, you should use a Gaussian pillbox (a rectangular prism). And, though no one ever says it, if you have any other situation you probably give up or go use the differential form and solve Laplace's equation (which is then a large focus in second courses on electrodynamics).
So why? We said Gauss' law is true for any Gaussian surface, why are we forced into making these specific choices for specific cases? And the answer is that we really aren't "forced," but if we want to actually find the electric field, any other choice will result only in headache, pain, and failure. Why is this?
In all these examples, the key to actually extracting the electric field from the flux is making an argument about symmetry which tells us that $\vec E\cdot d\vec A = Ed A$ where $E$ is the magnitude of the electric field, which we hope to be a constant over the surface, so we can then pull $E$ outside the integral and can then solve for it.
This is precisely what happens for a point particle with a Gaussian sphere. We use symmetry arguments to deduce that the electric field points only in the radial direction and depends only on the radial distance. This tells is that, for a sphere, $\vec E\cdot d\vec A = EdA$ with $E$ a constant everywhere on the sphere. If the electric field were not constant over the sphere, or if we used a different surface over which the electric field was not constant, we would not be able to pull it out of the integral and hence would be left with an integral equation for $\vec E$, which is no good. Indeed, it wouldn't even be enough to determine $\vec E$. For that we would really need to use the fact that Gauss' law holds for all surfaces.
This is always the down side of integral equations and the integral forms of equations. Unless you're dealing with a situation that has exceptional amounts of symmetry, so much so that you basically already know what the answer is supposed to be, integral equations aren't the best. There are far more techniques to solving their differential counterparts.
