Intuition behind Faraday's Law? Faraday's Law seems more like an observation than an explanation. Sure, a changing magnetic current causes emf, but why?
How does a changing magnetic field cause electrons to move in the direction of a wire? Assume that all you know is Bio-Savart's Law and $\vec{F} = q\vec{v} \times B$
I'm hoping for an explanation similar to how a battery generates a field that causes the electrons to move, which causes surface charge buildup and forces the electrons to move in the direction of the wire.
 A: The current is not directly due to the magnetic field, rather it is due to the electric field that is induced by the changing magnetic field. It is true that the electrons will experience a force from the magnetic field according to the Lorentz Force Law, but this force will always be perpendicular to the direction of motion and therefore will not produce a current. In this scenario, because the magnetic field is changing there will be an induced electric field in the wire, which is what will produce the current. 
To find the current we need to first find the emf, which is the line integral of the net force around the wire loop. The force from the magnetic field will always be perpendicular to the wire, so the only contribution to this line integral will be the contribution from the electric field. Faraday's Law tells us that the line integral of the electric field around the wire loop will be equal to the derivative of the flux of the magnetic field through the wire loop. Therefore, all we need to do to get the emf is calculate the derivative of flux of the magnetic field, and from the emf we can get the current.
Faraday's Law is the integral form corresponding to one of the four Maxwell Equations in differential form.
Starting with the following Maxwell Equation in differential form:
$$\nabla\times \overrightarrow{E} = -\frac{d\overrightarrow{B}}{dt} $$
taking the flux through any open surface $\Sigma$ on both sides yields
$$  \iint_{\Sigma}{(\nabla\times \overrightarrow{E}) \cdot \overrightarrow{dA}} = -\iint_{\Sigma}{\frac{d\overrightarrow{B}}{dt} \cdot \overrightarrow{dA}} $$
moving the time derivative outside the integral on the RHS
$$  \iint_{\Sigma}{(\nabla\times \overrightarrow{E}) \cdot \overrightarrow{dA}} = -\frac{d}{dt}\iint_{\Sigma}{\overrightarrow{B} \cdot \overrightarrow{dA}} $$
Applying Stokes' Theorem: 
$ \hspace{1pt}$  for any vector field $  \overrightarrow{v},\hspace{5 pt} \iint_{\Sigma}{(\nabla\times \overrightarrow{v}) \cdot \overrightarrow{dA}} = \oint_{\eth\Sigma}{\overrightarrow{v}\cdot\overrightarrow{dl}}  \hspace{10 pt}$
to the LHS yields
$$ \oint_{\eth\Sigma}{\overrightarrow{E}\cdot\overrightarrow{dl}} = -\frac{d}{dt}\iint_{\Sigma}{\overrightarrow{B} \cdot \overrightarrow{dA}} $$
which is Faraday's Law.
So we see that Faraday's Law is mathematically equivalent to one of the four Maxwell Equations in differential form, specifically it is the corresponding integral form. Since the two forms are mathematically equivalent, it doesn't really make sense to say that one form is more fundamental than the other. However, I think that the differential form is a little more enlightening than the integral form because it directly describes the relationship between the E & B fields, without any reference to paths or surfaces.
Now, if what you really want to know is why the differential form itself holds, in other words, why it is that $\nabla\times \overrightarrow{E} = -\frac{d\overrightarrow{B}}{dt}$, then that is a much harder question, and I'm not sure whether anyone knows the answer to it.
A: First, your force equation is wrong, as you're missing the electric field. Wait what electric field? That's the point! A changing magnetic field induces an electric field $\nabla\times E=-\frac{\partial B}{\partial t}$, and this "pushes" the current. Note that the applied magnetic field is perpendicular to the circuit/wire, so that at least part of the electric field is along the wire.
A: Faraday's law could not be explained from basic principles at the time of discovery. In that sense, it doesnt have an explanation. It was incorporated as a new law of nature, and included in what today are maxwell's equations.
A: Maybe the inverse statement could be explained by a knowledge that had existed before 1830, that is, how a rotational (curled) electric field can produce a time varying magnetic field? 
Before proceeding further let me just mention that a curled velocity field (say, of a fluid) would produce a tiny particle (at a given point in space) to rotate with a constant angular velocity. Likewise, a curled force field would make a particle to rotate with an angular acceleration. Thus, a curled electric field E would make, at a given point in space, an almost dimensionless charge to rotate with acceleration. Consequentially, the outer layers of the charge would make a circular motion with an increasing angular velocity (and with an increasing ordinary velocity along that circular path). According to the old Ampere law (that is, without the displacement current term) that would give rise to an increasing magnetic field right through the center of the imagined (almost) point charge. Thus, curl E would be equal to dB/dt, only without the intertwining rotational charge. However, maybe even there is something charged in the plain vacuum that rotates in curled electric fields (eg., some sort of bound electron positron pairs, etc.)? Therefore, a guy with an imagination like Maxwell had could had come onto the idea about the Faraday law even before the Faraday actually discovered it in 1830 (by only relying on the old Ampere law).   
A: Faraday's :aw of electromagnetic induction is basically an experimental result. As per the Faraday's Law, an induced emf is generated in a coil or loop if the loop / coil has changing magnetic flux linkage. Mind there that flux linkage through the coil must change to have an induced emf. If the circuit is complete then induced current shall flow. For detail and simulation of Faraday's Experiment, you must visit Faraday's Law of Electromagnetic Induction
A: I don't have the math all worked out yet, but hopefully this gives an idea on where to go. As an earlier answer mentioned, it ties into one of Maxwell's equations, $\nabla \times \vec{E} = -\frac{\partial B}{\partial t}.$ This can be proved from the Coulomb's Law, relativity, and the wave equation.
Coulomb's law is an "axiom", we assume it's true because all our data shows two charged particles exert forces on each other according to the equation $$\frac{q_1q_2\hat{r}}{4\pi\varepsilon_0|\vec{r}|^2}.$$
For a charge density $\rho$, this gives an electric field due to that bit of charge of $$\vec{E} = \frac{\rho\hat{r}}{4\pi\varepsilon_0|\vec{r}|^2}.$$
Using relativity you can rederive the Biot-Savart law. Manipulating this law gives
$$\vec{B} = \frac{\mu_0\vec{J}\times \hat{r}}{4\pi |\vec{r}|^2} = \frac{\vec{J}\times \hat{r}}{4\pi\varepsilon_0c^2|\vec{r}|^2} = \frac{\vec{v}\times \rho\hat{r}}{4\pi\varepsilon_0c^2|\vec{r}|^2},$$
$\rho \vec{v} = \vec{J}$. I think you can guess where this is going now. We have
$$\nabla \times \vec{E} + \frac{\partial B}{\partial t} = \left[\nabla + \frac{\vec{v}}{c^2}\frac{\partial}{\partial t}\right]\times\frac{\rho \hat{r}}{4\pi\varepsilon_0|\vec{r}|^2}.$$
We just need to show the bracketed portion is zero. Dotting with $\nabla$ gives $$\nabla\cdot\nabla + \frac{\nabla \cdot\vec{v}}{c^2\partial t} = \nabla\cdot\nabla + \frac{\partial}{c^2\partial t^2}$$
which should be zero from the wave equation or relativity.
