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. 395
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. 418