Here is one way to think about it:
When a charged particle travels in a magnetic field, it experiences a force. If the particle is stationary but the field is moving, then in the frame of reference of the field the particle should see the same force.
Now let's take a conductor wound into a coil. In order to increase the magnetic field inside, I could take a dipole magnet and move it close to the coil. As I do so, magnetic field lines cross the conductor, and generate a force on the charge carriers.
It is a convenient trick for figuring out "what goes where" to know that the induced current will flow so as to oppose the magnetic field change that generated it. In the perfect case of a superconductor, this "opposing" is perfect - this is the basis of magnetic levitation. For resistive conductors, the induced current is not quite sufficient to oppose the magnetic field, so some magnetic field is left.
The point is that the flowing of the current is instantaneous - it happens as the magnetic field tries to establish in the coil. So it's not "Apply field in coil. Coil notices, and generates an opposing field. " - instead, it is "Start to apply field in coil. Coil notices and prevents field getting to expected strength".
Not sure if this makes things any clearer...
I assume what is meant by Faraday's law of induction is what Griffiths refers to as the "universal flux rule", the statement of which can be found in this question. This covers both cases 1) and 2), even though in 1) it is justified by the third Maxwell equation1 and in 2) by the Lorentz force law.
The universal flux rule is a consequence of the third Maxwell equation, the Lorentz force law, and Gauss's law for magnetism (the second Maxwell equation). To the extent that those three laws are fundamental, the universal flux rule is not.
I won't comment on whether the universal flux rule is intuitively true. But the real relationship is given by the derivation of the universal flux rule from the Maxwell equations and the Lorentz force law. You can derive it yourself, but it requires you to either:
- know the form of the Leibniz integral rule for integration over an oriented surface in three dimensions
- be able to derive #1 from the more general statement using differential geometry
- be able to come up with an intuitive sort of argument involving infinitesimal deformations of the loop, like what is shown here.
If you look at the formula for (1), and set $\mathbf{F} = \mathbf{B}$, you see that
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{a} &= \iint_{\Sigma} \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a} + \iint_{\Sigma} \mathbf{v}(\nabla \cdot \mathbf{B}) \cdot \mathrm{d}\mathbf{a} - \int_{\partial \Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\
&= - \iint_{\Sigma} \nabla \times \mathbf{E} \cdot \mathrm{d}\mathbf{a} - \int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell \\
&= -\int_{\partial \Sigma} \mathbf{E} + \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell
\end{align*}
where we have used the third Maxwell equation, Gauss's law for magnetism, and the Kelvin--Stokes theorem. The final expression on the right hand side is of course the negative emf in the loop, and we recover the universal flux rule.
Observe that the first term, $\iint_\Sigma \dot{\mathbf{B}} \cdot \mathrm{d}\mathbf{a}$, becomes the electric part of the emf, so if the loop is stationary and the magnetic field changes, then the resulting emf is entirely due to the induced electric field. In contrast, the third term, $-\int_{\partial\Sigma} \mathbf{v} \times \mathbf{B} \cdot \mathrm{d}\boldsymbol\ell$, becomes the magnetic part of the emf, so if the magnetic field is constant and the loop moves, then the resulting emf is entirely due to the Lorentz force. In general, when the magnetic field may change and the loop may also move simultaneously, the total emf is the sum of these two contributions.
If you are an undergrad taking a first course in electromagnetism, you should know the statement of the universal flux rule, and you should be able to justify it by working out specific cases using the third Maxwell equation, the Lorentz force law, or some combination thereof, but I can't imagine you would be asked for the proof of the general case from scratch, as given above.
The universal flux rule only applies to the case of an idealized wire, modelled as a continuous one-dimensional closed curve in which current is constrained to flow, that possibly undergoes a continuous deformation. It cannot be used for cases like the Faraday disc. In such cases you will need to go back to the first principles, that is, the third Maxwell equation and the Lorentz force law. There is no shortcut or generalization of the flux rule that you can apply. You should be able to do this on an exam.
1 This equation is also often referred to as "Faraday's law" (which I try to avoid) or the "Maxwell--Faraday equation/law" (which I will also avoid here because of the potential to cause confusion).
Best Answer
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.