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).
"I think that magnetic flux through conductor remains constant as B is constant."
It's not the flux "through the conductor" that matters. It's the flux through the area swept out by the conductor. Imagine that the straight conductor (length $\ell$) is lying on a table, and that there is a uniform magnetic field acting downwards. (Actually there is : the vertical component of the Earth's field.) You then move the conductor across the table at speed v in a direction at right angles to itself. In time $\Delta t$ it sweeps out an area $\ell v \Delta t$
The flux through the swept out area is $$\Delta \Phi = (\ell v \Delta t)B$$
So according to Faraday's law, the induced emf is
$$\mathscr E=\frac {\Delta \Phi}{\Delta t}=\frac {(\ell v \Delta t)B}{\Delta t}=B\ell v$$
So we have recovered the result that you obtained from the magnetic Lorentz force. In my opinion the magnetic Lorentz force is more fundamental than Faraday's law when the emf is due to movement of conductors. However Faraday's law has the merit of spanning two types of electromagnetic induction: this one and the type due to changing flux through a stationary circuit, which depends on the electric field part of the Lorentz force.
Best Answer
yes, with can show it with Maxwell-faraday equation: $$∮_cE.dl= -\frac{d}{dt} ∬_sB.ds$$
We can suppose a contour with rectangular shape that our wire is one side of it. the wire side is moving and other side is constant so the area of contour is growing with V speed and if we write Maxwell-Faraday equation for this contour we have:
$$
emf=- \frac{d}{dt} ∬_sB.ds= -\frac{d}{dt} ∬_sB.dx.dy= -BL \frac{d}{dt} ∫dx=-BL \frac{d}{dt} (x)=-VBL $$