Deduce $\mathcal{E} = B l v$ by directly solving Maxwell's equations Consider the following situation:

We know this is a classic example frequently used when teaching Faraday’s Law, and the voltage/electromotive force (emf) $\mathcal{E}$ induced across terminals of the moving conductor (lighter gray bar in the picture) is
$$\mathcal{E} = B l v$$
where $B$ is the uniform magnetic field perpendicular to the area, $l$ is the length of the conductor and $v$ is the speed of conductor relative to the field. So far so good, but…
How can I deduce the formula by directly solving Maxwell's equations
$$\begin{aligned}
\nabla \cdot \mathbf{D} &= \rho_\text{f}\\
\nabla \cdot \mathbf{B} &= 0\\
\nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}} {\partial t}\\
\nabla \times \mathbf{H} &= \mathbf{J}_\text{f} + \frac{\partial \mathbf{D}} {\partial t}
\end{aligned}$$
?
By directly I mean calculate the electric field $\mathbf{E}$ first (by solving Maxwell's equations with proper initial conditions (i.c.) and boundary conditions (b.c.), using techniques taught in partial differential equation (PDE) course), and then calculate emf using the definition
$$\mathcal{E}=\oint_{C} \mathbf{E} \cdot \mathrm{d} \boldsymbol{ l }$$
 A: Starting from $\vec{\nabla}\times\vec{E} = -\partial \vec{B}/\partial t$, take a surface integral of both sides to find
\begin{align}
\iint_{S(t)} \vec{\nabla}\times\vec{E}\cdot d\vec{S} &= \iint_{S(t)} -\frac{\partial \vec{B}}{\partial t}\cdot d\vec{S}\\
\therefore\quad \oint_{\partial S(t)} \vec{E}\cdot d\vec{l} &= -\frac{d}{dt} \iint_{S(t)} \vec{B}\cdot \hat{S}\, dS - \oint_{\partial S(t)}\vec{v}\times\vec{B}\cdot d\vec{l}
\end{align}
where the LHS of the second line makes use of Stokes' theorem, while the RHS makes use of the Leibniz integral rule.
$\vec{B}\cdot \hat{S}=-B$ (magnetic field points into the page, while take area vector to point out) and $\iint_S dS = A(t) = l x(t)$ is the area of the (rectangular) surface enclosed by the loop, where $x(t)$ is the length of the sides that is varying with time.
So:
\begin{equation}
\mathcal{E} = \oint_{\partial S(t)} (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} = -\frac{d(-Blx(t))}{dt} = Bl\frac{dx(t)}{dt} = Blv
\end{equation}
N.B. emf is defined as the work done per unit charge,
\begin{equation}
\mathcal{E} = \frac{dW}{dq} = \frac{d}{dq}\int \vec{F}\cdot d\vec{l}=\frac{d}{dq}\int q(\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l} =\int (\vec{E}+\vec{v}\times\vec{B})\cdot d\vec{l}
\end{equation}
So as Puk has pointed out in the comments, this relies on knowing the Lorentz force. See e.g. Wikipedia for a discussion.
A: My answwer is you can't.
The maxwell equation $\nabla \times \bf E = -\partial \bf B/\partial t $
must be modified to
$\nabla \times \bf E = -\partial \bf B/\partial t + \nabla \times (\bf v \times \bf B) $. (cf. H.H. Skilling, "Fundamentals of Electric Waves", p. 82 eq. 194 )
The last term is based on the Lorentz force $ q \bf v \times \bf B $.
You have to be careful to associate time-varying areas with moving media.  Sometimes it doesn't work.  I have a WORD file showing such a situation but I'm a newbie at this forum & don't know how to upload that file as part of this answer.
A: The motional EMF in a conductor moving in magnetic field is really due to atomic lattice working on electrons, as these get pushed out of equilibrium into the lattice by external magnetic forces. It is a result of microscopic constraint forces keeping the electrons inside the conductor. So this is all about microscopic forces and Maxwell's equations for field are largely irrelevant - we already know the external magnetic field.
One can try to formulate microscopic theory of lattice and mobile electrons and of their interactions among themselves and with the external field. This would include Lorentz forces between all particles, external magnetic force, and forces of constraint, keeping the electrons bound to conductor. This is a hard problem.
Much easier is to visualize what happens in the inertial frame of the conductor. Using Lorentz transformations on the field tensor $F$, we find out that in the frame of the conductor there is external electric field perpendicular to both the magnetic field and to velocity of the conductor. Its magnitude is $Bv$ and it acts as electromotive agent on the mobile electrons. For conducting rod in the scenario you depicted, net work of the electric force per unit charge transported along the rod is $Blv$.
