You have derived emf already in the third equation. In the text after that, you seem to want to rephrase the result in terms of change of magnetic flux. That is possible, but you made an error in the step where you put equality in
$$
B \sin \theta \frac{d\mathbf S}{dt}= \frac{d|\mathbf B\times\mathbf S|}{dt}.
$$
The angle $\theta$ is between $\mathbf B$ and $\mathbf v$, not between $\mathbf B$ and $\mathbf S$, so the vector product is not correct there. Vectors $\mathbf B$ and $\mathbf S$ are actually parallel, so the dot product is already there.
An EMF from a source is defined as a force per unit charge line integrated about the instantaneous position of a thin wire so for an electromagnetic source:
$$\mathscr E=\oint_{\partial S(t_0)} \left(\vec E + \vec v \times \vec B\right)\cdot d \vec l.$$
Where $S(t_0)$ is a surface enclosed by the wire at time $t=t_0$ and the partial means the boundary, so $\partial S(t_0)$ is the instantaneous path of the wire itself at $t=t_0.$ The $\vec v$ is the velocity of the actual charges. Note this is not necessarily the work done on the charges if the wire is moving since the wire goes in a different direction than the charges go when there is a current.
Now, if the wire is thin and the charge stays in the wire and there are no magnetic charges we get $$-\oint_{\partial S(t_0)} \left(\vec v \times \vec B\right)\cdot d \vec l=\frac{d}{dt}\left.\iint_{\partial S(t)}\vec B(t_0)\cdot \vec n(t)dS(t)\right|_{t=t_0}$$
And regardless of magnetic charges or thin wires or whether charges stay in the wires we always get $$\oint_{\partial S(t_0)} \vec E\cdot d \vec l=\iint_{S(t_0)}\left.-\frac{\partial \vec B(t)}{\partial t}\right|_{t=t_0}\cdot \vec n(t_0)dS(t_0).$$
So combined together we get:
$$\mathscr E=\oint_{\partial S(t_0)} \left(\vec E + \vec v \times \vec B\right)\cdot d \vec l=-\left.\left(\frac{d}{dt}\Phi_B\right)\right|_{t=t_0}$$
The force due to the motion of the wire is purely magnetic, and the force due to the time rate of change of the magnetic field is purely electric. And the work done is an entirely different question than the EMF. The work happens for a motional EMF when a Hall voltage is produced.
So,is the former case of when the loop moves in a stationary magnetic field different?
A moving wire feels a magnetic force and magnetic forces can be a source term in an EMF.
Is electric field in the loop due to "motional emf" conservative?
Motional EMF is not caused by electric forces, it is caused by magnetic forces. Since magnetic forces depend on velocity, the word conservative does not even apply since the force depends on the velocity, not merely the path, and they don't do work.
And the book also,at one point, expresses electric field due to motional emf as a scalar potetnial gradient.
If the wire develops a Hall voltage due to the magnetic force, then the charge distribution for the Hall voltage would set up an electrostatic force, which is conservative.
In particular, if the magnetic field is not changing, then the electric field is conservative.
However,motional emf does sounds similar to induced emf.
When you compute the magnetic flux at two times the term $-\vec B \cdot \hat n dA$ can change for two reasons, a changing loop and a time changing magnetic field. You really get both effects from the product rule for derivatives. The one from the time changing magnetic field becomes equal to the circulation of the electric force per unit charge. The one from the time changing loop becomes equal to the circulation of the magnetic force per unit charge.
My question is,is E due to motional emf and induced E different or not,and why so?
The electric field is conservative if the magnetic field is not changing in time. And if the magnetic field is not changing in time, the EMF is due solely to the moving charges in the moving wire interacting with a magnetic field.
Best Answer
There are two things that lead to your confusion:
1) the integral $\oint_C q(\mathbf v\times \mathbf B)\cdot d\mathbf s = vBl$ along the circuit $C$ is not $work$ of magnetic force; it is called electromotive force of the circuit (emf). This is because $\mathbf v$ in this expression is not the velocity of the charged particle, but that of the element of conductor. The charged particle moves in a more complicated way.
The work of magnetic force per unit charge per displacement $d\mathbf r$ would be $$ (\mathbf u\times \mathbf B)\cdot d\mathbf r $$ where $d\mathbf r$ is displacement of the charged particle and $\mathbf u$ is its velocity. Since $\mathbf u = \frac{d\mathbf r}{dt}$ is parallel to $d\mathbf r$, the above product vanishes and the magnetic force does not work on the charged particle.
2) if magnetic force does not work on the charges, then why there is current and evolution of heat in the conductor in the first place?
In macroscopic theory, the description of such situations where current is due to nonelectrostatic force (such as current in the battery, or due to thermal gradients, or due to magnetic field, as in the rod) is via general notion of electromotive intensity $\mathbf E^*$. This is not electric field, but has the same units and its meaning is that the total force acting on the macroscopic charge is
$$ q(\mathbf E + \mathbf E^*). $$
One can derive approximate condition for $\mathbf E^*$ for the case of perfect conductor; the total force acting on the charge has to be zero and
$$ \mathbf E^* = - \mathbf E. $$
Various circumstances lead to various expressions for $\mathbf E^*$ in terms of other measurable quantities, for example in the Seebeck effect $\mathbf E^* = -k \nabla T$. In the case of magnetic forces, experiments are well explained by assuming that
$$ \mathbf E^* = \mathbf v \times \mathbf B, $$ but explaining why this works from the point of view of observer in the laboratory frame seems hard. It would require going into some microscopic theory of conduction and taking into account both the velocity of the rod $\mathbf v$ and the velocities of the individual charge carriers $\mathbf u_i$.
However, the above choice for $\mathbf E^*$ is well explained by the theory of relativity, if we look into the conduction from the point of view of obsever on the rod. There the electric field is composed of two parts, the electric field due to charges of the rod $\mathbf E_0'$ which from the theory of relativity is almost equal to $\mathbf E$, and the electric field due to external bodies $\mathbf E'_{ext}$, which from the theory of relativity is approximately equal to $\mathbf v \times \mathbf B$. So the total electric field in the frame of the rod is approximately $\mathbf E' = \mathbf E + \mathbf v\times \mathbf B$ and putting $\mathbf E' = \mathbf 0$ for ideal conductor of the rod and using the Ohm law for the rest of the circuit (resistance $R$), we can derive the required current $vBl/R$ and can motivate why $\mathbf E^* = \mathbf v\times\mathbf B$ works in the original frame.