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
Comment answer:
There is a subtle difference in Faradays law compared to the maxwell-faraday quation( you may not have known there are 2 to begin with)
Faraday law:
$\epsilon = -\frac{d\phi_{B}}{dt}$
$\epsilon = -\frac{d}{dt}\int \vec{B} \cdot \vec{da}$
describes both motional emf and transformer emf, ANY change in magnetic flux on a surface causes an EMF. this doesn't necessarily say whether or not the EMF is caused by an induced electric field, or if it is caused by the magnetic lorentz force. This equation COULD give rise to an emf( changing magnetic flux), due to a change in the boundary of the surface in question, or due to a changing B field($\frac{\partial \vec{B}}{\partial t})$ within that boundary
Whereas the maxwell-faraday equation $\int \vec{E} \cdot \vec{dl} = -\int \frac{\partial \vec{B}}{\partial t} \cdot \vec{da}$
only describes transformer emf, aka an emf caused by an induced electric field as a result of a changing magnetic field
In this situation there is only motional emf, as there is a NON changing magnetic field as its the field of a static wire. But, there is a moving loop, aka a changing magnetic flux caused by a change in the boundary of my surface. In this situation there is NO induced electric field. The EMF generated in this situation is not described by the maxwell-faraday equation. It is instead described by faradays law. With the physical mechanism for this EMF being the magnetic lorentz force on the charges in the boundary of my surface.
To find the emf I could compute the changing magnetic flux in my loop using FARADAYS law, to find the EMF. Or... as you have suggested. I Could use the lorentz force to calculate seperately the line integrals of all of the sides to find the total emf in the loop
Where
$\epsilon_{i} = \int_{a}^{b} (\vec{V}×\vec{B} ) \cdot \vec{dl}_{i}$
To mathematically understand why there is a difference between faradays law and the maxwell faradays equation. Is that the time derivative is in the inside of the integral. This forces the equation to completely ignore any change to the surface boundary