Electric fields produces a difference of potential on two points with different distances of the field source. Magnetic fields induces current on a closed loop if the loop is not on parallel in relation of the lines of field and the magnitude of the field does have to change (you have to have a flux).
If you have a magnetic field interfering on your circuit, you will have a constant inducted current flowing on it (which depends of its magnitude, the radius of your circuit and the tax of change of your field). So, the voltage drop will only depend of where is the point you are measuring, since it will depends of only the resistance between that two points. (Ohm's law - constant current)
If you have an electrical field interfering on your circuit, you will have a change of voltage on your circuit which will depends of the distance of the electrical field and also the distance between two points. So you have only to know the magnitude of the field at the two points you chose and the distance between them.
To consider both effects you have only to sum them.
God already stated all EM effects on four simple equations.
Also, we have an Electrical Engineering SE on which probably you will get more (and better) answers of related questions.
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
You're right. When you have a conservative $E$ field, you can define an electric potential $V$. And when you don't, you can't define such a potential (just as you can't define a potential energy $U$ for a nonconservative force $F$ - nonconservative forces still do work - but there is no associated potential energy function for such forces).
General Remarks
Electric potential is measured in volts and defined by $$V(x) = -\int_{\mathcal{O}}^x \vec{E}\cdot d\vec{l} \tag{1}$$ A voltage is defined by (in the textbooks) as a difference in potential $$\Delta V = V_b - V_a = -\int_a^b \vec{E}\cdot d\vec{l} \tag{2}$$
However this is too restrictive. In general, anything that takes the form $$\int \frac{\vec{F}}{q}\cdot d\vec{l} \tag{3}$$ can be called voltage and is measured in volts. So yes $\text{(2)}$ above is a voltage and even $\text{(1)}$ is a voltage if you like (as $V(x)$ is secretly $V(x) - V(\mathcal{O}) = V(x) - 0$ and therefore a difference in potential). But note that $\vec{F}$ can be anything. It doesn't have to be a conservative electric force. Again, anything that takes the form of $\text{(3)}$ is measured in volts and can be called a voltage. EMF takes the form $\text{(3)}$ [I explain this further down]. So too does potential difference. This is one reason why EMF and potential difference get mixed up: they take similar forms and hence both are measured in volts and can be called voltage (or induced voltage or whatever). And actually, this is great. If you are doing anything in the lab or talking to engineers, it doesn't really matter whether voltage means potential difference or EMF. It does, but all we really care about is energy. Voltage (equation $\text{(3)}$)is energy. EMF and potential difference/potential are energy. Energy is energy, whether it be EMF or potential difference. Voltage is just a general term for energy. If you are ever in a situation where you don't know whether to say "the EMF is 5 volts" or the "potential difference is 5 volts", just say "5 volts" or "the voltage/induced voltage/whatever is 5 volts" and you are safe.
What is EMF
In order to get current to flow around a circuit, you need some force pushing charges around the wire. Let's call this the driving force $\vec{F}$. EMF $\mathcal{E}$ is defined as
$$ \mathcal{E} = \oint \frac{\vec{F}}{q}\cdot d\vec{l} \tag{4}$$ where the integration is taken around the loop. There are two main forces that drive current around a circuit: a "source" force from say a battery and a conservative electric field which pushes charges around the wire. $\vec{F} = \vec{F}_s + q\vec{E}$. Therefore, $\text{(4)}$ can be written
$$\mathcal{E} = \int_a^b \frac{\vec{F}_s}{q}\cdot d\vec{l} $$ as $\vec{F}_s$ is usually confined to a section of the loop and $E$ is conservative so it integrates to 0 (started where we left off - once around the loop). $\vec{F}_s$ can be anything. It can be a chemical force, some temperature gradient thing, pressure on a crystal, a magnetic force, a nonconservative $E$ field, etc. So consider a battery. A conservative electric field goes from the positive terminal, around the loop, to the negative terminal, as well as from the positive terminal to the negative terminal inside the battery. Using the last equation, assuming the battery is ideal so that the chemical force is equal and opposite to the electric force,
$$ \mathcal{E} = \int_a^b \frac{\vec{F}_{\text{chemical}}}{q}\cdot d\vec{l} = -\int_a^b \vec{E}\cdot d\vec{l} = V$$ The EMF of the battery is equal to the potential difference across its terminals. But this does not mean that EMF is potential difference. It just happens to be so in this case. Most simple circuits turn out to be this way but realize again that EMF and potential difference are totally different. In the first place, you can't have a potential difference without an EMF generating that separation of charge in the battery. EMF generates a potential difference that happens to match the numerical value of the EMF (which you can think of as energy conservation). Then if you have a resistor connected to this battery, current $I = V/R$. I could also say $I = \mathcal{E}/R$ as they are numerically equivalent. But it's more appropriate in my opinion to use $I = V/R$ as the energy drop is coming as electric potential energy in a conservative $E$ field (which exists throughout the wire doing the pushing). Here we begin to see, as in the next section, that All circuits require an EMF to function.
Let's Look at your example
There is no such thing as an electric potential in your example. There's an $E$ field, but it's not conservative. Therefore, don't say potential. There is an EMF however. You can say there's a voltage or an induced voltage if you like (from the above discussion). But there's definitely not a potential difference/potential present. For this specific example, the EMF is given by
$$ \mathcal{E} = -\frac{d\phi}{dt} = \oint \vec{E} \cdot d\vec{l}$$ where the driving force is that nonconservative $E$ field. The current as you say is $I = \mathcal{E}/R$. Here again we see a true instance of the following: all circuits require an EMF. The idea that all points in a conductor are at the same potential is equivalent to saying that there is no $E$ field in a conductor. Note that this idea of there being no $E$ field in a conductor only holds for electrostatics + no time varying external magnetic fields (having a time varying $B$ prevents electrostatics anyways so saying "+ no time varying $B$" was redundant). Having an $E$ field in a conductor is completely fine. Turn on an $E$ field in a conductor. There is definitely an $E$ field in the conductor until electrostatics is reached. This is why conducting wires in simple circuits can have $E$ fields in them. This $E$ field is essential for driving current around even though it's through a conductor. It's just that the conductor can never reach electrostatics when it's part of a circuit. It desperately tries, but the battery prevents the wire from coming to a static situation. And with time-varying $B$ fields, nothing wrong with having an $E$ field in a conductor. When you stop varying $B$, you'll stop changing the $E$ field and things will reach statics. While you are changing the $B$ field, $E$ is changing with time. The conductor is trying to reach statics, but can never do so. So there will always be an $E$ present and hence the conductor won't be an equipotential (albeit, in simple circuits, you can take the wires to be equipotentials because $E$ is so so tiny).
Too Much Theory, What to know about EMF
From equation $\text{(4)}$, because of the closed line integral, EMF does not care about conservative forces while electric potential crucially depends on a conservative E field. EMF and potential are both instances of equation $\text{(3)}$, and therefore both tell you about energy. Potential is energy in a conservative E field. EMF is energy added to your circuit through "nonconservative" driving forces. In order for circuits to work, you need to pump energy into them so that charges will flow back down to low energy, making a circuit. EMF tells you how much energy driving forces give to a unit charge in one trip around the loop. Conservative forces don't give any net energy to a charge after one complete loop (started where you stopped). "Nonconservative" forces will give you some nonzero value to equation $\text{(4)}$. EMF tells you how much energy was added by driving forces and hence how much energy must be dropped by dissipative/"friction" forces in one trip around the loop. An EMF of 5 volts means 5 volts must be dropped by every unit of charge. If you have a battery providing 2 volts of EMF and a changing magnetic field providing 6 volts of EMF, 8 volts must be dropped
[by the way, If you ever see an example circuit out there with both a battery and a changing flux enclosed by the loop, more than likely their derived equations have wrong explanations. Right equation. Wrong explanation (which is basically just as bad as not knowing what you are doing). What they do is say $-d\phi/dt = \oint \vec{E}\cdot d\vec{l}$. This is Faraday's Law, true in anycase. But what actually does that $\vec{E}$ mean? It's the net $E$ field on your loop. In the case of a battery and a changing flux, that $E$ has both a conservative and a nonconservative component. The conservative component integrates to zero, leaving only the nonconservative $E$ providing the $-d\phi/dt$. Therefore, if you have this simple battery + changing flux + resistor circuit, $-d\phi/dt = I_{\phi}R$ where $I_{\phi}R$ is the integral of nonconservative $E$ around the loop. Now we can add a constant to each side of the equation. Since EMF battery $V_0 = I_0R$, we can say $V_0 - d\phi/dt = (I_{\phi} + I_0)R = IR$. Or if you want, you can write out $-d\phi/dt = I_{\phi}R + \oint \vec{E}_{\text{conserv}} \cdot d\vec{l} = I_{\phi}R - V_0 + I_0R$].