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A book i am referring to states:

"..When the loop moves toward the magnet,it is the magnetic force which drives the charge to flow.But what causes the induced current in a stationary loop when a magnet moves towards it?A magnetic field cannot exert force on a stationary conductor.Whenever a magnetic field is varying with time, an induced Electric field E is produced in any closed path in matter or empty space"

Then it goes on to state that this induced emf is non conservative because of a non-zero value of circular integral of E.dl.

So,is the former case of when the loop moves in a stationary magnetic field different? Is electric field in the loop due to "motional emf" conservative? And the book also,at one point, expresses electric field due to motional emf as a scalar potetnial gradient.

However,motional emf does sounds similar to induced emf. My question is,is E due to motional emf and induced E different or not,and why so?

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    $\begingroup$ Motional EMF is due to the magnetic field acting on the moving charges in the moving wire. Induced EMF is due to an actual nonconservative electric field that really exists even in empty space when magnetic fields are changing in time. They are as different as can be, and the electric ones always satisfy the rule for stationary wires, but if you want a corresponding Faraday's law for motional EMF you have to only apply it to thin wires that van keep the charges in the wire. It is a sometimes rule, not an always rule. The magnetic force can lead to a charge imbalance on moving wires. $\endgroup$
    – Timaeus
    Commented Sep 24, 2015 at 21:20
  • $\begingroup$ And that charge imbalance can cause a conservative electric field, but that field is just keeping the charges inside the wire. Now since the wire is moving you can do work on something and keep it in the wire. But an EMF is not defined as the work done per unit charge. In statics it happens to equal the work per unit charge, but that isn't the definition. $\endgroup$
    – Timaeus
    Commented Sep 24, 2015 at 21:21
  • $\begingroup$ As Timaeus said, and worth emphasizing that motinal emf is NOT by a motinal electric field. Many students confused on that. $\endgroup$
    – Jaseon Quanta
    Commented Mar 20, 2022 at 15:23

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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.

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Let's restate Faraday's Law of Induction carefully in integral form:

$\varepsilon = -\frac{d}{dt} \int_S B \bullet da = - \frac{d \Phi}{dt} $

Where C is a closed curve, and S is any smooth surface whose boundary is C.

So as you can see, no matter how the generation of electric field is interpreted (take the two scenarios in your question), the motional emf $\varepsilon$ is NEVER 0 unless the magnetic flux is not changing. In both of your scenarios, the magnetic flux is changing and in fact at the SAME rate! That is how they produce the same effect.

As Timaeus pointed out, in the case of stationary loop, there seems to be an electric field generated in the loop that drives the current. While in the case of moving loop, the magnetic force does the job. However, these interpretations are two sides of the same coin: relativity. From a relativistic point of view, magnetic fields are really born from principles of relativity + electric field. What is regarded as electrical effect in the stationary loop case transforms into a magnetic effect in the moving loop case. But since the two scenarios in your question are really just a switch between two inertial frames, there shouldn't be any difference in the outcome of the experiments (it would be weird if there were... Einstein would be shocked out of his grave). From that perspective it is also clear that the emf $\varepsilon$ should be the same in both scenarios.

So motional emf and induced emf are really just two descriptions of the same effect which is summarized by Faraday's law of induction. The fact that two views give the same result is a manifestation of the relativistic nature of magnetic fields.

For more information on deriving magnetism from electricity, and on field transformation laws, I refer you to chapters 5 and 6 of Edward Purcell's E and M textbook: http://www.amazon.com/Electricity-Magnetism-Edward-M-Purcell/dp/1107014026/ref=sr_1_2?ie=UTF8&qid=1443095756&sr=8-2&keywords=purcell+e+and+m

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  • $\begingroup$ When the loop moves in a constant magnetic field the emf around the loop is is due to the magnetic force on the moving charges (in the thin wire of the loop) as the loop moves. When the loop is stationary and the magnetic field changes the emf is due to the electric force from the nonconservative electric field that is associated with the changing magnetic field. The equation you wrote is for the case of the stationary loop. And an emf is not defined as the line integral of the electric field, if it were then you wouldn't get an emf in the case of a thin loop moving a uniform B field $\endgroup$
    – Timaeus
    Commented Sep 24, 2015 at 14:40
  • $\begingroup$ The equation I wrote works for either case I believe. In the stationary loop case, the magnetic flux changes because the moving magnet produces a changing magnetic field at the location of the loop. In the moving loop case, the magnetic flux changes because the loop travels through space, where magnetic field varies from point to point. And I believe emf is well defined as the line integral of E field in both cases. Regarding your last line, when there is a uniform B field, there is INDEED no emf because the magnetic forces on different charge carriers would cancel out. $\endgroup$
    – Zhengyan Shi
    Commented Sep 24, 2015 at 19:06
  • $\begingroup$ You believe wrong. The EMF is not defined as the line integral of the electric field. And the magnetic field can be both uniform and constant and a moving/deforming thin conducting loop can still feel an EMF. If you have a uniform B field and a conductor on a conducting rail that completes a conducting loop then you get an EMF as the area enclosed expands. If the conductors are thin enough (and the charges are compelled to stay in the conductor) then the EMF equals the change in magnetic flux. But that isn't the definition of an EMF. $\endgroup$
    – Timaeus
    Commented Sep 24, 2015 at 19:12
  • $\begingroup$ An EMF is the line integral of the force per unit charge in the instantaneous direction of the loop. The force can include magnetic forces when the loop is moving because then you can have a magnetic force in the direction the wire is instantaneously pointing. So magnetic field now and a moving wire, or current position of wire and time rate of change of magnetic flux. You can even say the electric EMF is proportional to the flux of $\partial \vec B/\partial t$ because it is true, and the magnetic EMF is solely due to magnetic field now and the velocity of the wire now. $\endgroup$
    – Timaeus
    Commented Sep 24, 2015 at 19:16
  • $\begingroup$ Oh sorry about the uniform B-field thing. I thought you were referring to globally uniform B-field in which no emf is generated... You are absolutely right in making the distinction between E and B fields clear in two reference frames, but my point was that fundamentally these fields are connected (by the Faraday tensor) in relativity and the distinction is really just an apparent effect. I fixed my answer based on some of the confusions you pointed out. Thank you very much! $\endgroup$
    – Zhengyan Shi
    Commented Sep 24, 2015 at 20:30

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