# PRoblem with differential form of Maxwell's third equation?

1. May 11, 2006

### leright

Why does it seem as if the standard differential form of Maxwell's third equation (Faraday's Law) for time varying fields not take into account motional EMF. The differential form simply says that the curl of E is equal to minus the time rate of change of B field. However, there could be a curl of E even in cases where B is constant with a time varying loop in a static B field, so the differential form fails in motional emf, correct? Why isn't the differential motional emf term added to the right hand side of Maxwell's third equation (in differential form)? If it were added in, Maxwell's third equation would work in all cases.

The integral form of Maxwell's third equation works fine in all cases, as long as the derivative is kept OUTSIDE of the flux integral. When it is brought in then there is trouble, and clearly two different answers will be obtained for the emf when comparing the derivative being inside of outside the integral (when the derivative is only applied to the B-field, and not ds). I feel as if this is a flaw in the logic in the derivation of the differential form of the third equation.

This is not independent research. I am just looking for answers to a question.

Thanks.

2. May 11, 2006

### leright

Any opinions?

He said that the differential motional emf term that I added was just an approximation at low velocities. Also, when Maxwell formulated the third equation he assumed the moving charge to be the reference frame. However, it seems that it wouldn't be too difficult to add in a term that describes the motional emf at a reference frame other than the bar.

The motional EMF term I added was integral(u x B)*dl. Stokes theorem can be applied to obtain integral(curl(u x B))*dS Therefore, the term curl(u x B) can be added to the right hand side of the differential form of the third equation.

Last edited: May 11, 2006
3. May 11, 2006

### vanesch

Staff Emeritus
No, the differential form doesn't fail, because BY DEFINITION, you have to calculate the curl (or the circular integral in the integral formula) in a time-fixed loop, that is, a curve which is time-independent wrt the coordinate frame in which you are expressing the E and B fields.

Why this apparent paradox ? The reason is that the E and B fields do not transform as vectors when you switch from a coordinate frame to another, moving frame (in Lorentz speak, a boost).
E and B aren't really vectors in fact (they only behave as vectors under translations and rotations of coordinates, but not under boosts). They are really elements of a 2-tensor.
So if you want to consider a "moving curve" in frame A, then you should first go to the frame B that is static wrt the curve, and then transform your E and B field from frame A into frame B. And now comes the miracle:

A pure B field in frame A transforms into a B field PLUS an E field in frame B. The extra E field is exactly what corresponds to the EMF.
Indeed, the B-field will be "moving" but that is just "changing" in the B frame, and hence there will be a changing B field and thus a curl of the E-field in the frame B.

4. May 11, 2006

### leright

I know that the curl is only calculated in a time fixed frame and I realize that if the derivative of B field in a time fixed frame is 0 then the B field does not induce an E field, and if the derivative of B field in a time fixed frame is NOT 0 then the B field DOES induce an E field. My issue is that an e-field CAN be induced by a constant B-field in the event of the motion of free charge (non-free space conditions) in the field.

I think that the commonly used differential form of Maxwell's equation assumes free space conditions where there is no moving free charge.

I also understand that the commonly used differential form of Maxwell's equation assumes that if there IS moving free charge the reference frame should be taken to be the free charge itself, and this is why Maxwell's equation in differential form "works", but it simply does not convey all possible imformation.

5. May 11, 2006

### leright

message has been deleted.

Last edited: May 12, 2006
6. May 11, 2006

### Hurkyl

Staff Emeritus

EMF is being induced -- not an E-field.

In the case of the time-varying magnetic field, the reason EMF gets induced is because the time-varying magnetic field induces an electric field, and that electric field starts pushing your (stationary) charges around.

In the case of motional emf, what happens is that the magnetic field itself starts pushing your (moving) charges around. It has absolutely nothing to do with the electric field.

In fact, my textbook specifically says that the integral form:

$$\oint \vec{E} \cdot d\vec{l} = - \frac{d \Phi_B}{dt}$$

is "valid only if the path around which we integrate is stationary."

7. May 11, 2006

### leright

In order for there to be an induced EMF (capacity to do work on charges) there MUST be an corresponding e-field. In order for there to be an induced EMF that can drive current around a closed path there MUST also be a "non-conservative" e-field produced, and the only way a non conservative e-field can be produced is from the energy stored in a b-field. The b-fields themselves do not do work. The energy in the b-fields INDUCE e-fields that can do work, and work over a closed path at that (conservation of energy is not violated since the energy allowing the e-field to do work over the closed path is coming from the b-field)

And in motional EMF, the magnetic field is not doing the work. The magnetic field is storing the ENERGY to produce an e-field that does the work. The force on a moving charge or current element is due to an induced e-field which comes from the energy in the b-field.

This is all getting more and more clear to me, and the more I think about it, the more things make sense. Before, the more I thought about it and the deeper I went, the more confused I got. I've reached the point where the deeper I get and the more I think about it, the more clearly things become.

Just because something has energy in some form does not mean that it itself can do work, even though energy is defined as the capacity to do work. The energy has to be converted to other forms of energy that can do the work. For instance, b-fields themselves cannot do work because the force is ALWAYS perpendicular to the direction of motion, and the b-field itself cannot increase the kinetic energy of a charge. However, the energy in the b-field induces an e-field which does the work for it. So, even if the energy itself cannot do work that doesn't imply that it isn't really energy. It can be converted to other forms of energy which can the work. So b-fields aren't DOING work, but they are facilitating work by inducing e-fields.

Last edited: May 11, 2006
8. May 11, 2006

### Hurkyl

Staff Emeritus
Work it out yourself:

(1) Electric field that is zero everywhere.
(2) Constant magnetic field.
(3) Conducting wire.

When the wire starts to move, you have moving charges in a magnetic field. Thus, those charges start to move along the wire, and you get current, even though the electric field is zero, and the magnetic field is constant.

Your entire line of thought is based on the fact that magnetic forces cannot do work, correct? So all that means is that something else is doing the work. Why do you assume it's the electric field? There's a much simpler explanation that doesn't involve rejecting established physical laws!

I wonder how that wire started moving...

9. May 11, 2006

### leright

ok, the b-field still is not doing WORK on the charges in the rod. When the rod moves in the constant magnetic field motional emf occurs. Motional emf is caused by an e-field that pushes the charges and that e-field is induced by the magnetic field. I am not rejecting any physical laws.

It is an electric field that is induced perpendicular to the magnetic field that causes the emf. The force is perpendicular to the magnetic field so it is not doing work...the induced e-field is doing the work, but the energy to produce the e-field is coming from the b-field.

Why is the force due to a magnetic field called Fsubm and when you divide Fsubm by Q it's called Esubm? Perhaps because it is an e-field doing the work. Also, take a look at the lorentz transformations (relationships between E and B fields with relativistic considerations) and you'll notice a UxB term when getting the e-field from b-field. The UxB is the force due to the B-field.

http://farside.ph.utexas.edu/teaching/em/lectures/node123.html - look at the determination of the perpendicular e-field due to the b-field.

Last edited: May 11, 2006
10. May 12, 2006

### Galileo

A moving (conducting) loop in a constant magnetic field (like a generator) will induce an EMF, but not because of Faraday's law. It's simply because in the loop you now have moving charges in a magnetic field which are pushed along by the Lorentz force and this causes the current. The magnetic force is not doing work ofcourse, but the person pulling the loop is.

11. May 12, 2006

### arunbg

I believe Galileo is right, the phenomenon is similar to Hall's effect which is attributed to Lorentz's forces.

However what I don't understand is that in many cases such as a rod rotating in a uniform magnetic field the induced emf across the ends of the rod can also be calculated by the eqn for motional "induced" emf
E = Blv (it is done so in my text) even though there is no change in net flux.I calculated the answer independently using lorentz forces and got the exact same answer which seemed the right approach to me.

Does this imply some kind of relation or is this simply a freak coincidence?

12. May 12, 2006

### leright

It is because of Faraday's law. Faraday's law says that the induced EMF is equal to the time rate of change in the flux through a conducting loop. This is what Faraday's law says. You can produce a change in the magnetic flux by changing the area of the loop, even in a constant B field.

The induced EMF can be because of a number of physical phenomena, but all of the physical phenomena is contained in Faraday's law.

13. May 12, 2006

### arunbg

There is no time rate of change of flux through a fixed conducting loop moving in a uniform magnetic field.

14. May 12, 2006

### Hurkyl

Staff Emeritus
My book doesn't use that notation, so I don't know what you mean by F_m and E_m.

But we're not doing a change of reference frame, so the Lorentz transformations are irrelevant. :tongue:

The Lorentz transformations say that the EMF induced by motion through a pure magnetic field in one frame has to have the same effect as the EMF induced by the electric field in some other reference frame where there is no motion.

But that does not mean that an electric field spontaneously appears in the original reference frame!

No -- there is no work being done in the production motional EMF. The energy to create the current comes from the kinetic energy of the wire. Something caused the conducting wire to start moving -- the magnetic field acts to change that motion into a current. If there was no friction, then given enough time and space, the magnetic field will convert all of the wire's kinetic energy into current, and the wire will stop moving.

You're rejecting the third of Maxwell's laws. And, it seems, the mathematical theorem that the integral form is true if and only if the differential form is true.

15. May 12, 2006

### leright

no no no....my whole point in this thread is that the differential form of the third equation FAILS to take into consideration motional EMF (and yes, motional EMF induces an E field with a curl), which is where there is a change in reference frame. That is the correction to the equation I am making, and for this you would need Lorentz transformations.

Last edited: May 12, 2006
16. May 12, 2006

### Staff: Mentor

If the loop rotates (as in a simple generator), the flux through the loop does change. For a fixed-area loop in a uniform magnetic field:

$$\Phi = AB \sin \theta$$

$$\frac{d \Phi}{dt} = AB \frac {d}{dt} (\sin \theta) = AB \cos \theta \frac {d\theta}{dt}$$

17. May 12, 2006

### leright

ok, well then take a time varying conducting loop!

18. May 12, 2006

### Hurkyl

Staff Emeritus
And my answer is that your point is based upon flawed assumptions.

The third equation does not fail to take into consideration motional EMF -- motional EMF does not act as a source of the electric field, and thus (correctly) does not appear in the equation.. :tongue:

This is true both in the differential and integral forms. The equation:

$$\oint_C \vec{E} \cdot d\vec{l} = - \frac{d \Phi_B}{dt}$$

is only valid for a fixed contour C. This equation cannot tell you anything about motional EMF.

If you do not understand this, then allow me to demonstrate some of the many problems that you will encounter, both mathematical and physical.

Suppose that

$$\oint_S (\nabla \times \vec{E}) \cdot d\vec{s} = - \frac{d \Phi_B}{dt}$$

is true for a time-varying surface S. Then we can consider two different time-varying surfaces, $S_1$ and $S_2$. Suppose that at t = 0 that $S_1 = S_2$. Then, the above equation tells us that the time derivative of the flux through $S_1$ is equal to the time derivative of the flux through $S_2$ -- but that can't be right, because those surfaces can be varying in vastly different ways!

Some physical problems: you want a velocity to appear in Maxwell's third equation. Well, velocity of what?

Let's suppose you're right -- some electric field does spontaneously appear due to the velocity (of something), and that's what causes motional EMF.

Well, what if our moving conducting wire passes right next to a stationary conducting wire. Well, since the moving conducting wire (somehow) causes there to be an electric field, then that should induce an EMF in the other wire in the same direction. Does that happen? No! In fact, once we account for all effects, don't we see the exact opposite?

If we had two conducting wires passing by each other moving at equal speeds, your argument would say that electric fields are induced... but since they're in opposite directions they cancel, and we see no motional EMF at all. But that's not what happens: we will see a current in both wires.

For the sake of completion, I will admit that you are right (but as far as I can tell, for the wrong reasons): motional EMF does induce an E-field with curl, but for reasons that can be entirely explained via Maxwells equations, differential or otherwise.

Motional EMF is, at the lowest level, nothing more than the acceleration of moving charged particles.
Accelerating charged particles (generally) result in nonstationary currents.
Because current acts as a source of the magnetic field, a nonstationary current gives rise to a time-varying magnetic field.
A time-varying magnetic field acts as a source of the electric field.
And in this way, motional EMF does induce an electric field.

But this is an electric field caused by motional EMF. The electric field has absolutely nothing to do with causing the motional EMF. Motional EMF is caused entirely by the magnetic field: moving charges in a magnetic field are deflected in a direction perpendicular to their motion.

(Of course, the electric field could indirectly cause motional EMF... say, by having the time variations of the electric field induce a magnetic field)

Last edited: May 12, 2006
19. May 13, 2006

### leright

OK, I see your point regarding point charges moving (non stationary current) which in turn produces a time variant magnetic field, which in turn produces an E field when in the presence of a B field. You may be correct on this statement, but I feel this only holds for moving CHARGES. Let us now consider a loop of conducting wire (but no current) that is rotated in free space with absolutely no B field present. This wire has no net charge OR current so as it moves it is not producing any b-field or changing b field. We know that no E field or EMF will be induced (or do we?) Let us assume for a moment that it doesn't produce a B field. It would be nice to have some experimental results handy claiming this is true. Someone should rotate an electrically neutral loop of wire somehow in a controlled environment (no B field, no E field, and in a vacuum) to and see if a B field is produced.

Now, let us move this same conducting loop of wire into a time invariant B field. We can rotate it in this B field or do whatever we want and produce a changing flux (not due to time variance of the B field, but time variance of the surface area of the loop....as we change the angle of the loop in the B field we are changing the magnetic flux in the B field even if the scalar area of the loop stays constant, since the angle between the surface vector of the loop and the time invariant B field changes we are changing the magnetic flux). We know from experiment that this changing FLUX induces a curl in E which produces EMF (or that this induces an EMF which in turn produces a curl in E, depending on how you look at it). However, there is no current or moving charges or anything, so that originally static field is still static. In this case we are changing ONLY the magnetic flux through the loop and not the B field. However, we know that there is an EMF induced by experiment, and if there is an EMF there is an E-field, PERIOD! NOWHERE in the differential form of Maxwell's third equation (and a changing B field doesn't pop out in any of the other equations in differential form, so your previous agument fails in this case) is this scenario accounted for, but the integral form covers it. The error (or assumption, perhaps) Maxwell made (or did Heaviside do this, I can't remember) when going from the integral form to the differential form was when he brought a partial derivative into the integral of B over a surface and didn't account for a changing area over time (ds). I don't know how many electrodynamics problems you've worked through, but there are COUNTLESS problems where the diffential form fails to yield the same result as the integral form. You can attempt some problems by computing the curl of the E field using the differential form and then integrating the curl over a surface to get the EMF and you DO NOT get the same answer as the integral form. This is a FACT, and is clear proof of the failure of the differential form of Maxwell's equation. Maxwell assumed that no motional EMF was taking place when he went from differential form to integral form. In fact, he assumed a lot of things. The equations are not wrong if this assumption is made clear whenever the equations are written down, but most textbooks FAIL to make this clear. In fact there are tons of websites out there with modified differential forms of maxwell's equations taking things like motional EMF, material space, etc into account (and even quantum effects if you want to get into QED). Trust me, the differential form of Maxwell's equation (3rd and 4th, really) has plenty of errors (or assumption that were made, so I don't insult Maxwell. :tongue: )

Your comment about about how the EMF causes the electric field, but an electric field does not cause the EMF I feel is incorrect, simply because of the way EMF is defined. I don't feel one necessarily causes the other, but I simply feel like they COMPLEMENT each other and occur simultaneously (an E field doesn't appear by itself in a time invariant frame and the EMF due to that E field appears in the next frame). If you have an EMF it is because of energy in a magnetic field allowing a non conservative E field to be produced. Remember, EMF is the closed path integral of a an E field. And I think it is pointless to argue over this (if the e-field produces emf, or if the emf produces the e-field). Under maxwell's equations, the condition of the field is taken in a time invariant frame so they are both really occuring simultaneously. This isn't what I am arguing about, really.

Last edited: May 13, 2006
20. May 13, 2006

### leright

http://www.wolfram-stanek.de/maxwell_equations.htm

here's a great website with all of the extensions on Maxwell's equations. Notice the ones on the top are the equations in question, and they fail in the case of moving objects, according to the website.

Last edited: May 13, 2006
21. May 14, 2006

### Hurkyl

Staff Emeritus
I don't think it is pointless, because as far as I know, you're wrong. And I'm optimistic enough to think that whichever one of us is wrong will be convinced in the end. :tongue:

You're confusing an equation with a definition -- and trying to apply the equation in circumstances in which it doesn't apply.

EMF is the "force" that causes a current to be induced -- nothing more, nothing less. It can be viewed as the integral of an electromotive force field. The term "EMF field" is not synonymous with "electric field"; it's just that they just happen to be equal in the case of a stationary loop.

I hope our disagreements are just this confusion over terminology. When I speak of the electric field, I'm talking about the fundamental field, and not some "effective" field that describes the amalgamation of multiple effects.

IMO, this entire thread is summed up in one statement:

Maxwell's third equation is not about the EMF field.​

You seem primarily interested in the EMF field -- and because Maxwell's equations are only equations for the electric and magnetic fields, you find them inadequate for your interests.

But I do not see that as a problem with Maxwell's equations. They (correctly) do what they were meant to do.

(Incidentally, I know very little about the macroscopic H and D fields -- I have only been interested in learning about the microscopic fields)

Bo ThidÃ©'s book derives a differential expression for the EMF field felt by a moving conductor:

$$\mathbf{E}^{\text{EMF}} = \mathbf{E} + \mathbf{v} \times \mathbf{B}$$

He derives it through a computation involving Faraday's law and a moving loop. (In particular, he does not use "Maxwell's generalized form" of Faraday's law, and even explicitly states that equation is only valid for a "generic stationary 'loop'")

It can be more quickly derived from the force law -- this is what I have been saying when I say that the magnetic field directly induces motional EMF. In the nonrelativistic case, the force on a moving charge is q(E+vxB), so the EMF field felt by that charge must be E+vxB.

(But, Bo had not yet introduced the force law when he did his derivation, so he couldn't have done it this way)

Last edited: May 14, 2006
22. May 14, 2006

### leright

EVERYWHERE I look on the internet and all the literature claims that EMF is produced by a special type of E-field (but it is still an E-field....it is just nonconservative E-field). I've never heard of this thing called an EMF field.

And I don't understand when you say Maxwell's third equation is not about an EMF field (if there is such a thing). Of course it is! It describes the curl of the E-field due to changing B fields (or with the motional EMF extension, movement of objects in the field). If you can consider something such as an "EMF field" it would be the total E-field (conservative + non-conservative parts), I would think, and Maxwell's equation gives the non-conservative part of the picture. Whenever change in magnetic flux induces an E-field it is a non-conservative E-field, which is what causes EMF

And I've used Maxwell's third integral equation for TONS of scenarios where the loop is time varying in many different applications (and so do text books when they work examples). the integral form DOES work in cases where the flux through a conducting loop is time varying but due to constant B field in a time varying loop. The differential form FAILS because it doesn't take into account a time varying loop. Did you look at the link I provided and the difference between the regular Maxwell's equations and the modified equations taking motion into account? The equation that takes motion into account has a total derivative in front of the B-field, and the equation that doesn't take motion into account has a partial derivative in front of the B-field.

Maxwell's original equations contained the partial derivative and not a total derivative because he assumed motion was not occuring.

Also, what is the E in the EMF field equation? It is the conservative part of the E-field. the VxB part is the non-conservative part. The EMF-field is the static E field + the time variant E field. That is all. It's all an E-field.

EDIT: OK, I admit that I made a terrible mistake. The EMF-field is the conservative E-field + the non-conservative E-field (if you want to call it an EMF-field). Maxwell's third equation describes the non-conservative part of the E-field and how it is produced. When you say it isn't about an EMF field then what IS it about? f you consider the EMF field to just be the non-conservative part of the E-field then OF COURSE the third equation is about the EMF field.

Last edited: May 14, 2006
23. May 14, 2006

### Hurkyl

Staff Emeritus
Yes -- and everything I found describes EMF in the setting of a stationary loop! If motional EMF is described, it was more-or-less as an addendum.

By definition, the electric field E is the vector value associated with a point in space that would cause a particle with charge q to experience a force qE.

I haven't succeeded in tracking down a general definition of the EMF field. But, it's essentially the field with the property that the net force on the charges in your circuit is qEEMF. Or, more accurately, the field such that integrating it around the loop gives you the EMF.

From the definitions of the fields, we see that there's no automatic reason they should be equal. In fact, there are some sources of EMF that are clearly not electromagnetic: for example, if I physically grab a charged ring and start rotating it, I have generated EMF in a mechanical way.

The phenomenon of motional EMF demonstrates that, even if we consider only electromagnetic sources, the EMF field is not a fundamental thing; the value of the EMF field is intimately tied to the motion of our conducting loop. So, unlike the electric field, the EMF field cannot be described as an independent entity.

No, the textbooks do not use Maxwell's third integral equation. They use Faraday's law:

EMF = -(d/dt) {magnetic flux through the loop}

It is a mistake to use Maxwell's third integral equation, because:
(1) Maxwell's third integral equation is only for fixed loops.
(2) The left hand side of Maxwell's third integral equation is not EMF.

No, it's not. E is the electric field: the force field that causes the force qE on a particle with charge q. B is the magnetic field: the force field that causes the force qvxB on a particle with charge q and velocity v.

Yes.

never forget the necessary relations for field transformations
(i.e. Lorentz' transformation) hidden in the left sides of the equations (1a) - (4a)​

if I read the page right, those equations are not for the electromagnetic field, but for a Lorentz transformation of the electromagnetic field -- the third equation is relating the electric field in one reference frame to the magnetic field in another frame.

24. May 14, 2006

### vanesch

Staff Emeritus
It is indeed a happy coincidence that the use of the integral form also works for non-time-stationary loops in certain circumstances (such as rotating loops in a homogeneous magnetic field), but in fact this is NOT a direct application of the integral form of Maxwell's third equation - although it surely takes on the same form.

As Hurkyl correctly pointed out, the EMF around a loop is the total force exerted on a charge around the loop integrated over the loop. In order to calculate this, one has to dispose many small test charges all oround the loop (co-moving with the loop) and calculate the force excerted on it. You then multiply with the (infinitesimal) distance, and sum over all the forces. If that loop is stationary in a frame, then the force acting on a test charge "fixed to the loop" is simply q.E, and the integral of this force around the loop is nothing else but the loop integral of the E-field, which is given by the integral form of Maxwell's third equation.
However, when the loop is not time-stationary, but moving or changing, then the (co-moving) testcharges are subjected to the total electromagnetic force, which is q(E + v x B), so THIS (as Hurkyl pointed out) is the quantity entering in the EMF. Here, v is the local velocity of the loop where the test charge is fixed.

And the "miracle" is now the following. The contribution of a test charge in a piece of loop dr is given by q dr.(E + v x B). But dr. (v x B), with v = ds/dt (the perpendicular motion of the piece of conductor) takes on the form: dr ds B / dt, which is dA B / dt, dA being the area whiped out by the piece of conductor dr in a time dt, and B dA being the magnetic flux whiped out by this piece of conductor, during time dt.

So we have that the contribution of the part v x B in the force law gives us, as a contribution of the EMF, simply d phi/ dt, the change of flux seen by the piece of wire due to its local motion.
And the contribution to the EMF due to the E-field is also d phi/ dt, but this time, the change in time of the flux in the stationary loop due to a changing B in time. So BOTH contributions take on the form d phi / dt, and hence the second contribution (from v x B) is AUTOMATICALLY taken into account when you calculate the total EMF due to a moving conductor loop. This was the "coincidence" I talked about.

And now you also understand why this works for the integral form, and not for the differential form: the differential form gives you only E, while the integral form automatically takes with it the contribution of v x B.

25. May 14, 2006

### leright

Tomorrow I am going to order a copy of Jackson's Classical Electrodynamics and start working through it to learn this stuff correctly. I am serious about learning this stuff, and I don't feel satisfied with what I got out of my ugrad coursework. Terminology like "EMF-field" is meaningless to me. If I understand correctly, the EMF-field is simply the field that models the TOTAL force that can act on a charge (not just the electric field), and can be from anything.

First, my textbook takes Faraday's law (Vemf is minus the rate of change in magnetic flux) and replace flux with the integral of B.ds. Then they take the derivative inside of the integral as a partial and proceed to apply stokes theorem on Vemf (rewriting integral(E.dl) as integral(curl(E).ds)) and then say that curl(E) = -partial(B)/partial(t). They DO say that this only holds for time invariant loops.

Also, whether or not the integral form holds for moving bodies or not depends on if you consider the integral form as being faraday's law with the flux simply replaced with integral(B.ds), or if you consider the integral form where they bring in the derivative into the integral. If you consider the integral form of the 3rd equation to be the former then the equation holds for moving objects. If you consider the integral form of the 3rd equation to be the latter then the equation DOES NOT hold. Unfortunately, I have seen the integral form of the 3rd equation being represented both ways on the internet.

Now, my textbook in chapter 8 says the mechanical force due to a magnetic field that acts on a moving charge is qV X B. I understand this. But then, in chapter 9, they went and defined a motional electric field Esubm, which was V X B. They then proceed to take the circulation integral of this around a closed path and call this the motional EMF (the EMF due to moving charge or loop in a constant B-field). They apply Stoke's and get the curl of E is the curl of (V X B). So, for a moving loop in an magnetostatic field curlE = curl(V X B).

The author then proceeds to claim that therefore for moving loops in time variant fields the total curl of E is the sum of the curl of U X B and negative the partial of B wrt t.

Hurkl also claimed that an E-field is not doing the work in the case of motional EMF, but the magnetic field is doing the work. Well, my book calls U X B the motional ELECTRIC field and determines the EMF by dotting this motional electric field over a closed path.

OK, so in the end I was CORRECT in my original post when I said that Maxwell's differential form of the third equation doesn't take into account motional EMF. I know for a fact that it doesn't. I now realize that the integral form also does not take into account motional EMF since the derivative is brought into the integral as a partial derivative and only B is differentiated. I was thinking that the integral form of the third equation the derivative was kept outside the integral, making it work in all cases. However, many sources on the internet show the integral for as having the derivative outside of the integral. In this case, then the integral for is the SAME THING AS FARADAYS LAW, and works in motional EMF cases also.

Another part of my confusion hinges on this motional electric field, which is Fsubm/q, or simply U X B. So should I be thinking about the force on moving charges moving in a magnetic field as being caused directly by the magnetic field, or should I be thinking about the force being caused directly by this so-called motional electric field, where the motional electric field is caused by the magnetic field. I think the latter is incorrect, but it seems as if all the literature disagrees, since they say that in a static magnetic field with a moving charge, there is a curl in E. However, a curl in E implies there is a time varying E field, and then therefore there would be a time varying magnetic field caused, and so on, which would imply an electromagnetic wave. However, we all know that a dc current in a magnetic field does not produce an electromagnetic wave (or does it?)

So, I really think the terminology was developed carelessly and there are lots of flaws.

Another thing I do not understand is the idea of a constant magnetic field in the, for example, x-direction. How can this be? If you have a constant B-field (at all points the magnitude and direction is the same) in the x-direction that would imply that it is a conservative B-field, which is bogus. The only way you could produce a constant B-field is with an infinite conducting place which is purely hypothetical. I see so many problems in textbooks that have constant B-fields in them. Are they just saying that the B-field is only constant in a particular region (center of a huge conducting plane, for instance?) I suppose this is ok, but it kinda bothers me a bit.

Also, one more question, in free space, maxwell's 4th equation says that if there is a changing D-field then there is a curl in H. LEt us now suppose that we have a D field whose derivative is a constant. If there is a constant derivative in the D field then there is a constant curl in H. This means that H only depends on x,y,z, but not on t. Therefore B is constant wrt time. Therefore E does not have a curl according to MAxwell's 3rd equation, correct? Therefore electromagnetic waves to dot propagate.

Last edited: May 14, 2006