Faraday's law and motional EMF paradox

[cnh1995]

Consider the diagram below (sorry for the quality). The circle is the front view of a cylindrical iron rotor (highly ferromagnetic, very low reluctance). The orange part is a single turn of a conductor (very high reluctance) wound around the rotor body. The grey lines are magnetic field lines from external magnetic poles, which you can see are crowded in the ferromagnetic iron body. The number of lines actually cutting the conductor is negligible compared to the those in the iron body because of very high reluctance of the conductor. (This is similar to what happens in a practical dc machine.) Now, keeping the magnetic field stationary, if the rotor is rotated through 90°, there will be a "change of flux" associated with the conductor loop. This should induce an emf in the loop as per Faraday's law E=dΦ/dt. But since the conductor is actually cutting a negligible number of magnetic field lines, the motional emf should be almost zero. What am I missing? Please help..

 

[Charles Link]

It's a little hard to follow your diagram, but just a guess is that the axis of rotation of the loop is horizontal (and not vertical) so that the ## d \Phi/dt ## is quite large.

 

[tech99]

Consider the diagram below (sorry for the quality). The circle is the front view of a cylindrical iron rotor (highly ferromagnetic, very low reluctance). The orange part is a single turn of a conductor (very high reluctance) wound around the rotor body. The grey lines are magnetic field lines from external magnetic poles, which you can see are crowded in the ferromagnetic iron body. The number of lines actually cutting the conductor is negligible compared to the those in the iron body because of very high reluctance of the conductor. (This is similar to what happens in a practical dc machine.) Now, keeping the magnetic field stationary, if the rotor is rotated through 90°, there will be a "change of flux" associated with the conductor loop. This should induce an emf in the loop as per Faraday's law E=dΦ/dt. But since the conductor is actually cutting a negligible number of magnetic field lines, the motional emf should be almost zero. What am I missing? Please help..
View attachment 104902

It looks me as if when the rotor turns through 90 degs, the conductor has passed through all the lines of force you have drawn. It seems to act like a much bigger loop but without an iron core.

 

[cnh1995]

It's a little hard to follow your diagram, but just a guess is that the axis of rotation of the loop is horizontal (and not vertical) so that the ## d \Phi/dt ## is quite large.

Yes, dΦ/dt is large and should induce an emf. But since the rotor is actually cutting "very few" magnetic field lines due to its high reluctance, motional emf should be almost 0. Hence, motional emf is not equal to dΦ/dt, which contradicts Faraday's law.

 

[cnh1995]

It looks me as if when the rotor turns through 90 degs, the conductor has passed through all the lines of force you have drawn. It seems to act like a much bigger loop but without an iron core.

How does the rotor "cut" all the magnetic lines? It has a very high reluctance, hence, most of the magnetic lines bypass the conductor and are crowded in the iron body.

 

[Charles Link]

How does the rotor "cut" all the magnetic lines? It has a very high reluctance, hence, most of the magnetic lines bypass the conductor and are crowded in the iron body.

If you are asking, does the conductor need to actually be in the (changing) magnetic field to acquire an EMF, the answer is no. It simply needs to surround the area that has the changing magnetic field.

 

[cnh1995]

If you are asking, does the conductor need to actually be in the (changing) magnetic field to acquire an EMF, the answer is no. It simply needs to surround the area that has the changing magnetic field.

I thought so in the beginning. But if the change of flux is taking place because of "movement" of the conductor loop, it has to physically "cut" the magnetic field lines. This is consistent with Gauss' law for magnetism.
https://www.google.co.in/url?sa=t&s...ggaMAA&usg=AFQjCNH9N69CnzBKuK7WS0vocM10Eop1Uw

 

[cnh1995]

Actually my original question stems from another thread of mine in the EE forum.
https://www.physicsforums.com/threads/dc-machine-magnetics.881877/
See #5 ,#6 and #10.

 

[Charles Link]

Actually my original question stems from another thread of mine in the EE forum.
https://www.physicsforums.com/threads/dc-machine-magnetics.881877/
See #5 ,#6 and #10.

I am going to need to look these over in detail, but they are interesting. I do expect there is a very logical answer supplied by the Faraday-Maxwell equation of your "link" in post #8.

 

[cnh1995]

I am going to need to look these over in detail, but they are interesting. I do expect there is a very logical answer supplied by the Faraday-Maxwell equation of your "link" in post #8.

Thanks a lot! I have been thinking on this for almost a month..

 

[Simon Bridge]

Please state Faraday's law and explain the problem clearly. Right now it looks like the metaphore of field lines is confusing things so try explaining the problem without them.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

 

[cnh1995]

Please state Faraday's law and explain the problem clearly.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

Well, I guess my diagram is hard follow.
The grey lines are magnetic field lines which are "bypassing" the orange conductor loop because of its high reluctance. So, there are very few field lines going "through" the conductor. But when the conductor loop is rotated through 90 degrees keeping the external field as it is, the loop will experience a change of flux. In the new position, the loop will be horizontal and "all" the flux will be linked with it while in the initial vertical position, flux linking with it was 0. This should induce an emf in the loop as per E=dΦ/dt.
But since there are actually very few lines going through the conductor, the the flux "cut" by the conductor will be very small. So motional emf Blvsinθ will be very small and since motional emf≠ dΦ/dt, this contradicts Faraday's law.

 

[Simon Bridge]

Still don't see the problem. The magnetic environment of the loop changes, therefore emf is induced. That is Faraday's law.
How many field lines are "cut" is irrelevant. Those lines are just a visual aid.

Perhaps the problem is not stated clearly, try restating without using the metaphore of cutting field lines.
Alternately, show me where the math does not work.

 

[cnh1995]

How many field lines are "cut" is irrelevant. Those lines are just a visual aid.

The wikipedia link in #8 says something different. Read "Faraday's law compared to Faraday- Maxwell equation" in that wiki article, especially the derivation.

 

[cnh1995]

How many field lines are "cut" is irrelevant.

The wikipedia link in #8 says something different. Read "Faraday's law compared to Faraday- Maxwell equation" in that wiki article, especially the derivation.

According to the wiki article I linked in #8, if the conductor is moving in a stationary field and the flux liking with the conductor is changing, the conductor must be "cutting" the field lines and hence, emf induced in the loop is given by E=Blvsinθ (motional emf) and it turns out from Gauss' law for magnetism, that the motional emf so induced is equal to the rate of change of flux linking with the loop i.e. motional emf=dΦ/dt.
Here in my problem, motional emf is very small since the conductor loop is "cutting" (that's the word used in that article too)
very few lines because of its high reluctance. Hence, motional emf≠ dΦ/dt. What will be the actual emf induced in the conductor loop in this situation? Motional emf formula predicts very small induced emf and Faraday's law predicts a larger emf because of large dΦ/dt. This is what I think is paradoxical. Did I make the problem clear?

 

[Simon Bridge]

Yeah - but the article uses it as a metaphor and also tells you what is actually happening.
You seem to be very resistant to ditching the metaphor: why?

In the other article I set up a simpler, illustrative, situation for you to work on, where you may expect "motional emf" yet the number of field lines "cut" is zero. You do not appear to have attempted it.

In your example in this thread ... try redrawing the diagram with the loop horizontal. Are there not field lines going through the loop that did not go through the loop before?

If you do not follow advise we cannot help you.

 

[cnh1995]

Sorry for replying late! My exams just got over.

If you do not follow advise we cannot help you.

I did attempt the example you set up earlier but I think that is not what I am asking in this thread.

Yeah - but the article uses it as a metaphor

I don't understand how it can be used as a metaphor when it's actually happening. The flux being cut by the conductor implies that there is a relative velocity between the flux and the conductor, which results in a force on the electrons of the conductor, given by F=q(v×B), where B is the magnetic field at the site of the conductor.

In the other article I set up a simpler, illustrative, situation for you to work on, where you may expect "motional emf" yet the number of field lines "cut" is zero.

Perhaps you mean something different by motional emf than I do. As per my understanding, motional emf is induced when a conductor is "moving" in a stationary magnetic field B with a velocity v and is given by induced electric field E=v×B. In the example you set up, there is no relative velocity between the flux and the conductor since the conductor is stationary and the flux is varying only with respect to time and not with respect to position. As per my book, that's what is called transformer emf.
In my original question, since the flux linking with the conductor loop is changing after 90 degree rotation, I agree that there must be an induced emf equal to dΦ/dt. But since the value of B at the site of the conductor is very small(because of its high reluctance), v×B will be negligible and hence, motional emf will be negligibly small. But according to Faraday's law, motional emf Blv must be equal to dΦ/dt.
I want to understand how this equality holds in this particular situation. I know it does, but how??

 

[Simon Bridge]

Oh I think I see - you are wondering how the EMF force via q(vxB) can arise when there is no magnetic field inside the conductor to act on the charges?

I think both examples in post #3 (other thread) illustrate how the law applies: in neither example does the field enter the conductor. You should also realize that motional and transformer emf are both manifestations of the same underlying principle.

 

[cnh1995]

you are wondering how the EMF force via q(vxB) can arise when there is no magnetic field inside the conductor to act on the charges?

Exactly!

You should also realize that motional and transformer emf are both manifestations of the same underlying principle.

I do know that. Emf induced=dΦ/dt, whatever may be the cause of change of flux. But D.J.Griffith and Feynman, in their books, say that "when a circuit moves in a magnetic field, the source of emf is the Lorentz force and when magnetic flux changes w.r.t. time in a stationary loop, concentric induced electric field lines are formed which cause the transformer emf. Though mathematically both the emfs are equal to dΦ/dt (the universal flux rule), the physical reasons are different. In my example, I just want to know if there is any q(vxB) force acting (on the electrons of the conductor) or not. If there is, how do I prove it equal to dΦ/dt? If there isn't, shall I treat it as an exception to the original equality of dΦ/dt and motional emf (which sounds weird)? I just want to know that if the electrons are in a very weak B field, what makes them move?

 

[Simon Bridge]

OK. The lynchpin to your problem here is to find out how the material excludes the magnetic field.
What is the mechanism that gives rise to this property?

It may help, also, to consider just where the free charges in a conductor actually live.

 

[cnh1995]

to find out how the material excludes the magnetic field.

As the loop moves, there is a change in permeability in space w.r.t. the field poles. Hence, magnetic lines rearrange themselves accordingly. Does this cause "cutting" of the field lines?

It may help, also, to consider just where the free charges in a conductor actually live

Surface charges on the conductor establish an electric field inside the wire and electrons inside the wire move along the field.

But considering the generator open circuited, the only movement of electrons will be due to the vxB force and current will be zero. So, free electrons are present in the entire volume of the conductor.
Is there any specific mechanism you are referring to?

 

[Simon Bridge]

As the loop moves, there is a change in permeability in space w.r.t. the field poles. Hence, magnetic lines rearrange themselves accordingly.

Are you saying that the material only excludes the magnetic field while the loop is moving?
I think you need to understand how this happens in terms of the charges inside the wire... however, I am not sure that "reluctance" is a useful concept for understanding induction.

Does this cause "cutting" of the field lines?

Please give up this language, it is useful for picturing some situations but it looks like it is confusing you in this case.

Surface charges on the conductor establish an electric field inside the wire and electrons inside the wire move along the field.

But considering the generator open circuited, the only movement of electrons will be due to the vxB force and current will be zero. So, free electrons are present in the entire volume of the conductor.

Hmmm... in the completely static case, no applied fields, the free charge resides about the outside of the conductor. DC current flows inside the conductor - generally considered to be through the whole volume, but a changing current experiences a skin effect where it is mostly outside.

Is there any specific mechanism you are referring to?

The usual models try to use a reference frame where internal forces are zero and modify the physics accordingly - which is why the carrier mass can differ from material to material. This is usually how you get some B field inside a conductor.
I think the short answer to your question is that the models you are using are incomplete.

I'm not sure how to best procede - I'll see if I can alert someone with more direct experience explaining this. For now - your best approach is to try to understand better the relationship between reluctance and induction.

 

[cnh1995]

For now - your best approach is to try to understand better the relationship between reluctance and induction.

All right! I will look more into this and post if anything is unclear. Thanks for your help so far!

 

[Dale]

Now, keeping the magnetic field stationary, if the rotor is rotated through 90°, there will be a "change of flux" associated with the conductor loop. This should induce an emf in the loop as per Faraday's law E=dΦ/dt. But since the conductor is actually cutting a negligible number of magnetic field lines, the motional emf should be almost zero.

The problem that you are having here is that your concern is based on a false premise. Like other posters, I am not a big fan of the "field lines" for any purpose other than textbook illustrations. However, if you like the concept and want to use it then you must correctly understand and use the rules that govern their behavior.

Magnetic field lines are closed loops, regardless of any material properties like reluctance. The density of lines is proportional to the flux. Consider a wire, a surface bounded by the wire, and a single line of flux. There are exactly three possibilities:

1 The line of flux does not intersect the surface, so there is no flux from that line.

2 The line of flux intersects the surface once, so there is flux from that line.

3 The line of flux intersects the surface twice, so there is no net flux from that line.

You cannot go from either case 1 or case 3 to case 2 without the wire cutting a line of flux. So if the flux changes then you are guaranteed that a line of flux cut the wire. This is contrary to your assertions that few lines are cut. In fact, many are cut, corresponding to the increase in flux.

 

[cnh1995]

This is contrary to your assertions that few lines are cut. In fact, many are cut, corresponding to the increase in flux.

Speaking in terms of magnetic flux density B, is the value of B same at the site of the conductor as it is in the iron core?
Does iron core not affect the magnetic field at the site of the conductor at all?

 

[Dale]

Speaking in terms of magnetic flux density B, is the value of B same at the site of the conductor as it is in the iron core?
Does iron core not affect the magnetic field at the site of the conductor at all?

Of course it affects the field, but that doesn't change the fundamental rules governing the field lines.

If you change flux then you cut field lines.

 

[cnh1995]

My apologies for replying so late!

Of course it affects the field, but that doesn't change the fundamental rules governing the field lines.

If you change flux then you cut field lines.

I am having trouble understanding this.
If the magnetic field B at the site of the conductor is very small (i.e. most of the field lines are bypassing the conductor due to its higher reluctance), the vxB force on the electrons of the conductor will also be very small. This means motional emf is very small and is not equal to dΦ/dt (which is quite large). What am I missing?

 

[Dale]

This means motional emf is very small and is not equal to dΦ/dt (which is quite large). What am I missing?

You are missing that this is not possible. If dΦ/dt is large then so is the EMF. I am not sure why you think that both are true, your drawing and description are not clear enough for me to understand, but one of the two statements is not correct. Either dΦ/dt is small or the EMF is large.

https://en.m.wikipedia.org/wiki/Faraday's_law_of_induction#Proof_of_Faraday.27s_law

 

[Charles Link]

The flux lines do not need to pass through the conductor to create an EMF in the conductor. Faraday's law says the EMF around a loop is equal to the ## -d \Phi/dt ## contained anywhere inside the loop. The border of the loop itself can have zero magnetic field and you still get the EMF. @cnh1995 To answer your question in post #28: The EMF is not a ## v \times B ## force. It depends upon a ## dB/dt ## that may be non-local. ...editing...Having an EMF around the loop means the loop contains an electric field ## E ##. ## \ ## EMF ## \mathcal{E}=\int E \cdot dl ##. This electric field ## E ## supplies a force on the electrons of the form ## F=e E ##.

 

[cnh1995]

The border of the loop itself can have zero magnetic field and you still get the EMF.

..only when the loop is stationary and the flux is varying w.r.t. time, just like in a transformer. Here, the conductor loop is moving in a stationary field and has a linear velocity v. The vxB force on the electrons of the conductor is responsible for creating the emf in such cases (which is called "motional emf"). I have posted a link in #8 about the same. Motional emf and dΦ/dt are always equal as @Dale has said earlier because when you change the flux by "moving the conductor", you "cut the field lines" . Here, emf is induced by the vxB force and mathematically, it turns out that motional emf and dΦ/dt are equal. Hence, in general we can say induced emf=rate of change of flux but the physical reasons are different when the "circuit moves in a steady field" and when the "flux varies in a fixed circuit".

your drawing and description are not clear enough for me to understand

Ok..Here's a quick description of the diagram..

The conductor (orange) is made of copper (high reluctance) and the rest of the core is made of iron(highly permeable, negligible reluctance). Now, if the magnetic poles are created as shown (using electromagnet), most of the field lines (grey) will bypass the conductor since it has a very high reluctance, and will be crowded in the rest of the core. So, magnetic field B at the site of the conductor(orange circles) is very weak.

Actually my original question stems from another thread of mine in the EE forum.
https://www.physicsforums.com/threads/dc-machine-magnetics.881877/
See #5 ,#6 and #10.

Isn't the B field at the site of the conductor(orange circles) much weaker than that in the rest of the core?

 

[Charles Link]

..only when the loop is stationary and the flux is varying w.r.t. time, just like in a transformer. Here, the conductor loop is moving in a stationary field and has a linear velocity v. The vxB force on the electrons of the conductor is responsible for creating the emf in such cases (which is called "motional emf"). I have posted a link in #8 about the same. Motional emf and dΦ/dt are always equal as @Dale has said earlier because when you change the flux by "moving the conductor", you "cut the field lines" . Here, emf is induced by the vxB force and mathematically, it turns out that motional emf and dΦ/dt are equal. Hence, in general we can say induced emf=rate of change of flux but the physical reasons are different when the "circuit moves in a steady field" and when the "flux varies in a fixed circuit".

Ok..Here's a quick description of the diagram..

The conductor (orange) is made of copper (high reluctance) and the rest of the core is made of iron(highly permeable, negligible reluctance). Now, if the magnetic poles are created as shown (using electromagnet), most of the field lines (grey) will bypass the conductor since it has a very high reluctance, and will be crowded in the rest of the core. So, magnetic field B at the site of the conductor(orange circles) is very weak.

Isn't the B field at the site of the conductor(orange circles) much weaker than that in the rest of the core?

Your more detailed description was very helpful. Near the poles of a cylindrical magnet or horseshoe shaped magnet (i.e. in the air outside of the material), the magnetic field outside the magnet can be nearly as large as it is inside the iron core of the magnet.

 

[Charles Link]

Additional comment on this one: There is a modified version of Faraday's law that applies when the surface of interest is also changing in size. Alternatively, the effect can be viewed as due to the magnetic field acting on the moving electrons in the conductor. Both methods give the same force, but in one case it is due to an electric field and the other case due to a magnetic field. Relativistically, I think the two go hand-in-hand and are part of the electromagnetic tensor. @vanhees71 I would welcome your input here-I think this problem is similar to a paper of yours that you referenced in a recent discussion of the effects of a changing magnetic field on the circuit equations. Your paper discussed the modified Faraday's law in the case of a changing boundary surface.

 

[Dale]

Isn't the B field at the site of the conductor(orange circles) much weaker than that in the rest of the core?

Is the conductor just a straight wire through the core or is it a loop? What is the difference between the orange circles and the rest of the conductor?

Whatever the geometry of the conductor, it is possible to make the field there much lower than the field nearby.

 

[Dale]

@vanhees71 I would welcome your input here-I think this problem is similar to a paper of yours that you referenced in a recent discussion of the effects of a changing magnetic field on the circuit equations. Your paper discussed the modified Faraday's law in the case of a changing boundary surface.

I am not sure if this is the @vanhees71 post you are thinking of, but I found it useful

https://www.physicsforums.com/threads/maxwells-curl-equations.831561/#post-5223098

 

[cnh1995]

Is the conductor just a straight wire through the core or is it a loop?

It is a loop. The diagram is the front view of a cylindrical iron rotor with a single turn of conductor (two orange circles are two slots).

Whatever the geometry of the conductor, it is possible to make the field there much lower than the field nearby.

Yes. The field at the site of the conductor is much weaker than the field nearby. How is vxB emf still equal to dΦ/dt?