1. Aug 20, 2016

### 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..

Last edited: Aug 20, 2016
2. Aug 20, 2016

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.

3. Aug 20, 2016

### tech99

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.

4. Aug 20, 2016

### cnh1995

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.

5. Aug 20, 2016

### cnh1995

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.

6. Aug 20, 2016

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.

7. Aug 20, 2016

### cnh1995

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.

8. Aug 20, 2016

### cnh1995

9. Aug 20, 2016

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.

10. Aug 20, 2016

### cnh1995

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

11. Aug 21, 2016

### Simon Bridge

12. Aug 21, 2016

### cnh1995

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.

13. Aug 21, 2016

### 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.

14. Aug 21, 2016

### cnh1995

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

15. Aug 21, 2016

### cnh1995

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?

16. Aug 24, 2016

### 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?

17. Aug 28, 2016

### cnh1995

Sorry for replying late! My exams just got over.
I did attempt the example you set up earlier but I think that is not what I am asking in this thread.
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.
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??

Last edited: Aug 28, 2016
18. Sep 2, 2016

### 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.

19. Sep 3, 2016

### cnh1995

Exactly!
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?

20. Sep 4, 2016

### 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.