How Does Faraday's Law Explain EMF Induction Outside a Solenoid?

[EMGuy101]

Some things to point out:

I know nothing about transformers.
Outside and inside there is an electric field E which can be anything but has to be generated by electrostatic charges, i.e. curl(E)=0.

E=-grad(V) - dA/dt
curl(E)= curl(-grad(V) - dA/dt), curl of grad(V)=0
curl(E)=-curl(dA/dt)
curl(E)=-d(curl(A)/dt
curl(E)=-dB/dt

but E=-dA/dt in this case not grad(V) since it is being generated by faradays law.
THUS, by saying curl(E)=0 does not mean E=-div(V) but more so that dB/dt = 0 NOT that dA/dt=0.

 

[Q-reeus]

I know nothing about transformers.

Well it's integral to your situation - which is all about transformer action. So check out a good article on transformers - Wikipedia a good place to start.

Let us simplify our problem, this way I should make it perfectly clear what bothers me. We imagine the problem to 2D. We have two regions in space, 1 and 2. Inside 1 the B-field is nonzero and outside B=0. Outside and inside there is an electric field E which can be anything but has to be generated by electrostatic charges, i.e. curl(E)=0.
We now imagine changing the B-field inside region 1. Clearly the changing B-field induces an electric field inside this region as described by:
∇xE = -dB/dt
Outside however nothing changes. B=0 and it remains so. Therefore -dB/dt=0 and ∇xE=0.

But as I kept trying to get through, ∇xE=0 relates to an intensive property of the field. Get used to that net emf involves integral form - as long as there is a threading time-changing flux through circuit of interest, there will be a net emf!

BUT! As the experiment and article stipulates. This is wrong. For some reason ∇xE is not equal to zero even in region 2!

See above!

I hope that I have now finally made it crystal clear what my problem is.

And I hope you finally get the distinction between differential and integral forms of Faraday's law, and which to apply to your situation.

And the solution is NOT(!) that B-field is only almost non-zero for all experiments performed in real life.

Agree entirely with that part. All a residual external B does - assuming some part of it 'reverse threads' the circuit of interest, is slightly reduce the net threading flux. If there is no 'reverse threading', it has zero impact on the situation.

 

[aaaa202]

But as I kept trying to get through, ∇xE=0 relates to an intensive property of the field.
And I hope you finally get the distinction between differential and integral forms of Faraday's law, and which to apply to your situation.

Maybe my problem is that I don't really know what is mean by an intensive property. Can you elaborate on that and how it relates to the distinction between differential and integral forms.

Emguy: I can only say that I think you are still completely misunderstanding the problem.

 

[Q-reeus]

Maybe my problem is that I don't really know what is mean by an intensive property. Can you elaborate on that and how it relates to the distinction between differential and integral forms.

Are you familiar with the idea of the gradient of a field or potential? It is an intensive property in that it involves a ratio of change/distance that persists at a point. Curl is just one step on from that - it involves comparing two such gradients at right angles to each other - see e.g. http://en.wikipedia.org/wiki/Curl_(mathematics)
The differential form of Faraday's law is really just an identity that equates a time-changing B field *at a point* to the associated curl of E field also *at that point*. An infinitesimal circuit there can have a substantial curl E acting even though the emf around such a circuit is vanishingly small. Shrinking the circuit to zero enclosed area will not reduce curl E but emf has gone to zero since so has the enclosed area. That's how such an intensive property acts. The integral form concerns itself with the net effect over a finite circuit = finite enclosed area, and that may or may not involve there being any flux intersecting the circuit itself. Net flux *threading* the circuit is all that matters to the extensive quantity of interest - line integral of E around the circuit. Must go.:zzz:

 

[harrylin]

I think I found an article describing this exact problem. I would have to say it agrees with me, and that the reasons for the voltmeters to change measurements is far more subtle than the reasons above. Do you disagree? If so, I have misunderstood you.
Emguy: I think you misunderstand the problem. The field is ZERO everywhere except inside the solenoid. Google the field of a solenoid.

What a coincidence. I found that same article at the same time, and as I just now saw this topic, I thought what a coincidence and I was going to suggest it to you, only to see that the coincidence was double! I think that it's a great article.

It's a bit mysterious that the induction takes place where the inducing field is practically zero; obviously there must be something else that is not part of the description and which makes it happen (perhaps corresponding to the vector potential?).

On a side note, I similarly found today interesting measurements of displacement current, which Maxwell did account for but which is often denied to exist.

 

[harrylin]

Practically 0 is not 0. Furthermore, something which is "practically 0" can still have an arbitrarily high rate of change. Your assertions are simply false, the field is non-zero, and the rate of change of the field is also non-zero.

It appears to me that the field outside the solenoid subtracts from the induction due to the field inside, because it is directed oppositely - what is measured is the induction from the field inside the solenoid minus the part between the outside of the solenoid and the loop. Thus the B field at the location of the wire can't explain Faraday's law.

 

[pervect]

http://www.astro.uvic.ca/~tatum/elmag/em09.pdf derives the magnetic vector potential for an infinite solenoid.

It appears to use cylindrical coordinates t,r,\phi, z.

Also \hat{z} represent a unit vector in the "z" direction, and \hat{\phi} a unit vector in the \phi direction, as is standard and used by the above reference.

If you work out the resulting electric and magnetic fields, you should get the following from the vector potential given by the reference.

Inside the solenoid: B = constant, in the \hat{z} direction.
Inside the solenoid E is proportional to both r and dI/dt, and is in the \hat{\phi} direction.

Outside the solenoid, B = 0
Outside the solenoid, E is proportional to dI/dt, but INVERSELY proportional to r.

It is left as an exercise to see how these results satisfy all of Maxwell's equations (or if I made a mistae, but I don't think I did).

Going through and working out how these solutions DO solve the equations should hopefully clear up the confusions in this thread. The integral form is more obvious, the easy integrals to perform are those about the center of the solenoid. We can see that inside the loop, the integral around the loop is proportional to r^2 and hence the area, which means it's proportional to both the product of the area and dI/dt, which seems exactly right.

Outside the loop, the integral of E around a loop is constant, which is also correct.

Computing the integral of E around a loop that's not centered seems messy, but I expect it will also give the correct answer of being proportional to the area. I haven't verified this personally, but if the magnetic vector potential given in the above reference is correct, it has to work that way.

The differential forms of the laws might be trickier, but by stokes theorem they are equivalnet. If there is any confusion over this, it might be helpful to review http://en.wikipedia.org/w/index.php?title=Stokes'_theorem&oldid=511213181, stokes theorem

The section on "underlying principle" is especially helpful, basically you can start by saying that if you divide the original area into pieces, you can find the line integral around the whole thing by adding together the line integral of the pieces - 4 in the picture - because the arrows for the repeated paths occur once in each direction in the interior, and cancel each other out, leaving only the sections of the path around the exterior, which contains the desired line integral.

Then you just need to show that the line integral over one of the small pieces is given by the the curl multipled by the area in the limit of a very small piece.

http://upload.wikimedia.org/wikipedia/commons/5/59/Stokes_patch.svgThus if I is fixed, there is no E field, only when I varies is there an E field.

 

[Q-reeus]

It's a bit mysterious that the induction takes place where the inducing field is practically zero; obviously there must be something else that is not part of the description and which makes it happen (perhaps corresponding to the vector potential?).

In terms of 'direct action of a field' at any point in the circuit, yes that corresponds, as stated in e.g. #7, #16, #27, to E = -∂A/∂t. In classical EM A is an auxiliary construct and only it's curl; ∇×A = B, or time-rate-of-change; -∂A/∂t = E (ignoring here the scalar potential part) is considered physically real. In QM things are different, as e.g. Aharanov-Bohm effect seems to establish beyond doubt, but I have neither the expertise nor interest to expand on that.

 

[Q-reeus]

It appears to me that the field outside the solenoid subtracts from the induction due to the field inside, because it is directed oppositely - what is measured is the induction from the field inside the solenoid minus the part between the outside of the solenoid and the loop. Thus the B field at the location of the wire can't explain Faraday's law.

Quite so - as per last part in my #32. Transformer manufacturers and designers understand that very well: Check out the 'E' amd 'I' core configurations here: http://en.wikipedia.org/wiki/Magnetic_core Both primary and secondary windings wrap around the central leg in those cores and link magnetically entirely according to the flux running through that central leg - the outer magnetic return path adds no extra coupling to the windings - but is extremely important in that it completes the magnetic circuit so as to make the effective magnetic susceptibility of the core very high.

 

[Q-reeus]

Going through and working out how these solutions DO solve the equations should hopefully clear up the confusions in this thread...

That there has been, but do you consider your entry here in any way opposed to my own bit in #34, rather than a more detailed re-packaging?
[The one misgiving I have with that illustration lifted from the Wiki site on Stoke's theorem, is that it could be taken as suggesting the strength of those peripheral vectors are indifferent to the size of the subdivided regions. That is not the case and the smaller the subdivided rectangular area, so smaller is the magnitude of the circulating field vector around the periphery. That follows geometrically from the ratio of circumference to enclosed area. Say for a circular enclosed region of radius r we have from Faraday's law emf ~ enclosed area ~ πr², but this occurs over a circuit length of 2πr. Hence the circulating field strength |E| = emf/length ~ πr²/(2πr) ~ 1/(2r). The article itself doesn't contradict that of course.]

Computing the integral of E around a loop that's not centered seems messy, but I expect it will also give the correct answer of being proportional to the area. I haven't verified this personally, but if the magnetic vector potential given in the above reference is correct, it has to work that way.

The whole point of the integral form of Faraday's law is that only the quantity of threading flux counts, so yes shifting the circuit off center makes no difference.

 

[vanhees71]

The cited source gives (without derivation however) the solution for the vector potential for a DC carrying solenoid (coil). This stationary situation has nothing to do with Faraday's law since in this case the equations for the electric and the magnetic components of the electromagnetic fields separate.

For the magnetic components for stationary currents, Maxwell's equations (in Heaviside-Lorentz units) simplify to
\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\mu \vec{j}.
The first equation tells you that there are no magnetic charges, and the second is Ampere's Law.

From the first equation you get, using Helmholtz's decomposition theorem for vector fields,
\vec{B}=\vec{\nabla} \times \vec{A}.
Plugging this into the second equation you get (in Cartesian coordinates!)
\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=\mu \vec{j}.
Since the vector potential, \vec{A} is only determined up to a gradient of an arbitrary scalar field ("gauge invariance"), you can pose the simplifying constraint (Coulomb gauge)
\vec{\nabla} \cdot \vec{A}=0.
Then Ampere's equation simplifies to
\Delta \vec{A}=-\mu \vec{j}.
In the case of the solenoid we have in cylinder coordinates
\vec{j}(\vec{x})=n I \delta(r-R) \vec{e}_{\varphi}.
For the vector potential we make the ansatz
\vec{A}=a(r) \vec{e}_{\varphi}.
The direction is suggested by the direction of the current density, and that the component in this direction depends solely on r by the symmetry of the situation (translation invariance in z direction and rotation invariance around the z axis).
Then the Coulomb-gauge condition is identically fulfilled since
\vec{\nabla} \cdot \vec{A}=0 with our ansatz.

We cannot directly evaluate \Delta \vec{A} in cylinder coordinates, but we have to take twice the curl. After some calculation (or using Mathematica ;-)) one gets
\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=-\frac{\mathrm{d}}{\mathrm{d} r} \left (\frac{1}{r} \frac{\mathrm{d}(r a)}{\mathrm{d} r} \right ) \vec{e}_{\varphi}.
Now this expression is, according to Ampere's Law 0 everywhere except at the singularity of the \delta distribution. Thus, we first solve the equation
\frac{\mathrm{d}}{\mathrm{d} r} \left (\frac{1}{r} \frac{\mathrm{d}(r a)}{\mathrm{d} r} \right )=0.
Successive integration leads to
a(r)=C_1 r + \frac{C_2}{r}.
To determine the integration constants we have to consider the \delta distribution, and these constant should be different for r<R and r>R. We can also assume without loss of generality that a(r) \rightarrow 0 for r \rightarrow \infty. That means that
a(r)=C_1 r \Theta(R-r)+\frac{C_2}{r} \Theta(r-R).
At r=R we must have a continuous, i.e.,
\frac{C_2}{R}=C_1 R,
and by integrating the differential equation, including the \delta-distribution on the right-hand side over an infinitesimal interval around r=R, we find
\left . -\frac{1}{r} \frac{\mathrm d}{\mathrm d}(r a) \right |_{r=R-0^+}^{r=R+0^+}=\mu n I.
This gives uniquely
a(r)=\frac{\mu n I}{2} \left [r \Theta(R-r)+\frac{R^2}{r} \Theta(r-R) \right].
For the magnetic field we find
\vec{B}=\vec{\nabla} \times [a(r) \vec{e}_{\varphi}]=\mu n I \Theta(R-r) \vec{e}_z
as is well known also from much more elementary considerations, using the integral form of Ampere's Law and the symmetries of the problem ;-).

 

[Dale]

It appears to me that the field outside the solenoid subtracts from the induction due to the field inside, because it is directed oppositely - what is measured is the induction from the field inside the solenoid minus the part between the outside of the solenoid and the loop. Thus the B field at the location of the wire can't explain Faraday's law.

The first sentence is correct, but the second sentence doesn't follow.

The field outside the solenoid does indeed point essentially in the opposite direction of the field inside the solenoid. So the further away from the solenoid you draw your ring the less flux through the surface. Since the further away from the solenoid the weaker the E field, this is exactly what you would expect from Faraday's law.

 

[Dale]

I completely understand your way of thinking. I actually thought the same initially but according to the article it does not seem to be the true reason, rather it is a problem of the topology of the region outside the solenoid - but I don't understand it too well.
Would you mind reading the 3 first pages of it, and explain to me what the conclusions are? If your idea is actually right after all, then I am happy because I understand what you think. I am just not unsure, on the basis of the article, that you are right.

So, I went over the paper in depth. The paper is essentially unrelated to your lab exercise for the reason that I objected to earlier: the solenoid in the lab exercise is finite, not infinite, and therefore the B is non-zero outside the solenoid. Furthermore, the topological reasoning expressed in the article does not apply for a finite solenoid since with a finite solenoid the space is simply connected topologically.

However, the paper does represent a very elegant solution to the THEORETICAL problem posed by an infinite solenoid. He shows essentially, that for the idealized infinite solenoid the topological features of the space allow for an E field which satisfies both the differential and integral forms of Faraday's law, even though at first glance they appear contradictory.

 

[harrylin]

[..]
The field outside the solenoid does indeed point essentially in the opposite direction of the field inside the solenoid. So the further away from the solenoid you draw your ring the less flux through the surface. Since the further away from the solenoid the weaker the E field, this is exactly what you would expect from Faraday's law.

Sorry, as I tried to make clear, the sign of the induced field is the opposite of what you suggest.

 

[Dale]

I'm not following you. Can you write down in equations or a drawing what you mean?

 

[Per Oni]

It looks to me that aaaa202 wants to visualise what is actually happening in say a toroid transformer. Q-eerus and others provide the perfect maths, but does that satisfy the visual picture of what is going on?
To make inroads towards that quest you have to go back to a much simpler coil. Start with a long straight wire and go through the following process.

First step: consider, no current no B field. Second: start a current wait till the maximum is flowing. Third: calculate B. Fourth: sit down and ponder, how did that field get there? Five: report your conclusions. Six: reverse process back to zero current, consider, where did the field go? What generated the back emf?

If you build up a picture which satisfies you then all you have to do is extend gradually this picture to a coil, to a toroid coil and lastly to a toroid transformer. I have a working model in my mind, which I can’t convey here due to the good rules of this forum. But it works for me. Physics deals with end products, not with visualisations no matter how valid they could be. (hint: I see a moving B field whenever dB/dt occurs).

 

[Q-reeus]

First step: consider, no current no B field. Second: start a current wait till the maximum is flowing. Third: calculate B. Fourth: sit down and ponder, how did that field get there? Five: report your conclusions. Six: reverse process back to zero current, consider, where did the field go? What generated the back emf?

If you build up a picture which satisfies you then all you have to do is extend gradually this picture to a coil, to a toroid coil and lastly to a toroid transformer. I have a working model in my mind, which I can’t convey here due to the good rules of this forum. But it works for me. Physics deals with end products, not with visualisations no matter how valid they could be. (hint: I see a moving B field whenever dB/dt occurs).

Trying to read between the lines here (esp. the last line), is your visualization centered around the notion of 'cutting of flux lines' by any chance?

 

[harrylin]

Hi Dalespam, I now see that we ended up in a series of misunderstandings. That's useless, so here's a retake (with more emphasis added):

[..] If the B-field is zero outside the solenoid, so is also the change of it in time and therefore the curl of E at those points. Where am I going wrong with this statement?

The B-field isn't really zero, just very small. But I don't think that's the point.

Indeed that was not the point, and EMGuy next went straight to the point in that post #4.
However, then the misunderstandings started:

[to aaaa] Not true.
http://en.wikipedia.org/wiki/Soleno...ctor_potential_for_finite_continuous_solenoid
You are starting from some false assumptions, and reaching erroneous conclusions. The field outside the solenoid in your lab experiment was NOT zero.

It appears to me that the field outside the solenoid subtracts from the induction due to the field inside, because it is directed oppositely - what is measured is the induction from the field inside the solenoid minus the part between the outside of the solenoid and the loop. Thus the B field at the location of the wire can't explain Faraday's law.

The B-field at the location of the wire is nearly zero and as I next pointed out, it's even in the wrong direction; that field is irrelevant (what matters is the enclosed field). The assumption that the B-field outside the coil is zero does not lead to an erroneous conclusion here.

 

[Dale]

The assumption that the B-field outside the coil is zero does not lead to an erroneous conclusion here.

The assumption that the B-field outside the coil is zero leads directly to the erroneous conclusion that the curl of the E field is 0 outside the coil.

 

[Per Oni]

Trying to read between the lines here (esp. the last line), is your visualization centered around the notion of 'cutting of flux lines' by any chance?

Yes it does.
If properly done it will give the correct direction of E-field in the same way as a traveling B field does. Maybe one day it will be thought at college/uni who knows? Note that when dI/dt exist there will be a wave front spreading out from a conductor having the same speed as the speed of light for the medium in which the conductor is located (think antenna). Perhaps you can see therefore that my picture is not that far fetched.

 

[harrylin]

The assumption that the B-field outside the coil is zero leads directly to the erroneous conclusion that the curl of the E field is 0 outside the coil.

Sorry, once more: the local curl of the E-field is irrelevant for the topic (see also post#22). That's where the OP went wrong, as most people explained starting with post #4.

 

[harrylin]

[..] the only place where the B-field is non zero is inside the solenoid itself. The circuit of the resistor was not a part of the inside of the solenoid. So -dB/dt would have to be zero for all points in space except inside the it. And that means no electric field can possibly have been induced from the varying magnetic field in the solenoid into the circuit of the resistor.
So how is the above experiment explained?

What perhaps was not clear, is that it can be misleading to say that the electric field is induced into the circuit.
In fact, the B-field induces an E-field everywhere around the field lines - not just where the circuit is. The induced E-field spreads out all around the solenoid.

Does that help?

 

[Q-reeus]

Yes it does.
If properly done it will give the correct direction of E-field in the same way as a traveling B field does. Maybe one day it will be thought at college/uni who knows? Note that when dI/dt exist there will be a wave front spreading out from a conductor having the same speed as the speed of light for the medium in which the conductor is located (think antenna). Perhaps you can see therefore that my picture is not that far fetched.

For sure there are situations where flux-linkage and flux-cutting can be used interchangeably, like that illustrated towards the bottom of the page here: http://www.thestudentroom.co.uk/showthread.php?t=1296165
On the other hand, take that toroidal (or any 'iron core') transformer case. No appreciable magnetic field ever 'cuts' the conductor windings wrapping around the magnetic core. I think historically there were many proponents of the idea that flux lines could in some way expand and shrink in and out of such magnetic cores. Just where those lines went to or came from was never quite explained. But it fell out of favor as giving physical reality to 'indestructible' lines, rather than viewing them as merely convenient visualizations of strength and direction for a continuous field that in the AC case, simply grows and diminishes in strength 'in place' - i.e. in the core. One situation where flux-cutting still seems to be in vogue is in explaining Faraday disk operation, although special relativity is nowadays seen as the underlying explanation.

 

[Dale]

Sorry, once more: the local curl of the E-field is irrelevant for the topic (see also post#22). That's where the OP went wrong, as most people explained starting with post #4.

Agreed.

 

[Per Oni]

On the other hand, take that toroidal (or any 'iron core') transformer case. No appreciable magnetic field ever 'cuts' the conductor windings wrapping around the magnetic core.

Thanks for your interesting reply. Good to know that in the past people were thinking up similar ideas then I do now.

I’m not sure why you single out iron core transformers. The same rules should apply for any type of transformer. Once one accepts the picture (visual aid) of a magnetic field spreading out from a single conductor it follows for all coils.

Of course the question is, is that a valid picture? Perhaps I could ask: is the picture of field lines valid? How many lines are there around a current carrying conductor? Why did Gauss need field lines but Coulomb did not?
Without Gauss we would perhaps still be in the stone age as far as em theory is concerned because his theory gives an action at a more precise location in space. But all could be explained and viewed as action at a distance.

 

[Q-reeus]

Thanks for your interesting reply.

Happy to provide.

I’m not sure why you single out iron core transformers. The same rules should apply for any type of transformer.

It's just that ferromagnetic core transformers are ubiquitous owing to the huge advantage of using that arrangement - enormous amplification of the relatively tiny magnetic field generated by the conducting windings. Anywhere from 100's up to ~ 1,000,000 times amplification. Also, by being confined to a closed magnetic circuit, the core field does not add unwanted appreciable EM interference to surrounding environment.

Once one accepts the picture (visual aid) of a magnetic field spreading out from a single conductor it follows for all coils.

Not necessarily. It does not follow that what applies to a straight wire carries over to all coil geometries. You get cancellation of magnetic field outside of an 'infinitely long' solenoid, or more practically, a toroidal geometry (analogy - electrostatic field of a charged infinitely long cylinder or torus appears only on the outside, none inside). So, if the primary windings are inside the secondary windings, none or very little field owing to the primary ever intersects the secondary windings. Taking advantage of the huge ferromagnetic core amplifications mentioned above, the kind of 'inverted toroid' geometries mentioned in #39 allow that very little of the net threading flux intersects either primary or secondary windings.

Of course the question is, is that a valid picture? Perhaps I could ask: is the picture of field lines valid? How many lines are there around a current carrying conductor? Why did Gauss need field lines but Coulomb did not?

The advantage of thinking in terms of field lines is that it allows a geometrically and visually neat way of explaining certain relationships. Simply by positing that electrostatic E field lines must begin and end on charge, we arrive at Gauss's law. But Coulomb's mathematical expression for field strength as a function of distance yields the same thing. There is never a need to use the concept of lines, but it can help. In the magnetic case it helps to know that expressed in terms of lines of B, they always form closed loops, given the absence of monopoles. But Biot-Savart expression just gives field strength and direction as a function of source current - no need for lines to enter the picture. And that's really where it's at. The source of both E and B field is always source currents - although, and this point I emphasize, in the case of ferromagnetic media as source, those currents, treated as a collection of Amperian loop currents, are formal and not real. It's a mistake then to think that time-changing B 'causes' an emf in a coil. There is an association between B and E, not a cause-effect relationship. Source of both is always currents - both actual moving charge, and formal as static or time-changing magnetization.

Without Gauss we would perhaps still be in the stone age as far as em theory is concerned because his theory gives an action at a more precise location in space. But all could be explained and viewed as action at a distance.

That last point is questionable when it comes to explaining EM waves. Field concept is there really indispensable, at least according to the overwhelming majority.

 

[andrien]

outside the solenoid,if one want to calculate induced electric field then considering a circular path whose radius R>r,where r is the radius of solenoid.we will have according to faraday law,

E.2∏R=-∂∅/∂t, where ∅=B.∏r²
clearly there is an induced electric field outside when B is varied because of faraday law.
however the region must include changing magnetic field location( the big circle must contain the changing B region).

 

[harrylin]

[..] I see a moving B field whenever dB/dt occurs).

That should give the same answer, but it's a different picture from that of the OP. Faraday's law works for two different cases (and the second is even a whole class, related to relative motion!).
- What the OP refers to is the picture of constant magnetic flux lines, but of varying intensity.
- You seem to picture moving flux lines, such as when a magnet is moved towards a conductor.

- http://en.wikipedia.org/wiki/Faraday's_law_of_induction

 

[Per Oni]

It's just that ferromagnetic core transformers are ubiquitous owing to the huge advantage of using that arrangement - enormous amplification of the relatively tiny magnetic field generated by the conducting windings. Anywhere from 100's up to ~ 1,000,000 times amplification. Also, by being confined to a closed magnetic circuit, the core field does not add unwanted appreciable EM interference to surrounding environment

The way I see the contribution of say a ferromagnetic core is that magnetic randomly aligned domains will be starting to line up. The only way their flux can go around the toroid is by traveling across the air gap (the inner ring) of the toroid. In doing so they will cut all windings.

You get cancellation of magnetic field outside of an 'infinitely long' solenoid, or more practically, a toroidal geometry (analogy - electrostatic field of a charged infinitely long cylinder or torus appears only on the outside, none inside). So, if the primary windings are inside the secondary windings, none or very little field owing to the primary ever intersects the secondary windings. Taking advantage of the huge ferromagnetic core amplifications mentioned above, the kind of 'inverted toroid' geometries mentioned in #39 allow that very little of the net threading flux intersects either primary or secondary windings

.
All the same cancellations will apply in my picture. The vast majority of field lines end up inside the toroid.

I can very well understand the objections to my non scientific approach, however, this model works for me.

That last point is questionable when it comes to explaining EM waves. Field concept is there really indispensable, at least according to the overwhelming majority

OK no problem accepting that.

 

[Q-reeus]

The way I see the contribution of say a ferromagnetic core is that magnetic randomly aligned domains will be starting to line up. The only way their flux can go around the toroid is by traveling across the air gap (the inner ring) of the toroid. In doing so they will cut all windings.

Yes it's proper to think in terms of individual domains aligning with the applied H field. But even without taking into account the mutual effect of adding all such domain fields - which leads to net cancellation exterior to the core, only some typically small fraction of any given domain's field lines intersect the windings. True regardless of domain orientation initially or finally. That becomes evident by studying the field of a dipole - e.g. http://en.wikipedia.org/wiki/Magnetic_dipole In terms of the domain's magnetic contribution to the -d∅/dt emf, all of it's field lines contribute [strictly true only if the core is driven to saturation]. So on that basis alone there is for your model a deficit to explain - no? [STRIKE]Another not so small factor is that, while some lines of a given domain may cut the windings, it's always overall a double-cut, that cancels - i.e. out+in = 0![/STRIKE] [Edit: Actually, thinking about it in terms of individual fictitious monopoles at each end of a magnetic dipole, there can be validity to the notion of line cutting. But given the overall exterior field cancellation, it's not one I'd use. But I do in hindsight concede it can have a certain use even in transformer core case]

OK no problem accepting that.

Glad we have some agreement.