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

[aaaa202]

Today we did an experiment with a solenoid through which we varied the current. Around this solenoid, was a different circuit (not coupled to the coil-circuit) where we measured the potential across a resistor. By varying the current my teacher showed how the potential over the resistor would vary.
Now faradays law essentially says that a time varying magnetic field induces a rotating electric field.
∇xE = -dB/dt
So I thought that the above experiment could be explained by the fact that the changing magnetic field induces an electric field which accounted for the electromotive force induced in the other circuit and thus any drop or increase in potential.
BUT! As my friend correctly stated, 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?

 

[Q-reeus]

Today we did an experiment with a solenoid through which we varied the current. Around this solenoid, was a different circuit (not coupled to the coil-circuit) where we measured the potential across a resistor. By varying the current my teacher showed how the potential over the resistor would vary.
Now faradays law essentially says that a time varying magnetic field induces a rotating electric field.
∇xE = -dB/dt
So I thought that the above experiment could be explained by the fact that the changing magnetic field induces an electric field which accounted for the electromotive force induced in the other circuit and thus any drop or increase in potential.
BUT! As my friend correctly stated, 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?

Curl of E is only non-zero where a B field exists also, but this does not mean the circulation of E requires magnetic field lines coincident with E field. The integral version of Faraday's law simply states the line integral of E around a closed path equals the negative of the time-changing rate of magnetic flux threading through such a closed path. Strictly speaking, for such transformer action situation, E is given at any point by minus the time-changing rate of A, the vector potential, and there can certainly be a non-zero A in the absence of a B at that point. Just check out the Wiki page http://en.wikipedia.org/wiki/Faraday's_law_of_induction

 

[aaaa202]

Come again please..

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?

 

[EMGuy101]

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

Take the integral form of Faradays law: i.e. that the surface integral of dB/dt is equal to the line integral of E around a closed contour which bounds that surface. Imagine then drawing a closed loop around your solenoid and any open surface which has this closed loop as its boundary. Is there is a non-zero flux of dB/dt going through that surface? If so, then the line integral of E around that loop (which is not inside of the solenoid) has a non-zero E.

 

[cmos]

Why should the B-field be zero outside the solenoid? You don't have an infinitely-long solenoid, do you?

 

[aaaa202]

Well suppose we had. Would the experiment still yield the same? Either way we can savely assume that the field is very small outside and thus the change in that should not induce a very strong electric field.

 

[Q-reeus]

Come again please..
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?

As per my #2, and #4, one has to distinguish between the differential property curl E as 'vorticity' of the field, which is an intensive property of E, and induced emf around a closed path, which is an extensive or 'global' property. Stick with the definition E = -∇phi -∂A/∂t and you won't go wrong.

 

[Dale]

BUT! As my friend correctly stated, the only place where the B-field is non zero is inside the solenoid itself.

This is only correct if you have an infinitely long solenoid.

 

[EMGuy101]

Well suppose we had. Would the experiment still yield the same? Either way we can savely assume that the field is very small outside and thus the change in that should not induce a very strong electric field.

Yes, see my post.

 

[cmos]

Well suppose we had. Would the experiment still yield the same? Either way we can savely assume that the field is very small outside and thus the change in that should not induce a very strong electric field.

If you had an infinitely long solenoid, then no, the experiment would not have the same result. Regardless of length, why do you assume the exterior B-field to be "very small?" If this were the case, then transformers wouldn't exist.

 

[aaaa202]

I can see that the flux of b through a closed area bounding the solenoid is non-zero. Thus you conclude that the integral of E around a closed path bounding the solenoid is non-zero - is that correctly understand.
Problem is just that doesn't get us very far. For where did this E-field generating a non-zero value for a closed line integral come from? It must come from the changing magnetic field. But that says ∇xE = -dB/dt
problem is just that the above would give zero. It seems the integral and differential form conflict with each other.

 

[EMGuy101]

If you had an infinitely long solenoid, then no, the experiment would not have the same result. Regardless of length, why do you assume the exterior B-field to be "very small?" If this were the case, then transformers wouldn't exist.

The magnetic field lines outside the solenoid tend to much "further" spaced apart than inside the solenoid (i.e. the flux outside << flux inside)

 

[EMGuy101]

I can see that the flux of b through a closed area bounding the solenoid is non-zero. Thus you conclude that the integral of E around a closed path bounding the solenoid is non-zero - is that correctly understand.
Problem is just that doesn't get us very far. For where did this E-field generating a non-zero value for a closed line integral come from? It must come from the changing magnetic field. But that says ∇xE = -dB/dt
problem is just that the above would give zero. It seems the integral and differential form conflict with each other.

I think you're getting a little bit confused. Think of Ampere's law. A magnetic field curls around some current. Imagine that current is due to a wire. Now imagine you are calculating the magnetic field far away from the wire. You still have a magnetic field even though there is no current close to where you are.

 

[EMGuy101]

I can see that the flux of b through a closed area bounding the solenoid is non-zero. Thus you conclude that the integral of E around a closed path bounding the solenoid is non-zero - is that correctly understand.
Problem is just that doesn't get us very far. For where did this E-field generating a non-zero value for a closed line integral come from? It must come from the changing magnetic field. But that says ∇xE = -dB/dt
problem is just that the above would give zero. It seems the integral and differential form conflict with each other.

Another thing, take the line integral of E around a closed loop. But now have the loop off to the side of the solenoid. That line integral will equal to zero because the - and + contributions will cancel each other out. BUT even though the line integral of E is zero, that doesn't mean that E itself is zero at any point. Does that help a little bit to give you a conceptual understanding?

 

[aaaa202]

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.

 

[Q-reeus]

I can see that the flux of b through a closed area bounding the solenoid is non-zero. Thus you conclude that the integral of E around a closed path bounding the solenoid is non-zero - is that correctly understand.

Yes. It is the manifestation of the integral form of Faraday's law - and that's the only one applicable to your case.

Problem is just that doesn't get us very far. For where did this E-field generating a non-zero value for a closed line integral come from? It must come from the changing magnetic field. But that says ∇xE = -dB/dt
problem is just that the above would give zero. It seems the integral and differential form conflict with each other.

No they don't. As stated in #7, the curl E version yields an intensive field property, valid 'at a point'. You want the net voltage around a circuit, then use the integral form, which only requires that some value of flux threads the circuit of interest. Note that this works perfectly in the case of a toroidal transformer configuration, which if properly constructed, has precisely zero magnetic field intersecting the secondary windings. Again, if in doubt always refer back to the definition for E, which here amounts to -∂A/∂t, as electrostatic fields are irrelevant to emf.

 

[aaaa202]

I find it hard to distinguish between the cases. If you look at the line integral around the solenoid as a sum of E*dr and have at each point that E is zero then WHY would the integral not yield a big, whopping zero?
Please take a look at the first page of the article attached above so you are sure what my problem is.
You are probably right, but I don't see it with your explanation.

 

[EMGuy101]

Just because curl(E) is zero doesn't mean that E is zero.

 

[EMGuy101]

Think of curl(E) as a line integral over an infinitesimally small loop. Thus curl(E) will be zero outside the solenoid because the sum of each of those infinitesimally small sides will equal to zero. Similarly you could deduce that curl(E) would be equal to zero over that infinitesimally small loop because if you use faradays integral form / Stroke's theorem you will see that there is no dB/dt going through the loop.

 

[Dale]

But that says ∇xE = -dB/dt
problem is just that the above would give zero.

Not true.

The field is ZERO everywhere except inside the solenoid. Google the field of a solenoid.

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.

 

[Q-reeus]

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.

A skim of that article makes it clear the circuit(s) considered introduce complexities not presented in your #1, which all assumed involved a simple case of measurement across the terminals of a secondary winding. The circuits in that article require application of Kirchoff's circuital laws in general, and obscure the simplicity of just dealing with application of Faraday's law in integral form. You agree transformers work, right?

 

[EMGuy101]

Not true.

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.

Something that might help you understand why this is true is magnetic field lines have to close on themselves (i.e. no such thing as a magnetic monopole). The only way for them to close on themselves is for them to loop back around outside the solenoid. Irregardless of this, if the magnetic field was zero outside (a theoretical situation) then you would still have an electric field outside the solenoid.

 

[aaaa202]

The article introduces exactly the problem that I introduced. Probably, I did not explain it very well though. Let me quote:
"Because curl E = 0 outside everywhere outside the solenoid it is tempting to conclude that the line integral between any two points is path independent. This is not true because of the topology of this region. Even though curl E vanishes in this region it is not simply connected and thus ∫Edr is not necessarily path independent."
Can anyone explain what the author tries to say with the region being not simply connected?
Dale: For a long solenoid the field is practically zero outside, and thus changes in dB are sufficiently small to be ignored. Have a look at the article posted if you have trouble understanding my problem.

 

[aaaa202]

Something that might help you understand why this is true is magnetic field lines have to close on themselves (i.e. no such thing as a magnetic monopole). The only way for them to close on themselves is for them to loop back around outside the solenoid. Irregardless of this, if the magnetic field was zero outside (a theoretical situation) then you would still have an electric field outside the solenoid.

I know that in reality the field outside will be there. It will just be extremely weak and not at all able to produce significant changes in the E-field. Indeed, as stated in the article, the solution to the problem is NOT that the B-field outside is always there but just extremely weak.
I don't see what you mean by "curl of E = 0 doesn't mean E is zero". That is not the point - of course there is an electric field outside the solenoid - but it is conservative. The whole idea of this experiment is to show that the electric field outside the solenoid is not conservative - something which DOES require curl E = 0!

 

[EMGuy101]

The article introduces exactly the problem that I introduced. Probably, I did not explain it very well though. Let me quote:
"Because curl E = 0 outside everywhere outside the solenoid it is tempting to conclude that the line integral between any two points is path independent. This is not true because of the topology of this region. Even though curl E vanishes in this region it is not simply connected and thus ∫Edr is not necessarily path independent."
Can anyone explain what the author tries to say with the region being not simply connected?
Dale: For a long solenoid the field is practically zero outside, and thus changes in dB are sufficiently small to be ignored. Have a look at the article posted if you have trouble understanding my problem.

Maybe you should take the time to write out EXACTLY what it is you do not understand. I am not sure what it is that is confusing you. I believe that quote is essentially saying that the E field outside the solenoid is not a conservative field (as one might assume if you say that the curl(E)=0).

 

[aaaa202]

what? curl(E)=0 => conservative field. But our field is not conservative. Please read the 2 first pages of the article, it shouldn't take that long and it describes exactly the problem I am trying to adress.

 

[Q-reeus]

The article introduces exactly the problem that I introduced. Probably, I did not explain it very well though. Let me quote:
"Because curl E = 0 outside everywhere outside the solenoid it is tempting to conclude that the line integral between any two points is path independent. This is not true because of the topology of this region. Even though curl E vanishes in this region it is not simply connected and thus ∫Edr is not necessarily path independent."
Can anyone explain what the author tries to say with the region being not simply connected?

Path independence (meaning 'simply connected' as applied here) is the property of a conservative potential - i.e. electrostatic case where E = -∇phi applies. Thus if there is a charged capacitor, no matter which path taken, moving from any point on one plate to any point on the other plate, the potential difference will be identical regardless of the path taken. In the case of transformer action where E = -dA/dt, net emf owes to a non-conservative field and is entirely dependent on the path taken. So, a secondary with N turns of wire circling the core where flux is concentrated, will read N times as much voltage across the terminals as for a single turn circuit. Those 'split circuits' in the article manifest the difference between these two basic situations.

 

[Dale]

Dale: For a long solenoid the field is practically zero outside, and thus changes in dB are sufficiently small to be ignored. Have a look at the article posted if you have trouble understanding my problem.

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.

 

[aaaa202]

I know nothing about transformers.
You have probably said the solution to my problem in the previous posts but I just don't understand it. So let us for a last time try to come to a conclusion - else I will have to ask my teacher instead but last time he said he couldn't explain it properly and referred me to read the article above.

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

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

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

 

[aaaa202]

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.

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.

 

[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&lt;R and r&gt;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.