## Main Question or Discussion Point

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?

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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

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 thats 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.

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

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 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
Mentor
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.

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.

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.

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.

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)

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.

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?

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.

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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.

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.

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

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
Mentor
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?

Not true.

Not true.

http://en.wikipedia.org/wiki/Solenoid#Magnetic_field_and_vector_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.

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.

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!

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).