Faraday’s Law Paradox

[PengKuan Em]

Faraday’s Law Paradox
9 October 2012

Faraday’s law defines how a varying magnetic field creates electric field. For the potential to be nonzero, EF must be nonzero. However, if we look at the electric field inside the conductor, it is not so. According to an electrostatic law, the force on the free electron must be zero; otherwise, the distribution of free electrons will change so that the electric field on the free electron becomes zero. On the surface, the electric field Es must be perpendicular to the surface. Considering such electric field in conductor, the potential between the point A and D is zero.

How can free electrons stay still against nonzero EMF? And how can EMF exist when the electric field on all free electrons is zero or perpendicular to the surface? No explanation exists now. So, there is a conflict between Faraday’s law and the electrostatic law. I call this conflict the Faraday’s law Paradox. Faraday’s law is the last major law of the electromagnetic theory to fall.


Please read the article at
http://pengkuanem.blogspot.fr/2012/10/faradays-law-paradox.html
http://independent.academia.edu/KuanPeng/Papers/1997018/Faradays_Law_Paradox

[ti...@physics.uq.edu.au]

On Tuesday, October 9, 2012 9:56:46 AM UTC+10, PengKuan Em wrote:
>
> For the potential to be nonzero, EF must be nonzero.

No. For the potential to change, EF must be non-zero.

> However, if we look at the electric field inside the conductor, it is not so.

No. It's common, quite usual, to have a non-zero electric field in a conductor.

> According to an electrostatic law, the force on the free electron must be zero; otherwise, the distribution of free electrons will change so that the electric field on the free electron becomes zero.

Yes, this is the case in _electrostatics_. This is not the case in general. The difference matters.

What is electrostatics? Electrostatics assumes that (a) E is constant in time, (b) J = 0, (c) E = 0 in conductors, treating all conductors as _perfect_ electrical conductors, and (d) the electric field E is fully described by the electrostatic potential Phi.

Note well that (d) requires curl(E)=0. So in the case presented in your linked article, (d) is not satisfied, and (b) is not satisfied. So, why should it be a great surprise to see that (c) is not satisfied, either?

> So, there is a conflict between Faraday’s law and the electrostatic law. I call this conflict the Faraday’s law Paradox.

So, a conflict between electrostatics, and non-electrostatics! Amazing!

Here's another similar paradox: connect a conducting wire across the terminals of a battery. A current flows in the wire! That means that E is non-zero in the conductor! How can that be, when electrostatics says E=0? You could call this the Battery Paradox.

Have you run out good paradoxes?

[Salmon Egg]

In article <56a959f7-bca6-4afd...@googlegroups.com>,
PengKuan Em <tita...@gmail.com> wrote:

> Faradayąs Law Paradox
> 9 October 2012
>
> Faradayąs law defines how a varying magnetic field creates electric field.

> For the potential to be nonzero, EF must be nonzero. However, if we look at
> the electric field inside the conductor, it is not so. According to an
> electrostatic law, the force on the free electron must be zero; otherwise,
> the distribution of free electrons will change so that the electric field on
> the free electron becomes zero. On the surface, the electric field Es must be
> perpendicular to the surface. Considering such electric field in conductor,
> the potential between the point A and D is zero.
>
> How can free electrons stay still against nonzero EMF? And how can EMF exist
> when the electric field on all free electrons is zero or perpendicular to the

> surface? No explanation exists now. So, there is a conflict between Faradayąs
> law and the electrostatic law. I call this conflict the Faradayąs law
> Paradox. Faradayąs law is the last major law of the electromagnetic theory to

> fall.
>
>
> Please read the article at
> http://pengkuanem.blogspot.fr/2012/10/faradays-law-paradox.html
> http://independent.academia.edu/KuanPeng/Papers/1997018/Faradays_Law_Paradox

The conductor that requires zero electric field is a perfect conductor,
Real conductors have resistance. The electric field inside this
"conductor" accelerates the carrier (often but not always electrons). In
a perfect conductor, the would accelerate forever. In real conductors
they bump into things and have to start all over again.

--

Sam

Conservatives are against Darwinism but for natural selection.
Liberals are for Darwinism but totally against any selection.

[bja...@iwaynet.net]

On 10/8/2012 7:56 PM, PengKuan Em wrote:
> Faraday’s Law Paradox
> 9 October 2012

> How can free electrons stay still against nonzero EMF?

And how can EMF exist when the electric field on all free

electrons is zero or perpendicular to the surface? No

explanation exists now. So, there is a conflict between

Faraday’s law and the electrostatic law. I call this

conflict the Faraday’s law Paradox. Faraday’s law is

the last major law of the electromagnetic theory to fall.

> Please read the article at
> http://pengkuanem.blogspot.fr/2012/10/faradays-law-paradox.html
> http://independent.academia.edu/KuanPeng/Papers/1997018/Faradays_Law_Paradox

And who says electrons stay still? Put a copper loop in an induced
field which produces an EMF and current flows. Duh!

So where is your paradox? You have an electrostatic E field which
applies and an induces E field that applies. The electrons obviously
experience BOTH E fields as a force that moves them. So if the
Electrostatic E field is zero then the force moving the electrons is the
EMF E field. And they move...it's called a current.

Didn't you try to have this discussion once before and also got nowhere
with it?

[PengKuan Em]

Le mardi 9 octobre 2012 11:18:48 UTC+2, bja...@teranews.com a écrit :
>
>
> And who says electrons stay still? Put a copper loop in an induced
>
> field which produces an EMF and current flows. Duh!
>

In my case, the wire in cut between A and D, there is not current. SO, the electrons are still

>
> So where is your paradox? You have an electrostatic E field which
>
> applies and an induces E field that applies. The electrons obviously
>
> experience BOTH E fields as a force that moves them. So if the
>
> Electrostatic E field is zero then the force moving the electrons is the
>
> EMF E field. And they move...it's called a current.

As the current is zero, there must be a counter E field that opposes exactly the EMF, because the electron distribution makes this counter E.

>
> Didn't you try to have this discussion once before and also got nowhere
> with it?

This the first Faraday's law discussion in 5 years. But I did discuss about other laws.

[PengKuan Em]

Le mardi 9 octobre 2012 04:05:08 UTC+2, (inconnu) a écrit :
> On Tuesday, October 9, 2012 9:56:46 AM UTC+10, PengKuan Em wrote:
>
> >
>
> > For the potential to be nonzero, EF must be nonzero.
>
>
>
> No. For the potential to change, EF must be non-zero.
>
>
>
> > However, if we look at the electric field inside the conductor, it is not so.
>
>
>
> No. It's common, quite usual, to have a non-zero electric field in a conductor.
>
>
>
> > According to an electrostatic law, the force on the free electron must be zero; otherwise, the distribution of free electrons will change so that the electric field on the free electron becomes zero.
>
>
>
> Yes, this is the case in _electrostatics_. This is not the case in general. The difference matters.
>
>
>
> What is electrostatics? Electrostatics assumes that (a) E is constant in time, (b) J = 0, (c) E = 0 in conductors, treating all conductors as _perfect_ electrical conductors, and (d) the electric field E is fully described by the electrostatic potential Phi.
>
>
>
> Note well that (d) requires curl(E)=0. So in the case presented in your linked article, (d) is not satisfied, and (b) is not satisfied. So, why should it be a great surprise to see that (c) is not satisfied, either?
>
>
>
> > So, there is a conflict between Faraday’s law and the electrostatic law. I call this conflict the Faraday’s law Paradox.
>
>
>
> So, a conflict between electrostatics, and non-electrostatics! Amazing!
>

The loop is cut between A and D ; there is not current. The conflict appears when current is 0.

>
> Here's another similar paradox: connect a conducting wire across the terminals of a battery. A current flows in the wire! That means that E is non-zero in the conductor! How can that be, when electrostatics says E=0? You could call this the Battery Paradox.
>

The is not current. In the battery case, if the wire is cut, the E field in the conductor is 0 and perpendicular to the surface.

>
> Have you run out good paradoxes?

This is a good paradox, only if one think carefully about it.

[PengKuan Em]

Le mardi 9 octobre 2012 10:56:09 UTC+2, Salmon Egg a écrit :
>
>
>
> > Faraday�s Law Paradox
>
> > 9 October 2012
>
> >
>
> > Faraday�s law defines how a varying magnetic field creates electric field.

>
> > For the potential to be nonzero, EF must be nonzero. However, if we look at
>
> > the electric field inside the conductor, it is not so. According to an
>
> > electrostatic law, the force on the free electron must be zero; otherwise,
>
> > the distribution of free electrons will change so that the electric field on
>
> > the free electron becomes zero. On the surface, the electric field Es must be
>
> > perpendicular to the surface. Considering such electric field in conductor,
>
> > the potential between the point A and D is zero.
>
> >
>
> > How can free electrons stay still against nonzero EMF? And how can EMF exist
>
> > when the electric field on all free electrons is zero or perpendicular to the
>

> > surface? No explanation exists now. So, there is a conflict between Faraday�s

>
> > law and the electrostatic law. I call this conflict the Faraday�s law
>

> > Paradox. Faraday�s law is the last major law of the electromagnetic theory to

>
> > fall.
>
> >
>
> >
>
> > Please read the article at
>
> > http://pengkuanem.blogspot.fr/2012/10/faradays-law-paradox.html
>
> > http://independent.academia.edu/KuanPeng/Papers/1997018/Faradays_Law_Paradox
>
>
>
> The conductor that requires zero electric field is a perfect conductor,
>
> Real conductors have resistance. The electric field inside this
>
> "conductor" accelerates the carrier (often but not always electrons). In
>
> a perfect conductor, the would accelerate forever. In real conductors
>
> they bump into things and have to start all over again.
>
>
>

There is not current because the wire is cut. In this case, the conductor is a perfect conductor.

[glen herrmannsfeldt]

Salmon Egg <Salm...@sbcglobal.net> wrote:

(snip)

> The conductor that requires zero electric field is a perfect conductor,
> Real conductors have resistance. The electric field inside this
> "conductor" accelerates the carrier (often but not always electrons).

Reasonably often holes. The metal most used for electric power
transmission (that is, long distance transmission lines) is aluminum.

Aluminum has both electron and hole bands that contribute to
conduction, with the electron band a little more at low
magnetic fields. As the magnetic field increases, the hole bands
become more important, such that the Hall coefficient goes
positive. I don't know what the fields are in long distance,
high voltage transmission lines, but there is still a lot of
hole conduction.)

> In a perfect conductor, the would accelerate forever. In real
> conductors they bump into things and have to start all over again.

There is also ion conduction in electrolytes and protons in
proton accelerators.

-- glen

[Jos Bergervoet]

On 10/9/2012 10:56 AM, Salmon Egg wrote:
> In article <56a959f7-bca6-4afd...@googlegroups.com>,
> PengKuan Em <tita...@gmail.com> wrote:
>

>> Faraday�s Law Paradox
>> 9 October 2012
>>
>> Faraday�s law defines how a varying magnetic field creates electric field.

>> For the potential to be nonzero, EF must be nonzero. However, if we look at
>> the electric field inside the conductor, it is not so. According to an
>> electrostatic law, the force on the free electron must be zero; otherwise,
>> the distribution of free electrons will change so that the electric field on
>> the free electron becomes zero.

...
...

> The conductor that requires zero electric field is a perfect conductor,
> Real conductors have resistance. The electric field inside this
> "conductor" accelerates the carrier (often but not always electrons).

And the resistance is not the only cause for nonzero
electric field. Even when there is no current at all,
there can be an E-field. Most clearly in semi-conductor
P-N junctions, where you have a "built-in potential"
often about 0.8 V.

But also in purely N-type material where there is
inhomogeneity in the density of states there will be
a drift field, which is described by a static potential
(the thermal voltage, kT/e, times the natural logarithm
of the density of donor states, i.e. the doping
concentration).

Strictly speaking, in any conductor where there is
inhomogeneity in the band structure there will be a
drift field. A zero E-field is only present, therefore,
in a current free, completely homogeneous conductor.

--
Jos

[Larry Harson]

On Tuesday, October 9, 2012 12:56:46 AM UTC+1, PengKuan Em wrote:
> Faraday’s Law Paradox
>
> 9 October 2012
>
>
>
> Faraday’s law defines how a varying magnetic field creates electric field. For the potential to be nonzero, EF must be nonzero. However, if we look at the electric field inside the conductor, it is not so. According to an electrostatic law, the force on the free electron must be zero; otherwise, the distribution of free electrons will change so that the electric field on the free electron becomes zero. On the surface, the electric field Es must be perpendicular to the surface. Considering such electric field in conductor, the potential between the point A and D is zero.
>
>
>
> How can free electrons stay still against nonzero EMF? And how can EMF exist when the electric field on all free electrons is zero or perpendicular to the surface? No explanation exists now. So, there is a conflict between Faraday’s law and the electrostatic law. I call this conflict the Faraday’s law Paradox. Faraday’s law is the last major law of the electromagnetic theory to fall.

It's not a paradox, but a useful question to test people's understanding of the physics rather than a blind application of Maxwell's equations.

Put a conductor in an external static electric field. There's now an internal static electric field inside the conductor which causes charge to move to the surface and generate an electric field that cancels the applied one.

The same sort of thing happens for an external induced electric field, but a non conservative electric field can't generally be cancelled by a conservative static electric field inside a volume, as discussed here a few weeks back. For a wire, the cancellation becomes more perfect as its dimensions approach one.

Larry.

[ti...@physics.uq.edu.au]

On Wednesday, October 10, 2012 2:14:18 AM UTC+10, PengKuan Em wrote:
> Le mardi 9 octobre 2012 04:05:08 UTC+2, (inconnu) a écrit :
>
> > So, a conflict between electrostatics, and non-electrostatics! Amazing!
>
> The loop is cut between A and D ; there is not current. The conflict appears when current is 0.

This was discussed very recently here.

https://groups.google.com/forum/?hl=en&fromgroups=#!topic/sci.physics.electromag/S3Y8w1haYwE

Essentially, your assumption that the current is zero is wrong. The induced field has curl(E) = non-zero, the field due to surface charges has curl(E) = 0 - they can't cancel over a volume.

The people who design transformers know about this. They design them to minimise the the currents.

> > Here's another similar paradox: connect a conducting wire across the terminals of a battery. A current flows in the wire! That means that E is non-zero in the conductor! How can that be, when electrostatics says E=0? You could call this the Battery Paradox.
>
> The is not current. In the battery case, if the wire is cut, the E field in the conductor is 0 and perpendicular to the surface.

Once equilibrium is reached, yes. This case is different from the induced field case, since here curl(E) = 0.

But still, with the wire not cut, you have a current in a conductor in steady-state. So, E is non-zero in the conductor. You wrote that the paradox was that electrostatics says that E=0 in conductors. If you say that the battery-and-wire isn't another example of the paradox, you were unclear.

Either way, it comes down to assuming that E=0 (as it would in electrostatics) is true when it isn't.

[Salmon Egg]

In article <4287060e-b940-4f12...@googlegroups.com>,

PengKuan Em <tita...@gmail.com> wrote:

> There is not current because the wire is cut. In this case, the conductor is
> a perfect conductor.

Then surface charges build up to cancel the E field. That is just the
same as what happens in a conductor placed into an electrostatic field,
The field distorts so as to contact the conducting surface normally.

What is the big deal?

--

[Szczepan Bialek]

"Larry Harson" <larry...@softhome.net> napisal w wiadomosci
news:450bd50a-61cf-4e61...@googlegroups.com...

On Tuesday, October 9, 2012 12:56:46 AM UTC+1, PengKuan Em wrote:
> Faraday’s Law Paradox
>
> 9 October 2012
>
>
>
> Faraday’s law defines how a varying magnetic field creates electric field.
> For the potential to be nonzero, EF must be nonzero. However, if we look
> at the electric field inside the conductor, it is not so. According to an
> electrostatic law, the force on the free electron must be zero; otherwise,
> the distribution of free electrons will change so that the electric field
> on the free electron becomes zero. On the surface, the electric field Es
> must be perpendicular to the surface. Considering such electric field in
> conductor, the potential between the point A and D is zero.
>
>
>
> How can free electrons stay still against nonzero EMF? And how can EMF
> exist when the electric field on all free electrons is zero or
> perpendicular to the surface? No explanation exists now. So, there is a
> conflict between Faraday’s law and the electrostatic law. I call this
> conflict the Faraday’s law Paradox. Faraday’s law is the last major law of
> the electromagnetic theory to fall.

<It's not a paradox, but a useful question to test people's understanding of
the physics rather than a blind application of Maxwell's equations.

<Put a conductor in an external static electric field. There's now an
internal static electric field inside the conductor which causes charge to
move to the surface and generate an electric field that cancels the applied
one.

Wrong. In Maxwell's definition the electric field is if the charge flows or
"want" to flow.
So the charges moved to ths surface do not cancel the applied one. They want
to flow but they can not.
S*

[PengKuan Em]

Le mercredi 10 octobre 2012 01:02:02 UTC+2, Larry Harson a écrit :
> On Tuesday, October 9, 2012 12:56:46 AM UTC+1, PengKuan Em wrote:
>
> > Faraday’s Law Paradox
>
> >
>
> > 9 October 2012
>
> >
>
> >
>
> >
>
> > Faraday’s law defines how a varying magnetic field creates electric field. For the potential to be nonzero, EF must be nonzero. However, if we look at the electric field inside the conductor, it is not so. According to an electrostatic law, the force on the free electron must be zero; otherwise, the distribution of free electrons will change so that the electric field on the free electron becomes zero. On the surface, the electric field Es must be perpendicular to the surface. Considering such electric field in conductor, the potential between the point A and D is zero.
>
> >
>
> >
>
> >
>
> > How can free electrons stay still against nonzero EMF? And how can EMF exist when the electric field on all free electrons is zero or perpendicular to the surface? No explanation exists now. So, there is a conflict between Faraday’s law and the electrostatic law. I call this conflict the Faraday’s law Paradox. Faraday’s law is the last major law of the electromagnetic theory to fall.
>
>
>
> It's not a paradox, but a useful question to test people's understanding of the physics rather than a blind application of Maxwell's equations.
>
>
>
> Put a conductor in an external static electric field. There's now an internal static electric field inside the conductor which causes charge to move to the surface and generate an electric field that cancels the applied one.
>

Yes, this is right for capacitors.

>
> The same sort of thing happens for an external induced electric field, but a non conservative electric field can't generally be cancelled by a conservative static electric field inside a volume, as discussed here a few weeks back. For a wire, the cancellation becomes more perfect as its dimensions approach one.
>
>

But induce E field cannot be canceled by conservative field. In my example, there is not current, the magnetic field increases constantly and the induced field is constant in the wire. If the E field in the wire is zero, the integration of it will be zero and non voltage appears.


>
> Larry.

[PengKuan Em]

Le mercredi 10 octobre 2012 02:25:33 UTC+2, Timo Nieminen a écrit :
> On Wednesday, October 10, 2012 2:14:18 AM UTC+10, PengKuan Em wrote:
>
> > Le mardi 9 octobre 2012 04:05:08 UTC+2, (inconnu) a écrit :
>
> >
>
> > > So, a conflict between electrostatics, and non-electrostatics! Amazing!
>
> >
>
> > The loop is cut between A and D ; there is not current. The conflict appears when current is 0.
>
>
>
> This was discussed very recently here.
>
>
>
> https://groups.google.com/forum/?hl=en&fromgroups=#!topic/sci.physics.electromag/S3Y8w1haYwE
>
>
>
> Essentially, your assumption that the current is zero is wrong. The induced field has curl(E) = non-zero, the field due to surface charges has curl(E) = 0 - they can't cancel over a volume.
>

The magnetic field increases constantly and the induced E field is constant around the wire. Like in a capacitor, even external E field is zero, there is not current inside.

>
>
> The people who design transformers know about this. They design them to minimise the the currents.
>

This is called the Eddy current which rises in alternate magnetic field. Here the M field is constantly increasing and no Eddy current is there.

> > > Here's another similar paradox: connect a conducting wire across the terminals of a battery. A current flows in the wire! That means that E is non-zero in the conductor! How can that be, when electrostatics says E=0? You could call this the Battery Paradox.
>
> >
>
> > The is not current. In the battery case, if the wire is cut, the E field in the conductor is 0 and perpendicular to the surface.
>
>
>
> Once equilibrium is reached, yes. This case is different from the induced field case, since here curl(E) = 0.
>
>
>
> But still, with the wire not cut, you have a current in a conductor in steady-state. So, E is non-zero in the conductor. You wrote that the paradox was that electrostatics says that E=0 in conductors. If you say that the battery-and-wire isn't another example of the paradox, you were unclear.
>

The induced voltage is V0 and is constant. At equilibrium, the electrons are still. But the integral of E is V0, thus, the E field is V0/l(length of the wire). We have then E=/=0 but electrons are still. Paradox.

[PengKuan Em]

Le mercredi 10 octobre 2012 03:55:09 UTC+2, Salmon Egg a écrit :
>
>
>
> > There is not current because the wire is cut. In this case, the conductor is
>
> > a perfect conductor.
>
>
>
> Then surface charges build up to cancel the E field. That is just the
>
> same as what happens in a conductor placed into an electrostatic field,
>
> The field distorts so as to contact the conducting surface normally.
>
>

The surface charges build up a E field. If you integrate E on the surface around a closed loop, you get 0. For example, from A to D on the internal side then from D to A on the external side. As from A to D is the same value then from D to A, the integral is 0 from A to D.

For induced E field, you have a voltage V0 that is not 0. So, the E field inside the wire is V0/l (length of the wire)

>
> What is the big deal?
>

The deal is we have 0 static voltage but V0 induced voltage.

[Timo Nieminen]

On Thursday, October 11, 2012 12:57:18 AM UTC+10, PengKuan Em wrote:
>
> The magnetic field increases constantly and the induced E field is constant around the wire. Like in a capacitor, even external E field is zero, there is not current inside.

No. Why do you way the current is zero? Look at transformer cores - they're made of laminae so as to minimise the currents.

You think a paradox exists because you assume that there is no current in the conductor. Yet induced currents are measured, and engineering things to minimise them is important. So, no paradox. See, e.g., http://en.wikipedia.org/wiki/Transformer#Cores

If your thought experiment disagrees with well-tested conventional theory, then perhaps conventional theory is wrong (probably the thought experiment is wrong, but sometimes people do succeed in poking holes in conventional theory). If your thought experiment disagrees with reality, then your thought experiment is wrong, and conflict with existing theory is expected.

[Larry Harson]

On Wednesday, October 10, 2012 8:19:41 AM UTC+1, Szczepan Bialek wrote:
> "Larry Harson" <larry...@soft.net> napisal w wiadomosci

Are you sure? His definition of an electric field was the force on a unit charge.

> So the charges moved to ths surface do not cancel the applied one. They want
>
> to flow but they can not.

So where does the cancellation of the applied field come from then?

> S*

[PengKuan Em]

Le jeudi 11 octobre 2012 02:13:49 UTC+2, Timo Nieminen a écrit :
> On Thursday, October 11, 2012 12:57:18 AM UTC+10, PengKuan Em wrote:
>
> >
>
> > The magnetic field increases constantly and the induced E field is constant around the wire. Like in a capacitor, even external E field is zero, there is not current inside.
>
>
>
> No. Why do you way the current is zero? Look at transformer cores - they're made of laminae so as to minimise the currents.
>

Maybe you think of alternate current and magnetic field. In my example, the magnetic field is a linear function of time, its time derivative is constant, so, Faraday's law gives a tension proportional to the rate of increase of magnetic field, that is, V constant. Because the wire is cut, the resistance is infinity, so I=V/R=0. The current is zero.

>
> You think a paradox exists because you assume that there is no current in the conductor. Yet induced currents are measured, and engineering things to minimise them is important. So, no paradox. See, e.g., http://en.wikipedia.org/wiki/Transformer#Cores
>

Here the resistance is infinity, the measured current is 0.

[Larry Harson]

J = \sigma E . For an ideal conductor, the conductivity \sigma = 1/p -> infinity. You can therefore say the electric field inside an ideal conductor is zero, providing the currents are *finite*. But any induced currents are going to be *infinite* unless the induced electric field can be cancelled someway.

For a practical conductor, \sigma will be very small and the induced currents very large unless they can be reduced by laminations. But even laminations can't get the induced currents down to zero exactly because the volumes are finite.

Larry.

[Timo Nieminen]

On Thursday, October 11, 2012 11:12:00 AM UTC+10, PengKuan Em wrote:
> Le jeudi 11 octobre 2012 02:13:49 UTC+2, Timo Nieminen a écrit :
> > On Thursday, October 11, 2012 12:57:18 AM UTC+10, PengKuan Em wrote:
> >
> > > The magnetic field increases constantly and the induced E field is constant around the wire. Like in a capacitor, even external E field is zero, there is not current inside.
> >
> > No. Why do you way the current is zero? Look at transformer cores - they're made of laminae so as to minimise the currents.
>
> Maybe you think of alternate current and magnetic field. In my example, the magnetic field is a linear function of time, its time derivative is constant, so, Faraday's law gives a tension proportional to the rate of increase of magnetic field, that is, V constant. Because the wire is cut, the resistance is infinity, so I=V/R=0. The current is zero.

AC versus linear ramp doesn't make a significant difference in this case.

The electrostatic "law" that you invoke to make the paradox paradoxical is E=0. Note that the relationship between E and current is J = conductivity * E. The overall resistance of the cut loop of wire is irrelevant. What matters are the current density J, the electric field E, and the conductivity. How does cutting the loop change the conductivity of the wire?

> > You think a paradox exists because you assume that there is no current in the conductor. Yet induced currents are measured, and engineering things to minimise them is important. So, no paradox. See, e.g., http://en.wikipedia.org/wiki/Transformer#Cores
>
> Here the resistance is infinity, the measured current is 0.

The net current around the loop is zero. That's measured and theoretical, in agreement. The current density in the wire (in the uncut portion) is not zero. Again, that's theoretical and measured, in agreement.

[Szczepan Bialek]

"Larry Harson" <larry...@softhome.net> napisał w wiadomości
news:5645a45b-fe08-43f0...@googlegroups.com...

On Wednesday, October 10, 2012 8:19:41 AM UTC+1, Szczepan Bialek wrote:
> "Larry Harson" <larry...@soft.net> napisal w wiadomosci
>
> news:450bd50a-61cf-4e61...@googlegroups.com...
>
>

> <It's not a paradox, but a useful question to test people's understanding
> of
>
> the physics rather than a blind application of Maxwell's equations.
>
>
>
> <Put a conductor in an external static electric field. There's now an
>
> internal static electric field inside the conductor which causes charge to
>
> move to the surface and generate an electric field that cancels the
> applied
>
> one.
>
>
>
> Wrong. In Maxwell's definition the electric field is if the charge flows
> or
>
> "want" to flow.

<Are you sure? His definition of an electric field was the force on a unit
charge.

You should agree that the force on the unit charge acts on the each next
unit charge.
E = 0 in conductors is the school simplification.

> So the charges moved to ths surface do not cancel the applied one. They
> want
>
> to flow but they can not.

So where does the cancellation of the applied field come from then?

No such cancellation. The force acts on each new charge injected into
conductor.
Do you think that Maxwell can be wrong?
S*

[PengKuan Em]

Take a wire of 0.1 mm of diameter and 1 m long, make it a circle without connecting the ends. Put it in a magnetic field with constant rate of variation. Will an electron move from one end to the other? Surely no, because it cannot get out of the end. It can do circles inside the conductor, the Eddy current, but this is not the subject.

[PengKuan Em]

Yes. The net current around the loop is zero means the overall tension is zero, thus we should measure zero tension between A and D in the figure 1.

[Timo Nieminen]

Yes? So, you agree there is no paradox? What are you saying "yes" to?

> The net current around the loop is zero means the overall tension is zero, thus we should measure zero tension between A and D in the figure 1.

Doesn't follow. You said yourself that around the loop, R = infinity. What is the product of infinity and zero?

(But that has very little to do with what you claimed as the paradox in the first place: the violation of the electrostatic "law" J=0 in the conductor. Are you changing what you're claiming to be paradoxical _again_?)

If you're happy to just approximate the cut wire as a thin wire with the only important thing being the net current (rather than the current density), then you at exactly the "battery paradox".

Connect a wire to the terminal of a battery - a current flows. Cut the wire, and the current stops. What is the potential difference V between the cut ends of the wire? Are you really trying to claim that because I=0, V=0? Try it, and measure V with a voltmeter.

[PengKuan Em]

Le jeudi 11 octobre 2012 14:48:40 UTC+2, Timo Nieminen a écrit :
> On Thursday, October 11, 2012 9:28:23 PM UTC+10, PengKuan Em wrote:
>
> > Le jeudi 11 octobre 2012 03:52:56 UTC+2, Timo Nieminen a écrit :
>
> > > The net current around the loop is zero. That's measured and theoretical, in agreement. The current density in the wire (in the uncut portion) is not zero. Again, that's theoretical and measured, in agreement.
>
> >
>
> > Yes.
>
>
>
> Yes? So, you agree there is no paradox? What are you saying "yes" to?

Yes to "The net current around the loop is zero"

> > The net current around the loop is zero means the overall tension is zero, thus we should measure zero tension between A and D in the figure 1.
>
>
>
> Doesn't follow. You said yourself that around the loop, R = infinity. What is the product of infinity and zero?
>

I was unclear, sorry. The B varies at constant rate and the wire reaches a equilibrium. So, the current in the wire zero, the electrons do not move, there is not arc between A and D.

The surface of the wire is an equipotential, thus, the tension between A and D is 0. But there is Faraday's tension measured between A and D. Why? Paradox.

[Larry Harson]

> Take a wire of 0.1 mm of diameter and 1 m long, make it a circle without connecting the ends. Put it in a magnetic field with constant rate of variation. Will an electron move from one end to the other? Surely no, because it cannot get out of the end. It can do circles inside the conductor, the Eddy current, but this is not the subject.- Hide quoted text -

When we connect the two ends, we know that any internal static charge
arrangement on the surface won't affect the path taken by the current.
So we can say the direction of the induced electric field determines
the path taken by the current which will be in a circular loop.

But when we disconnect the two ends, how are we to know what path the
induced current takes?

For that, we need to know the physics of what happens when a current
that was flowing is interrupted by the cut boundary. Obviously there
is an opposing electric field at the boundary to prevent the original
current from flowing across it, and this opposing field comes from the
build up of charge on the surface.

Don't you think that it will cause the original current to loop back
on itself within the wire?

Also the changing magnetic field enclosed by this current will lie
within the wire, meaning the current will be greatly reduced.

Larry.

[PengKuan Em]

Le jeudi 11 octobre 2012 19:39:56 UTC+2, Larry Harson a écrit :
>

> >
>


>
> > Take a wire of 0.1 mm of diameter and 1 m long, make it a circle without connecting the ends. Put it in a magnetic field with constant rate of variation. Will an electron move from one end to the other? Surely no, because it cannot get out of the end. It can do circles inside the conductor, the Eddy current, but this is not the subject.- Hide quoted text -
>
>
>
> When we connect the two ends, we know that any internal static charge
>
> arrangement on the surface won't affect the path taken by the current.
>
> So we can say the direction of the induced electric field determines
>
> the path taken by the current which will be in a circular loop.
>
>
>
> But when we disconnect the two ends, how are we to know what path the
>
> induced current takes?
>
>
>
> For that, we need to know the physics of what happens when a current
>
> that was flowing is interrupted by the cut boundary. Obviously there
>
> is an opposing electric field at the boundary to prevent the original
>
> current from flowing across it, and this opposing field comes from the
>
> build up of charge on the surface.
>

Do we need to know where the current passes? We just have to know that the surface is equipotential and A and D are on the same surface, that is, they are on the same equipotential surface. So, they have the same potential.

>
> Don't you think that it will cause the original current to loop back
>
> on itself within the wire?
>
>
>
> Also the changing magnetic field enclosed by this current will lie
>
> within the wire, meaning the current will be greatly reduced.
>
>
>
> Larry.

But, the surface has gradient of potential, due to Faraday. Hence, paradox.

[Timo Nieminen]

On Friday, October 12, 2012 1:08:39 AM UTC+10, PengKuan Em wrote:
> I was unclear, sorry. The B varies at constant rate and the wire reaches a equilibrium. So, the current in the wire zero, the electrons do not move, there is not arc between A and D.

No. The electrons DO MOVE! This can be, has been measured. Why do you think the electrons don't move? Just because electrostatics says so? This isn't an example of electrostatics.

The electrons do move. There will be circulating currents in the wire, even when the wire is cut. When the wire is cut, the circulation will be a loop within thin wire, not a loop around the whole wire. So, net current along the wire is zero, but current density will be non-zero.

> The surface of the wire is an equipotential, thus, the tension between A and D is 0.

Wrong assumption! This is NOT an electrostatic case. Here, curl(E) is non-zero, the electric field E is NOT -grad(potential).

So don't assume an equipotential.

> But there is Faraday's tension measured between A and D. Why? Paradox.

Potential doesn't tell you what the fields are. Potential difference doesn't tell you what the fields are.

Faraday's law tells you what curl(E) is. That, and the boundary conditions tell you E. That tells you that you have circulating eddy currents. As measured.

Theory agrees with measurement. Both disagree with your assumptions. Your asusmptions are wrong. But since theory agrees with measurment, no paradox.

[PengKuan Em]

Le vendredi 12 octobre 2012 00:01:44 UTC+2, Timo Nieminen a écrit :
> On Friday, October 12, 2012 1:08:39 AM UTC+10, PengKuan Em wrote:
>
> > I was unclear, sorry. The B varies at constant rate and the wire reaches a equilibrium. So, the current in the wire zero, the electrons do not move, there is not arc between A and D.
>
>
>
> No. The electrons DO MOVE! This can be, has been measured. Why do you think the electrons don't move? Just because electrostatics says so? This isn't an example of electrostatics.
>
>
>
> The electrons do move. There will be circulating currents in the wire, even when the wire is cut. When the wire is cut, the circulation will be a loop within thin wire, not a loop around the whole wire. So, net current along the wire is zero, but current density will be non-zero.
>

I agree that electrons move in closed tight loop within a cut wire. I'm happy to hear that this was measured, really.

Is it like the figure I made here ?
Current loop inside a thin wire
http://pengkuanem.blogspot.com/2012/10/current-loop-inside-thin-wire.html

Please indicate what is the tension integrated along the outside current and what is it along the inside current? Is it Faraday's law that gives it?

For equipotential surface, we will discuss later.

Pengkuan

[Larry Harson]

On Oct 11, 8:35 pm, PengKuan Em <titan...@gmail.com> wrote:
> Le jeudi 11 octobre 2012 19:39:56 UTC+2, Larry Harson a écrit :
>
>
>
>
>
>
>
> > > Take a wire of 0.1 mm of diameter and 1 m long, make it a circle without connecting the ends. Put it in a magnetic field with constant rate of variation. Will an electron move from one end to the other? Surely no, because it cannot get out of the end. It can do circles inside the conductor, the Eddy current, but this is not the subject.- Hide quoted text -
>
> > When we connect the two ends, we know that any internal static charge
>
> > arrangement on the surface won't affect the path taken by the current.
>
> > So we can say the direction of the induced electric field determines
>
> > the path taken by the current which will be in a circular loop.
>
> > But when we disconnect the two ends, how are we to know what path the
>
> > induced current takes?
>
> > For that, we need to know the physics of what happens when a current
>
> > that was flowing is interrupted by the cut boundary. Obviously there
>
> > is an opposing electric field at the boundary to prevent the original
>
> > current from flowing across it, and this opposing field comes from the
>
> > build up of charge on the surface.

> Do we need to know where the current passes?

Yes, we need to know the new path taken by the current, because we
can't assume its original path isn't affected by the cut at AD.

>We just have to know that the surface is equipotential and A and D are on the same surface, that is, they are on the same equipotential surface. So, they have the same potential.

Electrostatics is where all electric fields come from static charges,
and so excludes constant electric fields from changing currents. A
constant electric field from a magnetic field changing at a constant
rate isn't an electrostatic field just because it remains constant
with time. It's a constant non-convervative electric field, and
potential difference is meaningless here because you can't create a
potential function that's independent of path.

You can place a voltmeter across points AD, in which case it will
measure the voltage dropped across its internal resistance from the
induced current. But the potential difference between AD will be
different if you take some path through the wire. You have to be
careful about what you mean by "potential difference" in non-
conservative circuits.

> > Don't you think that it will cause the original current to loop back
>
> > on itself within the wire?
>
> > Also the changing magnetic field enclosed by this current will lie
>
> > within the wire, meaning the current will be greatly reduced.
>
> > Larry.
>

> But, the surface has gradient of potential, due to Faraday. Hence, paradox.- Hide quoted text -

Whether the electric field can be expressed as the gradient of a
potential is purely a property of the electric field and has nothing
to do with Faraday's law. Generally, E can be usefully written as Grad
phi only for electric fields generated by static charges because the
closed line integral of E is zero.

I think your main problem is that you think an induced constant
electric field is electrostatic just because its constant. It's a
constant non-conservative electric field which is different to an
electrostatic electric field which is always conservative, and
generated by the electric fields of static charge.

Larry.

[Timo Nieminen]

On Friday, October 12, 2012 9:25:25 AM UTC+10, PengKuan Em wrote:
> Le vendredi 12 octobre 2012 00:01:44 UTC+2, Timo Nieminen a écrit :
> > On Friday, October 12, 2012 1:08:39 AM UTC+10, PengKuan Em wrote:
>
> > > I was unclear, sorry. The B varies at constant rate and the wire reaches a equilibrium. So, the current in the wire zero, the electrons do not move, there is not arc between A and D.
>
> > No. The electrons DO MOVE! This can be, has been measured. Why do you think the electrons don't move? Just because electrostatics says so? This isn't an example of electrostatics.
> >
> > The electrons do move. There will be circulating currents in the wire, even when the wire is cut. When the wire is cut, the circulation will be a loop within thin wire, not a loop around the whole wire. So, net current along the wire is zero, but current density will be non-zero.
>
> I agree that electrons move in closed tight loop within a cut wire. I'm happy to hear that this was measured, really.

Measured by magnetic effects, and measured via heating.

> Is it like the figure I made here ?
> Current loop inside a thin wire
> http://pengkuanem.blogspot.com/2012/10/current-loop-inside-thin-wire.html

Yes. Just so.

> Please indicate what is the tension integrated along the outside current and what is it along the inside current? Is it Faraday's law that gives it?

Wrong concept. When you have E=-grad(potential), you can find the potential difference (i.e., "tension") between two points by integrating E.dl along a path connecting them.

When you don't have E=-grad(potential), you can't find the potential difference between two points by integrating E.dl along a path connecting them.

Here, you don't have E=-grad(potential), so integrating along the outside current doesn't give you a tension.

Faraday's law tells you the integral of E.dl along a closed loop. If the wire is thin, the outside integral and the inside integral will be approximately equal, and you can ignore the contribution from the two cut ends.

If you want to use potentials to deal with this kind of case, you can't just use the electrostatic potential (i.e., scalar potential). You need to use both the scalar potential and the vector potential A.

It is possible to find the scalar potential from E. But not by just integrating E.dl. You can find both the conservative and non-conservative parts of E; you can do this not just for electric fields, but for all sorts of fields. In principle, for any field where the Helmholtz decomposition into conservative and non-conservative parts works. For example, G. Pesce, G. Volpe, A.C.D. Luca, G. Rusciano, and G. Volpe. Quantitative assessment of non-conservative radiation forces in an optical trap. EPL (Europhysics Letters), 86:38002, 2009, and Pinyu Wu, Rongxin Huang, Christian Tischer, Alexandr Jonas, and Ernst-Ludwig Florin. Direct measurement of the nonconservative force field generated by optical tweezers. Physical Review Letters, 103(10):108101, 2009.

[PengKuan Em]

Le vendredi 12 octobre 2012 02:42:09 UTC+2, Larry Harson a écrit :

> > Do we need to know where the current passes?
>
>
>
> Yes, we need to know the new path taken by the current, because we
>
> can't assume its original path isn't affected by the cut at AD.
>

The path at equilibrium is constant.

>
> >We just have to know that the surface is equipotential and A and D are on the same surface, that is, they are on the same equipotential surface. So, they have the same potential.
>
>
>
> Electrostatics is where all electric fields come from static charges,
>
> and so excludes constant electric fields from changing currents. A
>
> constant electric field from a magnetic field changing at a constant
>
> rate isn't an electrostatic field just because it remains constant
>
> with time. It's a constant non-convervative electric field, and
>
> potential difference is meaningless here because you can't create a
>
> potential function that's independent of path.
>

When we integrate E.dl from A to D, we obtain a value, what is it? a tension, if not potential?

This tension is equal to Faraday's tension. And it is almost the same for all path inside the wire.

>
> You can place a voltmeter across points AD, in which case it will
>
> measure the voltage dropped across its internal resistance from the
>
> induced current. But the potential difference between AD will be
>
> different if you take some path through the wire. You have to be
>
> careful about what you mean by "potential difference" in non-
>
> conservative circuits.
>

OK, I will use the term tension at the place of potential.

>
> > > Don't you think that it will cause the original current to loop back
>
> >
>
> > > on itself within the wire?
>

This is exactly what our friend Timo Nieminen is saying above. I have drawn a figure here

Current loop inside a thin wire
http://pengkuanem.blogspot.com/2012/10/current-loop-inside-thin-wire.html

>

> Whether the electric field can be expressed as the gradient of a
>
> potential is purely a property of the electric field and has nothing
>
> to do with Faraday's law. Generally, E can be usefully written as Grad
>
> phi only for electric fields generated by static charges because the
>
> closed line integral of E is zero.
>

Let us use Tension instead.

>
>
> I think your main problem is that you think an induced constant
>
> electric field is electrostatic just because its constant. It's a
>
> constant non-conservative electric field which is different to an
>
> electrostatic electric field which is always conservative, and
>
> generated by the electric fields of static charge.
>

A induced E field pushes electron. If it is constant, it can be counter balanced by electrostatic force and is easier to handle. So, a any point in the wire, the induced E is constant, there must be a counter E to balance it since there electron stay still. What is this counter E? Will there be a counter E?

[PengKuan Em]

Le vendredi 12 octobre 2012 02:59:04 UTC+2, Timo Nieminen a écrit :
>
> > I agree that electrons move in closed tight loop within a cut wire. I'm happy to hear that this was measured, really.
>
>
>
> Measured by magnetic effects, and measured via heating.
>
>
>
> > Is it like the figure I made here ?
>
> > Current loop inside a thin wire
>
> > http://pengkuanem.blogspot.com/2012/10/current-loop-inside-thin-wire.html
>
>
>
> Yes. Just so.
>
>
>
> > Please indicate what is the tension integrated along the outside current and what is it along the inside current? Is it Faraday's law that gives it?
>
>
>
> Wrong concept. When you have E=-grad(potential), you can find the potential difference (i.e., "tension") between two points by integrating E.dl along a path connecting them.
>
>
>
> When you don't have E=-grad(potential), you can't find the potential difference between two points by integrating E.dl along a path connecting them.
>
>
>
> Here, you don't have E=-grad(potential), so integrating along the outside current doesn't give you a tension.
>

What is the quantity of the integral of E.dl along outside current? If it is not a tension?

>
> Faraday's law tells you the integral of E.dl along a closed loop. If the wire is thin, the outside integral and the inside integral will be approximately equal, and you can ignore the contribution from the two cut ends.
>

So, the integral of E.dl along the outside is Vout and along the inside is -Vout+eps. What will be the tension between A and D? Vout? Because if an electron travels along the outside current, it will get a energy E.l.

>
> If you want to use potentials to deal with this kind of case, you can't just use the electrostatic potential (i.e., scalar potential). You need to use both the scalar potential and the vector potential A.
>

Potential is only a name. What is important is the energy the electron gets.

[Don Kelly]

On 11/10/2012 8:28 PM, PengKuan Em wrote:
> Le vendredi 12 octobre 2012 02:42:09 UTC+2, Larry Harson a �crit :

And what happens if the conductor is cut and there is nowhere that the
charge carriers can go (assuming E is not high enough to cause emission)?
Do you have a distribution of the "free" charge carriers within the
conductor (assuming that the loop is cut) such as to provide an opposing
field? and balancing of forces on the charge carriers?
Take a look at semiconductor physics- or even back to the almost
prehistoric "triodelites" (related to coprolites).


--
Don Kelly
cross out to reply

[Timo Nieminen]

On Friday, October 12, 2012 1:46:30 PM UTC+10, PengKuan Em wrote:
> Le vendredi 12 octobre 2012 02:59:04 UTC+2, Timo Nieminen a écrit :
>
> > When you don't have E=-grad(potential), you can't find the potential difference between two points by integrating E.dl along a path connecting them.
> >
> > Here, you don't have E=-grad(potential), so integrating along the outside current doesn't give you a tension.
>
> What is the quantity of the integral of E.dl along outside current? If it is not a tension?

Work done on a unit charge as it follows that path.

Consider a mechanics problem. Take a curved frictionless surface with gravity acting downwards. The gravitational force on an object on the surface is proportional the gradient of the surface - the height of the surface tells you the gravitational potential.

You can integrate F.dl along a path, and find the potential difference (i.e., the height difference) between two points.

Now add a non-conservative force: friction. Now the force will be gravity + friction. What does integrating F.dl along a path tell you now? Consider a closed path, as well as open paths.

> > Faraday's law tells you the integral of E.dl along a closed loop. If the wire is thin, the outside integral and the inside integral will be approximately equal, and you can ignore the contribution from the two cut ends.
>
> So, the integral of E.dl along the outside is Vout and along the inside is -Vout+eps. What will be the tension between A and D? Vout? Because if an electron travels along the outside current, it will get a energy E.l.

The integral of E.dl does NOT give you a tension in this case. The E field is not conservative; it isn't the gradient of a scalar potential.

The integral along a given path will tell you the work done on an electron as it follows that path. NOT a potential.

> > If you want to use potentials to deal with this kind of case, you can't just use the electrostatic potential (i.e., scalar potential). You need to use both the scalar potential and the vector potential A.
>
> Potential is only a name. What is important is the energy the electron gets.

In this case, the energy the electron gets is path-dependent. The induced E field does work on the electron. The electron loses this energy due to the resistivity of the conductor.

If what is important is the energy the electron gets, just use Faraday's law to give you the integral along the electron's closed path. That tells you the work done as the electron goes around once, and the energy lost to heating the conductor.

[Timo Nieminen]

On Friday, October 12, 2012 1:28:18 PM UTC+10, PengKuan Em wrote:
>
> A induced E field pushes electron. If it is constant, it can be counter balanced by electrostatic force and is easier to handle.

No. An applied field E can only be counter-balanced by electrostatic forces if it is conservative. The induced E here is NOT conservative, and can't be counter-balanced this way.

That's why circulating currents flow in the wire. Because the induced E isn't counter-balanced. In the conductor, we have J = non-zero, and there is a non-zero current density.

[Salmon Egg]

In article <8bc29aca-0de7-45c7...@googlegroups.com>,

PengKuan Em <tita...@gmail.com> wrote:

> But induce E field cannot be canceled by conservative field. In my example,
> there is not current, the magnetic field increases constantly and the induced
> field is constant in the wire. If the E field in the wire is zero, the
> integration of it will be zero and non voltage appears.

Here is one last try before I ignore PengKuan posts entirely.

Figure out how to calculate current flow in a short circuited ring of
wire in a sinusoidal magnetic field (transformer), and why that current
value will change with temperature.

Good luck and probably goodbye.

[Jos Bergervoet]

On 10/11/2012 3:12 AM, PengKuan Em wrote:
...

> Maybe you think of alternate current and magnetic field. In my example,

> the magnetic field is a linear function of time, its time derivative is

> constant, so, Faraday's law gives a tension proportional to the rate of

> increase of magnetic field,

No! It does NOT. It gives E-field proportional to the rate of
increase, not "tension". (But at least you are not using the
term "EMF" anymore, which is good. Still, the only correct
statement is that the E-field is given, not anything else.
The law is:

rot E = -dB/dt

And it gives you the curl of E.

> that is, V constant.

No! It says nothing about V (which is meaningless anyway without also
knowing A, because of gauge invariance. You are making a lot of errors
here!)

> Because the wire is cut, the resistance is infinity,

No! There can be little loops of eddy current fitting
inside the wire diameter and the resistance for those
is definitely not infinity.

> so I=V/R=0.

No! I is NOT V/R. Correct is:

I = ( V + int dr dA/dt ) / R

You are (again) making the mistake of forgetting the
vector potential and assuming that the only cause for
a current can be the scalar potential V.

Mistakes, mistakes..

--
Jos

[PengKuan Em]

Le vendredi 12 octobre 2012 06:47:10 UTC+2, Timo Nieminen a écrit :
> On Friday, October 12, 2012 1:46:30 PM UTC+10, PengKuan Em wrote:
>
> > Le vendredi 12 octobre 2012 02:59:04 UTC+2, Timo Nieminen a écrit :
>
> >
>
> > > When you don't have E=-grad(potential), you can't find the potential difference between two points by integrating E.dl along a path connecting them.
>
> > >
>
> > > Here, you don't have E=-grad(potential), so integrating along the outside current doesn't give you a tension.
>
> >
>
> > What is the quantity of the integral of E.dl along outside current? If it is not a tension?
>
>
>
> Work done on a unit charge as it follows that path.
>
>
>
> Consider a mechanics problem. Take a curved frictionless surface with gravity acting downwards. The gravitational force on an object on the surface is proportional the gradient of the surface - the height of the surface tells you the gravitational potential.
>
>
>
> You can integrate F.dl along a path, and find the potential difference (i.e., the height difference) between two points.
>
>
>
> Now add a non-conservative force: friction. Now the force will be gravity + friction. What does integrating F.dl along a path tell you now? Consider a closed path, as well as open paths.
>

So, you consider that the integral of E is the work, heat:
integral (e.E.dl)=work
But this is not equal to the rate of variation of magnetic flux, that is Faraday's law:
integral (e.E.dl)=-e.d(phi)/dt

e being electron's charge.

[Larry Harson]

On Oct 12, 4:28 am, PengKuan Em <titan...@gmail.com> wrote:
> Le vendredi 12 octobre 2012 02:42:09 UTC+2, Larry Harson a écrit :

[snipped]

> > > > Don't you think that it will cause the original current to loop back
> > > > on itself within the wire?
>
> This is exactly what our friend Timo Nieminen is saying above. I have drawn a figure here

> Current loop inside a thin wirehttp://pengkuanem.blogspot.com/2012/10/current-loop-inside-thin-wire....

Looks good to me.

> > Whether the electric field can be expressed as the gradient of a
> > potential is purely a property of the electric field and has nothing
> > to do with Faraday's law. Generally, E can be usefully written as Grad
> > phi only for electric fields generated by static charges because the
> > closed line integral of E is zero.
>
> Let us use Tension instead.

"Work done" is better, I think.

> > I think your main problem is that you think an induced constant
>
> > electric field is electrostatic just because its constant. It's a
>
> > constant non-conservative electric field which is different to an
>
> > electrostatic electric field which is always conservative, and
>
> > generated by the electric fields of static charge.
>

> A induced E field pushes electron. If it is constant, it can be counter balanced by electrostatic force and is easier to handle. So, a any point in the wire, the induced E is constant, there must be a counter E to balance it since there electron stay still. What is this counter E? Will there be a counter E?- Hide quoted text -

Take a path in the volume where Line_integral (induced E) *dl = const.
Then around this same path Line_integral (static_E)*dl = 0. So there
is no way a static E can cancel an induced E *everywhere* around this
path. This is the proof that you're wrong to assume the induced E is
exactly balanced by a static E *everywhere*. Your conclusion should be
that the static E adds to the induced E so that there is a current
loop in the wire that gets reflected back at A and D.

What is the effect of the counter static E?

1. It must cancel completely the induced E normal to the cut surfaces
A and D because current can't flow across here.

2. It cancels most of the induced E inside the wire, with a small +dE
parallel to the outer surface to drive the current one way, and a
parallel -dE on the lower to drive it in the opposite direction. These
are the parts of the induced E it can't cancel out completely.

You can use Faraday's law to calculate the magnitude of this small
current since you know the approx area enclosed by it, and the
conductivity/resistance along the path.

Larry.

[PengKuan Em]

Le samedi 13 octobre 2012 01:23:49 UTC+2, Larry Harson a écrit :

> Then around this same path Line_integral (static_E)*dl = 0. So there
>
> is no way a static E can cancel an induced E *everywhere* around this
>
> path. This is the proof that you're wrong to assume the induced E is
>
> exactly balanced by a static E *everywhere*. Your conclusion should be
>
> that the static E adds to the induced E so that there is a current
>
> loop in the wire that gets reflected back at A and D.
>
>
>

> Larry.

"there is no way a static E can cancel an induced E *everywhere* around this path. " Exact. I was not saying that "the induced E is exactly balanced by a static E *everywhere*. " I said that if this were the case, then the integral of E around the loop would be zero, contradicting Faraday's law.

But if this is not the case, that is, no electrostatic E is created, then there is not movement of electrons and it should exist a force that maintains the electrons in place. This is not the case neither.

For example, in the middle of the wire, in a place between the forward and backward current, there is a place where the current is zero. The Faraday's law creates a E here but electrons do not move here.

[Timo Nieminen]

On Saturday, October 13, 2012 7:47:40 AM UTC+10, PengKuan Em wrote:
>
> So, you consider that the integral of E is the work, heat:
> integral (e.E.dl)=work
> But this is not equal to the rate of variation of magnetic flux, that is Faraday's law:
> integral (e.E.dl)=-e.d(phi)/dt e being electron's charge.

Why not? The E field in this case will consist of two components: a conservative one from surface charges on the conductor, and a non-conservative induced field.

The conservative field will not contribute to the work done around a closed path; this will depend only on the induced field. The curl of the induced field is what Faraday's law tells us.

Why do you think the work done around a closed path is not equal to the result using Faraday's law? Measurements of heating by induced currents agree with Faraday's law. For example, http://onlinelibrary.wiley.com/doi/10.1002/nme.796/abstract

Do you have some measurements of your own that lead you to believe that the heating isn't given by Faraday's law?

[Timo Nieminen]

On Saturday, October 13, 2012 9:42:46 AM UTC+10, PengKuan Em wrote:
>
> But if this is not the case, that is, no electrostatic E is created, then there is not movement of electrons and it should exist a force that maintains the electrons in place. This is not the case neither.

Why do you think there isn't an electrostatic field? Consider the uncut circular loop of wire - the current (and therefore the electrons) go around in a circle. An inward acceleration is required to make the electrons go in a circular path. The force producing this is due to the field of surface charges.

After you cut the loop, the force that causes electrons getting to the cut end to turn around and go back is due to the field of surface charges.

Not being able to cancel the induced field doesn't mean that no electrostatic field exists.

> For example, in the middle of the wire, in a place between the forward and backward current, there is a place where the current is zero. The Faraday's law creates a E here but electrons do not move here.

Along that line of E=0, the induced field is cancelled by the electrostatic field due to the surface charges. The conservative field due to the surface charges can't cancel the non-conservative induced field over a non-zero volume, but there's no problem doing so at points, along lines, or along sheets (points, lines, and sheets have zero volume).

[bja...@iwaynet.net]

On 10/13/2012 12:59 AM, Timo Nieminen wrote:


> After you cut the loop, the force that causes electrons getting to the cut end to turn around and go back is due to the field of surface charges.
>
> Not being able to cancel the induced field doesn't mean that no electrostatic field exists.
>
>> For example, in the middle of the wire, in a place between the forward and backward current, there is a place where the current is zero. The Faraday's law creates a E here but electrons do not move here.
>
> Along that line of E=0, the induced field is cancelled by the electrostatic field due to the surface charges. The conservative field due to the surface charges can't cancel the non-conservative induced field over a non-zero volume, but there's no problem doing so at points, along lines, or along sheets (points, lines, and sheets have zero volume).

But Timo, if one assumes an induce field ONLY in the theta direction
(circles) and also assume the wire is cut, we know that there is no
current in the wire. Now if we assume a line down the center of the
wire, you admit that on that line the movement of electric charges can
create a conservative statice E field that cancels the induced
non-conservative E field which is attempting to provide a force on the
charges pushing them round the circle.

So this says that if our wire is extremely thin, things work out fine.
But I suggest that with a fatter wire, the same thing holds true for ANY
line you choose within the wire! In fact I suggest that in this case
there would have to be cancellation at every point in the wire, because
were there not, then those charges would move under the force of the
induced field until they could move no more. And that point would be
when the induced field upon them is balanced by a static E field created
by the changes in charge density created by such movement. I don't agree
that the cancelling field is due only to surface charges.

[Larry Harson]

On Oct 13, 12:42 am, PengKuan Em <titan...@gmail.com> wrote:
> Le samedi 13 octobre 2012 01:23:49 UTC+2, Larry Harson a écrit :
>
> > Then around this same path Line_integral (static_E)*dl = 0. So there
>
> > is no way a static E can cancel an induced E *everywhere* around this
>
> > path. This is the proof that you're wrong to assume the induced E is
>
> > exactly balanced by a static E *everywhere*. Your conclusion should be
>
> > that the static E adds to the induced E so that there is a current
>
> > loop in the wire that gets reflected back at A and D.
>
> > Larry.
>
> "there is no way a static E can cancel an induced E *everywhere* around this path. " Exact. I was not saying that "the induced E is exactly balanced by a static E *everywhere*. " I said that if this were the case, then the integral of E around the loop would be zero, contradicting Faraday's law.
>
> But if this is not the case, that is, no electrostatic E is created, then there is not movement of electrons and it should exist a force that maintains the electrons in place. This is not the case neither.

Isn't it obvious that if the charge can't move across the surface from
the induced E, then it can only move along it?
And this build-up/depletion of charge on a surface creates an opposing
static E field?

> For example, in the middle of the wire, in a place between the forward and backward current, there is a place where the current is zero. The Faraday's law creates a E here but electrons do not move here.

The electrons don't move here because the static E has cancelled the
induced E along this part of the path. The interesting point is that
it makes no difference to Faraday's law if E:is the total E or just
the induced E:

loop_integral ( E_induced + E_static)*dl
= loop_integral (E_induced)*dl + loop_integral (E_static)*dl
= loop_integral (E_induced)*dl

Let's use the total E: We know that from A to D inside the wire that
loop_integral (E_induced + E_static)*dl = 0. From D to A across the
gap, loop_integral (E_induced + E_static)*dl = const. You may find it
surprising that two paths in parallel give different values for the
work done, but that's how nature is for non-conservative fields in
this case. For your uncut wire with a finite resistance R, you'll have
a current flowing in a loop:

I = 1/R loop_integral(E_induced)*dl

But note how the same current flows in opposite directions for two
identical halves in parallel.

Larry.

[Timo Nieminen]

On Saturday, October 13, 2012 8:04:02 PM UTC+10, bja...@teranews.com wrote:
> But Timo, if one assumes an induce field ONLY in the theta direction (circles) and also assume the wire is cut, we know that there is no current in the wire.

No net current. That is, the net flow of charge through a section of the wire is zero. (After the charge distribution and current has reached steady-state.)

This is quite possible, simple even, with J being non-zero. That means that the current along the inside of the wire is in one direction, and the current on the outside is in the other direction.

Non-zero currents densities mean non-zero E. Zero net current doesn't mean E=0.

> Now if we assume a line down the center of the wire, you admit that on that line the movement of electric charges can create a conservative statice E field that cancels the induced non-conservative E field which is attempting to provide a force on the charges pushing them round the circle. So this says that if our wire is extremely thin, things work out fine.

If the wire is extremely thin? Then the eddy currents are smaller. The integral of E.dl around the perimeter of the cut wire will be approximately 2*L*J/sigma. As the wire is made thinner, the flux through the wire (i.e., the flux enclosed by the integral around the perimeter of the wire) becomes small. So J becomes smaller.

Which is why thin laminae are used in transformer cores to reduce losses due to indeced currents.

Why, in the case of the wire, this is "working out fine", I don't see. Thick wire, thin wire, the pattern of current density is similar. Non-zero in both cases. No fundamental difference - only the numbers are different. For real life engineering, the numbers matter, and things are designed so that the numbers will be small when they should be small. But for a thought experiment like this, where we don't care how large the eddy currents and resulting heating are, why the smaller currents that result from a thin wire are "fine" is a mystery to me. They're still there, and still non-zero.

> But I suggest that with a fatter wire, the same thing holds true for ANY line you choose within the wire!

That there can be _a_ line (in this case, it will be sheet, not a line) along which E=0 doesn't mean that you get E=0 on all possible lines.

Next time you drain water out of a sink, watch the flow. At the (vertical) sides of the sink, the horizontal flow speed is zero. At the centre of the vortex above the drain, the horizontal speed is zero. That doesn't mean that the horizontal flow everywhere in the sink is zero.

What you suggest is not in agreement with experiment. What value is there in what you suggest?

Go ahead, give an example of a conservative (i.e., curl-free) field cancelling a divergence-free field over a non-zero volume.

> In fact I suggest that in this case there would have to be cancellation at every point in the wire, because were there not, then those charges would move under the force of the induced field until they could move no more.

What we see in experiment is that the charges move, and keep moving, under the influence of the induced field. Your "until they could move no more" is unnecessary, since the charges can keep moving, and do keep moving.

Again, what you suggest is not in agreement with experiment. What value is there in what you suggest?

> And that point would be when the induced field upon them is balanced by a static E field created by the changes in charge density created by such movement. I don't agree that the cancelling field is due only to surface charges.

(a) "At that point" is entirely imaginary - it doesn't happen. Over any non-zero volume of conductor [1], the current density will be non-zero - the charges _don't_ stop moving. Consequently, "when they stop moving" doesn't lead anywhere of interest!

(b) "Surface charges" is an approximation, but on the macroscopic scale, a good one. A nice student exercise is to, assuming an initial non-zero rho_0 in a conductor, how this changes over time. Try it. It might educate as to why "surface" is a good approximation.

(c) It doesn't matter for cancelling induced fields in this case. Surface or bulk charge density, the electrostatic field is still conservative, and can't cancel the induced field over a non-zero volume.

(d) From (c), the induced field also can't cancel the field due to any bulk charge density. Combine that with the result from (b), and you'll see why "surface" is still a good approximation even with induced fields present.

[1] There is an approximation here, of course. Here, we assume that the conductor is large enough so that the current can be described by a current density and the conductor can be described by its (scalar) conductivity. Make the wire out of a single chain of atoms, for example, and this fails.

[PengKuan Em]

When I said "integral (E.dl)=-d(phi)/dt", I meant
integral through the entire loop, ABCD. This integral must equal to -d(phi)/dt, Faraday's law, phi being the the magnetic flux through the whole loop, not the surface of covered by the wire.

But for you integral(e.E.dl)=work. When I do this integral through ABCD, E is very small and is not -d(phi)/dt. So, by using the tight loop of current inside the wire, the E does not give EMF of Faraday's law for the whole loop, the EMF measured between A and D.

[PengKuan Em]

Le samedi 13 octobre 2012 06:59:44 UTC+2, Timo Nieminen a écrit :
> On Saturday, October 13, 2012 9:42:46 AM UTC+10, PengKuan Em wrote:
>
> >
>
> > But if this is not the case, that is, no electrostatic E is created, then there is not movement of electrons and it should exist a force that maintains the electrons in place. This is not the case neither.
>
>
>
> Why do you think there isn't an electrostatic field? Consider the uncut circular loop of wire - the current (and therefore the electrons) go around in a circle. An inward acceleration is required to make the electrons go in a circular path. The force producing this is due to the field of surface charges.
>

It is to rule out this possibility in case someone thinks of EMF as an E without electrostatic consideration.

> > For example, in the middle of the wire, in a place between the forward and backward current, there is a place where the current is zero. The Faraday's law creates a E here but electrons do not move here.
>
>
>
> Along that line of E=0, the induced field is cancelled by the electrostatic field due to the surface charges. The conservative field due to the surface charges can't cancel the non-conservative induced field over a non-zero volume, but there's no problem doing so at points, along lines, or along sheets (points, lines, and sheets have zero volume).

This line of E=0 is the key. This line stretches from A to D, So, the integral along that line is 0 and we get
EMF=integral through ABCD along (E=0 line) of (E.dl) = 0
We do not get Faraday's law result for the circular loop .

[PengKuan Em]

Le samedi 13 octobre 2012 21:57:34 UTC+2, Larry Harson a écrit :
> On Oct 13, 12:42 am, PengKuan Em <titan...@gmail.com> wrote:
>
> The electrons don't move here because the static E has cancelled the
>
> induced E along this part of the path. The interesting point is that
>
> it makes no difference to Faraday's law if E:is the total E or just
>
> the induced E:
>
>
>
> loop_integral ( E_induced + E_static)*dl
>
> = loop_integral (E_induced)*dl + loop_integral (E_static)*dl
>
> = loop_integral (E_induced)*dl
>

> Larry.

I do not agree with you formula
EMF= loop_integral ( E_induced + E_static)*dl OK

loop_integral ( E_induced + E_static)*dl

=loop_integral (E_induced)*dl + loop_integral (E_static)*dl ,OK
but

loop_integral (E_induced)*dl + loop_integral (E_static)*dl

= loop_integral (E_induced)*dl not OK
Because
loop_integral (E_static)*dl=/=0.
Maybe you think E_static is conservative and its integral through closed path is zero.

Actually, it is not integrated through a close path. It is integrated from A to D, the remaining gap D to A is not in the path of the integral. So, the correct formula should be:
loop_integral ( E_induced + E_static)*dl (From A to D)
= loop_integral (E_induced)*dl (From Ato D) + integral (E_static)*dl (From A to D)
integral (E_static)*dl (CLOSED LOOP) =integral (E_static)*dl (From A to D) + integral (E_static)*dl (From D to A gap)=0

Then
integral (E_static)*dl (From A to D)
= -integral (E_static)*dl (From D to A, gap)= -EMF (gap)

For the induced E:
integral (E_induced)*dl (From A to D)=EMF

then
loop_integral ( E_induced + E_static)*dl (From A to D)
=EMF -integral (E_static)*dl (From D to A, Gap)=EMF-EMF=0

So, the integral along the E=0 line gives zero while the integral in the gap gives EMF

[Timo Nieminen]

On Sunday, October 14, 2012 12:51:36 PM UTC+10, PengKuan Em wrote:
> This line of E=0 is the key. This line stretches from A to D, So, the integral along that line is 0 and we get EMF=integral through ABCD along (E=0 line) of (E.dl) = 0 We do not get Faraday's law result for the circular loop .

Integral along ABCD isn't a closed loop, so Faraday's law doesn't tell you what the integral should be.

Integrate along ABCDA. The surface charges at D and A will give you a large field from D to A (i.e., in the space where the wire has been cut). Total integral will be what Faraday's law says it should be.

Easy to prove in the limit of a small cut: The total field in the conductor is the sum of the induced field and the electrostatic field. So, along the E=0 line in the conductor, E_electrostatic = -E_induced. But since E_electrostatic is conservative, the integral of E_electrostatic.dl around a closed path is zero. So, integral of E_electrostatic.dl from D to A is the integral of E_electrostatic.dl along ABCD.

[Salmon Egg]

One problem I see in most of the posts is not to consider the nature of
the conductor.

The ideal of a perfect conductor does no exist in physical form. A
superconductor ejects magnetic field from its interior when it goes from
being a normal conductor to a superconductor.

There is nothing in Maxwell's equations that tell what a conductor is.
Often, the concept of conductivity is introduced as an adjunct condition:
J = sigma E where J is current density and sigma quantifies the material
property of conductance.

With finite conductivity, there is a relaxation time (classical physics)
that measures the rate at which charge inside the material will reach
the surface.

There are quantum mechanical models for describing why conductivity is
finite. You have an empty space that gets filled with ions such as
copper ions. It forms a lattice of ions. The electrons that arise from
the neutral copper atoms are mobile.

If you could suddenly apply an electric field, the field would
completely penetrate the mostly empty space of the conductor. Forces
will be applied immediately to the ions and electrons of the conductor.
The ions will not move fast or far compared to the electrons. In times
of the order of the period of a light wave, electrons will move to
shield the interior from the external field. In doing so, the fastest
electrons will bounce of impurity atoms and other lattice defects. The
energy transfer will cause the lattice to vibrate and heat up. The
electrons can also bounce off of the vibrations (phonons). As long as an
emf is induced in a lattice, there WILL BE an electric field in the
conductor. The electrons will not be able to respond fast enough to
shield the interior rfrom the induced field.

Thus, too much time is being spent upon what happens in a perfect
conductor; a concept that is irrelevant.

This was composed late at night. Please o not expect perfect grammar and
spelling

[PengKuan Em]

Le dimanche 14 octobre 2012 08:01:56 UTC+2, Timo Nieminen a écrit :
> On Sunday, October 14, 2012 12:51:36 PM UTC+10, PengKuan Em wrote:
>
> > This line of E=0 is the key. This line stretches from A to D, So, the integral along that line is 0 and we get EMF=integral through ABCD along (E=0 line) of (E.dl) = 0 We do not get Faraday's law result for the circular loop .
>
>
>
> Integral along ABCD isn't a closed loop, so Faraday's law doesn't tell you what the integral should be.
>
>

This is not good argument because if the point A and D are plates of a capacitor, you will have a gap and you cannot get the EMF through the integral. The potential between the 2 plates ca only be known with the quantity of charges, that is not given by Faraday's law.

[PengKuan Em]

Le dimanche 14 octobre 2012 12:38:28 UTC+2, Salmon Egg a écrit :
> One problem I see in most of the posts is not to consider the nature of
>
> the conductor.
>
>
>
> The ideal of a perfect conductor does no exist in physical form. A
>
> superconductor ejects magnetic field from its interior when it goes from
>
> being a normal conductor to a superconductor.
>

I prefer superconductor concept because it is pure. Inside a superconductor wire, E=0. The tight loop current inside a wire is only a disturb that divert our attention and blur the focus.

So, the integral of E.dl through ABCD is 0. This is a result contrary to Faraday's law.

I do not assert that the experimental measure is 0, it is not. I just want to show that the law we use, integral of E.dl does not give Faraday's law's. This is a problem that no one can solve.

[Radi Khrapko]

Dear PengKuanem, you do not understand trivial things. It is a shame. In your situation the electric field is nonpotential. So, integral along the wire is zero, but integral between A and D equal d\Phi/dt

[Timo Nieminen]

Not a good argument? Faraday's law tells us what the integral would be about a CLOSED LOOP. You say that the integral along an OPEN LINE is different. So what? How is that any kind of paradox?

Besides, we do know E between D and A, well enough. Perhaps I was insufficiently clear in the final part of my post (which you cut without comment). Let me rephrase:

If the cut is small, the integral of E_induced.dl along ABCD is very close to the integral of E_induced.dl along ABCDA. This is given by Faraday's law. Call this I_faraday. The integral of E_total.dl along ABCD (along the E=0 line) is zero. Since E_total = E_induced + E_electrostatic, the integral of E_electrostatic.dl along ABCD = -I_faraday. Since the electrostatic field is conservative, the integral of E_electrostatic.dl along ABCDA is zero. Thus, the integral of E_electrostatic.dl along DA = I_faraday.

So, the integral of E_total.dl along ABCDA = I_faraday + -I_faraday + I_faraday = 0 + I_faraday = I_faraday.

QED.

[Don Kelly]

I think that there is some confusion here as to what loop is being
considered. In the case of a filamentary closed loop the only E causing
current is due to the flux enclosed- Faraday
Now consider a filamentary loop on the outside of the conductor and one
on the inside- There may or may not be some flux between these two
filaments and it is only this small flux, that is of concern- and only
so if the loops are cut.

p.S I don't know what you are calling ABCD

[Larry Harson]

On Oct 14, 4:14 am, PengKuan Em <titan...@gmail.com> wrote:
> Le samedi 13 octobre 2012 21:57:34 UTC+2, Larry Harson a écrit :
>
>
>
>
>
> > On Oct 13, 12:42 am, PengKuan Em <titan...@gmail.com> wrote:
>
> > The electrons don't move here because the static E has cancelled the
>
> > induced E along this part of the path. The interesting point is that
>
> > it makes no difference to Faraday's law if E:is the total E or just
>
> > the induced E:
>
> >    loop_integral ( E_induced + E_static)*dl
>
> > = loop_integral (E_induced)*dl + loop_integral (E_static)*dl
>
> > = loop_integral (E_induced)*dl
>
> > Larry.
>
> I do not agree with you formula
> EMF= loop_integral ( E_induced + E_static)*dl   OK
>  loop_integral ( E_induced + E_static)*dl
> =loop_integral (E_induced)*dl + loop_integral (E_static)*dl  ,OK
> but
> loop_integral (E_induced)*dl + loop_integral (E_static)*dl
> = loop_integral (E_induced)*dl not OK
> Because
> loop_integral (E_static)*dl=/=0.
>  Maybe you  think E_static is conservative and its integral through closed path is zero.

My loop_integral was meant to be closed anyway. I thought this was
obvious which is why I didn't bother to put it down as:

closed_loop_integral ( E_induced + E_static)*dl
= closed_loop_integral (E_induced)*dl + loop_integral (E_static)*dl
= closed_loop_integral (E_induced)*dl = emf

> Actually, it is not integrated through a close path. It is integrated from A to D, the remaining gap D to A is not in the path of the integral. So, the correct formula should be:
>   loop_integral ( E_induced + E_static)*dl  (From A to D)
>  = loop_integral (E_induced)*dl  (From Ato D) + integral (E_static)*dl  (From A to D)
> integral (E_static)*dl  (CLOSED LOOP) =integral (E_static)*dl  (From A to D) + integral (E_static)*dl  (From D to A gap)=0
>
> Then
> integral (E_static)*dl  (From A to D)
> = -integral (E_static)*dl  (From D to A, gap)= -EMF (gap)
>
>  For the induced E:
> integral (E_induced)*dl  (From A to D)=EMF
>
> then
>   loop_integral ( E_induced + E_static)*dl  (From A to D)
> =EMF  -integral (E_static)*dl  (From D to A, Gap)=EMF-EMF=0
>

> So, the integral along the E=0 line gives zero while the integral in the gap gives EMF- Hide quoted text -

OK. But I think it's simpler to just say:

closed_loop_integral (E_induced + E_static)*dl = emf
integral_A->D ( (E_induced + E_static) + integral_D->A ( (E_induced +
E_static) =emf
integral_D->A ( (E_induced + E_static) =emf

Larry.

[PengKuan Em]

Le lundi 15 octobre 2012 04:28:22 UTC+2, Don Kelly a écrit :
> On 13/10/2012 7:40 PM, PengKuan Em wrote:
>

> > Le samedi 13 octobre 2012 06:52:22 UTC+2, Timo Nieminen a �crit :

you are right.

>
> p.S I don't know what you are calling ABCD
>
>
>
> --
>
> Don Kelly
>
> cross out to reply

ABCD is the circular loop in the article Faraday’s Law Paradox
http://pengkuanem.blogspot.com/2012/10/faradays-law-paradox.html

[PengKuan Em]

Le lundi 15 octobre 2012 03:44:10 UTC+2, Timo Nieminen a écrit :
> On Monday, October 15, 2012 1:30:31 AM UTC+10, PengKuan Em wrote:
>
> > Le dimanche 14 octobre 2012 08:01:56 UTC+2, Timo Nieminen a écrit :
>
> > > On Sunday, October 14, 2012 12:51:36 PM UTC+10, PengKuan Em wrote:
>
> > >
>
> > > > This line of E=0 is the key. This line stretches from A to D, So, the integral along that line is 0 and we get EMF=integral through ABCD along (E=0 line) of (E.dl) = 0 We do not get Faraday's law result for the circular loop .
>
> > >
>
> > > Integral along ABCD isn't a closed loop, so Faraday's law doesn't tell you what the integral should be.
>
> > >
>
> > This is not good argument because if the point A and D are plates of a capacitor, you will have a gap and you cannot get the EMF through the integral. The potential between the 2 plates ca only be known with the quantity of charges, that is not given by Faraday's law.
>
>
>
> Not a good argument? Faraday's law tells us what the integral would be about a CLOSED LOOP. You say that the integral along an OPEN LINE is different. So what? How is that any kind of paradox?
>
>
>
> Besides, we do know E between D and A, well enough. Perhaps I was insufficiently clear in the final part of my post (which you cut without comment). Let me rephrase:
>

Sorry to cut without comment. I want to concentrate to one idea at once.

>
> If the cut is small, the integral of E_induced.dl along ABCD is very close to the integral of E_induced.dl along ABCDA. This is given by Faraday's law. Call this I_faraday. The integral of E_total.dl along ABCD (along the E=0 line) is zero. Since E_total = E_induced + E_electrostatic, the integral of E_electrostatic.dl along ABCD = -I_faraday. Since the electrostatic field is conservative, the integral of E_electrostatic.dl along ABCDA is zero. Thus, the integral of E_electrostatic.dl along DA = I_faraday.
>
>
>
> So, the integral of E_total.dl along ABCDA = I_faraday + -I_faraday + I_faraday = 0 + I_faraday = I_faraday.
>

I said that "This line of E=0 is the key. This line stretches from A to D, So, the integral along that line is 0 and we get EMF=integral through ABCD along (E=0 line) of (E.dl) = 0 We do not get Faraday's law result for the circular loop " was not a good argument because it has bad consequence.

Let us analyse your calculation
"...the integral of E_induced.dl along ABCDA. This is given by Faraday's law. Call this I_faraday. "
I_faraday=-d(phi/dt)

"The integral of E_total.dl along ABCD (along the E=0 line) is zero."

UB-UA=0,UC-UA=0,UD-UA=0 =>UB=UC=UD=UA

"E_total.dl along ABCDA = I_faraday."
Let A' be the terminal point of the integral which is at the same place than A, UA'-UA=I_faraday

I have drawn the curve of the potential here
http://3.bp.blogspot.com/-KfVmH57ZKsk/UHwlile-YbI/AAAAAAAAANs/dWmRC45krPE/s1600/curve+U+faraday.PNG
or in this page http://pengkuanem.blogspot.fr/2012/10/current-loop-inside-thin-wire.html
We see that the potential increases from UA to UA' through the gap, between D and A. This is contrary to the physical phenomenen.

[PengKuan Em]

Le lundi 15 octobre 2012 04:34:59 UTC+2, Larry Harson a écrit :
>
>
> OK. But I think it's simpler to just say:
>
>
>
> closed_loop_integral (E_induced + E_static)*dl = emf
>
> integral_A->D ( (E_induced + E_static) + integral_D->A ( (E_induced +
>
> E_static) =emf
>
> integral_D->A ( (E_induced + E_static) =emf
>
>
>
> Larry.

I have drawn the curve of the potential here
http://3.bp.blogspot.com/-KfVmH57ZKsk/UHwlile-YbI/AAAAAAAAANs/dWmRC45krPE/s1600/curve+U+faraday.PNG
or in this page http://pengkuanem.blogspot.fr/2012/10/current-loop-inside-thin-wire.html

We see that the potential increases from UA to UA' through the gap, between D and A. This is contrary to the physical phenomenon.

Your opinion is the same than Timo Nieminen's. Please read my explanation in the above post.

[Radi Khrapko]

This discussion is very strange. I cannot understand your problem. You cannot use the word "potential" because the E-field is nonpotential. I have drawn three lines of force roughly and four open paths: 1, 2, 3, 0 = ABCD. See http://khrapkori.wmsite.ru/ftpgetfile.php?id=104&module=files
R. Khrapko

[Larry Harson]

On Oct 15, 4:09 pm, PengKuan Em <titan...@gmail.com> wrote:
> Le lundi 15 octobre 2012 03:44:10 UTC+2, Timo Nieminen a écrit :
>
>
>
>
>
> > On Monday, October 15, 2012 1:30:31 AM UTC+10, PengKuan Em wrote:
>
> > > Le dimanche 14 octobre 2012 08:01:56 UTC+2, Timo Nieminen a écrit :
>
> > > > On Sunday, October 14, 2012 12:51:36 PM UTC+10, PengKuan Em wrote:
>
> > > > >  This line of E=0 is the key. This line stretches from A to D, So, the integral along that line is 0 and we get EMF=integral through ABCD along (E=0 line) of (E.dl) = 0 We do not get Faraday's law result for the circular loop .
>
> > > > Integral along ABCD isn't a closed loop, so Faraday's law doesn't tell you what the integral should be.
>
> > > This is not good argument because if the point A and D are plates of a capacitor, you will have a gap and you cannot get the EMF through the integral. The potential between the 2 plates ca only be known  with the quantity of charges, that is not given by Faraday's law.
>
> > Not a good argument? Faraday's law tells us what the integral would be about a CLOSED LOOP. You say that the integral along an OPEN LINE is different. So what? How is that any kind of paradox?
>
> > Besides, we do know E between D and A, well enough. Perhaps I was insufficiently clear in the final part of my post (which you cut without comment). Let me rephrase:
>
> Sorry to cut without comment. I want to concentrate to one idea at once.
>
>
>

[snipped]

> I have drawn the curve of the potential herehttp://3.bp.blogspot.com/-KfVmH57ZKsk/UHwlile-YbI/AAAAAAAAANs/dWmRC45...
> or in this pagehttp://pengkuanem.blogspot.fr/2012/10/current-loop-inside-thin-wire.html

> We see that the potential increases from UA to UA' through the gap, between D and A.

Your plot would be OK if potential was relabelled as "energy".
Potential and energy aren't the same thing as understood by physicists
and engineers, where potential is a scalar function, and taking its
gradient gives a vector field. I.e. Grad (phi) = E_static. As others
have pointed out to you, this can only be done for electrostatic
fields because closed_loop_integral(E_static) *dl = 0 is a necessary
condition. Your plot shows that this condition isn't true, so
therefore you can't use a potential function for your case. For a
start, A and A' are at the same point, yet with different values for
you "potential".

>This is contrary to the physical phenomenen.

What physical phenomena do you have in mind?

Think of your plot as showing the energy of charge as you take it
around the loop. The conclusion is that it gains energy which is
consistent with the physical fact that a constant current will flow
around the loop, despite it losing energy to the resistance of the gap
DA.

Larry.

[PengKuan Em]

I have said earlier that AD could be a capacitor. Here is the drawing
http://2.bp.blogspot.com/-_Ukhtt21tlk/UHyMb9ySTbI/AAAAAAAAAOE/j4Nw-WYs6mc/s1600/plate+AD.PNG
Or the page http://pengkuanem.blogspot.fr/2012/10/current-loop-inside-thin-wire.html

In the space between the plates A and D, the E field directs from D to A. So, the potential of D must be higher than that of A. This is a static one.

[Don Kelly]

It is as I thought-
Take a path as you show for the current and take the integral of E.dl
around this <closed> path. If this path in the wire encloses no flux,
this will be 0. Does that mean E is 0? -definitely not. If it encloses
flux, the integral will be non-zero and in the case of a cut wire there
will be net non-zero integral which can produce an eddy current. Faraday
applies.

Going along ABCD (assuming these are points along the wire)-is not going
along a closed path-Faraday doesn't apply
Going along ABCDA is a closed path where DA is the gap -Faraday
applies.. Since the gap has a very high impedance and the wire has a
very low impedance, one does expect a small potential difference between
points on the wire but the bulk of the potential difference is across
the gap (your own graph implies this). Faraday's Law doesn't tell you
how the induced voltage is distributed around the loop.

No paradox exists. No new discovery has been made.

[PengKuan Em]

This is the key.

I'm talking the ABCD and the gap with Timo and Larry. If you are interested, see the posts just above.

PK

[Timo Nieminen]

On Tuesday, October 16, 2012 1:09:31 AM UTC+10, PengKuan Em wrote:
>
> I have drawn the curve of the potential here
> http://3.bp.blogspot.com/-KfVmH57ZKsk/UHwlile-YbI/AAAAAAAAANs/dWmRC45krPE/s1600/curve+U+faraday.PNG
> or in this page http://pengkuanem.blogspot.fr/2012/10/current-loop-inside-thin-wire.html
> We see that the potential increases from UA to UA' through the gap, between D and A.

It is NOT a potential. The total field, E_total, is non-conservative, so it's integral doesn't give a potential (difference).

> This is contrary to the physical phenomenen.

And your measurement is ... ?

I know that it's contrary to your imagination, but the important point is whether or not it is contrary to reality. So: where is the measurement?

[PengKuan Em]

You have said once this already:"And your measurement is ... ?"
This is for putting you at a higher position than me and to make me shut up. It is not kind.
I do not have measurement, and then? Is Faraday's measurement not enough?

PK

[Timo Nieminen]

I have already, in this thread, referred to measurements which agree with Faraday's law. How is asking for what measurements you base your claim that the results of Faraday's law are not seen in reality?

You claim "contrary to reality". Yet people successfully measure what is predicted by Faraday's law. Where is the contrary measurement?

> I do not have measurement, and then?

If there is no measurement that you know of that disagrees with Faraday's law, on what basis do you claim "contrary to reality"?

(Your superluminal phase speeds in the near field of an antenna was similar - you claimed it can't happen, but it's known and measured, many times, for over a century.)

> Is Faraday's measurement not enough?

Are you claiming that Faraday's measurements disagree with Faraday's law? Be more specific!

[Timo Nieminen]

On Tuesday, October 16, 2012 12:21:55 PM UTC+10, Timo Nieminen wrote:
[Correction:]
> How is asking for what measurements you base your claim that the results of Faraday's law are not seen in reality "putting [me] at a higher position than [you], or asking anything more of you than I have provided?

... and usually people expect extraordinary claims to be supported by some evidence. You make the extraordinary claim on the basis of no evidence at all?

[Larry Harson]

> I have said earlier that AD could be a capacitor. Here is the drawinghttp://2.bp.blogspot.com/-_Ukhtt21tlk/UHyMb9ySTbI/AAAAAAAAAOE/j4Nw-WY...
> Or the pagehttp://pengkuanem.blogspot.fr/2012/10/current-loop-inside-thin-wire.html
>
> In the space between the plates A and D, the E field directs from D to A. So, the potential of D must be higher than that of A. This is a static one.- Hide quoted text -

I'm not sure what your problem is. The capacitor will become charged
with plate A negative, plate D positive as I would expect.

What do you think should happen?

Larry.

[Don Kelly]

I know this. I also know that the discussion has involved a lot of
unnecessary stuff and a lack of communication. However this doesn't
mean that Faraday's Law is invalid or that there is a paradox. It could
point to an error on your part.

What Timo ,Larry, and others are saying is that you are wrong. I agree
with them.

I see a serious and sophomoric error in the paragraph following Fig.1
in the "Faraday Paradox" blog. You apparently try an application of the
DC electrostatic version of Kirchoff's voltage law- Hey- if the voltage
between A and D going from A to D in one direction is 0, then going the
other direction it is also 0. Tain't so in the case of a
non-conservative system where Faraday's Law applies-even if the integral
of E.dl (which I am calling the induced voltage) is constant - the
distribution of this E.dl is not necessarily constant but depends on
the path. you are right- this is the key- but you are trying to fit it
in the wrong lock.


In your graph- you plot potential along the conductor- potential
(difference) with respect to what? Potential (difference) from A to B,
A to C, etc? Potential(difference) with respect to a remote reference?
What reference? What you appear to be saying, correctly, is that with
no current, the integral of E.dl within the conductor between A and D
is 0, and in the (conservative)electrostatic case, this is so whichever
way you go.
Now there is a big BUT- the situation is not conservative. A Bougerre
factor arises,courtesy of Messieurs Faraday and Maxwell recognizing
what is actually going on in the real world and trying to model it.

Having E=0 in the conductor does NOT mean that E across the gap is
also 0. In fact evidence that is easily obtained says otherwise.
You can experiment-and using alternating current vs a ramp won't make
any fundamental change except that it makes the experiment easier. You
can make it a thought experiment as well. If you don't know how to do
either, I can tell you. (hint-consider a resistance r in the conductor
and R across the gap-play with these). In the meantime, don't put your
fingers across the terminals of an open circuited HV transformer secondary.

[Radi Khrapko]

This discussion is awful. You cannot use the word "potential", because the E-field is nonpotential. In reality, there is no problem here. Please see http://khrapkori.wmsite.ru/ftpgetfile.php?id=104&module=files

[PengKuan Em]

Larry,

It would be better to make clear the used terms. I was using the term potential for the integral(E.dl), which is not understood as potential if the E has a non conservative field E(Faraday) as a component. I agree that this was confusing.

Let me explain with the term "energy" for the same thing. You said: "Your plot would be OK if potential was relabelled as "energy"." for the drawing
http://3.bp.blogspot.com/-KfVmH57ZKsk/UHwlile-YbI/AAAAAAAAANs/dWmRC45krPE/s1600/curve+U+faraday.PNG

So, e.UA,e.UB,e.UC,e.UD,e.UA' are the different energy at A,B,C,D,A' of an electron that has traveled through ABCDA'. The energy of the electron at D is e.UD=e.UA. This energy is computed by integrating along the wire.

According to Faraday's law, there will be a EMF across the gap DA. You said "The capacitor will become charged with plate A negative, plate D positive as I would expect." So, the plate D is charged positively, and an electron on the plate D has an energy e.UA+e.EMF. Do you agree with this energy?

We have thus two energy for the electron at the same place,
Energy=e.UA
for it having reached D from the wire, according to you "Your plot would be OK if potential was relabelled as "energy"."

But Energy=e.UA+e.EMF
for it being measured with Faraday's law method and also according to you: "The capacitor will become charged with plate A negative, plate D positive as I would expect."

Thus, this is a contradiction.

PK

[PengKuan Em]

Don,

Thank you for the explanation. In fact, I was proving what you are saying, but find it is inconsistent.

You said "In your graph- you plot potential along the conductor- potential (difference) with respect to what? Potential (difference) from A to B, A to C, etc? Potential(difference) with respect to a remote reference? "
I integrate from A, the original value of the integral is UA at A.

You said "What you appear to be saying, correctly, is that with no current, the integral of E.dl within the conductor between A and D is 0, and in the (conservative)electrostatic case, this is so whichever way you go."
You agree then the integral of E.dl is 0 from A to D, whether E is conservative or induced.

You said " Having E=0 in the conductor does NOT mean that E across the gap is also 0. In fact evidence that is easily obtained says otherwise."
I was also saying this: The voltage between D and A is EMF. This is given by Faraday's law, and I was not contradicting this value.

I am saying that, when a electron reaches D by passing through the wire, it has a voltage 0. But Faraday has measured it to be EMF. The difference between voltage=0 and voltage=EMF is the point of my claim. That is, the voltage, or energy (as Larry want to call) has two different values at the same place (See the post above to Larry). I am not claiming that EMF=0, as it appears to be understood here.

" You can experiment-and using alternating current vs a ramp won't make any fundamental change except that it makes the experiment easier. You can make it a thought experiment as well. If you don't know how to do either, I can tell you. (hint-consider a resistance r in the conductor and R across the gap-play with these). In the meantime, don't put your fingers across the terminals of an open circuited HV transformer secondary. "

Please do not try to kill me. There is other way to kill the discussion. Lol.

pk

[Larry Harson]

> Let me explain with the term "energy" for the same thing. You said: "Your plot would be OK if potential was relabelled as "energy"." for the drawinghttp://3.bp.blogspot.com/-KfVmH57ZKsk/UHwlile-YbI/AAAAAAAAANs/dWmRC45...

>
> So, e.UA,e.UB,e.UC,e.UD,e.UA' are the different energy at A,B,C,D,A'  of an electron that has traveled through ABCDA'. The energy of the electron at D is e.UD=e.UA. This energy is computed by integrating along the wire.
>
> According to Faraday's law, there will be a EMF across the gap DA. You said "The capacitor will become charged with plate A negative, plate D positive as I would expect." So, the plate D is charged positively, and an electron on the plate D has an energy e.UA+e.EMF. Do you agree with this energy?
>
> We have thus two energy for the electron at the same place,
> Energy=e.UA
>  for it having reached D from the wire, according to you  "Your plot would be OK if potential was relabelled as "energy"."
>
> But Energy=e.UA+e.EMF
> for it being measured with Faraday's law method and also according to you: "The capacitor will become charged with plate A negative, plate D positive as I would expect."
>
> Thus, this is a contradiction.
>

> PK- Hide quoted text -
>
> - Show quoted text -

There's no problem with the same point in space having two particles
with differing energies, since it means one has more kinetic energy
than the other. Of course if they're both static then their energies
are the same.

Previously I said your graphs would be OK if they were labelled
energy. But on reflection I think this is confusing because this
includes potential energy, and this can't be consistently defined in
your example of a non-conservative field. On the other hand, you can
label it as work-done.

What do we meant by potential energy for an electric field?

We mean that the electric field can change the kinetic energy of a
charged particle between different points. But if this change in
kinetic energy depends upon the path taken, then we can't assign a
difference in "potential energy" that depends only on the start and
end points.

So the conclusion is that the idea of "potential energy" cannot be
used for non-conservative fields, which is why it's never used in
these cases by scientists or engineers. Hence the contradiction arises
from you assuming that it's possible to assign a potential energy
value at each point for a non-conservative field, when you can't.

Larry.

[PengKuan Em]

Larry,

"On the other hand, you can label it as work-done." This is fine for me. So, the plot shows the work done on an electron from A to D, which is 0. What is the energy of electron on the back of the plate D? The energy at A, plus the work done. So, for an electron,
energy on the back of the plate D= energy at A + 0 =energy at A

What is the energy measured by Faraday's law on the plate D? energy at A + EMF

There are many electron in the conductor of the plate D and they are static. Some would have "energy at A" and some "energy at A + EMF"


Do you agree?

PK

[Larry Harson]

What do you mean by energy here?

As explained above, we have to confine ourselves to talking about
kinetic energy, because potential energy can't be defined at a point
in your example.

> There are many electron in the conductor of the plate D and they are static. Some would have  "energy at A" and some "energy at A + EMF"
>
> Do you agree?

If they're all static, then their kinetic energy is zero. A charge
being taken from D to A would be accelerated by the static electric
field, and end up with more kinetic energy at A than it had at the
start at A.

Larry.

[PengKuan Em]

Le mercredi 17 octobre 2012 03:10:07 UTC+2, Larry Harson a écrit :

> > Larry,
>
> >
>

> > "On the other hand, you can label it as work-done." This is fine for me. So, the plot shows the work done on an electron from A to D, which is 0. What is the energy of electron on the back of the plate D? The energy at A, plus the work done. So, for an electron,
>
> > energy on the back of the plate D= energy at A + 0 =energy at A
>
> >
>
> > What is the energy measured by Faraday's law on the plate D? energy at A + EMF
>
>
>
> What do you mean by energy here?
>

You do not like the term potential here because it is a non conservative field. OK. The charge before being accelerated from D had energy. What nature is the energy of? You said: "A charge being taken from D to A would be accelerated by the static electric field". You seem to think it is electrostatic, but not potential. Please give it a name.

>
>
> As explained above, we have to confine ourselves to talking about
>
> kinetic energy, because potential energy can't be defined at a point
>
> in your example.
>
>
>
> > There are many electron in the conductor of the plate D and they are static. Some would have "energy at A" and some "energy at A + EMF"
>
> >
>
> > Do you agree?
>
>
>
> If they're all static, then their kinetic energy is zero. A charge
>
> being taken from D to A would be accelerated by the static electric
>
> field, and end up with more kinetic energy at A than it had at the
>
> start at A.
>
>
>
> Larry.

Let call this energy the D energy. Has it the value "energy of A +EMF"?
Or the value "energy of A "? Because the travel through the wire gets 0 work.

PK

[Don Kelly]

No harm meant.

In a conservative system, the potential difference between A and D is
independent of the path -you get the same change in potential energy
whether you take the stairs or take an elevator to the top of a
building. This is not so for a non-conservative system and we are left
with the fact that the path does make a difference. One could deal with
a conductor with a given resistance r and a gap with a resistance R. We
have a closed circuit and we could say that integral of E.dl =ir +iR
This is imposing a restriction that is not implicit in Faraday's Law.
and if we said that the part loop integral E.dl in the conductor is ir
and in the gap it is iR then the the relative voltages Vconductor/Vgap =r/R
In other words ( and I credit Benj for putting this into my head) the
distribution of the induced voltage is determined by the characteristics
of the closed path. If you have an IR drop in part of the loop, then
there is a corresponding part of the total integral of E.dl in the same
location. In most cases, the problem is made conservative by putting an
induced voltage source in a part of the circuit where one either is not
interested in, or can't get at physically to make a measurement. Hence
we have nice motor and generator models as Thevenin sources.

There is another version of this problem given by Prof Lewin in his
lectures at MIT. This had a lot of people scratching their heads.

As for the electron moving to D - when it moved it had KE. When it got
there it lost this KE The distribution of electrons has been changed
and there is also now an energy containing E field across the gap. I
think that this is related to what Larry is saying.

[PengKuan Em]

Le mercredi 17 octobre 2012 05:51:53 UTC+2, Don Kelly a écrit :
> On 16/10/2012 8:49 AM, PengKuan Em wrote:
>

> > Le mardi 16 octobre 2012 04:43:19 UTC+2, Don Kelly a �crit :

Do you mean that my Faraday's law paradox is a known problem?

[Radi Khrapko]

среда, 17 октября 2012 г., 8:00:00 UTC+4 пользователь PengKuan Em написал:

> Do you mean that my Faraday's law paradox is a known problem?

No problem with Faraday's law

[bja...@iwaynet.net]

Consider the following apparatus: A solenoid that creates an increasing
magnetic field from an increasing current and a single circular turn of
wire at one end of it. The circle is broken. At the break one end is
attached a "ground" wire going to a voltmeter and the other wire from
the meter goes to a slider on the loop. It's all arranged so no voltages
are induced in these wires. (wires are all at right angles to the
magnetic fields)

People here seem to be implying a number of things.

1. There are eddy currents in the wire. This is not true to any
significant degree because the wire is in the direction of the E field
and the current ramp is steady so electrons only experience forces in
the direction of the wire at any point.

2. That wire resistance enters in. This is not true either. If there
were a current, we'd have a resistive drop of V=IR. But I = 0 hence V=0
and no voltage drops appear due to wire resistance.

3. That charges build up at the gap creating a large field at the gap.
This is true, because a large potential appears across that gap. And in
the gap (in air) one can calculate E.dL and find that potential using
the standard Electrostatic rules based on the charges sloshed to each
end of the wire.

4. But if we move our "slider" half way round the circle of wire, we now
do not read zero but actually HALF the gap voltage. The assumption has
been made that static electricity rules are applying, but they aren't.
Static electricity implies that there is NO field inside the wire. But
this is wrong. The solenoid is creating an E field EVERYWHERE in space
at it's end, including inside the wire! And as there are free electrons
inside the wire they experience a force and move around.


5. And if equilibrium is reached, each electron must experience no field
(because it's no longer moving). And this zero field for each electron
is created by the shift in positions of it's neighbors. But this does
NOT imply that E.dL integrated along a path inside the wire is zero even
though the charges have rearranged themselves so that the static fields
cancel the induced field at every point in the wire and each free
electron experiences no field.

But since all positive charges are fixed in the wire, the free negative
ones move so as cancel the induced E field. Hence there is a gradient in
the electron charge density in the wire from one end to the other which
represents the potential energy seen across the gap or at various places
along the wire.

The paradox is that while the total E field at each electron in the wire
is zero (it isn't moving) which sort of implies E is zero everywhere in
the wire, E.dL is NOT zero. In fact it is equal to the negative of the
conservative E.dL calculated in the gap. And in fact builds up from zero
to the gap potential as we travel around the wire. We can avoid the
non-conservative multi-valued problem by only going round the loop one
time. This is the typical trick in circuit theory.

In other words, by the time you've moved that slider half way round the
wire, it's picked up half the potential energy it would have gained by
simply crossing the gap.

But if a current flows then we are removing part of the charge
distribution which allows the induced E to dominate static E which in
turn produces a force on electrons entering the loop tending to force
them round to the other end. In other words a current.

Does this make sense?

[PengKuan Em]

Le mercredi 17 octobre 2012 07:06:56 UTC+2, bja...@teranews.com a écrit :
> On 10/16/2012 11:51 PM, Don Kelly wrote:
>
> > On 16/10/2012 8:49 AM, PengKuan Em wrote:
>

> >> Le mardi 16 octobre 2012 04:43:19 UTC+2, Don Kelly a �crit :

Great! Really.
At last someone has found the solution.
The problem is not in physics, but with physicists. They reject all questioning to established laws and get in trouble with old ones.

PK

[Don Kelly]

--------------

>
> Consider the following apparatus: A solenoid that creates an increasing
> magnetic field from an increasing current and a single circular turn of
> wire at one end of it. The circle is broken. At the break one end is
> attached a "ground" wire going to a voltmeter and the other wire from
> the meter goes to a slider on the loop. It's all arranged so no voltages
> are induced in these wires. (wires are all at right angles to the
> magnetic fields)
>
> People here seem to be implying a number of things.
>
> 1. There are eddy currents in the wire. This is not true to any
> significant degree because the wire is in the direction of the E field
> and the current ramp is steady so electrons only experience forces in
> the direction of the wire at any point.

No eddy currents unless the flux extends far enough so that some flux
passes through the cross section of the wire- then there is a
differential induced voltage.

>
> 2. That wire resistance enters in. This is not true either. If there
> were a current, we'd have a resistive drop of V=IR. But I = 0 hence V=0
> and no voltage drops appear due to wire resistance.

I don't recall mention ofany IR drop for I=0

>
> 3. That charges build up at the gap creating a large field at the gap.
> This is true, because a large potential appears across that gap. And in
> the gap (in air) one can calculate E.dL and find that potential using
> the standard Electrostatic rules based on the charges sloshed to each
> end of the wire.

OK- good

>
> 4. But if we move our "slider" half way round the circle of wire, we now
> do not read zero but actually HALF the gap voltage. The assumption has
> been made that static electricity rules are applying, but they aren't.
> Static electricity implies that there is NO field inside the wire. But
> this is wrong. The solenoid is creating an E field EVERYWHERE in space
> at it's end, including inside the wire! And as there are free electrons
> inside the wire they experience a force and move around.

Had a question here but you have answered it in your setup.

>
>
> 5. And if equilibrium is reached, each electron must experience no field
> (because it's no longer moving). And this zero field for each electron
> is created by the shift in positions of it's neighbors. But this does
> NOT imply that E.dL integrated along a path inside the wire is zero even
> though the charges have rearranged themselves so that the static fields
> cancel the induced field at every point in the wire and each free
> electron experiences no field.
>
> But since all positive charges are fixed in the wire, the free negative
> ones move so as cancel the induced E field. Hence there is a gradient in
> the electron charge density in the wire from one end to the other which
> represents the potential energy seen across the gap or at various places
> along the wire.
>
> The paradox is that while the total E field at each electron in the wire
> is zero (it isn't moving) which sort of implies E is zero everywhere in
> the wire, E.dL is NOT zero. In fact it is equal to the negative of the
> conservative E.dL calculated in the gap. And in fact builds up from zero
> to the gap potential as we travel around the wire. We can avoid the
> non-conservative multi-valued problem by only going round the loop one
> time. This is the typical trick in circuit theory.

OK -but the usual trick in circuit theory in my mind is to put the
source in a branch of a circuit loop that is not shared with any other
loop - This branch may have impedance but as long as you consider only
black box terminal conditions (i.e a Thevenin model) and don't look
inside the box- it works.

>
> In other words, by the time you've moved that slider half way round the
> wire, it's picked up half the potential energy it would have gained by
> simply crossing the gap.
>
> But if a current flows then we are removing part of the charge
> distribution which allows the induced E to dominate static E which in
> turn produces a force on electrons entering the loop tending to force
> them round to the other end. In other words a current.
>
> Does this make sense?

Yep

[Timo Nieminen]

On Wednesday, October 17, 2012 3:06:56 PM UTC+10, bja...@teranews.com wrote:
>
> The paradox is that while the total E field at each electron in the wire
> is zero (it isn't moving) which sort of implies E is zero everywhere in
> the wire, E.dL is NOT zero.

That would indeed be a great paradox, if E=0 but the integral of E.dl is non-zero.

Rather than inventing new types of integration rather radically different from boring conventional maths, a sounder application of mathematics would be to deduce from E.dl being non-zero that E is also non-zero. As experimentally measured, even.

[bja...@iwaynet.net]

I understand what you are saying about the problem with E.dL and E = 0.
And it IS a problem, but no "new math" is needed. In a nutshell the
problem goes like this. We know that there is a potential across the
gap. And indeed there is an E field in the gap. And indeed the integral
of E.dL works there because there the fields are all conservative (note
we are ignoring the INDUCED E field in the gap as small with the gap
short to E.dL is much smaller than the E static.dL.

But what about the rest of the loop? We DO know that charges move
through the wire to the ends of the loop. And we know that they build up
on the ends of the cut so we can do the E.dL thing above. All this is fine.

The question then is what about the REST of the wire. Is the E field
zero in the wire when no current is flowing? Yes, it MUST be! Because an
E field on an electron is a force and that would accelerate it! We know
that on average electrons anywhere in the wire are not moving (subject
to Kelly's comments about how one gets eddy currents) because there is
no current. And then as you note, mathematically E.dL integrated around
the wire would have to be zero! But it is not!

In fact, if one were to measure the potential from the gap to any point
in the loop (without drawing any current) one would discover that the
potential linearly increases around the loop until it reaches the full
gap potential at the gap! And oddly, because the induction is
non-conservative, if you go around the loop a second time you get twice
the potential! (Conservative field potentials are independent of path
choice)

So how can E.dL around the loop act as if E has a constant value and
yet, we know that E inside the wire must be zero because charges aren't
moving (no current). This takes some explaining!

I'm am not suggesting any "new maths". I'm suggesting a problem with the
old ones until the situation is explained.

[I don't want to get into circuit theory because it's only an
approximation to field theory, but there the way non-conservative fields
(batteries) are handled is the battery is DEFINED by the E.dL OUTSIDE
the battery where fields are conservative and potential calculations are
unquestioned. This works because the potential outside the battery
terminals must equal the potential inside the battery (MIT professor to
the contrary not withstanding) In this example it would be that the gap
potential would be used to define the loop potential!]

[Timo Nieminen]

On Thursday, October 18, 2012 3:00:42 PM UTC+10, bja...@teranews.com wrote:
>
> I understand what you are saying about the problem with E.dL and E = 0.
>
> And it IS a problem, but no "new math" is needed. In a nutshell the
> problem goes like this. We know that there is a potential across the
> gap.

> And indeed there is an E field in the gap. And indeed the integral
> of E.dL works there because there the fields are all conservative (note
> we are ignoring the INDUCED E field in the gap as small with the gap
> short to E.dL is much smaller than the E static.dL.

As has already been said by many others in this thread (including me).

As has also been said, this is exactly the situation one gets when one choose the line in the wire along which E=0.

But this was not all you were saying. As well as this (correct) thing, you were saying that (a) E=0 everywhere within the wire, and (b) there are no eddy currents. Those are wrong. And they aren't necessary to have the integral of E.dl within the wire equal to zero - it is sufficient to choose a line along which E=0 within the wire. Why add un-needed incorrect assumptions, when the correct result can be obtained without them?

[bja...@iwaynet.net]

I disagree that there are eddy currents inside the wire (at equilibrium
after charges have redistributed themselves) Eddy currents are the
results of imperfections in the field we have suggested so we are
assuming they can be zero. For some reason you think they are the answer
to everything.

And again, I have to point out that if there are no eddy currents, then
there are no currents in the wire, and thus, E MUST be zero inside the
wire or electrons would be moving (currents).

I do agree that INITIALLY when the Induced E is applied to the loop
electrons will quickly redistribute themselves inside the wire but they
also quickly come to a stop as all E fields are cancelled. I say you
are wrong in assuming that there are continuing eddy currents in the
wire. In just what direction are these current circulating? We can use
typical engineering practice to construct our conductive loop out of
many strands of ultra-fine wire which will minimize any cross-section
for eddy currents. But even though we reduce the eddy currents greatly
that has NO effect on induced potential around or even part way round
the loop.

And even if we allow eddy currents, they are in directions that both add
and subtract to E.dL hence do not contribute to the integral around the
loop.

I think eddy currents have nothing to do with this explantion.

[Timo Nieminen]

On Friday, October 19, 2012 8:41:59 AM UTC+10, bja...@teranews.com wrote:
>
> I think eddy currents have nothing to do with this explantion.

They (or their absence) certainly aren't needed for the explanation. They do have to do with the original claims by the OP that E=0 everywhere inside the wire. A claim which you are making as well.

The induced field has non-zero curl(E). The field due to the altered charge distribution in (i.e., on) the wire has curl(E)=0. How can they cancel over a non-zero volume?

Example, please, if you think it is possible.

Meanwhile, consider that eddy currents are _measured_. In real life, they exist. According to theory (and conventional mathematics, not your exotic new maths), they exist. You claim they don't exist (in the ideal case - are you blaming all measurements of them on "imperfections"?). In that case, the measurements, all of them, must be wrong. Either Faraday's law is wrong and the induced field has curl(E)=0, or conventional mathematics is wrong - which one are you claiming is wrong?

And since eddy currents have nothing to do with the explanation (of the OP's "paradox" as eventually clarified) (because you can choose a line in the wire with E=0 even with eddy currents), why begin the explanation with the (incorrect) claim that there are no eddy currents, that E=0 throughout the wire?

> In just what direction are these current circulating?

Given by Faraday's law. The integral version of Faraday's law tells us the integral of E.dl around a closed path. Note that the path has a direction (and the area bounded by the path has an orientation). That direction, and whether the integral of E.dl is positive or negative, tells us the direction of the eddy currents. (There are other ways. E.g., Lenz's law. Note that the eddy currents will result in loss of energy, and will appear as an increased resistance to the circuit driving the changing magnetic field producing the induced E field.)

[Don Kelly]

On 17/10/2012 10:01 PM, bja...@teranews.com wrote:
> On 10/17/2012 10:59 PM, Timo Nieminen wrote:
>> On Wednesday, October 17, 2012 3:06:56 PM UTC+10, bja...@teranews.com
>> wrote:
>>>
>>> The paradox is that while the total E field at each electron in the wire
>>> is zero (it isn't moving) which sort of implies E is zero everywhere in
>>> the wire, E.dL is NOT zero.
>>
>> That would indeed be a great paradox, if E=0 but the integral of E.dl
>> is non-zero.
>>
>> Rather than inventing new types of integration rather radically
>> different from
>
> boring conventional maths, a sounder application of mathematics would be
> to deduce f
>
> rom E.dl being non-zero that E is also non-zero. As experimentally
> measured, even.
>
> I understand what you are saying about the problem with E.dL and E = 0.
> And it IS a problem, but no "new math" is needed. In a nutshell the
> problem goes like this. We know that there is a potential across the
> gap. And indeed there is an E field in the gap. And indeed the integral
> of E.dL works there because there the fields are all conservative (note
> we are ignoring the INDUCED E field in the gap as small with the gap
> short to E.dL is much smaller than the E static.dL.

But while dl is small, the induced E may be relatively large. Certainly
when the gap is closed with whatever, a current flows. Then what is the
E.dl distribution?

>
> But what about the rest of the loop? We DO know that charges move
> through the wire to the ends of the loop. And we know that they build up
> on the ends of the cut so we can do the E.dL thing above. All this is fine.
>
> The question then is what about the REST of the wire. Is the E field
> zero in the wire when no current is flowing? Yes, it MUST be! Because an
> E field on an electron is a force and that would accelerate it! We know
> that on average electrons anywhere in the wire are not moving (subject
> to Kelly's comments about how one gets eddy currents) because there is
> no current. And then as you note, mathematically E.dL integrated around
> the wire would have to be zero! But it is not!
>
> In fact, if one were to measure the potential from the gap to any point
> in the loop (without drawing any current) one would discover that the
> potential linearly increases around the loop until it reaches the full
> gap potential at the gap! And oddly, because the induction is
> non-conservative, if you go around the loop a second time you get twice
> the potential! (Conservative field potentials are independent of path
> choice)

The trouble with measuring this potential is that it is damned hard to
do in this case as there is no loop containing a meter that doesn't
enclose flux---except in the case of a closed magnetic core where
leakage flux is negligable (This is generally a reasonable situation).
Now one can have, with care, a metering loop enclosing negligable flux.
Now there is a problem- the integral of E.dl will be 0 in this loop
but,in general, a current must flow for a voltage to be measured (A
nulling type electrostatic voltmeter might work ).

Have there been part loop measurements of voltage drops in which no
currents have been involved in either the original gapped loop or in
the metering loop?


In the case of constant d(flux)/dt and a gap resistance of R, and
conductor resistance r, it seems that any measurements between two
points on the conductor will reflect the Ir drop in that section and a
measurement across the "gap' reflects the IR drop in the gap. Now what
happens in the limit when the ratio R/r approaches infinity? It seems
that the gap voltage approaches the total integral of E.dl.

This seems to indicate that E inside the conductor, when no current is
flowing is 0. This is as far as one can go looking at a "circuit
approach" which, in this situation, is applicable.

Given that when there is no current, unrealizable except on paper, there
is a static field due to charge redistribution- both inside the
conductor and in the gap. In the conductor- this is not necessarily
uniform and if there is no E at any point, the induced E in the
conductor must be equal and opposite giving a net of 0. However, in the
gap- the static field and the induced field may be additive- too much
wine at present to check this out so this is all conjecture. In any
case, the source of any static field is the induced field and
measurements will not separate them. We are in an "if then maybe" region
In the view from outside-the simplest view is that the total induced
voltage is effectively (as we cannot reach the "ideal" situation)
across the gap.

We are trying to draw conclusions from an artificial model, realizable
only on paper. Another possible solution involves wee little men with
funny hats playing pool with electrons.

>
> So how can E.dL around the loop act as if E has a constant value and
> yet, we know that E inside the wire must be zero because charges aren't
> moving (no current). This takes some explaining!
>
> I'm am not suggesting any "new maths". I'm suggesting a problem with the
> old ones until the situation is explained.
>
> [I don't want to get into circuit theory because it's only an
> approximation to field theory, but there the way non-conservative fields
> (batteries) are handled is the battery is DEFINED by the E.dL OUTSIDE
> the battery where fields are conservative and potential calculations are
> unquestioned. This works because the potential outside the battery
> terminals must equal the potential inside the battery (MIT professor to
> the contrary not withstanding) In this example it would be that the gap
> potential would be used to define the loop potential!]

OK but-in the case of a battery- one knows its location in the
circuit. One know that there is a potential difference across its
terminals. This can also be done for a generator/motor or the secondary,
Consider a conductive rectangular loop with other conductive rectangular
loops on each side. Consider a changing field in the central loop. How
is the E.dl distributed? We cannot assume that it is in a specific
branch of the central loop as each of these is shared and the choice
of branch changes the final solution (a battery is known to be in a
specific branch- and to move it to a different branch can be done with
some topological work and replication (circuit theory is basically
topological). Fortunately, with the induced voltage- we can ignore
that and solve the (5) equations, 4 of which have a sum of voltages =0
and the 5th having the sum of voltages =the loop induced voltage.

With regard to eddy currents- my understanding is that, in the case
where the open ring is outside the field there will be no eddy currents
as the integral of E within the conductor is 0.
If there is flux through the cross section of the ring, then there will
be a differential voltage in loops within the ring and eddy currents
will exist because these internal loops now enclose flux.
My first previous comments, later corrected, were not right.

Hey, I'm having fun-possibly differentiating between the mind sets of
physicists, would be physicists, and engineers.

[Don Kelly]

Presence of eddy currents would eliminate the non-moving charge
argument. That is a different situation than what has been considered.

Also given by Faraday's Law is that if the path of integration doesn't
enclose any flux, there will be no induced voltage in this path to drive
any eddy currents. In the idealized case considered- the conductive
ring has a gap and is completely outside the field that it encloses.
So-if there is a closed conductive path that encloses a changing
magnetic flux- one will have eddy currents. Hence, in AC transformers
and machines the magnetic cores are laminated to minimize these
currents. The rotor of a DC machine is also laminated but the field
poles and yoke are not.

[Timo Nieminen]

On Friday, October 19, 2012 11:15:34 AM UTC+10, Don Kelly wrote:
>
> Presence of eddy currents would eliminate the non-moving charge
> argument. That is a different situation than what has been considered.

The OP's original statement of the problem was not entirely clear, but my reading of it is that B is (spatially) uniform, which means there will be eddy currents.

No problem. When the wire is cut, the eddy current is such that on the outside of the (cut) toroid, the current goes one way, and on the inside, the other way. In between, there is a sheet along which J=0, and hence E=0. Integrate along this, and all is the same as the no-eddy-current case.

Since it isn't necessary to assume no eddy currents, why do so? It's either restricting the problem or assuming incorrect things.

> Also given by Faraday's Law is that if the path of integration doesn't
> enclose any flux, there will be no induced voltage in this path to drive
> any eddy currents. In the idealized case considered- the conductive
> ring has a gap and is completely outside the field that it encloses.

Can do this idealisation for the thought experiment. Over the conducting ring, the field will be different in this case, but not in any way that fundamentally alters the problem. (E_induced will still be purely azimuthal, but will become smaller going outwards (as 1/r), rather than increasing outwards proportionally to r as in the uniform B case. No big deal.)

Doing it in the real world is a little harder, since you won't have B starting at z=-infinity going to z=infinity, but rather loops.

> So-if there is a closed conductive path that encloses a changing
> magnetic flux- one will have eddy currents. Hence, in AC transformers
> and machines the magnetic cores are laminated to minimize these
> currents. The rotor of a DC machine is also laminated but the field
> poles and yoke are not.

Yes. Reduce the eddy currents where they will be large, and ignore them when they are small enough so that you don't care. Laminated instead of one-piece doesn't come for free.

[bja...@iwaynet.net]

On 10/18/2012 9:03 PM, Don Kelly wrote:
> On 17/10/2012 10:01 PM, bja...@teranews.com wrote:

>> In fact, if one were to measure the potential from the gap to any point
>> in the loop (without drawing any current) one would discover that the
>> potential linearly increases around the loop until it reaches the full
>> gap potential at the gap! And oddly, because the induction is
>> non-conservative, if you go around the loop a second time you get twice
>> the potential! (Conservative field potentials are independent of path
>> choice)

> The trouble with measuring this potential is that it is damned hard to
> do in this case as there is no loop containing a meter that doesn't
> enclose flux---except in the case of a closed magnetic core where
> leakage flux is negligable (This is generally a reasonable situation).
> Now one can have, with care, a metering loop enclosing negligable flux.
> Now there is a problem- the integral of E.dl will be 0 in this loop
> but,in general, a current must flow for a voltage to be measured (A
> nulling type electrostatic voltmeter might work ).

Well, you are trying to make it a practical experiment, where I'm just
trying to suggest doable setups but without actually specifying
apparatus. But I suggest that even though apparatus isn't the question,
what I suggest CAN be done. Leads can be arrange along flux lines so no
flux is enclosed and even a primitive potentiometer draws no current at
balance. (though it does getting there).

> Have there been part loop measurements of voltage drops in which no
> currents have been involved in either the original gapped loop or in the
> metering loop?

I'm sure there have, though I have no specific references. The
alternative which is unviable would be that all the charges making
conservative potential in the gap is due ONLY to charges accumulated at
the ends of the wire! I don't think that flies.

> In the case of constant d(flux)/dt and a gap resistance of R, and
> conductor resistance r, it seems that any measurements between two
> points on the conductor will reflect the Ir drop in that section and a
> measurement across the "gap' reflects the IR drop in the gap. Now what
> happens in the limit when the ratio R/r approaches infinity? It seems
> that the gap voltage approaches the total integral of E.dl.

We've already noted there is no current, but to go with your approach,
in essence you have a circuit which is equivalent to a battery and a
resistor load. The load is the gap. The loop is the battery. The
potential across the gap is equal to the integral of E.dL across the
gap. But we also know that the potential at the "battery" terminals
generated by the battery (loop) must equal that potential even though in
the gap the field is conservative while in the loop it is non-conservative!

> This seems to indicate that E inside the conductor, when no current is
> flowing is 0. This is as far as one can go looking at a "circuit
> approach" which, in this situation, is applicable.

And again I say E inside the wire MUST be zero. If it is not, then
electrons will move. They only move until equilibrium is reached. Then
they can move no more! I think eddy currents are a side issue especially
if one arranges the loop outside the flux as you proposed.

Think of the static case. You have a conductor. Inside it are equal
negative and positive charges. They are uniformly distributed so they
cancel out. Now introduce some EXTRA charge into that mix. It produces E
fields. And those E fields move electrons. In fact they move to the
surface so that the field inside the conductor is once again zero. OK.
NOW add a linear E field throughout that volume! What happens? Clearly
electrons (positive charges are fixed in space) will feel a force and
move. They will move until they feel no more force (E=0) What will that
final distribution look like? Of course Timo will point out that our
linear induced field can't be linear because it has non-zero curl. True.
But it could be part of a circle with a very large radius.

> Given that when there is no current, unrealizable except on paper, there
> is a static field due to charge redistribution- both inside the
> conductor and in the gap. In the conductor- this is not necessarily
> uniform and if there is no E at any point, the induced E in the
> conductor must be equal and opposite giving a net of 0. However, in the
> gap- the static field and the induced field may be additive- too much
> wine at present to check this out so this is all conjecture. In any
> case, the source of any static field is the induced field and
> measurements will not separate them. We are in an "if then maybe" region
> In the view from outside-the simplest view is that the total induced
> voltage is effectively (as we cannot reach the "ideal" situation) across
> the gap.

Yes, now you get it! I already noted that the induced E adds to the
static E in the gap (the induced E is in the air) but we are ignoring it
for purposes of this discussion. It appears to be small and if the gap
is narrow, negligible.

But Timo also makes a very valid point with regard to a field with a
zero curl being able to cancel a field with a non-zero curl. Is that
possible even only over a limited volume? Note that this cancellation
is ONLY in the wire. The E field that results from the charge
redistribution inside the wire, produces who knows what kind of E field
outside the wire! I'm currently thinking that the fact that the charge
density is restricted to the boundary conditions of the volume of the
wire somehow plays a role in the cancellation of these two fields WITHIN
THAT VOLUME!

> We are trying to draw conclusions from an artificial model, realizable
> only on paper. Another possible solution involves wee little men with
> funny hats playing pool with electrons.

Come on Don! Save the "magic" solutions for cosmology and QM!

But obviously these should be very basic simple questions involving the
things much of our current electrical technology is based upon. How can
the answers be difficult?

> With regard to eddy currents- my understanding is that, in the case
> where the open ring is outside the field there will be no eddy currents
> as the integral of E within the conductor is 0.

Mine too.

> If there is flux through the cross section of the ring, then there will
> be a differential voltage in loops within the ring and eddy currents
> will exist because these internal loops now enclose flux.
> My first previous comments, later corrected, were not right.

Let's keep the flux out of the wire just for discussion.

> Hey, I'm having fun-possibly differentiating between the mind sets of
> physicists, would be physicists, and engineers.

Say, wait a minute. Since I'm the "official" crackpot here, I guess that
puts me into the would-be physicist category, right?

[Larry Harson]

On Oct 17, 4:03 am, PengKuan Em <titan...@gmail.com> wrote:
> Le mercredi 17 octobre 2012 03:10:07 UTC+2, Larry Harson a écrit :
>
>
>
>
>
> > > Larry,
>
> > > "On the other hand, you can label it as work-done." This is fine for me. So, the plot shows the work done on an electron from A to D, which is 0. What is the energy of electron on the back of the plate D? The energy at A, plus the work done. So, for an electron,
>
> > > energy on the back of the plate D= energy at A + 0 =energy at A
>
> > > What is the energy measured by Faraday's law on the plate D? energy at A + EMF
>
> > What do you mean by energy here?
>
> You do not like the term potential here because it is a non conservative field. OK.

OK, so by energy you mean potential energy.

>The charge before being accelerated from D had energy. What nature is the energy of?

It's you who's saying it has energy, and therefore you need to define
what you mean by it.

>You said: "A charge being taken from D to A would be accelerated by the static electric field". You seem to think it is electrostatic, but not potential. Please give it a name.

It's being accelerated by an electrostatic electric field and we can
assign a potential energy value to its positon within the air gap.

> > As explained above, we have to confine ourselves to talking about
>
> > kinetic energy, because potential energy can't be defined at a point
>
> > in your example.
>
> > > There are many electron in the conductor of the plate D and they are static. Some would have  "energy at A" and some "energy at A + EMF"
>
> > > Do you agree?
>
> > If they're all static, then their kinetic energy is zero. A charge
>
> > being taken from D to A would be accelerated by the static electric
>
> > field, and end up with more kinetic energy at A than it had at the
>
> > start at A.
>
> > Larry.
>
> Let call this energy the D energy. Has it the value "energy of A +EMF"?
> Or the value  "energy of A "? Because the travel through the wire gets 0 work.

You still haven't defined what you mean by energy.

For a start, it's a scalar quantity that is conserved in a closed
system where fields and particles don't cross the boundary. For a
closed electromagnetic system we can say:

total (mechanical_energy + electric_energy + magnetic_energy) = const

The static electric field contributes in part to the total
electric_energy of the system which we can write as a conservation
law, given that it's conservative:

total (electrostatic_potential_energy +
contribution_to_mechanical_kinetic_energy) = const

So here we can usefully say a charge at some position in the system
contributes to the potential energy of the system.

Energy is contained in the kinetic energy of particles and the energy
of the fields, the latter affected by how the particles are
positioned.

Larry.

[Don Kelly]

On 18/10/2012 7:57 PM, Timo Nieminen wrote:
> On Friday, October 19, 2012 11:15:34 AM UTC+10, Don Kelly wrote:
>>
>> Presence of eddy currents would eliminate the non-moving charge
>> argument. That is a different situation than what has been considered.
>
> The OP's original statement of the problem was not entirely clear, but my reading of it is that B is (spatially) uniform, which means there will be eddy currents.
>

OK - I have assumed that the flux was contained inside the inner radius
of the ring. In that case - no eddy currents. I also considered the case
where the flux was spatially uniform and extended past the outer radius
of the ring- as far as one wants. Then- there will be eddy currents. I
believe that I tried to deal with both cases.

> No problem. When the wire is cut, the eddy current is such that on the outside of the (cut) toroid, the current goes one way, and on the inside, the other way. In between, there is a sheet along which J=0, and hence E=0. Integrate along this, and all is the same as the no-eddy-current case.
>
> Since it isn't necessary to assume no eddy currents, why do so? It's either restricting the problem or assuming incorrect things.

If there are no eddy currents in the split ring- then it appears that
there are no currents at all in the ring and this appears to be where
all the fuss starts.

>> Also given by Faraday's Law is that if the path of integration doesn't
>> enclose any flux, there will be no induced voltage in this path to drive
>> any eddy currents. In the idealized case considered- the conductive
>> ring has a gap and is completely outside the field that it encloses.
>
> Can do this idealisation for the thought experiment. Over the conducting ring, the field will be different in this case, but not in any way that fundamentally alters the problem. (E_induced will still be purely azimuthal, but will become smaller going outwards (as 1/r), rather than increasing outwards proportionally to r as in the uniform B case. No big deal.)
>
> Doing it in the real world is a little harder, since you won't have B starting at z=-infinity going to z=infinity, but rather loops.

Agreed. The way around this and get the B totally enclosed is to have
the ring around the leg of a transformer core. Not perfect but it can be
quite good.

>
>> So-if there is a closed conductive path that encloses a changing
>> magnetic flux- one will have eddy currents. Hence, in AC transformers
>> and machines the magnetic cores are laminated to minimize these
>> currents. The rotor of a DC machine is also laminated but the field
>> poles and yoke are not.
>
> Yes. Reduce the eddy currents where they will be large, and ignore them when they are small enough so that you don't care. Laminated instead of one-piece doesn't come for free.

True -being able to cast a core would be nice but since typical
electrical steel for transformers and machines is relatively brittle and
shaping is a problem- lamination from sheet material also offers a
manufacturing benefit. There have been cores made with a glass-metal
material which has low hysteresis and eddy current losses. It is also
true that there are situations where lamination cannot be done- other
approaches are needed.

Incidentally, in a properly designed DC machine the fluxes in the pole.
and outer frame are all DC. In the rotor, there is a need for
laminations laminations.

[PengKuan Em]

I have got insulting critics such as “non sense, ignorant” and many unjustified “mistake, wrong …” here. In order to clarify the controversy, I have written what is accepted and what is controversial of my articles; see “Crux of the controversy about Faraday’s law” http://pengkuanem.blogspot.com/2012/10/crux-of-controversy-about-faradays-law.html

The point is that the voltage on the terminals A, D is OK, but the integral along the loop’s wire is not accepted. So, can you propose your version of computation?

Those who gave insulting judgment please give your justification with valid mathematical proof and show that you are good physicists and not only arrogant ones.

[Jos Bergervoet]

On 10/22/2012 11:16 PM, PengKuan Em wrote:
> I have got insulting critics such as �non sense, ignorant� and many unjustified �mistake, wrong �� here. In order to clarify the controversy, I have written what is accepted and what is controversial of my articles; see �Crux of the controversy about Faraday�s law� http://pengkuanem.blogspot.com/2012/10/crux-of-controversy-about-faradays-law.html
>
> The point is that the voltage on the terminals A, D is OK, but the integral along the loop�s wire is not accepted. So, can you propose your version of computation?

No-one here needs to tell you the correct way.
You can just study the accepted theories of
physics (especially gauge invariance!)

> Those who gave insulting judgment please give your justification with valid mathematical proof and show that you are good physicists and not only arrogant ones.

They don't need to do that at all! It is *you*
who needs to validate the controversial opinions
that you are posting here.

O, and please do not use "EMF". It is a crackpot
term!

--
Jos

[Larry Harson]

On Monday, October 22, 2012 10:16:43 PM UTC+1, PengKuan Em wrote:
> I have got insulting critics such as “non sense, ignorant” and many unjustified “mistake, wrong …” here. In order to clarify the controversy, I have written what is accepted and what is controversial of my articles; see “Crux of the controversy about Faraday’s law” http://pengkuanem.blogspot.com/2012/10/crux-of-controversy-about-faradays-law.html
>
>
>
> The point is that the voltage on the terminals A, D is OK, but the integral along the loop’s wire is not accepted. So, can you propose your version of computation?

Voltages are used in regions where it's value is independent of path. It's therefore confusing when you talk about voltages in non-conservative circuits unless you make clear which path you're using. But for now, I won't complain as long as you make the path clear. But really you should use work-done between A and B since this term isn't a problem.


The gap between A and D is small, so the electrostatic field will dominate the induced electric field within this region. We can therefore talk about the voltage between A and D for paths just in this region, because E_static*dl is independent of path here.

Inside the wire between A and D, the electrostatic and induced E are almost equal to one another since the total E is almost zero everywhere. But if we just use (E_electrostatic)*dl in the wire, then this will equal the voltage between A and D for paths in the gap.

So if you want to talk about the *voltage* between A and D independent of path, then only use E_static*dl. Hence when we take an electron once around the loop, then it's only E_induced*dl which increases its kinetic energy.

Moving onto your paper:

1. In the resistances, there is an internal static electric field as well as the induced E field. These accelerate the electrons until the average electric field of the ions that scatters the electrons, matches that of the accelerating E_total field. So you can say (E_static + E_induced)*dl = IR, without the minus as you've incorrectly put in.

2. You say: "Surprisingly, we have obtained 2 different values for the voltage between the same 2 points A and D, one positive and one negative.". If you remove the - sign and add, you will get the EMF which isn't surprising. Nor is it surprising that you've got two different values for the voltages because you've included the induced_E. If you only use E_static, then your voltages will be independent of path.

3. Please label your equations to make the discussion easier

Regards,

Larry.





Larry.

[PengKuan Em]

Le mercredi 24 octobre 2012 05:13:11 UTC+2, Larry Harson a écrit :
> Voltages are used in regions where it's value is independent of path. It's therefore confusing when you talk about voltages in non-conservative circuits unless you make clear which path you're using. But for now, I won't complain as long as you make the path clear. But really you should use work-done between A and B since this term isn't a problem.
>
> The gap between A and D is small, so the electrostatic field will dominate the induced electric field within this region. We can therefore talk about the voltage between A and D for paths just in this region, because E_static*dl is independent of path here.
>
> Inside the wire between A and D, the electrostatic and induced E are almost equal to one another since the total E is almost zero everywhere. But if we just use (E_electrostatic)*dl in the wire, then this will equal the voltage between A and D for paths in the gap.
> So if you want to talk about the *voltage* between A and D independent of path, then only use E_static*dl. Hence when we take an electron once around the loop, then it's only E_induced*dl which increases its kinetic energy.
>
> Moving onto your paper:
>
> 1. In the resistances, there is an internal static electric field as well as the induced E field. These accelerate the electrons until the average electric field of the ions that scatters the electrons, matches that of the accelerating E_total field. So you can say (E_static + E_induced)*dl = IR, without the minus as you've incorrectly put in.
>
> 2. You say: "Surprisingly, we have obtained 2 different values for the voltage between the same 2 points A and D, one positive and one negative.". If you remove the - sign and add, you will get the EMF which isn't surprising. Nor is it surprising that you've got two different values for the voltages because you've included the induced_E. If you only use E_static, then your voltages will be independent of path.
>
> 3. Please label your equations to make the discussion easier
>
> Regards,
>
>
>
> Larry.

Larry,

“If you only use E_static, then your voltages will be independent of path.”
This is what I have been waiting for: using (E_electrostatic)*dl only and let down E_induced*dl in the integral, because E_electrostatic is conservative and all paths are OK.

But this view introduces another polemic: to distinguish induced E field from electrostatic E field. In the paper ‘What do voltmeters measure?’ http://www.uvm.edu/~dahammon/Demonstrations/5ElectricityAndMagnetism/5bElectricFieldsAndPotential/5b10ElectricField/Faraday'sTeaser/Romer/Romer.pdf
the author criticized Shadowitz for making “an artificial division of the electric field into conservative and non conservative ones, a distinction not likely to be respected by real meters”

“If you remove the - sign and add, you will get the EMF which isn't surprising.”
The minus comes from the fact that the path of integration is ABCD where the work is done by the current, while in the path A to D through the gap the work done is against the current. If you want to make the sum of the work done in the load and nodes resistances by integrating, you have to follow the current’s direction through out the loop, that is, ABCDA or ADCBA. This way the 2 works done will have the same sign and its value EMF.

Following you suggestion, I have labeled all my equations.

Looking my work seriously but not as a blasphemy, finding solution outside the commonly accepted theory (to distinguish induced E field from electrostatic E field); you are really a good physicist.