Faraday disk based idea for generation
Dave
Of note, I see reference that the brushes are the stator, and the
rotor is the disk etc.
This concerns me, because it would mean that my ideas for a Faraday
disk based generator would not work.
See http://www.geocities.com/ling_the_monkey_boy/
With my (faulty?) logic, I assume that a net EMF would be generated
around the conductor loops in the drawings linked from the main page.
If not, why not?
François Guillet
"Dave" <sc...@ksc.th.com> a écrit dans le message de
news:fb3b045b.04060...@posting.google.com...
Current will not flow for the same reason it doesn't flow if you connect the
center of a Faraday disk to the rim via any conductor rotating with the
disk.
In the Lorentz Force F=qVxB, V is the speed of the charges relative to the
observer. But relative to any observer on the rotating conductor, V=0 so
F=0, so no emf can be felt.
A Faraday disk works due to the relative motion of 2 parts of a circuit in a
magnetic field thus one of them seeing emf of the other.
FG
Dave
> magnetic field thus one of them seeing emf of the other.
I agree.
But also, is it true that if two parts of a circuit are rotating in
different magnetic fields, then the emf of each part is also
different?
Dave
> center of a Faraday disk to the rim via any conductor rotating with the
> disk.
> In the Lorentz Force F=qVxB, V is the speed of the charges relative to the
> observer. But relative to any observer on the rotating conductor, V=0 so
> F=0, so no emf can be felt.
Sorry for two posts in answer, but I tried to think carefully about
what you said about the Lorentz force.
Please see my diagram on
http://www.geocities.com/ling_the_monkey_boy/HB/testvolt.html
What I think you are saying is that in case 3 (voltmeter moves with
the conductor), no voltage is seen by the voltmeter?
I'm not even sure about case 1. Maybe the conductor needs to move wrt
to magnet. But even in this case, if the conductor and voltmeter are
both moved wrt to magnet, then are you saying that no emf will be
shown on the voltmeter?
If so, this is quite a surprising result (for me, at least).
François Guillet
Only the voltmeter can say if there is emf or not. The voltmeter is the only
real "observer".
From the voltmeter's pointview in case 1 and 3, magnet and wire are always
at rest with it.
It doesn't a matter if they are in motion relative to other observers.
Thus case 1 = case 3, no emf.
In case 2, if the wire and the magnet moved around the voltmeter while
maintening constant the shape of the circuit, there will be no emf.
There would be emf only if there was a magnetic flux change through the loop
of the voltmeter circuit (it is a way to say that one section of the circuit
would move relative to the magnet).
FG
Dave
> real "observer".
> From the voltmeter's pointview in case 1 and 3, magnet and wire are always
> at rest with it.
> It doesn't a matter if they are in motion relative to other observers.
> Thus case 1 = case 3, no emf.
>
> In case 2, if the wire and the magnet moved around the voltmeter while
> maintening constant the shape of the circuit, there will be no emf.
> There would be emf only if there was a magnetic flux change through the loop
> of the voltmeter circuit (it is a way to say that one section of the circuit
> would move relative to the magnet).
>
> FG
Thanks for your reply, which is appreciated. I understand your
comments, and agree.
Now may I turn to a rotating system? With a rotating system, I think
there might be a possibility to do something that cannot be done with
a linear system. That is, I think: have a rotating conductor that has
no emf generated along it's length.
I have drawn cases 4 and 5, and I hope the attached text is self
explanatory.
Case 5 is the crucial point. Where have I gone wrong?
http://www.geocities.com/ling_the_monkey_boy/HB/rotatevolt.html
François Guillet
> "François Guillet" wrote >
> > Only the voltmeter can say if there is emf or not. The voltmeter is the
only
> > real "observer".
> > From the voltmeter's pointview in case 1 and 3, magnet and wire are
always
> > at rest with it.
> > It doesn't a matter if they are in motion relative to other observers.
> > Thus case 1 = case 3, no emf.
> >
> > In case 2, if the wire and the magnet moved around the voltmeter while
> > maintening constant the shape of the circuit, there will be no emf.
> > There would be emf only if there was a magnetic flux change through the
loop
> > of the voltmeter circuit (it is a way to say that one section of the
circuit
> > would move relative to the magnet).
> >
> > FG
>
> Thanks for your reply, which is appreciated. I understand your
> comments, and agree.
May be I was not clear about case 2: when I said "if the wire and the magnet
moved around the voltmeter while maintening constant the shape of the
circuit, there will be no emf.", I considered the voltmeter rotating also
around its axis (so it was the same as case 1 or 3).
> Now may I turn to a rotating system? With a rotating system, I think
> there might be a possibility to do something that cannot be done with
> a linear system. That is, I think: have a rotating conductor that has
> no emf generated along it's length.
> I have drawn cases 4 and 5, and I hope the attached text is self
> explanatory.
Case 4a and b: I agree with you.
> Case 5 is the crucial point. Where have I gone wrong?
I'll reply a bit later, question of time...
FG
> http://www.geocities.com/ling_the_monkey_boy/HB/rotatevolt.html
François Guillet
...
> Case 5 is the crucial point. Where have I gone wrong?
>
> http://www.geocities.com/ling_the_monkey_boy/HB/rotatevolt.html
Case 5 : how could the voltmeter rotate with the conductor and the magnet?
The circuit should cross the magnetic core somewhere.
In any case, if we assume the emf is due to the Lorentz force acting upon
the electrons, and the voltmeter rotates with the conductor, it follows
there will be no emf as the speed V of the electrons relative to the
voltmeter is 0, so F=qVxB=0 (V is relative to the observer, not to the
source of the magnetic field. A magnetic field can't be a referential
frame).
There is a near setup with a faraday disk: a circuit loop is made by
connecting a wire from the center of the disk to the rim. The circuit
rotates with the disk (no brush) and can include a voltmeter.
Even when shielding the wire to avoid it cuts the magnetic field lines, one
observe no emf.
In order to see emf, you must be in a referential frame from which you
observe the motion of a conductor in a B field. To get current, you must
connect a part of a circuit at rest in your referential frame to the moving
conductor.
If B is time varying, induction phenomenons appear but I suppose it is
outside of your project.
I hope this will help you.
FG
Dave
> >
> > http://www.geocities.com/ling_the_monkey_boy/HB/rotatevolt.html
>
> Case 5 : how could the voltmeter rotate with the conductor and the magnet?
> The circuit should cross the magnetic core somewhere.
The magnetic core rotates along with everything else.
> In any case, if we assume the emf is due to the Lorentz force acting upon
> the electrons, and the voltmeter rotates with the conductor, it follows
> there will be no emf as the speed V of the electrons relative to the
> voltmeter is 0, so F=qVxB=0 (V is relative to the observer, not to the
> source of the magnetic field. A magnetic field can't be a referential
> frame).
>
> There is a near setup with a faraday disk: a circuit loop is made by
> connecting a wire from the center of the disk to the rim. The circuit
> rotates with the disk (no brush) and can include a voltmeter.
> Even when shielding the wire to avoid it cuts the magnetic field lines, one
> observe no emf.
>
> In order to see emf, you must be in a referential frame from which you
> observe the motion of a conductor in a B field. To get current, you must
> connect a part of a circuit at rest in your referential frame to the moving
> conductor.
>
> If B is time varying, induction phenomenons appear but I suppose it is
> outside of your project.
Thanks, François. This has opened up all sorts of avenues of thought -
particularly a spiraling electron in a magnetic field (how does it
know it is travelling in a magnetic field?), and so on. But I don't
want to confuse the topic and would like to keep on track:
My first thought was "Wow, I never looked at it that way before". But
I have a concern, and would like to illustrate it as follows:
A closed magnetic path ( 'C' shape or similar) has a conductor in the
air gap. There is minimal field leakage. The conductor ends are looped
out of the room, across the hall, and into a far room down the
corridor. The other ends of the conductor are connected to a
voltmeter.
If I understand the theory above correctly, when I move the magnet and
air-gap conductor together, then the voltmeter observes somehow that
the air-gap conductor is moving in a magnetic field and shows a
voltage.
Now if I keep the magnet and air-gap conductor stationary, and move
the voltmeter in the other room, the same effect can be produced.
Without trying it, I have great difficulty in believing this would
happen.
So, with this in mind, I can only imagine that the relative speeds
that you mention can only apply if both conductors are in the same
magnetic field.
Have I understood you correctly?
Dave
> that you mention can only apply if both conductors are in the same
> magnetic field.
not "both conductors are in the same magnetic field". Is the relative
field intensity important?
François Guillet
> Thanks, François. This has opened up all sorts of avenues of thought -
> particularly a spiraling electron in a magnetic field (how does it
> know it is travelling in a magnetic field?), and so on.
May be it does not know, it only follows a geodesic path.
> But I don't
> want to confuse the topic and would like to keep on track:
>
> My first thought was "Wow, I never looked at it that way before". But
> I have a concern, and would like to illustrate it as follows:
>
> A closed magnetic path ( 'C' shape or similar) has a conductor in the
> air gap. There is minimal field leakage. The conductor ends are looped
> out of the room, across the hall, and into a far room down the
> corridor. The other ends of the conductor are connected to a
> voltmeter.
> If I understand the theory above correctly, when I move the magnet and
> air-gap conductor together, then the voltmeter observes somehow that
> the air-gap conductor is moving in a magnetic field and shows a
> voltage.
> Now if I keep the magnet and air-gap conductor stationary, and move
> the voltmeter in the other room, the same effect can be produced.
> Without trying it, I have great difficulty in believing this would
> happen.
>
> So, with this in mind, I can only imagine that the relative speeds
> that you mention can only apply if both conductors are in the same
> magnetic field.
>
> Have I understood you correctly?
Yes, and your objection is terrible for me :-(
My explanation worked well for the Faraday disk in all situations (rotating
or at rest: disk, magnet, voltmeter, brushes...).
I must admit it fails completly in the setup you described: we should
observe a voltage and we know we will not have it.
I don't think it is a question of "both conductors in the magnetic field"
because in the Faraday disk, only one conductor is in a magnetic field.
Well, back to point 0! I hope to get better ideas tomorrow and will sleep on
it.
FG
Dave
Hi François,
I have considered the subject further, and have prepared yet another
html page.
If I am wrong, at least it will allow us to break the problem down.
Please see the following link:
http://www.geocities.com/ling_the_monkey_boy/HB/truth_table.htm
My best guess is that a rotating system is it's own reference (i.e. a
rotating system knows it is rotating). Is that nonsense or not?
Best regards,
Dave
François Guillet
> I have considered the subject further, and have prepared yet another
> html page.
> If I am wrong, at least it will allow us to break the problem down.
> Please see the following link:
> http://www.geocities.com/ling_the_monkey_boy/HB/truth_table.htm
> My best guess is that a rotating system is it's own reference (i.e. a
> rotating system knows it is rotating). Is that nonsense or not?
> Best regards,
> Dave
Dave, I don't think it is a question of linear or rotating motion. In the
Faraday disk, one doesn't take account the pecularities of rotating motions
such as centripetal force or radial acceleration and considers only the
pseudo-linear speed of charges crossing the B field. Specific phenomena of
rotation are neglected.
Both in the linear and rotating motions there is no emf when the circuit is
made as a whole (without sliding contacts) and its area remains constant
during the motion in a uniform B field (if the area changes then the
magnetic flux change produces emf).
In this case, the speed of any part of the system is always null relative to
any other thus nothing can happen, all is at rest in the referential frame
of the system.
In your previous setting with the voltmeter in another room far from the
conductor and the magnet, and the voltmeter moving relative to the conductor
and magnet, here is my viewpoint:
1) if we consider the area of the circuit changes there will be an emf due
to the variation of the magnetic flux through it. If we move the voltmeter
creating this variation of the magnetic flux in the circuit, the emf will be
very very weak because the area remains about the same and also the most
part of the magnetic flux through the circuit area is near the magnet.
2) if we consider the area of the circuit is about constant, there will be
no emf, as we thought, for the following reason:
- the conductor moves in a magnetic field, we can admit there is the Lorentz
force applying to the electrons thus an emf is generated in the conductor.
- as the whole circuit is not inside the B field which is located near the
"conductor", the lines of field cut somewhere and in a reverse manner the
wires connecting the conductor to the voltmeter. Viewed from the voltmeter
the wires near the conductor are also moving, creating an opposing emf to
that of the conductor. This emf is opposite but equal in strength due the
fact the magnetic flux is conservative thus the voltmeter shows 0 V.
A possible objection is that the wires move along themself while the
conductor moves perpendicularly. It would not be relevant because the field
lines are not perpendicular to the wires thus there is a non null VxB
component.
A question remains open: why can't we shield a part of the circuit in order
to get emf from the other part.
FG
Dave
> "Dave" <sc...@ksc.th.com> a écrit dans le message de
> news:fb3b045b.04061...@posting.google.com...
> > Hi François,
> > I have considered the subject further, and have prepared yet another
> > html page.
> > If I am wrong, at least it will allow us to break the problem down.
> > Please see the following link:
> > http://www.geocities.com/ling_the_monkey_boy/HB/truth_table.htm
> > My best guess is that a rotating system is it's own reference (i.e. a
> > rotating system knows it is rotating). Is that nonsense or not?
> > Best regards,
> > Dave
>
> Dave, I don't think it is a question of linear or rotating motion. In the
> Faraday disk, one doesn't take account the pecularities of rotating motions
> such as centripetal force or radial acceleration and considers only the
> pseudo-linear speed of charges crossing the B field. Specific phenomena of
> rotation are neglected.
>
First of all, thanks for a very interesting discussion.
It seems to me that if we try to treat it as a linear system - charge
moving relative to B field - we then get problems with a rotating
magnet and conductor stationary, and reference frames (I saw the
previous dis-agreement in the 'Earth as a generator' thread). If it is
so simple, why couldn't everyone agree?
If we look at it as a rotating system, then if my 'truth table' is
correct, then it SEEMS as though it could be explained another way. It
doesn't mean that I'm right, just another way to visualise what is
going on. And in this case, it SEEMS that if a conductor is rotating
in a magnetic field, then emf is produced.
> Both in the linear and rotating motions there is no emf when the circuit is
> made as a whole (without sliding contacts) and its area remains constant
> during the motion in a uniform B field (if the area changes then the
> magnetic flux change produces emf).
I note "uniform magnetic field", and agree.
> In this case, the speed of any part of the system is always null relative to
> any other thus nothing can happen, all is at rest in the referential frame
> of the system.
>
In a uniform magnetic field, in this case, nothing will happen. But
the expanation of this in terms of relative speed is something I am
not sure about. A linear system cannot see if it is moving, so the
statement "all is at rest.." is OK. However, a rotating system knows
when it is rotating, therefore the statement "all is at rest...." may
not be correct. I think there might be a difference.
> In your previous setting with the voltmeter in another room far from the
> conductor and the magnet, and the voltmeter moving relative to the conductor
> and magnet, here is my viewpoint:
>
> 1) if we consider the area of the circuit changes there will be an emf due
> to the variation of the magnetic flux through it. If we move the voltmeter
> creating this variation of the magnetic flux in the circuit, the emf will be
> very very weak because the area remains about the same and also the most
> part of the magnetic flux through the circuit area is near the magnet.
Yes, I agree with this. But we can see that, although the air-gap
conductor and magnet are moving relatively at the same speed in the
opposite direction, they have no effect on the voltmeter.
>
> 2) if we consider the area of the circuit is about constant, there will be
> no emf, as we thought, for the following reason:
> - the conductor moves in a magnetic field, we can admit there is the Lorentz
> force applying to the electrons thus an emf is generated in the conductor.
> - as the whole circuit is not inside the B field which is located near the
> "conductor", the lines of field cut somewhere and in a reverse manner the
> wires connecting the conductor to the voltmeter. Viewed from the voltmeter
> the wires near the conductor are also moving, creating an opposing emf to
> that of the conductor. This emf is opposite but equal in strength due the
> fact the magnetic flux is conservative thus the voltmeter shows 0 V.
I agree. This is similar to the case when the whole system is at rest
in the lab, but travelling as the earth spins.
> A possible objection is that the wires move along themself while the
> conductor moves perpendicularly. It would not be relevant because the field
> lines are not perpendicular to the wires thus there is a non null VxB
> component.
>
Yes, of course.
> A question remains open: why can't we shield a part of the circuit in order
> to get emf from the other part.
I think it depends how you shield. If the shield concentrates the
field by causing all the normally far reaching field to be guided
locally, and the wires go through the shield, then the interaction of
the wires and field at the shield should be carefully checked.
You have mentioned about shielding the wires from a Faraday disk: do
you have any details of the experiment?
Dave
François Guillet
> First of all, thanks for a very interesting discussion.
> It seems to me that if we try to treat it as a linear system - charge
> moving relative to B field - we then get problems with a rotating
> magnet and conductor stationary, and reference frames (I saw the
> previous dis-agreement in the 'Earth as a generator' thread). If it is
> so simple, why couldn't everyone agree?
> If we look at it as a rotating system, then if my 'truth table' is
> correct, then it SEEMS as though it could be explained another way. It
> doesn't mean that I'm right, just another way to visualise what is
> going on. And in this case, it SEEMS that if a conductor is rotating
> in a magnetic field, then emf is produced.
I'm a bit confused with your 'truth table'. I don't understand how you move
each part.
- How could you rotate B or move linearly B with C at rest? What happens to
the wires?
- When A at rest and B moves linearly, B will go outside of the field of the
magnet so we have a non null dphi/dt, it's not at all the same case as
rotation so the comparisson is not possible.
- In the rotating setup, V is connected to the ends of B. The emf due to the
Lorentz force is the same from the center to each end of B. The voltmeter
should never show emf because the potential at each end of B is the same.
> In a uniform magnetic field, in this case, nothing will happen. But
> the expanation of this in terms of relative speed is something I am
> not sure about. A linear system cannot see if it is moving, so the
> statement "all is at rest.." is OK. However, a rotating system knows
> when it is rotating, therefore the statement "all is at rest...." may
> not be correct. I think there might be a difference.
Please see here on the Faraday disk:
http://lakeweb.com/Faraday/images/text1_1.gif
The viewpoint of an observer on the rotating disk is that the Lorentz force
applies to the portion of the circuit external to the disk, and the
viewpoint of an observer in the laboratory is that the Lorentz force applies
to the disk.
But both agree on the voltmeter reading.
It is also said that the ratio Lorentz force/Centripetal force is 10^9 so we
can neglect rotation effects.
...
> You have mentioned about shielding the wires from a Faraday disk: do
> you have any details of the experiment?
Imagine a wire connecting the rim of the disk to a capacity and a large
resistance in series connected to the center of the risk (and fastened near
the center), all rotating.
If the wire is not shielded, there will be the same Lorentz force applying
to the electrons in the wire as in the disk. The capacity will show no
difference of potential at its terminals.
If the wire was shielded, the Lorentz force in the wire should be null or
less than that in the disk, so the capacity should charge through the
resistance. When we stop the disk, the capacity discharges through the
resistance but as the value of the resistance is large, we have time to
measure a voltage.
I have tried myself by passing the wires through ferrite toroids. What I
have measured is 0.
FG
>
> Dave
Dave
> each part.
> - How could you rotate B or move linearly B with C at rest? What happens to
> the wires?
> - When A at rest and B moves linearly, B will go outside of the field of the
> magnet so we have a non null dphi/dt, it's not at all the same case as
> rotation so the comparisson is not possible.
Consider only the time when the conductor is moving across the face of
the magnet.
If we only rotate, or move the conductor a little, then the wires will
simply twist or move slightly.
> - In the rotating setup, V is connected to the ends of B. The emf due to the
> Lorentz force is the same from the center to each end of B. The voltmeter
> should never show emf because the potential at each end of B is the same.
You are correct for A,B,C rotating. If C is stationary, and wires
twisting, then even on your reference page emf should be observed by
V.
I think that the Voltage at each end of B is not the same, even if all
are rotating. Is it so that any loop of conductor in a field in free
air, even if the field is local to a magnet and has a high gradient,
will have a net emf of zero around the loop? But the emf need not be
produced symmetrically, and wouldn't be if the field has a gradient.
The reference example asssumes constant B everywhere.
Because V itself does not produce an emf (not rotating), and Conductor
and wires do not form a complete loop (loop emf not equal to 0), then
it seems possible that a net emf could be observed.
>
> > In a uniform magnetic field, in this case, nothing will happen. But
> > the expanation of this in terms of relative speed is something I am
> > not sure about. A linear system cannot see if it is moving, so the
> > statement "all is at rest.." is OK. However, a rotating system knows
> > when it is rotating, therefore the statement "all is at rest...." may
> > not be correct. I think there might be a difference.
>
> Please see here on the Faraday disk:
> http://lakeweb.com/Faraday/images/text1_1.gif
> The viewpoint of an observer on the rotating disk is that the Lorentz force
> applies to the portion of the circuit external to the disk, and the
> viewpoint of an observer in the laboratory is that the Lorentz force applies
> to the disk.
> But both agree on the voltmeter reading.
> It is also said that the ratio Lorentz force/Centripetal force is 10^9 so we
> can neglect rotation effects.
Tricky, but I think I can follow this. So far, I think the 'truth
table' is still correct? Of course, in the truth table set-up, V is
far away from the magnet.
> > You have mentioned about shielding the wires from a Faraday disk: do
> > you have any details of the experiment?
>
> Imagine a wire connecting the rim of the disk to a capacity and a large
> resistance in series connected to the center of the risk (and fastened near
> the center), all rotating.
> If the wire is not shielded, there will be the same Lorentz force applying
> to the electrons in the wire as in the disk. The capacity will show no
> difference of potential at its terminals.
> If the wire was shielded, the Lorentz force in the wire should be null or
> less than that in the disk, so the capacity should charge through the
> resistance. When we stop the disk, the capacity discharges through the
> resistance but as the value of the resistance is large, we have time to
> measure a voltage.
> I have tried myself by passing the wires through ferrite toroids. What I
> have measured is 0.
This reminds me of the strange question that I always had about
electric motors:
When we put a conductor in the rotor, with soft iron around it. The
magnetic field is led AWAY from the conductor and through the area
beside the conductor. Very little magnetic field passes through the
conductor. So why does the conductor generate the same force as though
the magnetic field were passing directly through it?
I think the same effect could possibly be acting at the ferrite cores.
So you may believe that you are shielding the wire, but in fact, not
having any effect.
Dave
arthur
Hello,
I permet to neter your discussion.
François is a good friend from fr.sci.physique.
Hope I can help...
> My first thought was "Wow, I never looked at it that way before". But
> I have a concern, and would like to illustrate it as follows:
>
> A closed magnetic path ( 'C' shape or similar) has a conductor in the
> air gap. There is minimal field leakage. The conductor ends are looped
> out of the room, across the hall, and into a far room down the
> corridor. The other ends of the conductor are connected to a
> voltmeter.
ok
> If I understand the theory above correctly, when I move the magnet and
> air-gap conductor together, then the voltmeter observes somehow that
> the air-gap conductor is moving in a magnetic field and shows a
> voltage.
ok.
From a referentiel at rest a rotating magnet generates a E field.
This is explained properly by relatitivy and this can be calcuted with
lorentz transformation of fields even out of the context of RR : E' =
E - v X B
In your manip : total force acting on electrons in the wire remains
ZERO like in the referentiel of the wire (or the magnet).
Why, because from the referentiel at rest, there is a E field
producing a force on them equal and opposite to q. v X B.
In fact, it is a pity that we have been learned Lorentz force was q
vXB.
Lorentz Force is q.E + q.vXB. Only this full expressions obeys to
relativity (perfectly to special relativity, quite good to galilean
one).
I hope what I wrote is clear...
> Now if I keep the magnet and air-gap conductor stationary, and move
> the voltmeter in the other room, the same effect can be produced.
> Without trying it, I have great difficulty in believing this would
> happen.
ok.
I have already answered this question to François in another forum and
I am gonna say the contrary right now ;-)
I add this to my former answer : *assuming we don't have to take into
account* the variation of phi in the loop due to the movement of the
voltmeter, we still measure nothing.
How to justify this and this is the point I missed in my discussion
with François. When we look at it in the referentiel at rest, the
magnet is rotating *on itself*. That means that flux lines moves BUT
right but phi does not change its local value. The the referentiel of
the magnet, when it will look at the volmeter moving, this last one
will have a movement such that phi does not change its local value
neither. no variation of phi. no emf.
I love ELM :-)
François Guillet
transformation of the fields. A moving B field transforms into E field and
vice versa. Please have a look at Arthur's post.
The Lorentz force is F=qE+qVxB. From an observer seeing a charge at speed V
in a B field, the force F=qVxB.
For the charge moving in the B field, the field transforms into an E field
and from the charge's pointview the force F=qE is the same than F=qVxB from
the observer's pointview.
Nevertheless it is not yet clear for me how one can distinguish a charge
moving in a uniform B field and a charge at rest, when the observer is in
the referential of the charge.
FG
"Dave" <sc...@ksc.th.com> a écrit dans le message de
news:fb3b045b.04061...@posting.google.com...
arthur
> Dave, I think we both were wrong because we neglected the Lorentz
> transformation of the fields. A moving B field transforms into E field and
> vice versa. Please have a look at Arthur's post.
>
> The Lorentz force is F=qE+qVxB. From an observer seeing a charge at speed V
> in a B field, the force F=qVxB.
> For the charge moving in the B field, the field transforms into an E field
> and from the charge's pointview the force F=qE is the same than F=qVxB from
> the observer's pointview.
>
> Nevertheless it is not yet clear for me how one can distinguish a charge
> moving in a uniform B field and a charge at rest, when the observer is in
> the referential of the charge.
>
Hi,
see fr.sci.physique but in a few words :
simply in referring to what create the B field, not just the B field.
a rotating uniform B is "completely" different from a uniform stating
B.
a rotating magnet is seen as rotating from in observer point of view
EVEN if B still remains constant where it is measured due to the fact
B generated by the magnet is uniform.
Dave
Thanks for your comments, Arthur.
I need a little time to study this.
But in the meantime, I have a question:
When a Faraday disk is rotating while fixed to the magnet, then there
are two points of view.
From the stator point of view, the emf is generated on the rotor.
From the rotor point of view, the emf is generated on the stator.
If emf is physically produced because the electrons move (or want to
move) to one end of the conductor, then in which part of the circuit
is this happening?
arthur
> When a Faraday disk is rotating while fixed to the magnet, then there
> are two points of view.
I am not sure to understand.
-----
I assume you mean the Faraday disk is rotating and the magnet is fixed
> From the stator point of view, the emf is generated on the rotor.
from the stator point of view (magnet) Lorentz Force makes electron
move on the disk (rotor) -> ok with you.
> From the rotor point of view, the emf is generated on the stator.
from the rotor point of view, nothing moves but the E field coming
*from* the magnet generates the movement of electrons *on the disk*
(rotor)
> If emf is physically produced because the electrons move (or want to
> move) to one end of the conductor, then in which part of the circuit
> is this happening?
from both point of view, movement of electrons in the disk are due to
Lorentz force (once B the other time E)
-----
If the disk and the magnet are linked together, what happens ?
both their point of view is the same given they are in the same
refentiel. They both see a voltmeter with 2 wires connected to the
disk having a "strange" movement of rotation on itself.
they see wires moving in a constant E field and qvB part of Lorentz
force generate the emf.
-----
did I understand your question properly ?
Dave
> I assume you mean the Faraday disk is rotating and the magnet is fixed
>
> > From the stator point of view, the emf is generated on the rotor.
>
> from the stator point of view (magnet) Lorentz Force makes electron
> move on the disk (rotor) -> ok with you.
>
> > From the rotor point of view, the emf is generated on the stator.
>
> from the rotor point of view, nothing moves but the E field coming
> *from* the magnet generates the movement of electrons *on the disk*
> (rotor)
>
> > If emf is physically produced because the electrons move (or want to
> > move) to one end of the conductor, then in which part of the circuit
> > is this happening?
>
> from both point of view, movement of electrons in the disk are due to
> Lorentz force (once B the other time E)
>
> -----
>
> If the disk and the magnet are linked together, what happens ?
>
> both their point of view is the same given they are in the same
> refentiel. They both see a voltmeter with 2 wires connected to the
> disk having a "strange" movement of rotation on itself.
>
> they see wires moving in a constant E field and qvB part of Lorentz
> force generate the emf.
>
> -----
>
> did I understand your question properly ?
No, maybe I wasn't clear enough.
I was referring to the disk when it is rotating with the magnet (both
are the 'rotor'), and the 'stator' is the non-moving circuit created
by using brushes as in the link
http://lakeweb.com/Faraday/images/text1_1.gif
The reference states that an observer in either point of view (i.e.
rotating with the magnet and disk, or stationary with the brushes and
wires), sees the emf generated in the other part.
My question is where do the electrons actually experience force? It
cannot be both sides at the same time. If electrons experience force,
do they get slightly closer together at one end of the conductor
(making charge)? Then it seems that in one reference frame, they move,
but in another reference frame, they do not.
Very confusing.
arthur
ok... This is not an easy problem.
Firstly, what I wrote before when I looked at magnet and disk linked
together was wrong because I forgot one force.
Secondly, I think your remark according to which "emf" must appear at
the same place is judicioux and I expect well illustrated in this
explanation.
------------------
// 1. (fixed) circuit loop point of view. //
/ 1.1 what happends in the disk. /
We have a circuit comprising the galvanometer, 2 wires fixed and a
*disk*.
In front of this : a rotating magnet.
From the fixed observer point of view, charges in the disk are
influenced by Lorentz force F = q.E + q.v X B with 2 contributions :
- moving charges from the disk in a B field
- E generated by the magnet (cfr former posts on the topic)
total forces is **zero** because E = - v.B
démo1 : immediate via Lorentz tranformations of ELM field.
démo2 : let's forget the loop circuit, let's place ourselves in the
referentiel of the disk. Lorentz force is zero because there is no
mouvement of the disk (q.v X B = 0) and there is no E field from a
immobil magnet (q.E = 0).
a force = to zero is perceived equal to zero in all referentiel.
***therefore, no emf is generated in the disk***
(Which is the contrary of what is written I think in the reference you
gave us).
/ 1.2 what happens in the wire and in the galvanometer /
First they rae fixed and therefore q.v X B = 0 in Lorentz force
/1.2.1/ assume B is in all space // to the axis of the magnet.
as a consequence, E is in all space radial to the magnet.
and therefore :
in the wires, no emf is produced (only a radial force on charges)
in the galvanometer, an emf is produced due to radial E field
/1.2.2/ let's assume B is diverge as much as we are away from the
magnet.
as a consequence, E is not radial any more but has an axial component.
in the wires, an emf is produced.
in the galvanomter, an emf is produced too due to "remaining radial E
field
/1.2.3/
note that it can be demonstrated that whatever hypothesis (1.2.1 or
1.2.2) the effect in the galvanometer will be the same.
// 2. disk point of view //
/2.1/ what happens in the disk.
In the disk, nothing can happen given it is fixed relatively to the
magnet.
v = 0 ; E = 0 -> F = 0
/2.2/ what happens in the wire and in the galvanomter.
There is just a B field in space, no E field given magnet is still.
The wires and the galvonometer are rotating in front of the magnet.
We get exactly the same results as in the former case according to the
hypothesis we consider about the B field from the magnet.
/2.2.1/ B infinetely parallel
In the wires, Lorentz force produce only a radial force and therefore
no axial force an no emf.
In the galvanometer, we have an radial force due to Lorentz and
therefore an emf.
/2.2.2/ real configuration of B.
In the wires, Lorentz B has a radial component that produce an axial
force and as a consequence an emf.
In the galvanometer, we have a less important axial component of B and
a less important radial force and emf than in case /2.2.1/
// 3 conclusions //
In the described experiment the emf is generated in *wires* of the
circuit but not the rotating disk. This, whatever if we analyse the
problem from the disk or from the laboratory.
You remark is perfectly right.
Dave
Arthur,
Well... what a wonderful explanation! Thank you very much.
May I make an observation based on your conclusion?
The circuit works *because* the rotating disk produces NO Lorentz
force in a magnetic field, compared to the Lorentz force which is
produced in the wires. The disc is important for this reason.
By the way, I set up a 40mm Neodymium magnet and Aluminium disc (all I
had available) in my lathe, yesterday, rotating together. I could not
measure any significant mV or mA by using my DVM probes as brushes,
but I did see a few mV difference if I reversed the leads to check for
DVM zero errors. (??).
The magnet is so strong that if I dropped the aluminium disc on to it,
the eddy currents cushion the fall.
Once again, thanks. I will need to go away for a little while and
ponder over this.
Dave
arthur
Hi Dave,
thank you for the experiments !
> Arthur,
> Well... what a wonderful explanation! Thank you very much.
> May I make an observation based on your conclusion?
>
> The circuit works *because* the rotating disk produces NO Lorentz
> force in a magnetic field, compared to the Lorentz force which is
> produced in the wires. The disc is important for this reason.
With a disk not fixed to the magnet this would not work.
If this is what you mean, I agree with you.
Eg if the disk is fixed in the lab and in front of a rotating magnet
emf produced by lorentz force (qE) in the disk compensates
emf produced lorentz force (qvB) in the wires.
But without disk at all, this would work with brushes in direct
connection with the magnet.
Role played by the disk would be played by the surface of the magnet.
> By the way, I set up a 40mm Neodymium magnet and Aluminium disc (all I
> had available) in my lathe, yesterday, rotating together. I could not
> measure any significant mV or mA by using my DVM probes as brushes,
> but I did see a few mV difference if I reversed the leads to check for
> DVM zero errors. (??).
Unless mistake in my calcul Vm = Emf = 1/2 B.w.R^2
I assume you parameters were :
R = 0.02 m
B ~ 1 T
w ~ 1000 rd/min ~ 100 rad/s
is it right ?
if so, Emf ~ 20 mV
NB:
I don't understand how you can say that you see a
few mV difference if you don't measure any significant mV... ???
In other words : if "V1 - V2 = a few mV" then "V1 or V2 is > a few mV"
Pmb
"Dave" <sc...@ksc.th.com> wrote in message
news:fb3b045b.04061...@posting.google.com...
FYI - When the Faraday disk is used to generate current it is also called a
"homopolar generator". They are low voltage high current generators.
Warning: The homopolar generator is loved by crackpots. They don't
understand how it works and that has lead many of them to make bogus claims
that it generates free-energy and thus violates the principle of the
conservation of energy. Bruce DePalma was the leader of those crackpots.
Pmb
Dave
> sc...@ksc.th.com (Dave) wrote in message news:<fb3b045b.04061...@posting.google.com>...
>
> Hi Dave,
>
> thank you for the experiments !
>
> > Arthur,
> > Well... what a wonderful explanation! Thank you very much.
> > May I make an observation based on your conclusion?
> >
> > The circuit works *because* the rotating disk produces NO Lorentz
> > force in a magnetic field, compared to the Lorentz force which is
> > produced in the wires. The disc is important for this reason.
>
> With a disk not fixed to the magnet this would not work.
> If this is what you mean, I agree with you.
Yes, this is what I mean. The disk is important because it is fixed to
the magnet.
One thought: If I had several disks which rotate with the magnet, each
one insulated from the other, then I could set up a series of loops
where each loop increments the emf. Since the emf in the wires is
generated, but the emf in the disk is zero, then I could simply add
several 'turns'?
Each disk would have brushes, and the brushes would connect to fixed
wires which loop around the next disk and connect to it via another
set of brushes.
> Eg if the disk is fixed in the lab and in front of a rotating magnet
> emf produced by lorentz force (qE) in the disk compensates
> emf produced lorentz force (qvB) in the wires.
>
> But without disk at all, this would work with brushes in direct
> connection with the magnet.
> Role played by the disk would be played by the surface of the magnet.
>
> > By the way, I set up a 40mm Neodymium magnet and Aluminium disc (all I
> > had available) in my lathe, yesterday, rotating together. I could not
> > measure any significant mV or mA by using my DVM probes as brushes,
> > but I did see a few mV difference if I reversed the leads to check for
> > DVM zero errors. (??).
>
> Unless mistake in my calcul Vm = Emf = 1/2 B.w.R^2
>
> I assume you parameters were :
>
> R = 0.02 m
> B ~ 1 T
> w ~ 1000 rd/min ~ 100 rad/s
>
> is it right ?
>
> if so, Emf ~ 20 mV
>
> NB:
> I don't understand how you can say that you see a
> few mV difference if you don't measure any significant mV... ???
> In other words : if "V1 - V2 = a few mV" then "V1 or V2 is > a few mV"
I need to try again, because my 1.5 yr old daughter was up to mischief
in my workshop, and I couldn't concentrate. The DVM was showing 100mV
unconnected (noise); 1mV approx (shorted); variable mV 5-10(connected
to spinning disk) - but I wasn't sure because of noise; not able to
see any mV; reversing the leads gave me the impression of a reversal
in mV reading. I need a better instrument (moving coil meter less
noisy).
Significant, for me, means no doubt as opposed to wishful thinking.
Dave
Dave
> "homopolar generator". They are low voltage high current generators.
>
> Warning: The homopolar generator is loved by crackpots. They don't
> understand how it works and that has lead many of them to make bogus claims
> that it generates free-energy and thus violates the principle of the
> conservation of energy. Bruce DePalma was the leader of those crackpots.
>
> Pmb
Thanks Pmb,
Yes, now I see where the reaction is coming from.
In my reply to Arthur's last post, I mentioned an idea for increasing
the output voltage using more disks.
Using all this new information (for me), I will need to review the
statements made on my feeble effort of a web site.
Dave
Pmb
You're welcome. Did I show you my work on this stuff? I focused in on the
rotating magnet part of the generator. The rotating magnet becomes charged
(total charge remains zero). There is a separation of charge on such a
magnet. See - http://www.geocities.com/physics_world/em/rotating_magnet.htm
The physics is quite interesting. One thing people miss on relativity of
moving bodies is that charge density is not invariant. I.e. its quite
possible that there can be zero charge density in one frame and a non-zero
charge density in another frame (bilge missed this aspect of relativistic
electrodynamics a while back in another thread).
The relativity of moving bodies is quite an interesting branch of
relativity.
Pmb
Dave
> > Thanks Pmb,
> > Yes, now I see where the reaction is coming from.
> > In my reply to Arthur's last post, I mentioned an idea for increasing
> > the output voltage using more disks.
> >
> > Using all this new information (for me), I will need to review the
> > statements made on my feeble effort of a web site.
>
> You're welcome. Did I show you my work on this stuff? I focused in on the
> rotating magnet part of the generator. The rotating magnet becomes charged
> (total charge remains zero). There is a separation of charge on such a
> magnet. See - http://www.geocities.com/physics_world/em/rotating_magnet.htm
>
You didn't, but you have now!
I see some similarity between your equations 15,16,17 and Arthurs
equation for the emf generated by a rotating conductor in a magnetic
field.
The charge separation and the emf due to rotation in a magnetic field
have to cancel out, according to Arthur's post.
> The physics is quite interesting. One thing people miss on relativity of
> moving bodies is that charge density is not invariant. I.e. its quite
> possible that there can be zero charge density in one frame and a non-zero
> charge density in another frame (bilge missed this aspect of relativistic
> electrodynamics a while back in another thread).
>
> The relativity of moving bodies is quite an interesting branch of
> relativity.
>
> Pmb
I have to follow a trail of references in order to understand half of
what you guys talk about. Lorentz force - contraction - tensor - etc
etc.
Dave
arthur
> One thought: If I had several disks which rotate with the magnet, each
> one insulated from the other, then I could set up a series of loops
> where each loop increments the emf. Since the emf in the wires is
> generated, but the emf in the disk is zero, then I could simply add
> several 'turns'?
> Each disk would have brushes, and the brushes would connect to fixed
> wires which loop around the next disk and connect to it via another
> set of brushes.
emf is produced in the wires.
if all wires are next to each other with the configuration you suggest
this should work
> I need to try again, because my 1.5 yr old daughter was up to mischief
> in my workshop, and I couldn't concentrate. The DVM was showing 100mV
> unconnected (noise); 1mV approx (shorted); variable mV 5-10(connected
> to spinning disk) - but I wasn't sure because of noise; not able to
> see any mV; reversing the leads gave me the impression of a reversal
> in mV reading. I need a better instrument (moving coil meter less
> noisy).
> Significant, for me, means no doubt as opposed to wishful thinking.
LOL LOL LOL
arthur
Wonderfull !!
I have developped exactly the same on fr.sci.physique on different
posts treating about the "neutrality of a wire"... I read your website
carefully and I come back to you.
arthur
> You're welcome. Did I show you my work on this stuff? I focused in on the
> rotating magnet part of the generator. The rotating magnet becomes charged
> (total charge remains zero). There is a separation of charge on such a
> magnet. See - http://www.geocities.com/physics_world/em/rotating_magnet.htm
you have demonstrated the origin of the E field I am talking about and
that can be found via Lorentz transformation of EM fields.
you stopped at the polarisation P of the "coils" but these dipoles are
responsible of a E radial field.
question : how to explain the E field not for a rotating magnet but
for a rotating coil ?
> The physics is quite interesting. One thing people miss on relativity of
> moving bodies is that charge density is not invariant.
exact.
> I.e. its quite
> possible that there can be zero charge density in one frame and a non-zero
> charge density in another frame (bilge missed this aspect of relativistic
> electrodynamics a while back in another thread).
>
> The relativity of moving bodies is quite an interesting branch of
> relativity.
the relativity of *charged* moving bodies !
Another question :
When there is a current in a wire electrons are mowing while the atoms
(+) are at rest. The global charge density is not zero but in fact
negative...
conclusion : there is a E field generated when there is current in a
wire... Who *didn't* miss this ?
Question : can this have a consequence of the experimental value
deduce for vacuum permeability ?
Pmb
"arthur" <cee...@hotmail.com> wrote
> question : how to explain the E field not for a rotating magnet but
> for a rotating coil ?
There is no E-field for a rotating coild.
> the relativity of *charged* moving bodies !
Whether a moving body is charged or not depends on you're frame of reference
>
> Another question :
>
> When there is a current in a wire electrons are mowing while the atoms
> (+) are at rest. The global charge density is not zero but in fact
> negative...
In the rest frame of a wire that wire has a net charge density of zero.
That's the total charge density. i.e. the charge density of "+" charges is
the same as the charge density of "-" charges yielding a zero charge
density. This is not true in other frames of reference.
>
> conclusion : there is a E field generated when there is current in a
> wire... Who *didn't* miss this ?
>
> Question : can this have a consequence of the experimental value
> deduce for vacuum permeability ?
Not that I know of.
Pmb
arthur
> "arthur" <cee...@hotmail.com> wrote
>
> > question : how to explain the E field not for a rotating magnet but
> > for a rotating coil ?
>
> There is no E-field for a rotating coild.
This seems obvious. But this is contradictory to Lorentz equations of
EM field transformation.
In theory, the fact that B is produced by a magnet or by a coil
doesn't change anything in term of these transformations ?
A comment ?
> > the relativity of *charged* moving bodies !
>
> Whether a moving body is charged or not depends on you're frame of reference
1 point for you ! I forgot matter was composed of charges !
now 1 point for me : as you showed very well on your website the total
"charge" of the body doesn't change only charge density. "Whether a
moving body is charged on not DO NOT depends on frame of reference..."
Agreed ?
> >
> > Another question :
> >
> > When there is a current in a wire electrons are mowing while the atoms
> > (+) are at rest. The global charge density is not zero but in fact
> > negative...
>
> In the rest frame of a wire that wire has a net charge density of zero.
> That's the total charge density. i.e. the charge density of "+" charges is
> the same as the charge density of "-" charges yielding a zero charge
> density. This is not true in other frames of reference.
ok but what about the fact that a wire crossed by a current generate a
E field ?
Pmb
"arthur" <cee...@hotmail.com> wrote in message
news:f4fb6cc0.04062...@posting.google.com...
> "Pmb" <som...@somewhere.com> wrote in message
news:<ZIOdncaqF_l...@comcast.com>...
> > "arthur" <cee...@hotmail.com> wrote
> >
> > > question : how to explain the E field not for a rotating magnet but
> > > for a rotating coil ?
> >
> > There is no E-field for a rotating coild.
>
> This seems obvious. But this is contradictory to Lorentz equations of
> EM field transformation.
Why do you think that? Suppose you had a single current loop which was
rotating. While the field would be that of a magnetic dipole you can't treat
it as if it were a magnetic dipole in all cases. Think of the current loop
as the supperposition of two charged rings on top of each other. Do you
think that charge density of the rotating ring is the same or different than
the same ring not rotating? Think that one through and I believe the physics
will become clear.
>
> In theory, the fact that B is produced by a magnet or by a coil
> doesn't change anything in term of these transformations ?
>
> A comment ?
>
>
> > > the relativity of *charged* moving bodies !
> >
> > Whether a moving body is charged or not depends on you're frame of
reference
>
> 1 point for you ! I forgot matter was composed of charges !
>
> now 1 point for me : as you showed very well on your website the total
> "charge" of the body doesn't change only charge density. "Whether a
> moving body is charged on not DO NOT depends on frame of reference..."
>
> Agreed ?
If you are refering to total charge then yes. I agree.
> ok but what about the fact that a wire crossed by a current generate a
> E field ?
Please clarify. That statement is unclear to me. Please give an explicitly
example. Thanks.
Pmb
arthur
> Please clarify. That statement is unclear to me. Please give an explicitly
> example. Thanks.
>
Hi,
I am happy to have met somebody who agrees with this vision of ELM :
no a new theory but a will to go deeper into understanding of ELM.
I lack time to discuss properly with you.
I come back this week-end with more time and clarified explanations.
Arthur
sophie duhem
> "Pmb" <som...@somewhere.com> wrote in message news:<ZIOdncaqF_l...@comcast.com>...
>
>>"arthur" <cee...@hotmail.com> wrote
>>
>>
>>>question : how to explain the E field not for a rotating magnet but
>>>for a rotating coil ?
>>
>>There is no E-field for a rotating coild.
>
>
> This seems obvious. But this is contradictory to Lorentz equations of
> EM field transformation.
>
> In theory, the fact that B is produced by a magnet or by a coil
> doesn't change anything in term of these transformations ?
>
> A comment ?
>
>
>
>>>the relativity of *charged* moving bodies !
>>
>>Whether a moving body is charged or not depends on you're frame of reference
>
>
> 1 point for you ! I forgot matter was composed of charges !
>
> now 1 point for me : as you showed very well on your website the total
> "charge" of the body doesn't change only charge density. "Whether a
> moving body is charged on not DO NOT depends on frame of reference..."
>
> Agreed ?
Sorry to be intrusive.
The total charge of a body is a relativistic invariant:
it's total charge does not depend on any reference frame.
Here, you have an infinite wire, who has an infinite charge
, theerefore you have not that problem.
>
>>>Another question :
>>>
>>>When there is a current in a wire electrons are mowing while the atoms
>>>(+) are at rest. The global charge density is not zero but in fact
>>>negative...
>>
>>In the rest frame of a wire that wire has a net charge density of zero.
>>That's the total charge density. i.e. the charge density of "+" charges is
>>the same as the charge density of "-" charges yielding a zero charge
>>density. This is not true in other frames of reference.
>
>
> ok but what about the fact that a wire crossed by a current generate a
> E field ?
A constant current in a wire does not generate an E field in the
wire's ref frame. It does so in any ref frame in movement relative
to the wire. The classic explanation exhibited by Pmb shows
it clearly. Let me summarize it so we can kno< if we agree on it.
Firstly, an infinite static line of charge generates a radial
field: e=2lambda0/r, where lambda0 is the line density in the
wire rest frame R'.
Now suppose that this line moves with velocity v relative to R.
Bercause of length contraction the line density in R is
lambda=gamma.lambda0, and there it constitutes a current
i=gamma.lambda0.v
Let us locate the location of the static line charge with
the x' axis of R'. The only non-vanishing component of the em
field at the point P(0,r,0) in R' is e'2=2lambda0/r.
Transforming the field into R we find that the only non
vanishing components are e2=2gamma.lambda0/r and
b3=2gamma.lambda0.v/cr (I have chosen gaussian units: we are
on an english forum :-)).
The strength of the b-field is only a fraction v/c of the e field.
The e-force is much greater, but in a real wire it is neutralized
by the fact that a real wire corresponds to two charge distributions:
a fixed positive line charge (ions) and a negative one travelling
at speed -v.
Before the current is switched on, we can think of
the ions as a row of chairs on which the free electrons sit.
When current is flowing, electrons play musical chairs, always
moving to the next chair since there can be no build-up of
charge. This means that the elctrons are now as far apart
in motion as are the stationary ions. Hence the respective
line densities of ions and electrons are equal and opposite
+/- lambda, and the current is given by i=lambda.v.
The electric fields cancel exactly, while the magnetic
field is given as before by 2i/c.r.
Consider now a charge moving with velocity u parallel
to the wire. It experiences a force in the direction
uxb: radially towards or away from the wire. In it's
rest frame, where it can be only affected by electric
fields, it sees two moving lines of positive and negative charge.
Without length contraction, it would always see the
two lines as having equal and opposite line densities, and
so never feel a force. But in fact, because of length
contraction, these two charge densities will differ in absolute
value in any frame but the rest frame of the wire.
It is this difference which provides the electric field
that causes the charge to accelerate in it's rest frame.
This is a direct manifestation of length contraction,
arising from a speed difference of a few millimeters
by second (typical drift velocity of electrons in metals).
For those who hace access to it, this is very well written
in Rindlers "Relativity", a wonderfull book.
Pmb
"sophie duhem" <s.d...@infonie.fr> wrote in message
news:cb8mb4$172$1...@news.tiscali.fr...
If a straight and infinitely long current carrying wire is uncharged in the
wires rest frame then it will be charged in a frame which is moving along
the length of the wire. This in and of itself is a violation of the
conservation of charge but since this is not a physically realizable
situation its not really a problem. If, however, the wire is only part of a
loop then one side will be have "+" charges on it and the other side will
have "-" charges on it and the total of those charges will be zero.
(Note: Charge density is the time component of a 4-vector)
> A constant current in a wire does not generate an E field in the
> wire's ref frame.
Of cours it is possible for the wire to be charged in its rest frame. Just
dump excess charge on the wire and in its rest frame it will be charged.
Exactly! A similar situtation holds true for magnets. Magnitizaion in one
frame is a combination of magnitization and polarization in another frame.
Polarization in one frame is a combination of polarization and magnetization
in another frame.
> For those who hace access to it, this is very well written
> in Rindlers "Relativity", a wonderfull book.
Very true. That a current carrying wire which is neutral in its rest frame
is not neutral in a frame moving parallel to the wire is also proved in the
Feynman lectures.
See - "The Feynman Lectures on Physics - Volume II," Feynman, Leighton and
Sands, Addison Wesley, 1977, pages 13-7 to 13-12
Pmb
sophie duhem
>>
>>
>>Sorry to be intrusive.
>>The total charge of a body is a relativistic invariant:
>>it's total charge does not depend on any reference frame.
>>Here, you have an infinite wire, who has an infinite charge
>>, theerefore you have not that problem.
>
>
> If a straight and infinitely long current carrying wire is uncharged in the
> wires rest frame then it will be charged in a frame which is moving along
> the length of the wire. This in and of itself is a violation of the
> conservation of charge but since this is not a physically realizable
> situation its not really a problem. If, however, the wire is only part of a
> loop then one side will be have "+" charges on it and the other side will
> have "-" charges on it and the total of those charges will be zero.
>
> (Note: Charge density is the time component of a 4-vector)
>
>
does not arise whith the unphysical infinite straight line,
thanks for putting it clearly.
arthur
> Sorry to be intrusive.
> The total charge of a body is a relativistic invariant:
> it's total charge does not depend on any reference frame.
> Here, you have an infinite wire, who has an infinite charge
> , theerefore you have not that problem.
yes. we all three agree on that.
and many others ;-)
> > ok but what about the fact that a wire crossed by a current generate a
> > E field ?
>
> A constant current in a wire does not generate an E field in the
> wire's ref frame. It does so in any ref frame in movement relative
> to the wire. The classic explanation exhibited by Pmb shows
> it clearly. Let me summarize it so we can kno< if we agree on it.
Yes that is ture... for an infinite wire.
Exact.
arthur
> Very true, this is exactly what I meant when I said that the problem
> does not arise whith the unphysical infinite straight line,
> thanks for putting it clearly.
but I complete by stating that
in the frame of a loop where a current circulates
even if it is globally neutral (total Q = 0)
it is not obvious or trivial that the E field all around is 0...
I made the calculus last WE but must check results...
---
Just to give the thoughts at the basis of the idea :
in the piece of such a loop, there are moving electrons and
fixed atoms (+). All in equal number.
(Total Q is therefore 0).
the E field associated to such an atom is :
kQ/r^2
but the E field associated with a moving electron is :
kQ/r^2 (1-B²)/(1-B².sin²(th))^3/2
(all terms must be defined but are well known)
Integrated on all the charges of the loop why would these two terms
compensate ?
They do not.
Pmb
"sophie duhem" <s.d...@infonie.fr> wrote
> A constant current in a wire does not generate an E field in the
> wire's ref frame. It does so in any ref frame in movement relative
> to the wire. The classic explanation exhibited by Pmb shows
> it clearly.
One way to understand this is by considering how a current can be
established in the wires rest frame. Let the lattice of "+" ions which make
up the wire define the wire and rest frame. Let the "-" charges be at rest.
Now accelerate the electrons along the wire starting at the same time, have
the same acceleration and then stop accelerating at the same time as
measured in the rest frame. The "-" charges will be the same distance apart
as they were when they weren't moving. Now look at this in the frame of
reference which is moving in the direction that the "-" charges end up at
rest in. Since the charges started accelerating at the same time in the
wire's rest frame they will not accelerate at the same time in the moving
frame. This is the reason the charge density changes in one frame and not
the other.
Pmb
Pmb
"sophie duhem" <s.d...@infonie.fr> wrote
> A constant current in a wire does not generate an E field in the
> wire's ref frame. It does so in any ref frame in movement relative
> to the wire. The classic explanation exhibited by Pmb shows
> it clearly.
One way to understand this is by considering how a current can be