Charge moving with a constant linear velocity....

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A charge moving with constant linear velocity does produce a magnetic field, contrary to the initial claim. This is explained through Maxwell's equations and Lorentz transformations, which show that the electromagnetic field is frame-dependent, meaning observers in different frames will perceive the presence or absence of magnetic fields differently. The discussion emphasizes that while a charge at rest only generates an electric field, a moving charge generates both electric and magnetic fields in the observer's frame. The concept of absolute motion or aether is dismissed, as special relativity does not support a preferred reference frame. Experimental evidence for these principles is acknowledged as lacking for single charged particles, yet macroscopic observations in electrical engineering align with theoretical predictions.
  • #31
Maciej Orman said:
But electrons in wire never move with constant linear velocity... What's constant is the average velocity...
So what? If each electron has instantaneous velocity ##v_i<<c## then its contribution to the magnetic field is proportional to ##v_i##. Add up the total fields and the result is proportional to ##\sum v_i## and hence proportional to the average velocity.

I'm modelling the wire as much thinner than the distance at which you measure the field. Feel free to integrate over a cylindrical wire if you like.
 
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  • #32
Ibix said:
So what?
Magnetic field generated by charge moving with accelerated motion is well known... One can generate large magnetic field just by waving an electret which has low charge in uC... The Rowland disk experiment shows that... Still, electret produces no magnetic field when with constant linear velocity motion due to 500 m/s orbital motion of the Earth...
 
  • #33
Maciej Orman said:
Still, electret produces no magnetic field when with constant linear velocity motion due to 500 m/s orbital motion of the Earth...
Motion of the Earth with respect to what? The charge's motion with respect to your sensors is what matters for whether there is a magnetic field detected or not.
 
  • #34
Maciej Orman said:
OK, we have the model (still parameter descriptions missing) and now all we need is the experimental evidence... Or is there any at all? Also the above model should be solved for magnetic field not a force...

I think that's a good point and I think a little bit of research is needed to answer your question about experimental evidence. But i would like to add a couple of points that may be relevant.

Firstly, your Van de Graaf belt is analogous to a straight current carrying wire. Using ordinary electric currents eg from a power pack you can easily detect the field around wire geometries such as solenoids. But it's most difficult to detect it around a straight wires, simply because that geometry gives the weakest field. To give a convincing demonstration you have to use quite high currents and depending on what you use that can lead to blown fuses or molten wires.

Secondly, although the Van de Graaf can produce high voltages the currents can be low, so any magnetic field will be difficult to detect.

I suggest that you do some research, find out what a typical current carried by a V de G belt is and then Biot and Savart to estimate the B field due to that current. I suspect that your magnetometer might not be sensitive enough to detect That field.
 
  • #35
Dadface said:
I think that's a good point and I think a little bit of research is needed to answer your question about experimental evidence. But i would like to add a couple of points that may be relevant.

Firstly, your Van de Graaf belt is analogous to a straight current carrying wire. Using ordinary electric currents eg from a power pack you can easily detect the field around wire geometries such as solenoids. But it's most difficult to detect it around a straight wires, simply because that geometry gives the weakest field. To give a convincing demonstration you have to use quite high currents and depending on what you use that can lead to blown fuses or molten wires.

Secondly, although the Van de Graaf can produce high voltages the currents can be low, so any magnetic field will be difficult to detect.

I suggest that you do some research, find out what a typical current carried by a V de G belt is and then Biot and Savart to estimate the B field due to that current. I suspect that your magnetometer might not be sensitive enough to detect That field.
But one can arrange the belt to move around the circle thus making the motion of the charge accelerate and magnetic field can be easily detected even with a compass... See the Rowland disk experiment...
 
  • #36
Maciej Orman said:
But one can arrange the belt to move around the circle thus making the motion of the charge accelerate and magnetic field can be easily detected even with a compass... See the Rowland disk experiment...

I'm not disputing the results of the Rowland disc experiment. In fact I don't recall having seen a demonstration of it before. What I'm suggesting is that you search for the relevant data and do some calculations. I for one would be interested to see how it all works out. For example how does a typical current from a Rowland disc compare to a typical current from a straight V de G belt?
 
  • #37
Maciej Orman said:
Yes, but I need simple equation shown magnetic field intensity at some point near the charge and on the right factor of velocity times factor of charge magnitude thus showing that magnetic field is proportional to charge and velocity...
It's a lot to type. So let me keep it short and skip the primes for the frame, where the charge is moving.

First of all you get in the frame, where the particle moves with velocity ##v## in ##x^1## direction ##u^{\mu}=\gamma(1,v,0,0)## and thus
$$\rho^2=(u \cdot x)^2-\vec{x}^2=\gamma^2(x^1-v t)^2+(x^2)^2 + (x^3)^2,$$
so that
$$(A^{\mu})=(q u^{\mu} \phi(\rho))=\frac{\gamma q}{4 \pi \rho} (1,\beta).$$
The fields follow by taking the derivatives:
$$\vec{E}=-\vec{\nabla} A^0 - \dot{\vec{A}} = \frac{q \gamma}{4 \pi \rho^{3/2}} (x^1-v t,x^2,x^3)$$
and
$$\vec{B}=\vec{\nabla} \times \vec{A} = -\gamma \vec{v} \times \vec{\nabla} \phi(\rho)=\vec{v} \times \vec{E}=\frac{\gamma q v}{4 \pi \rho^{3/2}} (0,-x^3,x^2).$$
 
  • #38
Dadface said:
I'm not disputing the results of the Rowland disc experiment. In fact I don't recall having seen a demonstration of it before. What I'm suggesting is that you search for the relevant data and do some calculations. I for one would be interested to see how it all works out. For example how does a typical current from a Rowland disc compare to a typical current from a straight V de G belt?
I have searched for over three months, now and cannot find anything relevant...
 
  • #39
vanhees71 said:
It's a lot to type. So let me keep it short and skip the primes for the frame, where the charge is moving.

First of all you get in the frame, where the particle moves with velocity ##v## in ##x^1## direction ##u^{\mu}=\gamma(1,v,0,0)## and thus
$$\rho^2=(u \cdot x)^2-\vec{x}^2=\gamma^2(x^1-v t)^2+(x^2)^2 + (x^3)^2,$$
so that
$$(A^{\mu})=(q u^{\mu} \phi(\rho))=\frac{\gamma q}{4 \pi \rho} (1,\beta).$$
The fields follow by taking the derivatives:
$$\vec{E}=-\vec{\nabla} A^0 - \dot{\vec{A}} = \frac{q \gamma}{4 \pi \rho^{3/2}} (x^1-v t,x^2,x^3)$$
and
$$\vec{B}=\vec{\nabla} \times \vec{A} = -\gamma \vec{v} \times \vec{\nabla} \phi(\rho)=\vec{v} \times \vec{E}=\frac{\gamma q}{4 \pi \rho^{3/2}} (0,-x^3,x^2).$$
Thank you, Ibix already provided the formulae and it looks that the field would be significant at relativistic speeds...
 
  • #40
Ibix said:
Motion of the Earth with respect to what? The charge's motion with respect to your sensors is what matters for whether there is a magnetic field detected or not.
It is irrelevant the only difference is with relative motion there will be a pulse but without motion there will be continuous indication of magnetic field...
The observer has no influence on the charge in motion...
 
  • #41
Ibix said:
Motion of the Earth with respect to what? The charge's motion with respect to your sensors is what matters for whether there is a magnetic field detected or not.
Sun, for example.. Surely you are not disputing the orbital motion of the Earth...
 
  • #42
Marciej
I think a typical V de G current can be taken as a couple of mA (please check on this)

Approximating the belt to a straight wire you can use Biot and Savarts law to estimate the value of the field at any particular distance from the belt. Try googling "magnetic field around a straight wire."

It would then be interesting to compare the field you calculate to the Earth's field.
 
  • #43
Well the point is that in such cases ##\vec{j}=\rho \vec{v}##, i.e., the electric current is a convection current. What sounds trivial today was not in the early days of electromagnetism, i.e., the 19th and (very) early 20th century, and thus people made all kinds of experiments with moving charges. The great puzzle indeed was the "electrodynamics of moving bodies", and that's why Einstein's famous paper about SR of 1905 is titled in this way.

From the experimental side I only know the Rowland experiment (conducted in Berlin) and the Röntgen-Eichenwald experiment, where the convection current was created by moving a polarized dielectric (conducted in Gießen). The analogon with moving magnetization creating an electric field are the unipolar induction experiments a la Faraday.
 
  • #44
Dadface said:
Marciej
I think a typical V de G current can be taken as a couple of mA (please check on this)

Approximating the belt to a straight wire you can use Biot and Savarts law to estimate the value of the field at any particular distance from the belt. Try googling "magnetic field around a straight wire."

It would then be interesting to compare the field you calculate to the Earth's field.
I use CST Studio for computing electromagnetic fields both static and dynamic... https://www.cst.com/products/cstems
 
  • #45
Maciej Orman said:
But one can arrange the belt to move around the circle thus making the motion of the charge accelerate and magnetic field can be easily detected even with a compass... See the Rowland disk experiment...
I haven't found a complete analysis of the experiment from a quick search. But from what I have found, it appears to be showing that the magnetic field due to a ring of charge spinning about its axis is the same as the magnetic field due to a current flowing in a circular loop of wire. All the analyses of the magnetic field of a current in a circular loop that I have seen work by treating it as a series of infinitesimal straight wires and adding up the fields from the straight wires. This is completely consistent with the notion that the only thing that matters is the linear speed of the charge with respect to the sensor.
Maciej Orman said:
It is irrelevant the only difference is with relative motion there will be a pulse but without motion there will be continuous indication of magnetic field...
I don't know what you're trying to say here.
Maciej Orman said:
The observer has no influence on the charge in motion...
True. But the motion of the observer has an effect on how the electromagnetic field breaks down into an electrostatic field and a magnetic field. Or, more precisely, two observers at the same location but in motion relative to one another will measure different electric and magnetic fields; they combine into the same electromagnetic field tensor, however.
Maciej Orman said:
Sun, for example.. Surely you are not disputing the orbital motion of the Earth...
I'm disputing that it's relevant to this question. If you had a sensor on the Sun sufficiently sensitive to detect the fields of a charge on the Earth then the relative motion of the Earth and the Sun would be relevant.
 
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  • #46
Maciej Orman said:
Yes, but I need simple equation shown magnetic field intensity at some point near the charge and on the right factor of velocity times factor of charge magnitude thus showing that magnetic field is proportional to charge and velocity...
Maciej Orman said:
Yes, but I need simple equation shown magnetic field intensity at some point near the charge and on the right factor of velocity times factor of charge magnitude thus showing that magnetic field is proportional to charge and velocity...
Maciej Orman said:
A belt of Van De Graaff generator carries excess charge with constant linear velocity but I have never seen anybody detecting magnetic field around it...

By assuming the current in the belt of a Van de Graaf is a few mA and that the belt is a reasonable analogue of a straight wire, I estimate the field at a distance r from the belt to be of the order of 4 (10-10)/r. Tesla. That means that even if your magnetometer or compass can get to a distance of 1cm from the wire that field would still be about one thousand times weaker than the Earth's field. I think that's difficult to measure but not impossible.
(Please check my calculation)
 
  • #47
Dadface said:
(Please check my calculation)
It's comparable to my answer treating it as a flat sheet of charge in motion.
 
  • #48
Dadface said:
By assuming the current in the belt of a Van de Graaf is a few mA and that the belt is a reasonable analogue of a straight wire, I estimate the field at a distance r from the belt to be of the order of 4 (10-10)/r. Tesla. That means that even if your magnetometer or compass can get to a distance of 1cm from the wire that field would still be about one thousand times weaker than the Earth's field. I think that's difficult to measure but not impossible.
(Please check my calculation)
Yes, but you are assuming normal operation in which charge is transferred to the sphere... I should have provided more detail description: In the proposed experiment there is only a belt with one metal and one glass roller and only a ground comb for charging the belt and there is no top comb to remove charge and transfer outside the belt, so on each revolution belt gets charge until dielectric break down thus maximum charge on the belt reaches a magnitude of several orders larger than in normal operation of the VDG...
 
  • #49
Maciej Orman said:
Yes, but you are assuming normal operation in which charge is transferred to the sphere... I should have provided more detail description: In the proposed experiment there is only a belt with one metal and one glass roller and only a ground comb for charging the belt and there is no top comb to remove charge and transfer outside the belt, so on each revolution belt gets charge until dielectric break down thus maximum charge on the belt reaches a magnitude of several orders larger than in normal operation of the VDG...
Also see how easy it is to observe magnetic field when motion is accelerated using the same method of tribioelectric charging: http://www.6moons.com/audioreviews/furutech8/hero4.jpg
 
  • #50
Maciej Orman said:
Yes, but you are assuming normal operation in which charge is transferred to the sphere... I should have provided more detail description: In the proposed experiment there is only a belt with one metal and one glass roller and only a ground comb for charging the belt and there is no top comb to remove charge and transfer outside the belt, so on each revolution belt gets charge until dielectric break down thus maximum charge on the belt reaches a magnitude of several orders larger than in normal operation of the VDG...

Maciej Orman said:
Also see how easy it is to observe magnetic field when motion is accelerated using the same method of tribioelectric charging: http://www.6moons.com/audioreviews/furutech8/hero4.jpg

If you can create a greater rate of charge flow ( current) then you will get a bigger magnetic field. But aren't you suggesting that an accelerated charge is needed as in, for example, Rowlands experiment?
 
  • #51
Dadface said:
If you can create a greater rate of charge flow ( current) then you will get a bigger magnetic field. But aren't you suggesting that an accelerated charge is needed as in, for example, Rowlands experiment?
Not suggesting, magnetic field due to accelerated excess charge is basic knowledge... What is missing is magnetic field due to constant linear velocity motion in straight line of excess charge experimental evidence... Logic suggests that since constant linear motion can only be visible in relative to other bodies then magnetic field created due to such motion and proportional to the velocity cannot exist since it would define an absolute motion and absolute speed measuring method...
 
  • #52
You don't need acceleration. A steady current also gives a magnetic field. Usually you start with that simpler special case when learning electrodynamics, namely with electrostatic and magnetostatic fields. What you get in addition when charges are accelerated (i.e., charge-current distributions become time-dependent) is electromagnetic radiation in addition to the static field around any charge.
 
  • #53
vanhees71 said:
You don't need acceleration. A steady current also gives a magnetic field. Usually you start with that simpler special case when learning electrodynamics, namely with electrostatic and magnetostatic fields. What you get in addition when charges are accelerated (i.e., charge-current distributions become time-dependent) is electromagnetic radiation in addition to the static field around any charge.
Steady current? You mean DC current in wire? If so, then such represent accelerated motion of electrons and such motion is never in straight line...
 
  • #54
Maciej Orman said:
...maximum charge on the belt reaches a magnitude of several orders larger than in normal operation of the VDG...
Any actual numbers on the charge and velocity of the VDG versus charge and tangential velocity of the Rowland disc?

Maciej Orman said:
Logic suggests that since constant linear motion can only be visible in relative to other bodies then magnetic field created due to such motion and proportional to the velocity cannot exist..
No, it just suggests that this magnetic field would be frame dependent.
 
  • #55
Maciej Orman said:
Not suggesting, magnetic field due to accelerated excess charge is basic knowledge...
Reference, please. Note that simply citing the Rowland disc is not sufficient, since the results are entirely explicable in terms of the linear speed of the charges (see p98-9 of https://books.google.co.uk/books?id...B#v=onepage&q=rowland disc experiment&f=false).

Note that the numbers quoted there suggest that Rowland was able to detect an incredibly small magnetic field, but there is no suggestion that the acceleration was relevant.
Maciej Orman said:
Logic suggests that since constant linear motion can only be visible in relative to other bodies then magnetic field created due to such motion and proportional to the velocity cannot exist since it would define an absolute motion and absolute speed measuring method...
...or, the magnetic field could be frame-dependent.
 
  • #56
A.T. said:
Any actual numbers on the charge and velocity of the VDG versus charge and tangential velocity of the Rowland disc?

No, it just suggests that this magnetic field would be frame dependent.
Only at relativistic speeds...
 
  • #57
Maciej Orman said:
Steady current? You mean DC current in wire? If so, then such represent accelerated motion of electrons and such motion is never in straight line...
I've given the formula for a particle in unaccelerated uniform motion. In the reference frame, where the velocity doesn't vanish, the electromagnetic field has both electric and magnetic components. You don't need accelaration of charges to get a magnetic field. That's all I'm saying, and it follows straight-forwardly from Maxwell theory!
 
  • #58
A.T. said:
Any actual numbers on the charge and velocity of the VDG versus charge and tangential velocity of the Rowland disc?
According to the link in my previous post, Rowland had a 20cm diameter ring charged to 10-7C spinning at 50 revs/s. And was able to detect a magnetic field from it, which suggests epic experimental skills to me.
 
  • #59
Ibix said:
...or, the magnetic field could be frame-dependent.
The electromagnetic field is described by a 2nd-rank tensor and thus frame-independent. The split into electric and magnetic components of course is frame-dependent.
 
  • #60
vanhees71 said:
I've given the formula for a particle in unaccelerated uniform motion. In the reference frame, where the velocity doesn't vanish, the electromagnetic field has both electric and magnetic components. You don't need accelaration of charges to get a magnetic field. That's all I'm saying, and it follows straight-forwardly from Maxwell theory!
Yes, except this is only a theory and if such was true we would have a method of absolute speed measuring using excess charge body as a sensor...
 

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