Charge moving with a constant linear velocity....

[Maciej Orman]

Charge moving in constant linear velocity does not produce magnetic field...
If not, please provide an explanation...

 

[Orodruin]

Charge moving in constant linear velocity does not produce magnetic field...

Yes it does.

If not, please provide an explanation...

Your question is unclear. You labeled this thread "A". Which part of Maxwell's equations do you not understand?

 

[vanhees71]

The most simple way to get the electromagnetic field (which of course has non-zero electric and magnetic components) is to use Lorentz invariance. Just solve Maxwell's equations for the charge at rest (leading of course to a Coulomb field) and then do a Lorentz boost. You'll get both electric and magnetic field components! Of course, it's easier to first use the four-potential and then take the derivatives to get the field-strength tensor.

 

[Maciej Orman]

OK, so if I place a magnetometer probe next to the running belt fo Van De Graff generator I should detect magnetic field, or should I? Also if constant linear velocity of a charge generate magnetic field then it means that absolute motion can be measured and the concept of aether is reality... Still one needs experimental confirmation but I cannot find any...

 

[vanhees71]

Of course not! There's no aether. Contrary to the Newtonian spacetime model special relativity (i.e., the Minkowskian spacetime model) precisely makes the description of electromagnetism possible without assuming a preferred reference frame (aka. "restframe of the aether" from the ancient days before Einstein).

Of course if there's a moving charge in a reference frame, the observer at rest wrt. this frame will observe both electric and magnetic field components. In a frame, where the charge stays at rest, he/she will observe only electric field components. The electromagnetic field is of course invariant, i.e., frame independent (it's an antisymmetric tensor field, the Faraday tensor).

 

[Maciej Orman]

Of course not! There's no aether. Contrary to the Newtonian spacetime model special relativity (i.e., the Minkowskian spacetime model) precisely makes the description of electromagnetism possible without assuming a preferred reference frame (aka. "restframe of the aether" from the ancient days before Einstein).

Of course if there's a moving charge in a reference frame, the observer at rest wrt. this frame will observe both electric and magnetic field components. In a frame, where the charge stays at rest, he/she will observe only electric field components. The electromagnetic field is of course invariant, i.e., frame independent (it's an antisymmetric tensor field, the Faraday tensor).

Does that mean that if charge has magnetic field around it, it (the magnetic field) indicates an absolute motion?

 

[Maciej Orman]

Just like to point that an abstract quantities like kinetic energy can change wrt to an observer but magnetic field is a physical quantity and does not change due to relative motion of an observer thus correct logic deduction suggests that constant linear velocity of charge particle does not produce magnetic field...
Also absence of magnetic field around charged particle would indicate an absolute stop motion... Feather more all objects with excess charge located on the Earth
would generated their own magnetic field due to the orbital motion of around 500 m/s...

 

[Ibix]

Does that mean that if charge has magnetic field around it, it (the magnetic field) indicates an absolute motion?

No. Whether a charge has a magnetic field or not is a frame-dependent question. If it has a magnetic field then it is in motion in that frame. If it does not have a magnetic field it is not in motion in that frame.

It always has an electromagnetic field. In some frames this may have a magnetic component or be purely electrostatic.

 

[Ibix]

magnetic field is a physical quantity and does not change

This is incorrect, as I wrote above.

 

[Maciej Orman]

No. Whether a charge has a magnetic field or not is a frame-dependent question. If it has a magnetic field then it is in motion in that frame. If it does not have a magnetic field it is not in motion in that frame.

It always has an electromagnetic field. In some frames this may have a magnetic component or be purely electrostatic.

Laws of physics must hold across all frames of motion and cannot be frame dependent...

 

[Maciej Orman]

This is incorrect, as I wrote above.

It would mean that charged particle in motion produces magnetic field which intensity is observer dependent thus completely hiding magnetic field for moving observer and showing magnetic field for non moving observer... That would be totally illogical...

 

[Ibix]

Laws of physics must hold across all frames of motion and cannot be frame dependent...

True. Whether or not a charge possesses a magnetic component to its electromagnetic field is not a law of physics. The relevant laws of physics are Maxwell's equations, which have the same form in all frames.

It would mean that charged particle in motion produces magnetic field which intensity is observer dependent thus completely hiding magnetic field for moving observer and showing magnetic field for non moving observer... That would be totally illogical...

If you think that then you need to revise your notion of what constitutes "logical". It's trivial to show that it happens by carrying out a Lorentz transform on the electromagnetic field strength tensor.

 

[Maciej Orman]

True. Whether or not a charge possesses a magnetic component to its electromagnetic field is not a law of physics. The relevant laws of physics are Maxwell's equations, which have the same form in all frames.
If you think that then you need to revise your notion of what constitutes "logical". It's trivial to show that it happens by carrying out a Lorentz transform on the electromagnetic field strength tensor.

Logic also suggests that mathematical models have no effect on physics of reality thus incorrect models will not be supported by experimental evidence...
In short if you can point to real experiment where charge moving in constant linear velocity produces magnetic field then that will prove that your Lorentz transform correctly represents mathematical model of the phenomena in question...

 

[Ibix]

In short if you can point to real experiment where charge moving in constant linear velocity produces magnetic field

A straight wire carrying a constant current.

 

[vanhees71]

Indeed the electromagnetic field,
$$\boldsymbol{F}=F^{\mu \nu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{e}_{\nu}$$
is a tensor field and thus invariant under Lorentz transformations. There's no way to infer an absolute coordinate frame, i.e., there's no aether in special relativity. That's how special relativity has been discovered by Lorentz, Fitzgerald, Poincare, and finally Einstein.

 

[Maciej Orman]

Indeed the electromagnetic field,
$$\boldsymbol{F}=F^{\mu \nu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{e}_{\nu}$$
is a tensor field and thus invariant under Lorentz transformations. There's no way to infer an absolute coordinate frame, i.e., there's no aether in special relativity. That's how special relativity has been discovered by Lorentz, Fitzgerald, Poincare, and finally Einstein.

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...

 

[Ibix]

OK, we have the model (still parameter descriptions missing) and now all we need is the experimental evidence... Or is there any at all?

A straight wire carrying a constant current.

 

[Maciej Orman]

A straight wire carrying a constant current.

There is not a single electron moving in constant linear velocity in a wire carrying constant current...

 

[vanhees71]

Now let's finally do the calculation, which hasn't occurred in this thread. The most simple way is to just do a Lorentz boost of the four-potential.

So let $x^{\mu}$ inertial coordinates, where the charge is at rest sitting in the origin of the spatial frame. Then, in (1+3) notation we have simply a Coulomb field, which is described in Lorenz gauge by the potentials
$$A^0(x)=\frac{q}{4 \pi |\vec{x}|}, \quad \vec{A}(x)=0. \qquad (1)$$
The four-potential in Lorenz gauge transforms as a four-vector, i.e.,
$$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\rho} A^{\rho}(x).$$
To make our life easier, instead of doing the combersome transformation we rather write everything in a covariant way. The situation is described in a covariant way by the charge $q$ of the particle (which is a scalar) and the particles' constant velocity, which is described covariantly by the four-velocity $(u^{\mu})=(\gamma,\gamma \vec{v})$, where $\gamma=1/\sqrt{1-\vec{v}^2}$ (I set $c=1$ for convenience).

Thus, the building blocks for our field, given in the particles's restframe, where $(u^{\mu})=(1,0,0,0)$, are only $x^{\mu}$ and $u^{\mu}$. It's clear that $A^{\mu} \propto u^{\mu}$ and finally we need $\vec{x}^2$. It's easy to project out the time component from the four vector $x^{\mu}$:
$$(x^{\mu}-u^{\mu} u \cdot x)=(0,\vec{x})$$
and thus
$$-\vec{x}^2=[x-u(u \cdot x)]^2=x^2-(u \cdot x)^2.$$
So finally we have
$$A^{\mu}=\frac{q}{\sqrt{(u \cdot x)^2-x^2}} u^{\mu}.$$
In the reference frame where the particle moves with three-velocity $v$ we have
$$u^{\prime \mu}=\gamma (1,\vec{v})$$
and
$$(u \cdot x)^2-x^2=\gamma^2 (t'-\vec{v} \cdot \vec{x})^2-t^{\prime 2}+\vec{x}^{\prime 2}.$$
With this you get the field components in the primed frame by taking the usual differential operators,
$$\vec{E}'=-\dot{\vec{A}}'-\vec{\nabla} A^{\prime 0}, \quad \vec{B}'=\vec{\nabla}' \times \vec{A}'.$$

 

[vanhees71]

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...

Hm, that's a good question. I'm not aware of an experiment which has ever measured the electromagnetic field of a single charged body moving with constant velocity. Of course the measurable macroscopic fields used in everyday electrical engineering all follow Maxwell's equations, and thus we can be pretty sure that everything works also for this special case.

 

[Maciej Orman]

Hm, that's a good question. I'm not aware of an experiment which has ever measured the electromagnetic field of a single charged body moving with constant velocity. Of course the measurable macroscopic fields used in everyday electrical engineering all follow Maxwell's equations, and thus we can be pretty sure that everything works also for this special case.

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...

 

[Maciej Orman]

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...

Rowland disk experiment proves that accelerated static charge on a disk creates magnetic field...

 

[vanhees71]

The Röntgen-Eichenwald experiment proves that also a moving electric polarization leads to a magnetic field either...

 

[Maciej Orman]

Now let's finally do the calculation, which hasn't occurred in this thread. The most simple way is to just do a Lorentz boost of the four-potential.

So let $x^{\mu}$ inertial coordinates, where the charge is at rest sitting in the origin of the spatial frame. Then, in (1+3) notation we have simply a Coulomb field, which is described in Lorenz gauge by the potentials
$$A^0(x)=\frac{q}{4 \pi |\vec{x}|}, \quad \vec{A}(x)=0. \qquad (1)$$
The four-potential in Lorenz gauge transforms as a four-vector, i.e.,
$$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\rho} A^{\rho}(x).$$
To make our life easier, instead of doing the combersome transformation we rather write everything in a covariant way. The situation is described in a covariant way by the charge $q$ of the particle (which is a scalar) and the particles' constant velocity, which is described covariantly by the four-velocity $(u^{\mu})=(\gamma,\gamma \vec{v})$, where $\gamma=1/\sqrt{1-\vec{v}^2}$ (I set $c=1$ for convenience).

Thus, the building blocks for our field, given in the particles's restframe, where $(u^{\mu})=(1,0,0,0)$, are only $x^{\mu}$ and $u^{\mu}$. It's clear that $A^{\mu} \propto u^{\mu}$ and finally we need $\vec{x}^2$. It's easy to project out the time component from the four vector $x^{\mu}$:
$$(x^{\mu}-u^{\mu} u \cdot x)=(0,\vec{x})$$
and thus
$$-\vec{x}^2=[x-u(u \cdot x)]^2=x^2-(u \cdot x)^2.$$
So finally we have
$$A^{\mu}=\frac{q}{\sqrt{(u \cdot x)^2-x^2}} u^{\mu}.$$
In the reference frame where the particle moves with three-velocity $v$ we have
$$u^{\prime \mu}=\gamma (1,\vec{v})$$
and
$$(u \cdot x)^2-x^2=\gamma^2 (t'-\vec{v} \cdot \vec{x})^2-t^{\prime 2}+\vec{x}^{\prime 2}.$$
With this you get the field components in the primed frame by taking the usual differential operators,
$$\vec{E}'=-\dot{\vec{A}}'-\vec{\nabla} A^{\prime 0}, \quad \vec{B}'=\vec{\nabla}' \times \vec{A}'.$$

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]

The Röntgen-Eichenwald experiment proves that also a moving electric polarization leads to a magnetic field either...

Yes, but this is not an excess electric charge moving in constant linear velocity...

 

[Maciej Orman]

The Röntgen-Eichenwald experiment proves that also a moving electric polarization leads to a magnetic field either...

It was Rowland who first shown this effect to Hertz and all that is charge accelerated in circular motion produces magnetic field...

 

[Ibix]

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...

Have an infinite flat sheet of charge lying in the x-z plane. The electric field is purely in the y-direction and is given by $E_y=\sigma/2\epsilon_0$ where $\sigma$ is the charge density. That means $F^{ij}=0$ for all i and j except $F^{02}=-F^{20}=\sigma/2c\epsilon_0$. Lorentz transforming to a frame where the sheet is moving at speed $v$ gives a magnetic field in the z direction of $B_z=\sigma v/2c^2\epsilon_0$. (Beware sign errors in that.)

I can't find an authoritative source for the charge density on the belt of a van de Graaff generator. http://www.chegg.com/homework-help/questions-and-answers/belt-van-de-graaff-generator-carries-surface-charge-density-7-00-mc-m-2-belt-0500-m-wide-m-q7847099 citing 7mC/m², and this one (also a homework question) gives 65μC/m². Taking the higher figure and its cited belt speed of 21m/s gives a field of around 9x10^-8T - about 0.2% the strength of the Earth's magnetic field. I'm sure it's possible to detect that, but it's not going to be done casually.

 

[Maciej Orman]

Have an infinite flat sheet of charge lying in the x-z plane. The electric field is purely in the y-direction and is given by $E_y=\sigma/2\epsilon_0$ where $\sigma$ is the charge density. That means $F^{ij}=0$ for all i and j except $F^{02}=-F^{20}=\sigma/2c\epsilon_0$. Lorentz transforming to a frame where the sheet is moving at speed $v$ gives a magnetic field in the z direction of $B_z=\sigma v/2c^2\epsilon_0$. (Beware sign errors in that.)

I can't find an authoritative source for the charge density on the belt of a van de Graaff generator. http://www.chegg.com/homework-help/questions-and-answers/belt-van-de-graaff-generator-carries-surface-charge-density-7-00-mc-m-2-belt-0500-m-wide-m-q7847099 citing 7mC/m², and this one (also a homework question) gives 65μC/m². Taking the higher figure and its cited belt speed of 21m/s gives a field of around 9x10^-8T - about 0.2% the strength of the Earth's magnetic field. I'm sure it's possible to detect that, but it's not going to be done casually.

Using the above equation for charge of 1C and speed of 1m/s I calculated field of 5.55556 E-18 so for non relativistic speeds large charge o 1C has no detectable magnetic field... Thank you it is the answer I have been looking for...

 

[Ibix]

Using the above equation for charge of 1C and speed of 1m/s I calculated field of 5.55556 E-18 so for non relativistic speeds large charge o 1C has no detectable magnetic field... Thank you it is the answer I have been looking for...

That formula isn't applicable to point charges. Both E and B fields vary with position in that case. But for a point charge the B field at any point will be weaker by a factor of at least $\gamma v/c^2$ than the E field measured at the same point, so the same comments apply. It's only when you get a large number of electrons in motion (like in a wire) that the magnetic field is detectable.

 

[Maciej Orman]

That formula isn't applicable to point charges. Both E and B fields vary with position in that case. But for a point charge the B field at any point will be weaker by a factor of at least $\gamma v/c^2$ than the E field measured at the same point, so the same comments apply. It's only when you get a large number of electrons in motion (like in a wire) that the magnetic field is detectable.

But electrons in wire never move with constant linear velocity... What's constant is the average velocity...

 

[Ibix]

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.

 

[Maciej Orman]

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...

 

[Ibix]

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.

 

[Dadface]

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.

 

[Maciej Orman]

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...

 

[Dadface]

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?

 

[vanhees71]

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
$$\vec{B}=\vec{\nabla} \times \vec{A} = -\gamma \vec{v} \times \vec{\nabla} \phi(\rho)=\vec{v} \times \vec{E}=\frac{\gamma q v}{4 \pi \rho^{3/2}} (0,-x^3,x^2).$$

 

[Maciej Orman]

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...

 

[Maciej Orman]

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...

 

[Maciej Orman]

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...

 

[Maciej Orman]

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...

 

[Dadface]

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.

 

[vanhees71]

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.

 

[Maciej Orman]

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

 

[Ibix]

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.

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.

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.

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.

 

[Dadface]

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...

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...

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)

 

[Ibix]

(Please check my calculation)

It's comparable to my answer treating it as a flat sheet of charge in motion.

 

[Maciej Orman]

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...

 

[Maciej Orman]

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

 

[Dadface]

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

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?