Charge moving with a constant linear velocity....

[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?

 

[Maciej Orman]

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...

 

[vanhees71]

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.

 

[Maciej Orman]

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...

 

[A.T.]

...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?

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.

 

[Ibix]

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.

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.

 

[Maciej Orman]

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...

 

[vanhees71]

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!

 

[Ibix]

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.

 

[vanhees71]

...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.

 

[Maciej Orman]

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...

 

[Ibix]

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...

No you would not. For the reason I've given twice, and vanhees71 and A.T. have given at least once each.

 

[Maciej Orman]

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.

No, triboelectric charging method creates volume charge for which there is no method of measurement... Thus suggesting that the charge is much larger than estimated...

 

[Ibix]

No, triboelectric charging method creates volume charge for which there is no method of measurement... Thus suggesting that the charge is much larger than estimated...

Reference please.

 

[vanhees71]

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=MbSUqqzFDocC&pg=PA98&lpg=PA98&dq=rowland+disc+experiment&source=bl&ots=0OzuvkesWW&sig=G_oqn3m8UNNQmnz97pz9_tQXVyQ&hl=en&sa=X&ved=0ahUKEwiI3dfXo_LVAhWGC8AKHbvyCWsQ6AEIDTAB#v=onepage&q=rowland disc experiment&f=false).

The point of Rowland's experiment (and other similar experiments in the late 19th century) was to proof that electric currents are due to moving charges ("electrons"). Nowadays that sounds trivial, because according to our present understanding all electric currents (and magnetizations) are due to elementary particles, but at this time the very nature of electric charges and currents was not that clear.

If I understand the excerpt from the textbook right, you have a charged ring of radius $R$. The corresponding current density is
$$\vec{j}(\vec{x})=\rho \vec{v}=\frac{q}{2 \pi R} \delta(\varrho-R) \delta(z) \omega \varrho \vec{e}_{\varphi},$$
where $(\varrho,\varphi,z)$ are standard cylinder coordinates. The corresponding magnetic field can be calculated using the vector potential (in Coulomb gauge),
$$\vec{A}(\vec{x})=\frac{1}{c} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{\vec{j}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|},$$
and finally the magnetic field as
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
The integral can be found in some good textbooks (like Jackson), leading to elliptic functions.