Rotating a spherical conductor in an Electric field

[tim9000]

Hi,

I recently came across the familiar image of a metal sphere in an electric field:

https://i.stack.imgur.com/x58Ia.jpg

I noted how the free-charges on the surface of the sphere align with the electric field lines as opposite charges are attracted.

Then I wondered, 'what if the sphere was spun with a radial velocity around an axis going into the page?'
This would be a moving charge as seen from the rotating conductors reference perspective. As the charges on the surface always wanted to align with the respective electric field. But does this count as an electric current? And thus have a magnetic field induced around the conductor? Because it seems like a paradox that the electric charge on the surface of the conductor is stationary with respect to an outside point, so how could it create a magnetic field?

I tried consulting:
http://www.maxwells-equations.com/maxwells-equations.gif

However, I wasn't able to comprehend how this could guide me in forming a conclusion.

So, would this rotating sphere create a magnetic field going into and out of the page?

Cheers

 

[mfb]

It is easier to study this in an inertial (non-rotating) reference frame, where you have a current (via positive and negative charges rotating differently). It induces a magnetic field. The contribution that cancels the external magnetic field is not rotating, but the remaining object is positively charged and rotating.

Edit: This is wrong, see later posts.

 

[tim9000]

It is easier to study this in an inertial (non-rotating) reference frame, where you have a current (via positive and negative charges rotating differently). It induces a magnetic field. The contribution that cancels the external magnetic field is not rotating, but the remaining object is positively charged and rotating.

Hey, thanks for the reply mfb.
Unfortunately I'm a complete simpleton, and try as I might, I don't really understand.
Okay, so inertial reference frame: we're in the lab watching the ball rotate as described. However, does this not mean that there is no electric current? Because it is only the ball that is spinning, not the external electric field around it (I hope I made that clear enough previously).
I don't understand what you mean by contribution which cancels the external magnetic field or how the object could have any [net] charge. Sorry.

Note: I would expect that if were in the non-inertial reference frame of the ball that we would see a current.

Thanks!

 

[vanhees71]

This sounds like a very interesting but tough problem. Indeed you induce a charge distribution ##\rho##, and through the rotation you'll get a current distribution ##\vec{j}=\rho \vec{v}## (it's called a convection current) and thus also a magnetic field.

 

[tim9000]

This sounds like a very interesting but tough problem. Indeed you induce a charge distribution ##\rho##, and through the rotation you'll get a current distribution ##\vec{j}=\rho \vec{v}## (it's called a convection current) and thus also a magnetic field.

I guess it's kind of like the Electric version of a Faraday generator (where there is a stationary magnetic field and a rotating disc).

So from what you said I'd assume that it doesn't matter which reference frame the charge is moving in, a magnetic field will be generated and seen in all reference frames. However, it's this vector that I'm still skeptical in: ##\vec{j}=\rho \vec{v}## because from the inertial reference frame the charge distribution doesn't have a velocity (as far as I can imagine).

Thanks!

 

[mfb]

Ah wait, I forgot the symmetry of the setup. You have negative charge accumulating at one side, but positive charge accumulating at the other side.
Ignore my previous post. I wouldn't expect a magnetic field unless the sphere has an overall net charge.

@vanhees71: In our inertial reference frame the charge distribution should be static.

 

[tim9000]

Ah wait, I forgot the symmetry of the setup. You have negative charge accumulating at one side, but positive charge accumulating at the other side.
Ignore my previous post. I wouldn't expect a magnetic field unless the sphere has an overall net charge.

@vanhees71: In our inertial reference frame the charge distribution should be static.

Exactly what I thought.
Yeah, so no magnetic field.

Thanks again!

 

[tim9000]

A friend of mine said that this video was sophistry:



So I suppose this thought experiment we've been discussing is proof of that?

 

[Vanadium 50]

Ignore my previous post

Whew!

If the resistance is truly 0, you can view this as spinning the sphere, but leaving the free charges where they are. If the resistance is small and non-zero, the charges are dragged with the metal, but since there is still no net charge, there shouldn't be a net current, so no net field.

 

[Gigaz]

You don't need a net charge to have a current. A wire with some current inside is not charged but still produces a magnetic field.

There is a magnetic field, but the situation is actually the other way round from what we have in such a wire. In a wire, the electrons follow the potential gradient. In the rotating sphere, the atomic nuclei are bound by the crystal lattice and rotate, whereas the conduction electrons are fixed by the electric field.

 

[Vanadium 50]

but since there is still no net charge

Replace with "net motion of charge".

 

[tim9000]

You don't need a net charge to have a current. A wire with some current inside is not charged but still produces a magnetic field.

There is a magnetic field, but the situation is actually the other way round from what we have in such a wire. In a wire, the electrons follow the potential gradient. In the rotating sphere, the atomic nuclei are bound by the crystal lattice and rotate, whereas the conduction electrons are fixed by the electric field.

Replace with "net motion of charge".

I agree and think we're all totally on the same page. So are these results in-line with the explanation of magnetism in the video? And what are your thoughts on the validity of this explanation?

Thanks!

 

[vanhees71]

I guess it's kind of like the Electric version of a Faraday generator (where there is a stationary magnetic field and a rotating disc).

So from what you said I'd assume that it doesn't matter which reference frame the charge is moving in, a magnetic field will be generated and seen in all reference frames. However, it's this vector that I'm still skeptical in: ##\vec{j}=\rho \vec{v}## because from the inertial reference frame the charge distribution doesn't have a velocity (as far as I can imagine).

Thanks!

It has! To a very good approximation you can take ##\vec{v}=\vec{\omega} \times \vec{r}## since the conduction currents are negligible here.

I'd also not try to calculate this in the rotating body-rest frame by the way.

 

[Vanadium 50]

I don't comment on videos. I think that it is the responsibility of the person citing the video to extract the central argument and to clearly write it down - not to make the people who are trying to help spend their time trying to fish it out. To be blunt, if it's not worth the poster's time to do this, how much less valuable is mine?

In your picture, the charge distribution is the same whether you rotate the sphere or not. That's the key.

 

[shiv222]

Ah wait, I forgot the symmetry of the setup. You have negative charge accumulating at one side, but positive charge accumulating at the other side.
Ignore my previous post. I wouldn't expect a magnetic field unless the sphere has an overall net charge.

@vanhees71: In our inertial reference frame the charge distribution should be static.

I think as charges move there will be current in one direction due to positive charges on one edge of sphere and due to negative charges on other end. While integral of current around the sphere will be zero, there will be local fields around each current.

 

[tim9000]

It has! To a very good approximation you can take ##\vec{v}=\vec{\omega} \times \vec{r}## since the conduction currents are negligible here.

I'd also not try to calculate this in the rotating body-rest frame by the way.

I'm confused now, because isn't the whole point that the electrons DON'T have a radial velocity ω? Because they're hanging on the surface permanently facing the positive side or (higher potential) of the external electric field? And thus not moving around the centre of the sphere in space or in time from an outside perspective.

I don't comment on videos. I think that it is the responsibility of the person citing the video to extract the central argument and to clearly write it down - not to make the people who are trying to help spend their time trying to fish it out. To be blunt, if it's not worth the poster's time to do this, how much less valuable is mine?

In your picture, the charge distribution is the same whether you rotate the sphere or not. That's the key.

That's fair-enough, I shouldn't have assumed you'd be interested enough to sit and watch it; I appreciate your frankness.
The thesis of the video is that a positively charged cat is moving with the same velocity parallel to an electric current, so via special relativity the cat sees a negative electric repulsive force (because from the cat's perspective the moving positively charged nuclei of the wire is moving and has a greater spatial density, and the stationary electrons have a more stretched-out distribution). Which the video asserts explains the perceived 'magnetic' attraction or repulsion.
I see this as being analogous to the thought-experiment of this thread, and thus if it were true then a "magnetic field" would be created from the positive atoms of our rotating sphere. Given that we've previously concluded that it is not the case, I am lead to believe that the video is incorrectly asserting what a 'magnetic' field is.

When you say "In your picture, the charge distribution is the same whether you rotate the sphere or not" I want to clarify, I thought that picture was indicating that the electrons are hanging on the surface permanently facing the positive side or (higher potential) of the external electric field and the positive valance shell holes are on the side of the sphere facing the lower potential of the external electric field? So there is a charge distribution across the surface of the sphere, but it's not moving in space or time.

I think as charges move there will be current in one direction due to positive charges on one edge of sphere and due to negative charges on other end. While integral of current around the sphere will be zero, there will be local fields around each current.

I'm having trouble visualising what you mean, because from the perspective of the electric field I don't think the charges are moving. On what axis is the negative charge flowing locally?Thanks all

 

[vanhees71]

I'm confused now, because isn't the whole point that the electrons DON'T have a radial velocity ω? Because they're hanging on the surface permanently facing the positive side or (higher potential) of the external electric field? And thus not moving around the centre of the sphere in space or in time from an outside perspective.

Why? I forgot the exact reference, but there is a very nice American Journal of Physics article on the homopolar generator with an experimental verification that the electrons in a conductor are indeed moving with the body as a whole. I think one should be able to treat this problem as a static one. Since after some transient state, i.e., after a sufficiently long time you should simply have a static charge-current distribution (as long as ##\vec{\omega}## is time independent of course). If I find the time at this weekend, I'll give it a try. It looks like a very interesting static problem (maybe something to puzzle about for our students in the E&M lecture in the upcoming fall semester ;-))).

 

[Vanadium 50]

So there is a charge distribution across the surface of the sphere, but it's not moving in space or time.

That's right. That's why there is no magnetic field. (Again, this relies on R=0, a perfect conductor; a physical sphere is different)

 

[tim9000]

Why? I forgot the exact reference, but there is a very nice American Journal of Physics article on the homopolar generator with an experimental verification that the electrons in a conductor are indeed moving with the body as a whole. I think one should be able to treat this problem as a static one. Since after some transient state, i.e., after a sufficiently long time you should simply have a static charge-current distribution (as long as ##\vec{\omega}## is time independent of course). If I find the time at this weekend, I'll give it a try. It looks like a very interesting static problem (maybe something to puzzle about for our students in the E&M lecture in the upcoming fall semester ;-))).

That's right. That's why there is no magnetic field. (Again, this relies on R=0, a perfect conductor; a physical sphere is different)

Hi Vanhees and Vanadium,

That is something that I didn't consider, due to Resistance the electrons would have a slow drift velocity, possibly smaller than that of an applied radial velocity of the sphere, and so would get dragged along with the conductor (sphere) as they tried to move along the electric field lines. This would decrease the maximum charge density. However, I doubt this could look like an electric current externally, so as you say, it's still a static charge distribution.Regarding the lack of magnetic field of the sphere (of which I concur). If you wouldn't mind, I do feel that this outcome flys in the face of the explanation of magnetism via Special Relativity, the hypothesis of which I critiqued (from the aforementioned video):

That's fair-enough, I shouldn't have assumed you'd be interested enough to sit and watch it; I appreciate your frankness.
The thesis of the video is that a positively charged cat is moving with the same velocity parallel to an electric current, so via special relativity the cat sees a negative electric repulsive force (because from the cat's perspective the moving positively charged nuclei of the wire is moving and has a greater spatial density, and the stationary electrons have a more stretched-out distribution). Which the video asserts explains the perceived 'magnetic' attraction or repulsion.
I see this as being analogous to the thought-experiment of this thread, and thus if it were true then a "magnetic field" would be created from the positive atoms of our rotating sphere. Given that we've previously concluded that it is not the case, I am lead to believe that the video is incorrectly asserting what a 'magnetic' field is.

I disagree with this video because by this logic: An external object that was negatively charged would see the spinning positive charges of the sphere as being of a greater density in space (due to special relativity length contraction) than the static or slower-moving electrons, and thus experience a magnetic attraction, which I find implausible to say the least. I'm not saying that this does not occur, but I do not accept that it is the explanation for magnetism, does anyone else agree/disagree?

Thanks very much!

 

[vanhees71]

That's right. That's why there is no magnetic field. (Again, this relies on R=0, a perfect conductor; a physical sphere is different)

Indeed, to do the calculation you need Ohm's Law in the form ##\vec{j}_{\text{cond}}=\sigma(\vec{E}+\vec{v} \times \vec{B}/c)=0## inside the rotating sphere to have stationary conditions. As I said before, this implies that you can with good approximation assume that ##\vec{j}=\vec{j}_{\text{convect}}=\rho \vec{\omega} \times \vec{r}##. I'll try to do the calculation later.

 

[Amin-1983]

This sounds like a very interesting but tough problem. Indeed you induce a charge distribution ##\rho##, and through the rotation you'll get a current distribution ##\vec{j}=\rho \vec{v}## (it's called a convection current) and thus also a magnetic field.

I think the formula you wrote is not applicable in this case. Because, the voltage throughout the conductive sphere is the same. So, the voltage difference is zero and we will not have any current.