Magnetic field from Van De Graff

[arydberg]

Is there a magnetic field generated from the charges on a moving Van De Graff belt?

If yes how come they do not move at relativistic speeds.

If no doesn't this represent a current and doesn't a current produce a magnetic field.

 

[bcrowell]

Is there a magnetic field generated from the charges on a moving Van De Graff belt?

Yes. Any moving charge generates a magnetic field.

If yes how come they do not move at relativistic speeds.

Why would you expect that they would?

 

[HallsofIvy]

Is there a magnetic field generated from the charges on a moving Van De Graff belt?

As BCrowell said, yes, of course.

If yes how come they do not move at relativistic speeds.

A very strange question. You just said the charges are on the Van De Graff belt and it is not moving at relativistic speeds.

[If no doesn't this represent a current and doesn't a current produce a magnetic field.

 

[arydberg]

I assumed relativistic velocities are necessary because just as space and time take on different values depending on the velocity of the viewing frame so do electric and magnetic fields change values depending on being viewed from a moving frame.

 

[pervect]

If you look at the expression for magnetic force, F = q v x I, you can rewrite it as

I = (rho)*A*v', where rho is the density of charge per meter's in the wire and v' is the velocity of the charges in the wire. Typically v' is very small, by the way. So you get something like

F is proportional to q rho A v v', where v is the velocity of your test charge and v' is the velocity of the charge in the current.

If v <<c and v' <<c, one might expect this expression to be small, but it turns out to be easily measurable in practice, in part because the electromagnetic field is so strong.

 

[arydberg]

I am assuming that the very existence of a magnetic field is a relativistic effect. This idea comes from Lectures in Physics by Feynman ( section 13-6 and 13 -7)
If this is so then is there a difference between the field formed by two Van De Graff belts one carrying half the charge of the second but running at twice the speed. Of course the equation says there is no difference as the moved charge per second is the same but it is hard to imagine relativistic effects arising from such slow velocities.

 

[PAllen]

I am assuming that the very existence of a magnetic field is a relativistic effect. This idea comes from Lectures in Physics by Feynman ( section 13-6 and 13 -7)

It is, but that doesn't mean you need relativistic velocities. All ordinary magnetism comes from charge flow rates extremely small compared to c. The strength is due to the amount of charge, and the strength of E/M interaction. You don't feel how incomprehensibly strong the coulomb field would be for the amount of charge in ordinary matter because of neutrality.

For example, if magically 1 kg of pure protons in a 1 foot ball materialized near you, I think you would essentially instantly be turned into plasma from the combination of electrons stripped from you, and rapidly expanding protons tearing through you.

 

[arydberg]

OK

Thanks for the help.

 

[arydberg]

OK so if i have a Wimshurst machine with charged spinning disks as shown here

http://www.scientificsonline.com/wi...nerator.html?gclid=CN3NiKqy26kCFSN5gwodJH-naw

And if we assume each metallic element is 10 pF and the voltage is 10 KV then the charge per plate is Q = C X V or 10 X exp -12 x 10 exp 3 = 100 exp -9 coulombs

assuming 500 rpm or 3.14159 feet x 500 = 1570 feet per minute or 26 feet per second or 8 meters per second

now the current = q x velocity = 1 x exp -9 coulombs x 8 meters per second = 8 * exp -7 amps or about 1 micro amp.

So can we detect this magnetic field?

 

[bcrowell]

So can we detect this magnetic field?

I suggest you try estimating it numerically before you try to plan how to measure it. The method of detection is going to depend on the order of magnitude of the effect you're trying to measure. Find the magnetic field of a straight wire carrying a micro-amp of current, at the relevant distance. I suspect it will be much too small to measure by any practical method that you have available.