Moving Electric Charge Creating a Magnetic Field


by Eldgar
Tags: electric charge, magnetic field
Eldgar
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#1
Jan7-10, 12:19 PM
P: 11
Does a moving charge create a magnetic field?

At first the answer was obvious to me, since I = Q/t then if a charge is moved it is simillar to an electric current, and electric currents create magnetic fields.

However in a conductor, an electric current consists of electrons moving past relatively stationary protons.
So I know there is a physical difference between moving a charge and an electric current.
I am just wondering what the differences are,
my guess is that the faster a charged particle moves the electric field get weaker and the magnetic field gets stronger. But i haven't been able to find anything helpful for my understanding.

If anyone knows of any explantions or concepts to help understand what happens it would be appreciated.
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espen180
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#2
Jan7-10, 12:49 PM
P: 836
Quote Quote by Eldgar View Post
Does a moving charge create a magnetic field?
Yes.

Quote Quote by Eldgar View Post
At first the answer was obvious to me, since I = Q/t then if a charge is moved it is simillar to an electric current, and electric currents create magnetic fields.
It's not only similar to an electric current, moving charge is the definition of electric current. So they are the same, only in a conductor, you typically have a whole bunch of electrons moving together, so their magnetic fields add together.
Bob S
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#3
Jan7-10, 01:08 PM
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See my post #25 in
http://www.physicsforums.com/showthr...c+force&page=2

and this attachment
http://www.physicsforums.com/attachm...6&d=1259869448

showing that the Coulomb field gets weaker, and the magnetic field gets stronger, as the velocity approaches c, where the two opposing forces cancel.
Bob S

Firefox123
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#4
Jan7-10, 01:52 PM
P: 189

Moving Electric Charge Creating a Magnetic Field


Quote Quote by Bob S View Post
See my post #25 in
http://www.physicsforums.com/showthr...c+force&page=2

and this attachment
http://www.physicsforums.com/attachm...6&d=1259869448

showing that the Coulomb field gets weaker, and the magnetic field gets stronger, as the velocity approaches c, where the two opposing forces cancel.
Bob S
I just looked over that paper you wrote....I need to sit down and go through it, but from a first read I think your approach was excellent.

I enjoyed your calculation....very straightforward and well written.
Born2bwire
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#5
Jan8-10, 01:37 AM
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You can look up the Liénard-Wiechert potentials in most textbooks, like Jackson. They will give you the electric and magnetic fields for a charge of arbitrary trajectory. But you will find that you do not get an increase in the magnetic field versus the electric field. Once you have a moving charge, the radiation is electromagnetic waves. The electric and magnetic fields maintain a constant relationship between them. So if the electric field decreases, the magnetic field decreases by the same amount. This can be seen explicitly with either the potential or the field equations.
Bob S
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#6
Jan8-10, 10:46 AM
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In charged particle beams, the charge density of the beam produces a radial electric field that produces a repulsive force on individual particles. See Eqn (4) in Karlheinz Schindl's paper
http://cas.web.cern.ch/CAS/Loutraki-...ndl/paper1.pdf
Simultaneously, the current of the moving beam produces a magnetic field that produces an attractive force on individual particles. See eqn (8). These two opposing forces are combined in Eqn (9) to (11), cancelling each other as β → 1. The cancellation is proportional to 1/γ2.

Space charge forces, both Coulomb and magnetic, are well known in charged particle beams, and cause emittance blow-up in low-β, high-current beams. The two forces cancel for very relativistic beams.
Bob S
Eldgar
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#7
Jan8-10, 09:03 PM
P: 11
Ok, i think i have a better understanding of the concepts, though i dont really understand
all of those formulas.
If i were to keep it simple and just consider two moving point charges with the same charge
and velocity.
would the forces be?

Electric Force
Fe=k(q1 x q2)/d2

Magnetic Force
Fm = (k(q1 x q2)/d2)(v2/c2)

Net Force
FN = (k(q1 x q2)/d2)(1-v2/c2)
Born2bwire
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#8
Jan8-10, 11:17 PM
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Quote Quote by Eldgar View Post
Ok, i think i have a better understanding of the concepts, though i dont really understand
all of those formulas.
If i were to keep it simple and just consider two moving point charges with the same charge
and velocity.
would the forces be?

Electric Force
Fe=k(q1 x q2)/d2

Magnetic Force
Fm = (k(q1 x q2)/d2)(v2/c2)

Net Force
FN = (k(q1 x q2)/d2)(1-v2/c2)
The easiest thing would be to just choose the reference frame that moves with the charges. In that case, the force would be just the usual Coulomb force. But for the lab frame, the magnetic force that you have is incorrect. If we were to take the lab frame, then we would need to calculate the magnetic field of each charge and apply its resulting Lorentz force on the other charge. If the velocities are non-relativistic you could just use the Biot-Savart law otherwise you use the appropriate relativistic forms, like Jefimenko's equations. You would also need to take into account the vector cross products of the field and velocity vectors since force is a vector.
Eldgar
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#9
Jan9-10, 02:24 AM
P: 11
the problem i face with Biot-Savart law and briefly looking at Jefimenko's equations from wikipedia.
http://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law

is finding the current or current density of a small charge such as an electron, or a point charge
based on its velocity.
The way i see it if dealing with a point charge that has any velocity will have infinite current.

I was thinking I = (Q x v)/d where d is the diameter, and V is the Velocity, but this didn't seem to make sense when i looked at other equations.
Born2bwire
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#10
Jan9-10, 02:44 AM
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Oh yeah my bad. I just realized I wasn't seeing the velocity dependence in the equations from Jackson, he takes his cross product of the electric fields using the directional vector for the retarded position, which has a term of v/c tucked away in there when you convert the vector from the retarded position to the instantaneous position. So the magnetic fields, for a constant moving charge, are proportional to the electric fields by the velocity of the charge. It would get a bit messy in general for a charge of arbitrary trajectory since the cross product is taken using the retarded position (which for an accelerating charge would be not be directly dependent on the velocity since the retarded position would follow an arbitrary path in time).

For Biot Savart, it's fairly easy, you just treat the point charge as an infinitesimal current element. But actually I found what I wanted to reference in the first place, tucked away further down the page are the equations for a particle of constant velocity. There are added terms that arise if the particle is accelerating, which is why I usually reference the Lienerd-Wiechert potentials (though I haven't found an expression for the fields from them for a point charge on wikipedia though you can find them in Griffiths, Jackson, etc.). Jefimenko's equations aren't terribly useful for this problem, it would be annoying to work it out from them, I just forgot that the field equations I wanted to reference were buried on the Biot-Savart page. So just look further down and they have the field equations derived from Maxwell's equations directly (not from Biot-Savart though).

Also, since the current element that the charge represents in space is localized, you would also want to have a dirac delta in your current statement. This is how they lose the integral when they use the Biot-Savart law to find the magnetic field from a charge for the non-relativistic expresion on the Wiki page.
Bob S
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#11
Jan9-10, 12:17 PM
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The easiest way to calculate the vector magnetic and electric field components of a relativistic constant-velocity point charge might be just writing down the individual vector components of the isotropic electric field at a distance x=r·sin(θ) from a charge at rest, and Lorentz-transforming them to the electric and magnetic components in the moving frame (including Lorentz-contraction z' = z/γ) using the four equations at the bottom of:
http://pdg.lbl.gov/2009/reviews/rpp2...-relations.pdf
This would avoid the problem of the point-charge singularity in the Biot-Savart Law.
Bob S
issabella10
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#12
Jan15-10, 09:16 PM
P: 1
Actually, I have already learned that electrical building like a Mig Welding when I was in high school. But now that I am in college, I already forgot about it so I study again about it and found out that when an electrical charge is moving or an electric current passes through a wire, a circular magnetic field is created.


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