Electric field of a current loop

[mathperson]

What is the electric field outside a steady state current loop?
Why is this not discussed in textbooks?

 

[Dale]

What is the electric field outside a steady state current loop?

0 (provided the usual case of an uncharged wire)

Why is this not discussed in textbooks?

Because 0 is boring, even for textbooks.

 

[DaTario]

and what about inside? Does the battery provide an electric field with rotational character?

Best Regards

DaTario

 

[Dale]

and what about inside?

That depends on the properties of the wire itself. There will be no e-field inside a perfect conductor, there will be an e-field inside a resistor.

 

[cabraham]

J = sigma*E, or alternately E = rho*J.

If the conductor is perfect, sigma is infinite, rho is 0, so that E = 0.

Make sense to you?

Claude

 

[mathperson]

even though the negative charge is accalerating toward the center of the loop, and the velocity vector cross the acceleration vector is non-
zero?
p.s. as i am not a physics person, would you include the definitions of terms, so i can look them up?
p.p.s. i don't think the question i trivial, even if the answer is.
The result of the michaelson-morley experiment was essentially "0"

 

[Dale]

Magnetic fields produce uniform circular motion of free charges, not electric fields. However, don't forget that the drift velocity in a wire is on the order of mm/s or even um/s. And electrons have very little mass. So the required centripetal force is incredibly small.

 

[mathperson]

Let me try again. Consider a charged metal ring in the x y plane. It generates an electric field. Now consider the same charged ring rotating rapidly about the z axis. Is the new electric field different from the original one? Since this is a steady state situation, it would seem that Maxwell's time-varying eq

 

[Phrak]

You seem have gotten the treatment. By now you may have figured out that an uncharged ring having a constant current generates no electric field.

A charged ring has an electric field with equipotentials looping around the ring in some fashion. You might find a picture of this in Wikipedia.

A charged rotating ring will be the simple sum of the two. It has charge and current. Its field will be exactly the same as the charged ring. The motion of the charge will have no additional effect.

 

[Meir Achuz]

The motion of the charge will have no additional effect.

That is not quite true. The relativistic corrections to the E field of a rotating charged ring are of the same order of magnitude as the B field. A rotating charged ring is not the same as a ring with a current.

 

[Phrak]

That is not quite true. The relativistic corrections to the E field of a rotating charged ring are of the same order of magnitude as the B field. A rotating charged ring is not the same as a ring with a current.

What relativistic corrections?

 

[Meir Achuz]

The E and B fields of a moving electric charge are the Lienard-Wiechert fields.
See an advanced EM text.

 

[Phrak]

The E and B fields of a moving electric charge are the Lienard-Wiechert fields.
See an advanced EM text.

No need for that. A Wikipedia article has http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential" .

The Lienard-Wiechert potential addresses a charge in arbitrary motion incorportating a ratarded time into the 4-vector potential; not relevant to a steady state current density where retarded time will effectively cancel, leaving the 4-vector potential unchanged.

 

[Meir Achuz]

Yes, but a rotating charged ring is not the same as a steady state current.
A good example of a rotating charged ring is a circular accelerator which radiates.

 

[Dale]

The current in an accelerator is in bunches, it is not a uniform current. That is not the same as what you described in post 8.

 

[Meir Achuz]

Dale: You are thinking of a phasotron or something like the LHC when it runs. A betatron or cyclotron is a bit old fashioned, but has a relatively uniform spatial distribution. The flow of individual charges in any accelerator is not a current, but is a flow of individual charges. This would be equivalent, in its E field, to that in post #8, and would be different than for the ring at rest.

 

[Phrak]

We know that a a rotating charged ring is not the same as a steady state current. The first is charged, the second is not.

Let me see if I have this right. Are you claiming that the field solution of a ring that is charged and rotating in not the same as the superposition of a static ring of charge and a ring of current, classically treated?

 

[Dale]

clem, the long and the short of it is Maxwell's equations. I believe that you want to consider a steady-state setup where all of the d/dt terms are 0. That leaves only the curl of B equal to the current and the divergence of E equal to the charge. That is it.

 

[mathperson]

trying again: A charged particule going in a circle generates an electric field in part due to its acceleration. Yet the acceleration from a classical current loop does not contribute to the electric field? What hypotheses are used to decide this either way? Isn't a classical wire a superposition of an equal amount of + & - charge continuously distributed along a curve?

 

[Meir Achuz]

Are you claiming that the field solution of a ring that is charged and rotating in not the same as the superposition of a static ring of charge and a ring of current, classically treated?

Yes......

 

[Meir Achuz]

The motion of charged particles is in many textbooks. It does not matter if you have one particle or billions. One particle in motion or many are not the same as a current.
That is why you can't use the fields of a current for one or many moving moving particles.

 

[Dale]

The motion of charged particles is in many textbooks. It does not matter if you have one particle or billions. One particle in motion or many are not the same as a current.
That is why you can't use the fields of a current for one or many moving moving particles.

This is completely incorrect. A single moving charge is in fact a current. For a uniform ring of current the magnitude of the current is given by J DiracDelta(x²+y²-R²,z). For a single moving point charge the magnitude of the current is given by qv DiracDelta(x-X,y-Y,z-Z). In the limit of a continuum of charges moving in a ring given by X²+Y²=R² and Z=0 it is clear that the above two expressions are the same. And the description for both are handled by Maxwell's equations.

 

[Dale]

trying again: A charged particule going in a circle generates an electric field in part due to its acceleration. Yet the acceleration from a classical current loop does not contribute to the electric field? What hypotheses are used to decide this either way? Isn't a classical wire a superposition of an equal amount of + & - charge continuously distributed along a curve?

Note the key word "superposition" that you used above. Don't forget that you can get both constructive and destructive interference in superposition. These radiation components from anyone point charge are eliminated by the principle of superposition by destructive interference from the contributions of all of the other charges.

Radiation requires a changing multipole moment. This system has exactly one non-zero moment, the magnetic dipole moment, and it's static.

The argument that a single accelerating charge radiates neglects a very important fact: this is not a single accelerating charge. Suppose I had a rotating plate, and I place a charge at 0 degrees. This plate radiates. Now, I put another charge at 180, and now I no longer have any electric dipole radiation. I do still have quadrupole radiation, so I place charges at 90 and 270. Now the highest surviving moment is electric octopole, which can be zeroed out with charges every 45 degrees. In the limit of a continuous ring of charge, there is no radiation: radiation from any point on the ring is canceled by radiation from other points on the ring.

This has nothing to do with superconductivity. It's true if you simply glued charges to a ring. It's all in Chapter 9 of Jackson.

Now, you might ask, "yes, but we know charges aren't continuous: they're discrete - doesn't that wreck your argument?" Yes, but they are really, really close to [continuous]: you have millions of electrons, so it all cancels except for the million-pole terms, and those radiate very, very little power. Possibly the age of the universe per photon.

 

[mathperson]

trying again, again: In Griffiths's Intro. to Electrodynamics 2nd ed. page 424 he refers to the acceleration field of a (classical) charge. If a charge going in a circle generates such a field, then wouldn't an arc of charge do the same? Why should the acceleration field disappear because the arc is extended full circle?
Also on page 489 he treats a wire a a superposition of + & - continuous line charges with no acceleration. I'm asking about a similar situation with centripital acceleration and constant angular speed.
Again, I ask, What are the hypotheses? Simply saying "Maxwell's equations" in a steady state situation seems too vague to me.
Thanks for your patience.

 

[Dale]

If a charge going in a circle generates such a field, then wouldn't an arc of charge do the same?

Certainly not, what would make you think that? It would be astounding indeed for a linear system to have the same output for a "point" input as for an arc of points. Like finding that 1+1=1 instead of 2.

Also on page 489 he treats a wire a a superposition of + & - continuous line charges with no acceleration.

Yes, this is the correct approach. The ring can be treated as a superposition of a continuous circle of point charges in uniform circular motion. In terms of the principle of superposition, do you understand the concept of destructive interference that I mentioned in my previous post and that Vanadium described implicitly the quote?

 

[mathperson]

Hi again! Griffiths page 417:"In Maxwell's electrodynamics, formulated as it is in terms of charge and current densities, a point charge must be regarded as the limit of an extended charge..."
Divide a circle of charge into small enough arcs so that each arc may be regarded as a point charge. If the circle is rotating, each point has a velocity cross acceleration vector pointing in the same direction. Apply eq(9.107) page 424 (electric field of a point charge equation) to each point charge and take the vector sum.
Although I don't know enough to follow the superposition argument, my guess is that it shows that there is no time varying electric field.
Consider a ring of - charge superimposed on a ring of + charge of equal value. If both rings are stationary, there is no net field anywhere. If the - ring rotates and the + one doesn't, the above argument shows there is an electrostatic field.

 

[Dale]

Divide a circle of charge into small enough arcs so that each arc may be regarded as a point charge. If the circle is rotating, each point has a velocity cross acceleration vector pointing in the same direction. Apply eq(9.107) page 424 (electric field of a point charge equation) to each point charge and take the vector sum.
Although I don't know enough to follow the superposition argument, my guess is that it shows that there is no time varying electric field.

Yes.

Consider a ring of - charge superimposed on a ring of + charge of equal value. If both rings are stationary, there is no net field anywhere. If the - ring rotates and the + one doesn't, the above argument shows there is an electrostatic field.

How so?

 

[mathperson]

In the equation there is an (r cross (v cross a)) term (script r in the text) where:
r=vector from a point charge q to a test charge Q,
v=velocity vector of q,
a=acceleration vector of q.
For each q on the rotating ring, v cross a is a vector parallel to the z axis. Put Q on the x-axis far enough from the ring so that each r points in approximately the same direction. The triple product will have a component parallel to the y axis. Add.