# Electric field of a current loop

1. Sep 21, 2009

### mathperson

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

2. Sep 21, 2009

### Staff: Mentor

0 (provided the usual case of an uncharged wire)
Because 0 is boring, even for textbooks.

3. Sep 21, 2009

### DaTario

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

Best Regards

DaTario

4. Sep 22, 2009

### Staff: Mentor

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.

5. Sep 22, 2009

### 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

6. Sep 22, 2009

### 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"

7. Sep 23, 2009

### Staff: Mentor

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.

8. Oct 3, 2009

### 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

9. Oct 4, 2009

### 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.

10. Oct 4, 2009

### clem

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.

11. Oct 4, 2009

### Phrak

What relativistic corrections?

12. Oct 4, 2009

### clem

The E and B fields of a moving electric charge are the Lienard-Wiechert fields.

13. Oct 4, 2009

### Phrak

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

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.

Last edited by a moderator: May 4, 2017
14. Oct 5, 2009

### clem

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.

15. Oct 5, 2009

### Staff: Mentor

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.

16. Oct 5, 2009

### clem

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.

17. Oct 5, 2009

### 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?

18. Oct 5, 2009

### Staff: Mentor

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.

19. Oct 6, 2009

### 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?

20. Oct 6, 2009

### clem

Yes...................

21. Oct 6, 2009

### clem

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.

22. Oct 6, 2009

### Staff: Mentor

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.

23. Oct 6, 2009

### Staff: Mentor

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 any one point charge are eliminated by the principle of superposition by destructive interference from the contributions of all of the other charges.

24. Oct 6, 2009

### 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.