E-field from Time Varying B-field of Current Loop

Click For Summary

Discussion Overview

The discussion revolves around calculating the electric field (E-field) generated by a time-varying magnetic field (B-field) from a current loop. Participants explore various theoretical approaches, including the application of Faraday's law and Ampere's law, while considering the implications of harmonic variations in the fields.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a model for the B-field from a current loop and proposes that the E-field should also vary harmonically with a phase delay, leading to the equation derived from Faraday's law.
  • Another participant suggests using Ampere's law and Ohm's law to relate the magnetic field to the electric field, noting that in a superconducting loop, the E-field would cancel out.
  • A participant clarifies their interest in the E-field outside the current loop, emphasizing the need for fields resulting from the time-varying B-field.
  • One participant introduces the concept of using the magnetic potential to derive the E-field, referencing the relationship between the magnetic vector potential and the E-field.
  • Another participant provides a formula for the E-field in the presence of time-varying magnetic fields and charges, indicating the need for a scalar potential.
  • A later reply confirms that the scalar field must be zero everywhere, aligning with the reasoning about the absence of charges.

Areas of Agreement / Disagreement

Participants express differing views on the methods to calculate the E-field, with no consensus reached on the most effective approach. Some participants focus on the implications of harmonic fields, while others emphasize the conditions under which the E-field is derived.

Contextual Notes

Participants mention challenges related to the mathematical complexity of the derivatives involved and the potential for conflicting results from computational tools like Mathematica.

Who May Find This Useful

This discussion may be of interest to researchers and students working on electromagnetic theory, particularly those exploring the interactions between time-varying fields and their implications in various physical contexts.

luke
Messages
25
Reaction score
0
I am trying to calculate the E-field from a varying B-field. The B-field is from a current loop.

[tex]\vec{B}(\vec{r},t)=\vec{B_{0}}(\vec{r})Sin(\omega t)[/tex]

where [tex]\vec{B_{0}}=[/tex] http://www.netdenizen.com/emagnet/offaxis/iloopoffaxis.htm"

Now since B is varying with time there should be an electric field. Since B is harmonic E should also vary harmonically but with a phase delay. I am assuming that omega is small enough and the length scales are small enough that this can be treated quasi-static. If I assume E takes this form...
[tex]\vec{E}(\vec{r},t)=\vec{E_{0}}(\vec{r})Cos(\omega t)[/tex]

then faraday's equation gives
[tex]\nabla \times \vec{E_{0}}=-\omega \vec{B_{0}}[/tex]
I tried to solve for E this way but I am not sure if it is possible to do analytically since it involves elliptic integrals. I attempted to integrate using Mathematica but that didn't work either. I can go into more detail on this approach but I do not feel like it will work.

The other way I attempted was to us ampere's equation. This would give [tex]\nabla \times \vec{B_{0}}=-\frac{\omega}{c^{2}} \vec{E_{0}}[/tex]
This seemed like it would be nice and easy but it didn't work out either. Here is what I did.
[tex]\nabla \times \vec{B_{0}}=\frac{1}{r}(\frac{\partial B_{z}}{\partial \theta}-r\frac{\partial B_{\theta}}{\partial z})\hat{e_{r}} +<br /> (\frac{\partial B_{r}}{\partial z} - \frac{\partial B_{z}}{\partial r})\hat{e_{\theta}} +<br /> \frac{1}{r}(\frac{\partial}{\partial r} (r B_{\theta}) - \frac{\partial B_{r}}{\partial \theta} )\hat{e_{z}}[/tex]
well we already know [tex]\vec{B_{\theta}}=0[/tex] and that there is no dependence on [tex]\theta[/tex] for [tex]\vec{B_{r}}[/tex] and [tex]\vec{B_{z}}[/tex] so
[tex]\nabla \times \vec{B_{0}}= (\frac{\partial B_{r}}{\partial z} - \frac{\partial B_{z}}{\partial r})\hat{e_{\theta}}[/tex]
I first tried to take these derivative by hand but after I had worked it all out (it gets kind of long) I decided to let Mathematica do it and see what it gave. I attached the notebook. Well Mathematica give 0 as the derivative. Right off the bat that does not seem right since looking that the vector field it is obvious that there should be a theta component in the curl. Also this would suggest that there is no E-field from a changing B-field which isn't right either.

I do not really want to finish working through the derivatives by hand since they are messy and this point I will have no idea if I am right since if they cancel then it agrees with mathematica but seems unphysical and if they don't cancel then it would probably but to messy to know if it is right.

Any ideas on how to determine the E-field in this case? I would be exterminly happy also if someone was to just point to the literature where this has already been treated. I had a hard time searching for it since I don't know the proper keywords to use.

This problem is a part of the research that I am working on. I will be using these fields in a particle tracing code. It has really had me stuck for a while.

Thank you very much for any help or suggestions. I am willing to provide any clarifications that are necessary.
 

Attachments

Last edited by a moderator:
Physics news on Phys.org
So you have a loop with ac current and you wish to find the E field. Off the cuff, here is how I'd approach it. Ampere's law, AL, relates the magnetic field with the current, B as a function of I. Ohm's law, OL, relates the current density with the electric field, J = sigma*E, where J is in amp/m^2, and sigma is conductivity in mho/m. Then there is Faraday's law, FL.

The boundary condition that must be met is that the E field must comply with OL. The line integral of the E field around the loop equals the voltage around the loop along a particular path, and OL is V=I*R. You should be able to solve it. If the loop is superconducting, then the E field generated by the ac B field exactly cancels the E field of the generating source, hence the net is zero, in compliance with OL.

Have I helped? BR.
 
I believe what you are referring is the E-field in the loop. The E that is responsible for the current. But I am not interested in the inside of the current loop. Maybe I am misunderstanding what you are suggesting but I do not feel like it would produce the E-field in the regions about the current loop.

I just want the fields that result from the time varying B.

A recent thought that I have was that I could use the magnetic potential to get E.
[tex]\vec{B}=\nabla \times \vec{A}[/tex]
I pointed out before that since the fields are harmonic in time then
[tex]\nabla \times \vec{E_{0}}=-\omega \vec{B_{0}}[/tex]
so
[tex]\nabla \times \vec{E_{0}}=\nabla \times (-\omega \vec{A})[/tex]
so
[tex]\vec{E_{0}} = -\omega \vec{A} + \nabla f[/tex] where f is a scalar field
right? Well the magnetic potential is given for a current loop in Jackson EM. I am not sure how I would begin to determine [tex]\nabla f[/tex]. Looking at the fields with Matlab it seems that [tex]\nabla f = 0[/tex] but I have no way to show it.

Any suggestions?
 
E = - dA/dt. This relation is for time varying magnetic fields with no charges present. With charges present, the E field is E = - grad V.

If both charges and time varying magnetic fields are present:

E = - grad V - dA/dt.
 
Ah ok good. That is what I had figured. I was later able to reason that my scalar field must be zero everywhere since it would correspond with charges. So it is good that you pointed it out because now I know I was thinking about it correctly. Thank you.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K