Cylindrical symmetric magnetic field

Click For Summary

Homework Help Overview

The problem involves understanding the behavior of a cylindrical symmetric magnetic field and its relationship to Stokes' theorem, specifically how the magnetic field decreases with distance from the axis of the cylinder.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Stokes' theorem and the implications of cylindrical symmetry on the magnetic field. Questions arise regarding the constancy of the magnetic field along the integration path and the directionality of current density in relation to the area element.

Discussion Status

Participants are actively exploring the implications of cylindrical symmetry on the magnetic field and questioning the assumptions related to the constancy of the field along the path of integration. Some guidance has been offered regarding the relationship between the magnetic field and current density.

Contextual Notes

There is an ongoing discussion about the assumptions related to the direction of current density and the area element in the context of cylindrical symmetry, which may affect the interpretation of the integral equations involved.

Maike
Messages
12
Reaction score
0

Homework Statement


Suppose the magnetic field line pattern is cylindrical symmetric. Explain with Stokes theorem that the field decreases like 1/r (with r the distance from the axis of the cylinder).

Homework Equations


Stokes theorem

The Attempt at a Solution


I was thinking of a circular loop around the axis. The line integral around this loop is B*2*pi*r. But I don't really know what I can say about the curl of B if you only know about the cylindrical symmetry.

Thanks in advance for helping me out!
 
Physics news on Phys.org
Integrating the Maxwell equation for ## \nabla \times B ##: ## \ ##
## \int \nabla \times B \cdot dA=\int \mu_oJ \cdot dA=\mu_o I ##. By Stokes theorem ## \int \nabla \times B \cdot dA=\oint B \cdot dl ##. One question to ask at this point-why is it that ## \oint B \cdot dl =B 2 \pi r ##? i.e. How do we know that ## B ## is constant along the path of the integral?
 
Thanks very much for your answer!
I think B is constant along the path because of the cylindrical symmetry. Or is that a wrong conclusion?
Are you sure by the way that ∫J⋅da = I in this case? How can you know that J and da point in the same direction? Or doesn't it matter?
 
Maike said:
Thanks very much for your answer!
I think B is constant along the path because of the cylindrical symmetry. Or is that a wrong conclusion?
Are you sure by the way that ∫J⋅da = I in this case? How can you know that J and da point in the same direction? Or doesn't it matter?
Yes, that is correct. ## B_{\phi} ## is constant along the circular path. ## J ## often points along the z-direction in problems with cylindrical symmetry, but it isn't a strict requirement.
 
  • Like
Likes   Reactions: Maike

Similar threads

Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
905
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
5
Views
2K
  • · Replies 46 ·
2
Replies
46
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K