Cylindrical symmetric magnetic field

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SUMMARY

The discussion focuses on the application of Stokes' theorem to analyze the behavior of a cylindrical symmetric magnetic field, specifically demonstrating that the magnetic field strength, B, decreases with distance from the axis of the cylinder as 1/r. Participants utilized the integral form of Maxwell's equations, particularly the relationship between the curl of B and the line integral around a circular path. The conclusion drawn is that B remains constant along the circular path due to cylindrical symmetry, allowing for the simplification of the integral to B * 2πr.

PREREQUISITES
  • Understanding of Stokes' theorem
  • Familiarity with Maxwell's equations
  • Knowledge of cylindrical symmetry in magnetic fields
  • Basic calculus for evaluating line integrals
NEXT STEPS
  • Study the implications of Stokes' theorem in electromagnetism
  • Explore the derivation of magnetic field equations in cylindrical coordinates
  • Learn about the physical significance of curl in vector fields
  • Investigate the relationship between current density J and magnetic fields in cylindrical systems
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Students of electromagnetism, physicists studying magnetic fields, and educators seeking to explain the principles of cylindrical symmetry in magnetic field theory.

Maike
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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!
 
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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.
 
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