Boundary conditions ##\vec{B}## and ##\vec{H}##

In summary, the conversation discusses the boundary conditions for the magnetostatics problem, specifically for the components of the vectors B and H. It is mentioned that the curls of the vectors are not necessary to consider, and the focus is on the surface current model. The problem can be solved using two different methods, and using Stokes theorem, the parallel components of H on the entire surface can be determined.
  • #1
happyparticle
456
21
Homework Statement
Find boundary conditions ##\vec{B}## and ##\vec{H}## for a cylinder of radius a and length 4a and ##\vec{M} = M\hat{z}## on the axis of the cylinder
Relevant Equations
##\vec{\nabla} \cdot \vec{B} = 0##
##\vec{\nabla} \cdot \vec{H} = - \vec{\nabla} \cdot \vec{M}##
##\vec{\nabla} x \vec{B} = \mu_0 \vec{J}##
##\vec{\nabla} x \vec{H} = \mu_0 \vec{J}_f##
When asking for boundary conditions I'm wondering if this is enough in this situation to give
##\vec{\nabla} \cdot \vec{B} = 0 , B_{2\perp} - B_{1 \perp} = 0##
##\vec{\nabla} \cdot \vec{H} = - \vec{\nabla} \cdot \vec{M}, H_{2\perp} - H_{1 \perp} = - (M_{2\perp} - M_{1 \perp})##
##\vec{\nabla} \times \vec{B} = \mu_0 \vec{J}, B_{2\||} - B_{1 \||} = \mu_0 \vec{K} \times \hat{n}##
##\vec{\nabla} \times \vec{H} = \mu_0 \vec{J}_f , H_{2 \||} - H_{1 \||} = \vec{K}_f \times \hat{n}##
 
Physics news on Phys.org
  • #2
It probably isn't necessary to include anything about the curls of the vectors, because that is a surface current type treatment that is completely separate from the pole model of magnetostatics. Meanwhile, what do you know about ## M_{2 \, perpendicular} ## ? =if it is outside of the material? For the ## M_{in \, perpendicular} ##, it is ## -M ## on the left endface, and ## +M ## on the right endface.

Edit: Scratch part of that=I think they may be looking for boundary conditions everywhere on the surface, (they aren't completely clear here=do they want just the on-axis conditions? ), and it may be useful to employ the surface current model to get the conditions for the parallel components.

It may be worth mentioning that this problem has two very standard ways of solving it, and Legendre's method with boundary conditions are not needed to solve it. See https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/
 
Last edited:
  • Like
Likes happyparticle
  • #3
a follow-on: It still needs a little work, but let's get started using ## \nabla \times \vec{H}=\vec{J}_{free}=0 ##. Using Stokes theorem on this, what can you say about the parallel components of ## \vec{H} ## anywhere on the entire surface, i.e. ## H_{out \, parallel} ## and ## H_{in \, parallel} ##?
 
  • #4
Finally find it, thank you.
 
  • Like
Likes Charles Link

FAQ: Boundary conditions ##\vec{B}## and ##\vec{H}##

What are boundary conditions for ##\vec{B}## and ##\vec{H}##?

Boundary conditions for ##\vec{B}## and ##\vec{H}## refer to the conditions that must be satisfied at the interface between two different media, such as air and a magnet, for the electromagnetic fields to be continuous and well-defined.

What are the types of boundary conditions for ##\vec{B}## and ##\vec{H}##?

The two types of boundary conditions for ##\vec{B}## and ##\vec{H}## are the tangential and normal conditions. The tangential condition states that the tangential components of ##\vec{B}## and ##\vec{H}## must be continuous across the interface, while the normal condition states that the normal component of ##\vec{B}## must be continuous and the normal component of ##\vec{H}## must have a discontinuity equal to the surface current density.

How do boundary conditions for ##\vec{B}## and ##\vec{H}## affect electromagnetic wave propagation?

Boundary conditions for ##\vec{B}## and ##\vec{H}## play a crucial role in determining the behavior of electromagnetic waves at interfaces. They determine the reflection and transmission coefficients of the waves, as well as the polarization of the transmitted wave.

Can boundary conditions for ##\vec{B}## and ##\vec{H}## be simplified in certain cases?

Yes, in some cases, boundary conditions for ##\vec{B}## and ##\vec{H}## can be simplified. For example, in the absence of surface currents, the normal condition reduces to the requirement that the normal component of ##\vec{H}## is continuous, while the tangential condition reduces to the requirement that the tangential components of ##\vec{B}## and ##\vec{H}## are continuous.

How do boundary conditions for ##\vec{B}## and ##\vec{H}## differ from those for ##\vec{E}## and ##\vec{D}##?

Boundary conditions for ##\vec{B}## and ##\vec{H}## are different from those for ##\vec{E}## and ##\vec{D}## because they involve different physical quantities. While the boundary conditions for ##\vec{B}## and ##\vec{H}## are based on the continuity of the fields, the boundary conditions for ##\vec{E}## and ##\vec{D}## also take into account the presence of surface charges and bound charges at the interface.

Back
Top