- #1
JustinLevy
- 895
- 1
By "static" in this thread, I mean the charge and current density is constant in time (so things can be moving, but the distribution cannot).
We can build an arbitrary eletrostatic field with a "eletric monopole" density.
Since Del.B=0, can we build an arbitrary magnetostatic field with just a "magnetic dipole" density?
--
Since we also need the direction of the dipoles, I guess this density it would be a vector field.
Based on analogy with magnetostatics in materials where [tex]B = \mu_0(H + M)[/tex] and M can be considered the "dipole density", it seems like this is possible because I could just consider _every_ current a "bound current" instead of a "free current" and push everything onto M.
Correct?
Or is there something that prevents considering all "free currents" as "bound currents"?
In particular, I can't figure out how to derive an M which alone would recreate an arbitrary magnetostatic field.
I have:
[tex] \nabla \times \mathbf{H} = 0[/tex]
[tex] \nabla \times \mathbf{B} = \mu_0 \nabla \times (\mathbf{H} + \mathbf{M})[/tex]
[tex] \nabla \cdot \mathbf{B} = \mu_0 \nabla \cdot (\mathbf{H} + \mathbf{M})=0[/tex]
So while I can find the curl in terms of B
[tex] \nabla \times \mathbf{M} = \frac{1}{\mu_0}\nabla \times \mathbf{B}[/tex]
I can't figure out how to get the divergence in terms of B
[tex] \nabla \cdot \mathbf{M} = - \nabla \cdot \mathbf{H}[/tex]
because H and M depend on each other. It looks like there is no unique solution.
We can build an arbitrary eletrostatic field with a "eletric monopole" density.
Since Del.B=0, can we build an arbitrary magnetostatic field with just a "magnetic dipole" density?
--
Since we also need the direction of the dipoles, I guess this density it would be a vector field.
Based on analogy with magnetostatics in materials where [tex]B = \mu_0(H + M)[/tex] and M can be considered the "dipole density", it seems like this is possible because I could just consider _every_ current a "bound current" instead of a "free current" and push everything onto M.
Correct?
Or is there something that prevents considering all "free currents" as "bound currents"?
In particular, I can't figure out how to derive an M which alone would recreate an arbitrary magnetostatic field.
I have:
[tex] \nabla \times \mathbf{H} = 0[/tex]
[tex] \nabla \times \mathbf{B} = \mu_0 \nabla \times (\mathbf{H} + \mathbf{M})[/tex]
[tex] \nabla \cdot \mathbf{B} = \mu_0 \nabla \cdot (\mathbf{H} + \mathbf{M})=0[/tex]
So while I can find the curl in terms of B
[tex] \nabla \times \mathbf{M} = \frac{1}{\mu_0}\nabla \times \mathbf{B}[/tex]
I can't figure out how to get the divergence in terms of B
[tex] \nabla \cdot \mathbf{M} = - \nabla \cdot \mathbf{H}[/tex]
because H and M depend on each other. It looks like there is no unique solution.