# How to obtain the magnetic field of the following set up [closed]

This is the experimental set up: A ferromagnetic cylindrical disk (of irrelevant dimensions) has either sides set at different electric potentials (left side at 0 volts and right side at 1 volt). A radio frequency current is passed through the disk and flows between these two regions. I have been able to use mathematica to determine the electric field the disk exhibits but am unsure of how to obtain the corresponding magnetic field. For the sake of the problem let the current be 1 mA with frequency 1 Ghz. Let the disk have diameter 10 microns and height 0.5 microns.

R1 =
ImplicitRegion[x^2 + y^2 <= 1 ∧ Abs[z] <= 0.1,
{{x, -1, 1}, {y, -1, 1}, {z, -0.1, 0.1}}]

DBC1 = {
DirichletCondition[u[x, y, z] == 0,
(x^2 + y^2 <= 1 ∧ -1 <= x <= -0.8 ∧ Abs[y] <= 0.6 ∧ Abs[z] == 0.1)],
DirichletCondition[u[x, y, z] == 1,
(x^2 + y^2 <= 1 ∧ 0.8 <= x <= 1 ∧ Abs[y] <= 0.6 ∧ Abs[z] == 0.1)]
}

pot =
NDSolveValue[{Laplacian[u[x, y, z], {x, y, z}] == 0, DBC1},
u, Element[{x, y, z}, R1]]

Efield = -Grad[pot[x, y, z], {x, y, z}]


Any tips on how I would go about obtaining the magnetic field from the information I have would be a great help.

## closed as unclear what you're asking by Jens, MarcoB, ilian, m_goldberg, C. E.Jul 22 '15 at 3:40

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• To get the magnetic field, you need the current density as a function of position, not just the current. This isn't a Mathematica problem, though. – Jens Jul 21 '15 at 17:05
• I'm voting to close this question as off-topic because the issue it raises is not a Mathematica issue but a physics issue. That it is formulated in terms of Mathematica is not sufficient to make it an appropriate question for Mathematica.SE. – m_goldberg Jul 21 '15 at 21:47
• Yes, I think that is a good idea. – Michael ponds Jul 23 '15 at 8:09