5

As a preamble, I would not expect that splitting $E$ into real/imaginary parts is very profitable. Normally, block 2x2 systems are motivated because one block of unknowns is "easier" to solve than the other in some sense (better conditioned? smaller in cardinality? etc). This is not the case for time-harmonic Maxwell, I think you'd be better off ...


3

Using the typical expansion functions (1-forms/edge-elements for E, and 2-forms/facet-elements for B) the formulations are basically the same after spatial discretization and you'd expect more or less the same accuracy. I do think they express slightly different opinions about time integration. The mixed E/B formulation nudges you in the direction of ...


3

Your particle is a rounded proton (mass m = 2e-27 kg instead of 1.672e-27 kg). The equation of motion is $$ \dot x=v,~~~ m\dot v = q\,v\times B, $$ where $B=(0,0,B_z)$ with $B_z=4T=4N/(m\,A)$ and $q=1e=1.602·10^{-19} C$, $C=A\,s$ This then gives for the acceleration m=2e-27 e_charge = 1.6e-19 q=+1*e_charge Bz = 4 ax = q/m*vy*Bz; ay = -q/m*vx*Bz; az = 0 For ...


2

This is the code I came up with. I still dont feel confident about the results, mostly because I didn't find the usual amount of errors yet. import numpy as np from scipy.special import kn, iv import matplotlib.pyplot as plt global size, mu_0, p, a, q, current, order mu_0 = 4 * np.pi * 1e-7 size = 10 order = 5 # order of bessel functions considered p = 1.8 ...


1

Disclaimer: I'm outside my area of expertise here. Due to continuity constraints, when solving electromagnetic simulations with FEM, one uses edge based vector basis functions. After solving, I would expect you to construct your solution field $$\vec{H}(x) = \sum_i H_i \vec{N}_i(x)$$ or, element-wise $$\vec{H}(x) = \sum_i H^e_i \vec{N}^e_i(x) \quad\forall x \...


1

The problem is that, first, the vector field has a singularity at the origin. Moving straight towards the singularity results in a catastrophe in finite time. Second and related, that the speed and all stiffness measures (higher derivatives) increase significantly the closer you get to the origin. So while one would expect a Kepler ellipse (after setting the ...


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