Closed ZedThree closed 1 year ago
I don't think this is actually the correct fix, and instead we should have $BR = B{pa} Z / (R a \rho)$ which is the derivative of the existing $\psi$ times a scaling factor
The first fix was the one I had in mind. \psi is indeed normalised, so no need for the factor of two.
However, the default density fit needs to be changed such that it's linear in rho (that is, linear in distance in meters from the magnetic axis).
This week has been madly busy, I'll be able to chat next week if you like.
Closing this; #74 adds ConstantCurrentDensityField
which has the original behaviour, and CircularCrossSectionField
has $\psi = \rho / a$ and consistent field components from that.
For
find_B_method = "analytical"
, circular cross-section geometry, $\psi(R, 0)$ is linear, but so is $B_R(R, 0) \simeq \frac{d\psi}{dR}$.Instead we need $\psi \simeq \rho^2$ so that the magnetic field components come out linear.
@valerian-chen $\psi$ is the normalised flux, right? So we don't need an extra
B_p_a / 2
in there to makeB_R
etc correct?Unfortunately, this does change the results a bit, so the golden answer test needs updating.