Magnetic Fields and Magnetic Diagnostics for Tokamak Plasmas

Magnetic fields and tokamak plasmas
Alan Wootton
R
Ψ = 2πRA φ
= MI = 2π ∫ B z
RdR
1.39
0
That is, for the system we are considering, the poloidal flux at a position R is simply the vertical
field B z integrated across a circle of radius R. Note that sometimes in the literature the flux
function ψ = Ψ/(2π) is used.
As an example we show in Figure 1.10 the poloidal flux ψ produced by a single circular
filament (see the section on vector potentials for the derivation of A φ ) of radius R 0 = 1 m, current
I = 1/ µ0 . Results are shown in the plane of the coil (z = 0). The exact results are shown as the
solid line. Also shown are two approximate solutions; the very near field solution and the far
field solution. Near the current (“near field”) we can write
ψ = RA φ
≈ µ 0 IR 0
2π
⎡ ⎛ ⎛
ln⎜
8R 0⎞
⎞ ⎛
⎜ ⎟ − 2 ⎟ − ⎜ ρ ⎞
⎟ cos( ω ) ⎛ ⎛
ln⎜
8R 0
⎞ ⎞ ⎤
⎜ ⎟ −1 ⎟
⎢
⎣ ⎝ ⎝ ρ ⎠ ⎠ ⎝ R 0
⎠ 2 ⎝ ⎝ ρ ⎠ ⎠ ⎥
⎦
1.40
The very near field solution is the zero order in ρ/R 0 term, i.e. ψ 0
= µ 0 IR 0
2π
from the loop (“far field”) we can write
⎛ ⎛
ln⎜
8R 0
⎞ ⎞
⎜ ⎟ − 2 ⎟ . Far
⎝ ⎝ ρ ⎠ ⎠
MR 2
ψ =
( R 2 + z 2
) 3 2
1.41
We see that neither the very near or far field solutions are good for distances of about half a coil
radius from the coil itself. However, if we use the full expansion expression including terms of
order ρ/R 0 (Equation 1.40) then the results are very near to the exact solution. This is shown in
Figure 1.11.
ψ
far field
exact
very near field
R (m)
19