Magnetic Fields and Magnetic Diagnostics for Tokamak Plasmas

Magnetic fields and tokamak plasmas
Alan Wootton
where the total current I flows inside the volume V, the line element dl is along the direction of
the total current I, and r is the distance from the line element to the point of interest. A useful
example for us is the vector potential of a circular filament. This is used to represent windings
(vertical field, shaping, ohmic heating) on the tokamak, and elements of the plasma current itself.
P
r
dl
z
Y
dl
R 0
r
R 0
−φ
φ
R
X
Figure 1.6. The geometry used to evaluate the vector potential of a circular filament.
Consider a circular filament of radius R 0 , with current I in the φ direction.
A φ must be
independent of φ, so choose the point of interest P in the (X,z) plane of Figure 1.6, where φ = 0.
Pairing equidistant elements dl shown in thickened lines at ±φ we see the resultant is normal to
(R,z). Therefore only consider the component dl φ of dl in the direction normal to the plane (R,z);
dl φ = R 0 cos(φ)dφ. The radial distance r from the point P to the element is given by
r 2 = z 2 + R 2 0
+ R 2 − 2R 0
Rcos( φ). Then
π
A φ
= µ Idl φ
4π ∫ = µI R 0
cos(φ )dφ
r 2π ∫ 1.16
0 R 2 0
+ R 2 + z 2 − 2R 0
Rcos(φ)
( ) 1 2
Far from the loop (i.e. a small loop) we have r 0 = (R 2 +z 2 ) 1/2 >> R 0 , and the integral becomes
A φ
= µI
2π
≈ RR 2
0µI
4r 3
R 0
cos(φ) ⎛
⎜ 1 + RR cos(φ)
0
⎞
⎟ dφ
2
r 0
⎝ r 0
⎠
1.17
= µ ( M × r)
4πr 3
π
∫
0
Here we have written the magnetic moment of the loop M = πR 0 2 I , directed upwards.
If the loop is not small, then let φ = π + 2θ, so dφ = 2 dθ and cos(φ) = 2 sin 2 (θ) - 1, and we obtain
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