Magnetic Fields and Magnetic Diagnostics for Tokamak Plasmas
Magnetic Fields and Magnetic Diagnostics for Tokamak Plasmas
Magnetic Fields and Magnetic Diagnostics for Tokamak Plasmas
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<strong>Magnetic</strong> fields <strong>and</strong> tokamak plasmas<br />
Alan Wootton<br />
Energy associated with toroidal fields W 1<br />
B<br />
The energy associated with poloidal currents is written as<br />
W B 1<br />
= L I 2<br />
1 1<br />
2 + L I 2<br />
1e 1e<br />
2<br />
+ M 1<br />
I 1<br />
I 1e<br />
6.0.5<br />
Here I 1 is the poloidal current in the plasma, <strong>and</strong> I 1e is the poloidal current in the toroidal field<br />
coil (subscript e <strong>for</strong> external). I 1 is that poloidal current flowing in the plasma edge which<br />
produces a toroidal field equal to the difference between the internal toroidal field B φi <strong>and</strong> the<br />
external toroidal field B φe . By definition we have<br />
L 1<br />
I 1<br />
2<br />
( ) 2 V<br />
B<br />
2 = φi<br />
− B φe<br />
6.0.6<br />
2µ 0<br />
( B φi<br />
− B φe )B φe<br />
V<br />
M 1<br />
I 1<br />
I 1e<br />
=<br />
6.0.7<br />
µ 0<br />
Now the circuits I 1 <strong>and</strong> I 1e are perfectly coupled, so that L 1 = M 1. The field B 1 = µ 0 I 1 /(2πR), <strong>and</strong><br />
so<br />
B 1<br />
2<br />
M 1<br />
= L 1<br />
= 1 ⎛<br />
2<br />
I 1<br />
∫ dV = µ 0<br />
R − R 2 − a 2<br />
µ ⎝<br />
0<br />
V<br />
( ) 1 2<br />
⎞<br />
⎠ ≈ µ a 2<br />
0<br />
2R<br />
6.0.8<br />
<strong>for</strong> skin currents. To get the <strong>for</strong>ces we will need only the functional dependencies, namely<br />
∂L 1<br />
∂a = 2 L 1<br />
a<br />
∂M 1<br />
∂a = 2M 1<br />
a<br />
∂L 1<br />
∂R = − L 1<br />
R<br />
∂M 1<br />
∂R = − M 1<br />
R<br />
6.0.9<br />
6.0.10<br />
The <strong>for</strong>ces will be computed at constant current. For example, the part of the <strong>for</strong>ce due to<br />
∂/∂R(L 1 I 1 2 /2) is then written as (I 1 2 /2)∂/∂R(L 1 ) = -(I 1 2 /2)L/R = -(L 1 I 1 2 /2)(1/R). Using Equation<br />
6.0.6 this becomes (I 1 2 /2)∂/∂R(L 1 ) = V/(2Rµ 0 ). Doing this <strong>for</strong> each component in<br />
Equation 6.0.5 gives<br />
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