280 Chapter 9. MHD equilibria
showing that the functional form of Eq.(9.34) is consistent with Ampere’s law
∮
B·dl=μ 0 I.
The poloidal magnetic field is
Bpol=
1
2 π
(∇ψ×∇φ). (9.36)
Integration of the poloidal magnetic field over the area of a circle of radiusrwith center at
axial locationzgives
∫r
0
Bpol·ds=
∫r
0
1
2 π
∇ψ×∇φ·zˆ 2 πr′dr′=ψ(r,z); (9.37)
thusψ(r,z)is the poloidalflux at locationr,z.The concept of poloidalflux depends on
the existence of axisymmetry so that a circle of radiusrcan always be associated with a
locationr,z.
Axisymmetry also provides a useful relationship between toroidal and poloidal vectors.
In particular, the curl of a toroidal vector is poloidal since
∇×Btor=
μ 0
2 π
∇I×∇φ (9.38)
and similarly the curl of a poloidal vector is toroidal since
∇×Bpol=∇×(Brrˆ+Bzzˆ)=ˆφ
(
∂Br
∂z
−
∂Bz
∂r
)
. (9.39)
The curl of the poloidal magnetic field is a Laplacian-like operator onψsince
∇φ·∇×Bpol=∇·(Bpol×∇φ)=∇·
(
1
2 π
[∇ψ×∇φ]×∇φ
)
=−
1
2 π
∇·
(
1
r^2
∇ψ
)
,
(9.40)
a relationship established using the vector identity∇·(F×G)=G·∇×F−F·∇×G.
Because∇×Bpolis purely toroidal andφˆ=r∇φ,one can write
∇×Bpol=−
r^2
2 π
∇·
(
1
r^2
∇ψ
)
∇φ. (9.41)
Ampere’s law states that∇×B=μ 0 J. Thus, from Eqs. (9.38) and (9.41) the respective
toroidal and poloidal currents are
Jtor=−
r^2
2 πμ 0
∇·
(
1
r^2
∇ψ
)
∇φ (9.42)
and
Jpol=
1
2 π
∇I×∇φ. (9.43)
We are now in a position to evaluate the magnetic force in Eq.(9.22). After decomposing
the magnet field and current into toroidal and poloidal components, Eq.(9.22)becomes
∇P=Jpol×Btor+Jtor×Bpol+Jpol×Bpol. (9.44)
TheJpol×Bpolterm is in the toroidal direction and is the only toroidally directed term on
the right hand side of the equation. However∂P/∂φ=0 because all physical quantities
are independent ofφand soJpol×Bpolmust vanish. This implies
(∇I×∇φ)×(∇ψ×∇φ)=0 (9.45)