3.7 Non-adiabatic motion in symmetric geometry 97
z B
r,zis flux
linked by this
circle
Figure 3.12: Azimuthally symmetricflux surface
Thusψis constant along the symmetry axisr= 0;for convenience we choose this
constant to be zero. Hence,ψ(r,z,t)is precisely the magneticflux enclosed by a circle of
radiusrat axial locationz.We can also use the vector potentialAto calculate the magnetic
flux through the same circle and obtain
∫
B·ds=
∫
∇×A·ds=
∮
A·dl=
∫ 2 π
0
Aθrdθ= 2πrAθ. (3.150)
This shows that thefluxψand the vector potentialAθare related by
ψ(r,z,t) = 2πrAθ. (3.151)
No other component of vector potential is required to determine the magnetic field and so
we may setA=Aθ(r,z,t)ˆθ.
The currentJ=μ− 01 ∇×Bproducing this magnetic field is purely azimuthal as can be
seen by considering therandzcomponents of∇×B.The actual current density is
Jθ = μ− 01 r∇θ·∇×B
= μ− 01 r∇·(B×∇θ)
= −
r
2 πμ 0
∇·
(
1
r^2
∇ψ
)
= −
r
2 πμ 0
[
∂
∂r
(
1
r^2
∂ψ
∂r
)
+
1
r^2
∂^2 ψ
∂z^2