Fundamentals of Plasma Physics

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

]

(3.152)