9.2 Vacuum magnetic fields 2659.2 Vacuum magnetic fields
The simplest non-trivial magnetic field results when the magnetic field is produced by
electric currents located outside the volume of interest and there are no currents in the
volume of interest. This type of magnetic field is called a vacuum field sinceit could exist
in a vacuum. Because there are no local currents, a vacuum field satisfies
∇×Bvac=0. (9.2)Since the curl of a gradient is always zero, a vacuum field must be the gradientof some
scalar potentialψ, i.e., the vacuum field can always be expressed asBvac=∇ψ. For
this reason vacuum magnetic fields are also called potential magnetic fields. Because all
magnetic fields must satisfy∇·B=0,the potentialψsatisfies Laplace’s equation,
∇^2 ψ= 0. (9.3)Hence the entire mathematical theory of vacuum electrostatic fields canbe brought into
play when studying vacuum magnetic fields. Vacuum electrostatic theory showsthat if ei-
therψor its normal derivative is specified on the surfaceSbounding a volumeV,then
ψis uniquely determined inV. Also, if an equilibrium configuration has symmetry in
some direction so that the coefficients of the relevant linearized partial differential equa-
tions do not depend on this direction, the linearized equations may be Fourier transformed
in this ‘ignorable’ direction. Vacuum is automatically symmetric in all directions and Pois-
son’s equation reduces to Laplace’s equation which is intrinsically linear. The linearity and
symmetry causes Laplace’s equation to reduce to one of the standard equations of math-
ematical physics. For example, consider a cylindrical configuration with coordinatesr,θ,
andzand suppose that this configuration is axially and azimuthally uniform so that bothθ
andzare ignorable coordinates;i.e., the coefficients of the partial differential equation do
not depend onθor onz. Fourier analysis of Eq.(9.3) implies thatψcan be expressed as
the linear superposition of modes varying asexp(imθ+ikz).For each choice ofmandk,
Eq. (9.3) becomes
∂^2 ψ
∂r^2
+
1
r∂ψ
∂r−
(
m^2
r^2
+k^2)
ψ= 0. (9.4)Definings=kr,this can be recast as
∂^2 ψ
∂s^2+
1
s∂ψ
∂s−
(
1 +
m^2
s^2)
ψ= 0, (9.5)a modified Bessel’s equation. The solutions of Eq.(9.5) are the modified Besselfunctions
Im(kr)andKm(kr). Thus, the general solution of Laplace’s equation here is
ψ(r,θ,z) =∑∞
m=−∞∫
[am(k)Im(kr) +bm(k)Km(kr)]eimθ+ikzdk (9.6)where the coefficientsam(k),bm(k)are determined by specifying eitherψor its normal
derivative on the bounding surface. Analogous solutions can be found in geometries having
other symmetries.
This behavior can be viewed in a more general way. Equation (9.3) states thatthe sum
of partial second derivatives in two or three different directions is zero, so at least one of
these terms must be negative and at least one term must be positive. Since negativeψ′′/ψ
corresponds to oscillatory (harmonic) behavior and positiveψ′′/ψcorresponds to exponen-
tial (non-harmonic) behavior, any solution of Laplace’s equation must be oscillatory in one