Precis of “Dielectric (and Magnetic) Image Methods”
Dr. Kirk T. McDonald
Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544
(November 21, 2009; updated February 7, 2020)
Annotated by Dr. Edward P. Sayre, P.E.
North East Systems Associates, Inc.
This is an annotated precis of the McDonald paper covering the applicability of image
theory to evaluate the fiels within planar slabs of permeable material. In “classical” 1
electrodynamics, one characterizes permeable material presence in macroscopic media by
the density M of magnetic moments. Noting that B = µ(H + M) and that ∇ ・ B = 0, one
can define an effective magnetic charge density according to ρm,eff = −∇ ・ M. Isolated
effective magnetic charges do not exist and the effective magnetic charge density is only
an alternative representation of Amperian currents in bulkmatter. 2
We will consider image methods for permeable media in two-dimensional cases with a
line conduction current I parallel to the z-axis and intersecting the x-axis.
X
Hy
I
a
µ’
Z
µ”
Y
Line Current I in Two Permeable Media Slabs
Figure 1
Away from the line current, ∇ × H = 0, so we could write H = −∇ ΦM where ΦM is a
scalar potential. However, when conduction currents are present it seems better to work
with the vector potential A, such that B = ∇ ×A. We recall that for a current I along the zaxis in a medium of relative permeability μ’, the magnetic field is
1
The units for this paper have been transformed to MKS formulation for more convenient enginnering applications.
L.D. Landau, E.M. Lifshitz and L.P. Pitaevskii, Electrodynamics of Continuous Media, 2nd ed. (Butterworth-Heinemann, 1984),
http://physics.princeton.edu/~mcdonald/examples/EM/landau_ecm2.pdf
2
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B = μ_H = μ’I θ/2πr,
so the vector potential is
A = −(μ’I /2 π)* ln r z
where: θ is taken as the angular unit vector in a cylindrical coordinate system, and
z is the unit vector parallel to the Z axis.
The approach taken by McDonald is to consider a trial image solution of a line current in
the media spaces half media spaces µ’ and µ” as shown in Figure 1.
Consider the case of a medium with relative permeability μ” in the half space x < 0 with a
line current I at (x,y) = (a, 0), in the half space region x > 0 which has a relative
permeability μ’. The classical image method assumes that the vector potential in the
region x > 0 is that due to the original current I at (a, 0) plus an image current I’ at (−a,
0). However, this solution is only valid when I and I’ are equal and opposite.
Additionally, McDonald posits that because the permeable material μ” will not suffice for
the equal and opposite condition, the potential in the half region x < 0 requires the
original image currents plus an additional current I” also at (a, 0).
Based on these assumptions, assume there are two currents, I, I” which satisfy the usual
image solution for x > 0 if the region x < 0 were a good conductor. The extra (third)
current I” is suggested to account for the finite value of the permeability µ” in the region,
x < 0. The application of boundary conditions allows the relationships between the three
line current unknowns to be determined.
According to the suggested image method, the vector potential at x > 0 can be written
as,
Az (x > 0) = −μ’(I/2π)*ln[(x − a)2 + y2] – μ’(I’/2π)*ln[(x + a)2 + y2] ,
and that for x < 0 as,
Az(x < 0) = −μ’((I + I”)/2π)*ln[(x − a)2 + y2].
At the boundary plane, x=0, the vector potentials are continuous, and
A’z(x = 0) = A”z(x = 0)
Resulting in the relationship between the image currents:
I’ = I”
Also, application of the tangential magnetic field Hy boundary condition across the
boundary between the two media requires that,
Hy (x=0+) = Hy (x=0-)
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where: Hy is the tangential magnetic field component in near vicinity of the
(x=0, y) plane.
As noted earlier, taking the curl of the vector potential and dividing the result by the
relative permeability gives the magnetic field.
Hy (x = 0+) = By (x = 0+)/μ' = −1* ∂Az (x = 0+) = Hy (x = 0-) =−1* ∂Az (x = 0-)
µ’ ∂x
µ" ∂x
By direct substitution,
µ’(I - I’) = µ”(I + I”)
From the continuity of the vector potential for x = 0, and the equality between the Ycomponents of the magnetic field in the near vicinity of the media planar boundary, we
obtain,
I’ = I” = I * (µ’ - µ”)/ (µ’ + µ”)
Giving in the region x > 0,
By (x > 0) = (µ’I/π) *{(x – a)/((x – a)2 + y2) + ((µ’ - µ”)/( µ’ + µ”) * ( x + a)/((x + a)2 + y2)}
And, in the region x < 0,
By (x < 0) = (2µ’I/π) * [(µ”* µ’)/( µ’ + µ”)] * [ (x – a)/((x – a)2 + y2)]
To complete the analysis, the vertical (X) component of the magnetic flux density will be
determined. Applying the boundary condition for the magnetic flux density,
Bx (x = 0+) = Bx (x = 0-)
Results in the following in the region x > 0,
Bx (x > 0) = (µ’I/π) *{y/((x – a)2 + y2) + ((µ’ - µ”)/( µ’ + µ”) * y/((x + a)2 + y2)}
And, in the region x < 0,
By (x < 0) = (2µ’I/π) * [(µ”* µ’)/( µ’ + µ”)] * [ (y/((x – a)2 + y2)]
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Conceptual and Graphical Interpretation of Results
In general, image solutions replace the materials with vacuum properties and the material
effects are expressed in equivalent sources with the material properties as noted in the
field solutions.
In the region x > 0 the vector potential Az and potential and B field are given as if due to
an effective line current μ’I at x = a and an effective image current μ’I” at x = −a, both in
vacuum. The geometry is illustrated in Figure 2.
X
X > 0 Field
Point
xf
(xf, yf)
Line Current
Source
x =a
Z
yf
Imaged Line
Current Source
x = -a
Y
Illustration of Line Current and Image Current for x > 0
Figure 2
The vector potential Az and the components of the B field in the region x < 0 are as if that
region were vacuum and the original line current I were replaced by an effective current
μ_(I’ + I”) = 2Iμ’μ”/(μ’ + μ”).
X
X < 0 Imaged Line
Current Source
x =a
Z
yf
xf
Y
(xf, yf)
X < 0 Field
Point
Illustration of Image Current for x < 0
Figure 3
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