Radio Science Advances in Shielding properties of a ... · Professur fur Theoretische Elektrotechnik und Numerische Feldberechnung, Helmut-Schmidt-Universit¨ at, Universit¨ ¨at

Advances in Radio Science, 3, 119–123, 2005SRef-ID: 1684-9973/ars/2005-3-119© Copernicus GmbH 2005

Advances inRadio Science

Shielding properties of a conducting bar calculated with a boundaryintegral method

L. O. Fichte, S. Lange, Th. Steinmetz, and M. Clemens

Professur fur Theoretische Elektrotechnik und Numerische Feldberechnung, Helmut-Schmidt-Universitat, Universitat derBundeswehr Hamburg, PO. Box 700822, 22008 Hamburg, Germany

Abstract. A plane rectangular bar of conducting and per-meable material is placed in an external low-frequency mag-netic field. The shielding properties of this object are in-vestigated by solving the given plane eddy current problemfor the vector potential with the boundary integral equationmethod. The vector potential inside the rectangle is governedby Helmholtz’ equation, which in our case is solved by sep-aration. The solution is inserted into the remaining bound-ary integral equation for the exterior vector potential in thedomain surrounding the bar. By expressing its logarithmickernel as a Fourier integral the overall solution inside andoutside the bar is calculated using analytical means only.

1 Introduction

Subject of this investigation are the shielding properties of arectangular bar of constant conductivityκi and permeabil-ity µi . The space outside the bar has the arbitrary con-stant permeabilityµa and conductivityκa=0. The vectorrq=xqex+yqey is pointing at the lower left corner of the bar,the dimensions of it beinga andb, as shown in Fig. 1. Wename the cross-section of the bar and its contourC=∂.

An exciting loop consists of two thin conductors, whichcarry the currentsi(t) and−i(t), with i(t)=<Iejωt

, I be-ing the phasor describing the complex current . They arelocated at

re = ±xeex .

The angular frequencyω is considered to be low enough sothat displacement currents can be neglected:J

∂∂t

D, whereJ is the current density inside the bar andD the electric flux.

The z-directed dimensions of both bar and loop are sup-posed to be infinite.

2 Differential equations for the vector potential

As all exciting currents are oscillating at one single fre-quency, all fields will show the same time dependency andcan be expressed by their complex phasors.

Correspondence to:L. O. Fichte([email protected])

Introducing a complex vector potentialA as B=curlA,where B denotes the magnetic flux, and investigating itsproperties, we find that an additional term has to be addedto the vector potential if we consider points inside the bar.The additional term can be expressed as the time domain in-tegral of the gradient of a complex electrical scalar potentialφi(z) by∫

(gradφi)dt = Cez,

whereC denotes an unknown complex constant. The vec-tor potentialAi inside the bar can be redefined by using theBuchholz convention

A∗

i = Ai +

∫grad8idt = Ai + Cez

from which the fields can be computed with

B i = curlA∗

i and Ei = −jωA∗

i .

From Maxwell’s equations we find that the modified vec-tor potentialinside the bar is governed by Helmholtz’ equa-tion, which for complex fields takes the form

1A∗

i − jωµiκiA∗

i = 0. (1)

Outside the bar the equation

1Aa = −µaJ e, (2)

is valid for the vector potentialAa. The exciting current den-sity J e can be expressed as

J e = δ1(x − xe) − δ1(x + xe)δ1(y) · Iez.

The symbolδ1(x−xo) denotes the one-dimensional Diracfunction; accordingly,δ2 and δ3 are the two- and three-dimensional Dirac functions.

Since we allow onlyz-directed exciting currents, all vectorpotentials will also be exclusivelyz-directed. They can bedescribed by theirz-componentsAaz andA∗

iz, respectively.The boundary conditions forAaz andA∗

iz are

Aaz = A∗

iz − C ∧1

µa

no · gradAaz =1

µi

no · gradA∗

iz,

whereno is a unit vector normal to the boundary.

120 L. O. Fichte et al.: Shielding properties of a conducting bar calculated with a boundary integral method

SHIELDING PROPERTIES OF A CONDUCTING BAR

CALCULATED WITH A BOUNDARY INTEGRAL METHOD

Lars Ole Fichte, Sebastian Lange, Thorsten Steinmetz, Markus ClemensProfessur fur Theoretische Elektrotechnik und Numerische Feldberechnung

Helmut-Schmidt-UniversitatUniversitat der Bundeswehr Hamburg

PO. Box 700822, D-22008 Hamburg, GermanyTel.(+49)40-6541 2460, Fax.(+49)40-6541 3764

ABSTRACT: A plane rectangular bar of con-ducting and permeable material is placed in an ex-ternal low-frequency magnetic field. The shieldingproperties of this object are investigated by solvingthe given plane eddy current problem for the vec-tor potential with the boundary integral equationmethod. The vector potential inside the rectangle isgoverned by Helmholtz’ equation, which in our caseis solved by separation. The solution is inserted intothe remaining boundary integral equation for theexterior vector potential in the domain surround-ing the bar. By expressing its logarithmic kernelas a Fourier integral the overall solution inside andoutside the bar is calculated using analytical meansonly.

1. INTRODUCTION

Subject of this investigation are the shielding prop-erties of a rectangular bar of constant conductivityκi and permeability µi. The space outside the barhas the arbitrary constant permeability µa and con-ductivity κa = 0. The vector ~rq = xq~ex + yq~ey ispointing at the lower left corner of the bar, the di-mensions of it being a and b, as shown in Fig.1. Wename the cross-section of the bar Ω and its contourC = ∂Ω.

An exciting loop consists of two thin conductors,which carry the currents i(t) and −i(t), with i(t) =<Iejωt, I being the phasor describing the com-plex current . They are located at

~re = ±xe~ex.

The angular frequency ω is considered to be lowenough so that displacement currents can be ne-glected: ~J ¿ ∂

∂t~D, where ~J is the current density

inside the bar and ~D the electric flux.

The z-directed dimensions of both bar and loop aresupposed to be infinite.

2. DIFFERENTIAL EQUATIONS FORTHE VECTOR POTENTIAL

As all exciting currents are oscillating at one single

PSfrag replacements

x, x0

y, y0

−I0 +I0

xc, x0c

yc, y0c

a

b

conducting bar

~rq

Figure 1: Screening bar and exciting loop

frequency, all fields will show the same time de-pendency and can be expressed by their complexphasors.

Introducing a complex vector potential ~A as ~B =curl ~A, where ~B denotes the magnetic flux, and in-vestigating its properties, we find that an additionalterm has to be added to the vector potential if weconsider points inside the bar. The additional termcan be expressed as the time domain integral ofthe gradient of a complex electrical scalar potentialϕi(z) by

w(gradϕi)dt = C~ez,

where C denotes an unknown complex constant.The vector potential ~Ai inside the bar can be re-defined by using the Buchholz convention

~A∗i = ~Ai +w(gradϕi)dt = ~Ai + C,

from which the fields can be computed with

~Bi = curl ~A∗i and ~Ei = −jω ~A∗i .

From Maxwell’s equations we find that the modi-fied vector potential inside the bar is governed by

Fig. 1. Screening bar and exciting loop.

3 Derivation of Boundary Integral Equation (BIE)

In Green’s second theorem∫∫τ

∫(8 19−9 18)dτ=

= vF=∂τ

(8grad9 − 9grad8) · nodF,

we insert the single component of the vector potentialAa as9 and a kernel functionK satisfying1K=−δ3 as8. As aresult, we get an integral representation valid forAa:

Aa = − vF=∂τ

(Aa gradK − KgradAa) · nodF+

+µaI

+∞∫−∞

K

∣∣∣xo=±xe

yo=0

dzo.

The integrand of the surface integral at the right hand sideof the above equation does not depend onz. Hence thez-directed integration is only affecting the kernel functionK,for which the elementary Kernel function

K =1

4πr,

is inserted. Thereby we can define a new kernel functionG20as

G20=

+∞∫−∞

Kdzo=1

4π

+∞∫−∞

dzo√(x−xo)2+ (y−yo)2+ (z−zo)2

= −1

2πln

(√(x − xo)2 + (y − yo)2

ρo

)= −

1

2πln

(ρ

ρo

),

with ρ=

√(x−xo)2+(y−yo)2 andρo=const. While we are

aware that the above integral is divergent, it can be solved byperforming a suitable renormalization.

The functionG20 satisfies1G20=−δ2, as is shown inEhrich et al.(2000).

Helmholtz’ equation, which for complex fields takesthe form

∆ ~A∗i − jωµiκi~A∗i = 0. (1)

Outside the bar the equation

∆ ~Aa = −µa~Je, (2)

is valid for the vector potential ~Aa. The excitingcurrent density ~Je can be expressed as

~Je = δ1(x− xe)− δ1(x+ xe)δ1(y) · I~ez.

The symbol δ1(x−xo) denotes the one-dimensionalDirac function; accordingly, δ2 and δ3 are the two-and three-dimensional Dirac functions.

Since we allow only z-directed exciting currents, allvector potentials will also be exclusively z-directed.They can be described by their z-components Aazand A∗iz, respectively. The boundary conditions forAaz and A∗iz are

Aaz = A∗iz − C ∧1

µa

~no · gradAaz =1

µi

~no · gradA∗

iz,

where ~no is a unit vector normal to the boundary.

3. DERIVATION OF BOUNDARY INTE-GRAL EQUATION (BIE)

In Green’s second theorem

ww

τ

w(Φ∆Ψ−Ψ∆Φ)dτ =

=

F=∂τ

(Φ gradΨ−ΨgradΦ) · ~no dF,

we insert the single component of the vector po-tential Aa as Ψ and a kernel function K satisfying∆K = −δ3 as Φ. As a result, we get an integralrepresentation valid for Aa:

Aa = −

F=∂τ

(Aa gradK −K gradAa) · ~no dF+

+ µaI

+∞w

−∞

K∣

∣

∣

xo=±xe

yo=0

dzo.

The integrand of the surface integral at the righthand side of the above equation does not depend onz. Hence the z-directed integration is only affectingthe kernel function K, for which the elementaryKernel function

K =1

4πr,

PSfrag replacements

xc, x0c

~n1 = ~ey ~n2 = −~ex

~n4 = ~ex ~n3 = −~ey

yc, y0c

Ω

a

b

xq

yq

Figure 2: Normals on boundary scetions

is inserted. Thereby we can define a new kernelfunction G20 as

G20=

+∞w

−∞

Kdzo=1

4π

+∞w

−∞

dzo√

(x−xo)2+ (y−yo)2+ (z−zo)2

= −1

2πln(

√

(x− xo)2 + (y − yo)2

ρo) = −

1

2πln(

ρ

ρo),

with ρ =√

(x− xo)2 + (y − yo)2 and ρo = const.While we are aware that the above integral is di-vergent, it can be solved by performing a suitablerenormalization.

The function G20 satisfies ∆G20 = −δ2, as is shownin [3].

If we use the new kernel and define the influence ofthe exciting loop as

Ae := µaIG20|xo=±xe

yo=0

,

we can write the integral representation for Aa as

Aa = −z

C=∂Ω

(Aa gradG20−G20 gradAa) ·~no ds+Ae.

We use the cross-section of the bar as integrationdomain Ω with its facet normals defined as shownin Fig. (2).

By taking into account the boundary conditions dis-cussed above, we get an integral representation forthe vector potential outside the bar:

Aa = −z

C=∂Ω

[(Wi − C) gradG20−

−G20µa

µi

gradWi] · ~no ds+Ae.

(3)

Fig. 2. Normals on boundary scetions.

If we use the new kernel and define the influence of theexciting loop as

Ae := µaIG20|xo=±xe

yo=0

,

we can write the integral representation forAa as

Aa = −

∮C=∂

(Aa gradG20 − G20 gradAa) · no ds + Ae.

We use the cross-section of the bar as integration domain

with its facet normals defined as shown in Fig. (2).By taking into account the boundary conditions discussed

above, we get an integral representation for the vector poten-tial outside the bar:

Aa =

−

∮C=∂

[(A∗

i −C) gradG20−G20µa

µi

gradA∗

i ] · no ds+Ae. (3)

It can be shown that∮C=∂

(C gradG20) · no ds = 0,

so we can omit this term in Eq. (3).In a last step we take the normal derivative ofAa on each

of the four contour sections of and move the observationpoint to the very same section, i.e.

(1) y = yq , (3) y = yq + b,(2) x = xq + a, (4) x = xq .

As a result, we get four independent integral equationsµa

µi

no · gradA∗

i = no· grad (4)∮

C=∂

(A∗

i gradG20−G20µa

µi

gradA∗

i ) no ds+Ae

∣∣∣x=xq∨x=xq+a,y

x,y=yq∨y=yq+b

Conventionally, one would derive another boundary inte-gral equation, this one valid forA∗

i , using

G20i=1

4jH1

0(kρ

ρ0) with k2

=jωκµi

L. O. Fichte et al.: Shielding properties of a conducting bar calculated with a boundary integral method 121

PSfrag replacements

x/m

y/m

−0.1 +0.1

−0.1

+0.1

~B

Figure 3: Quadratic bar in homogenous field

Figure 4: Abs(A∗i ) for bar in homogenous field

As an additional verification, different values of ω, κand µri =

µi

µ0

are considered. We calculated the av-erage of the absolute of A∗i over the cross-section ofthe bar Ω as functions of . By this we can estimatewhich amount of the outside field permeates intothe conducting area. The average of A∗i has beennormalized on the value A0 = 1 · 10−6 V s

Ain Figs.

(5) and (6) and on A1 = 1 · 10−7 V sA

in Fig. (7).

4.2 SINGLE EXCITING LINE

In a next step, one conducting bar of quadraticcross-section with a = b = 0, 1m is placed at~rq = (−0.1,−0.1)m. One single exciting line, carry-ing I = 100 A, is placed at (0.0,−0.15) m, as shownin Figure (8).

Figure (9) displays the resulting lines of the mag-netic field for ω = 50 1

sand κ = 57 · 106 A

Vm, µri = 1;

the field attenuation ∆Aa (and thereby the shield-

PSfrag replacements

A/A

0

ω/(2π) / 1

s

Figure 5: A/A0 as a function of ω, κ, µ const.

PSfrag replacements

A/A

0

κ / 106 A

Vm

Figure 6: A/A0 as a function of κ, ω, µ const.

PSfrag replacements

A/A

1

µri

Figure 7: A/A0 as a function of µi, ω, κ const.

Fig. 3. Quadratic bar in homogenous field.

PSfrag replacements

x/m

y/m

−0.1 +0.1

−0.1

+0.1

~B




µi

µ0


Ain Figs.

(5) and (6) and on A1 = 1 · 10−7 V sA

in Fig. (7).




sand κ = 57 · 106 A

Vm, µri = 1;


PSfrag replacements

A/A

0

ω/(2π) / 1

s


PSfrag replacements

A/A

0

κ / 106 A

Vm


PSfrag replacements

A/A

1

µri


Fig. 4. Abs(A∗i ) for bar in homogenous field.

as a kernel functionHanson and Yakovlev(2002). The func-tion H1

0 is the Hankel function of the first kind and orderzero. The resulting system of two integral equations couldbe solved to yield the solutions for the two unknown func-tionsAa andA∗

i .

Here, we find a solution to Eq. (1) by separation, using aproduct of functions depending only on one coordinate forthe descrition ofA∗

i .

A∗

i (xc, yc) =

∞∑n=1

[v1ncosh(βn(b−yc)) + v3n cosh(βnyc)

]cos(αnxc)

+[v2n cosh(βn(a−xc)) + v4n cosh(βn xc)

]cos(αnyc),

αn =nπ

a, βn =

√α2

n + jωκµ,

αn =nπ

b, βn =

√α2

n + jωκµ. (5)

Then, the unknown constantsvin, i=1, ..., 4 will have tobe determined.

We can now insert the result forA∗

i into Eq. (4). Using the

PSfrag replacements

x/m

y/m

−0.1 +0.1

−0.1

+0.1

~B




µi

µ0


Ain Figs.

(5) and (6) and on A1 = 1 · 10−7 V sA

in Fig. (7).




sand κ = 57 · 106 A

Vm, µri = 1;


PSfrag replacements

A/A

0

ω/(2π) / 1

s


PSfrag replacements

A/A

0

κ / 106 A

Vm


PSfrag replacements

A/A

1

µri


Fig. 5. A/A0 as a function ofω, κ, µ const.

PSfrag replacements

x/m

y/m

−0.1 +0.1

−0.1

+0.1

~B




µi

µ0


Ain Figs.

(5) and (6) and on A1 = 1 · 10−7 V sA

in Fig. (7).




sand κ = 57 · 106 A

Vm, µri = 1;


PSfrag replacements

A/A

0

ω/(2π) / 1

s


PSfrag replacements

A/A

0

κ / 106 A

Vm


PSfrag replacements

A/A

1

µri


Fig. 6. A/A0 as a function ofκ, ω, µ const.

orthogonality of the cosine functions

2π∫0

cos(nξ) cos(pξ)dξ =

0 n 6= p

π n = p 6= 02π n = p = 0,

we can isolate one coefficientvim on the left hand side ofEq. (4). Expanding the right hand side into a series of cosinefunctions by multiplication with cos(αmxc) (or cos(αmyc),respectivly) and computing its integral along one of the con-tour sections we get

µa

µi

vimdim =

4∑i=1

∞∑n=1

[γinmvin] + bm, m ∈ N.

If we take only the firstN series elements into account,this equation can be written as a matrix equation

0v = b, (6)

with matrix 0 and vectorsv, b of finite dimensions. Equa-tion (6) represents a system of linear equations from whichthe 4N unknown coefficientsvin can be computed.

The values of the coefficientsγinm, dim andbm, needed forthe solution of the matrix equation, can be obtained analyti-cally (seeFichte et al., 2004).

122 L. O. Fichte et al.: Shielding properties of a conducting bar calculated with a boundary integral method

PSfrag replacements

x/m

y/m

−0.1 +0.1

−0.1

+0.1

~B




µi

µ0


Ain Figs.

(5) and (6) and on A1 = 1 · 10−7 V sA

in Fig. (7).




sand κ = 57 · 106 A

Vm, µri = 1;


PSfrag replacements

A/A

0

ω/(2π) / 1

s


PSfrag replacements

A/A

0

κ / 106 A

Vm


PSfrag replacements

A/A

1

µri

Figure 7: A/A0 as a function of µi, ω, κ const.Fig. 7. A/A0 as a function ofµ, ω, κ const.

PSfrag replacements

x/ m

y/ m

0.1 +0.3

0.1

0.3

0.05

Figure 8: Quadratic bar and single line

Figure 9: Magnetic field at 50 Hz

ing effect of the conducting bar) is plotted in Figure(10). The resulting plots are slightly non-symmetricbecause of roundoff-errors.

Figs. (11) and (12) show the lines and the attenua-tion of the magnetic field for the same configurationand a significantly lower frequency of ω = 5 1

s. Con-

ductivity and permeability are κ = 57·106 A

Vm, µri =

1.

5. CONCLUSIONS

We applied a hybrid method to determine the fieldproperties of a rectangular conducting bar in thefield of an exciting loop. While Helmholtz’ equationgoverning the vector potential inside the bar hasbeen solved by separation, we obtained an integral

Figure 10: Field attenuation at 50 Hz


Fig. 8. Quadratic bar and single line.

Once the values of these coefficients are known, they canbe used to express the vector potential outside the shieldingbar as a series of known functionsAk(x, y), k=1, ..., 4:

Aa(x, y) =

M∑n=1

v1nA1(x, y) + v2nA2(x, y)+

+v3nA3(x, y) + v4nA4(x, y).

This solution for the vector potentialAa is used to deter-mine the shielding properties of the bar by

1A =Aa

Ae.

4 Numerical results

4.1 Verification of method

The applied method involves rarely used special functions,i.e. the exponential integral function for complex arguments,E1(z) (seeAbramowitz and Stegun(1970), pp. 227 for de-tails), which is not included in commericial numerical soft-ware packages. A focus of our paper is on the validation ofthe model. Therefore, in a first step, we use the presentedmethod on a conducting bar which has been placed into an

PSfrag replacements

x/ m

y/ m

0.1 +0.3

0.1

0.3

0.05





s. Con-


Vm, µri =

1.

5. CONCLUSIONS




Fig. 9. Magnetic field at 50 Hz.

PSfrag replacements

x/ m

y/ m

0.1 +0.3

0.1

0.3

0.05





s. Con-


Vm, µri =

1.

5. CONCLUSIONS





homogenous magnetic field – see Fig. (3 – and compare theresults with the expected behaviour.

The following Fig.4 shows the absolute value of the mod-ified vector potential inside the plate.

As an additional verification, different values ofω, κ andµri=

µiµ0

are considered. We calculated the average of the ab-solute ofA∗

i over the cross-section of the bar as functionsof . By this we can estimate which amount of the outsidefield permeates into the conducting area. The average ofA∗

ihas been normalized on the valueA0=1·10−6 V s

Ain Figs.5

and6 and onA1=1·10−7 V sA

in Fig. 7.

4.2 Single exciting line

In a next step, one conducting bar of quadratic cross-sectionwith a=b=0, 1 m is placed atrq=(−0.1, −0.1) m. Onesingle exciting line, carryingI=100 A, is placed at (0.0,−0.15) m, as shown in Fig.8.

Figure9 displays the resulting lines of the magnetic fieldfor ω=501

s and κ=57·106 AVm , µri=1; the field attenuation

L. O. Fichte et al.: Shielding properties of a conducting bar calculated with a boundary integral method 123

PSfrag replacements

x/ m

y/ m

0.1 +0.3

0.1

0.3

0.05





s. Con-


Vm, µri =

1.

5. CONCLUSIONS






representation for the vector potential in the exte-rior space from Green’s second theorem. Couplingthese two equations for points on the contour of thebar resulted in a system of linear equations. Thesolution to this system delivered the coefficients de-scribing all fields. This method has been used tocalculate the shielding properties of a conductingbar placed in the field of a exciting loop.

References

[1] Abramowitz, M.; Stegun, I.A.: Handbook ofMathematical Functions, New York 1970

[2] Hanson, G.W., Yakovlev, A.B.; Operator The-ory for Electromagnetics, Springer, New York2002

[3] Ehrich, M.; Fichte, L.O.; Luer, M.: Contri-bution to Boundary Integrals by the Singu-larity of Kernels satisfying Helmholtz’ Equa-tion CJMW’2000 China-Japan Joint Meetingon Microwaves, Nanjing

[4] Fichte, L.O.; Ehrich, M.; Kurz, S.: An Ana-lytical Solution to the Eddy Current Problemof a Conducting Bar EMC 2004 InternationalSymposium on Electromagnetic Compatibility,Sendai Conference Proceedings CD-ROM

Fig. 12. Field attenuation at 5 Hz.

1Aa (and thereby the shielding effect of the conducting bar)is plotted in Fig.10. The resulting plots are slightly non-symmetric because of roundoff-errors.

Figures11and12show the lines and the attenuation of themagnetic field for the same configuration and a significantlylower frequency ofω=51

s. Conductivity and permeabilityareκ=57·106 A

Vm , µri = 1.

5 Conclusions

We applied a hybrid method to determine the field proper-ties of a rectangular conducting bar in the field of an excitingloop. While Helmholtz’ equation governing the vector po-tential inside the bar has been solved by separation, we ob-tained an integral representation for the vector potential inthe exterior space from Green’s second theorem. Couplingthese two equations for points on the contour of the bar re-sulted in a system of linear equations. The solution to thissystem delivered the coefficients describing all fields. Thismethod has been used to calculate the shielding properties ofa conducting bar placed in the field of a exciting loop.

References

Abramowitz, M. and Stegun, I. A.: Handbook of MathematicalFunctions, New York 1970.

Ehrich, M., Fichte, L. O. and Luer, M.: Contribution to Bound-ary Integrals by the Singularity of Kernels satisfying Helmholtz’Equation CJMW’2000 China-Japan Joint Meeting on Mi-crowaves, Nanjing, 2000.

Fichte, L. O., Ehrich, M., and Kurz, S.: An Analytical Solution tothe Eddy Current Problem of a Conducting Bar EMC 2004 Inter-national Symposium on Electromagnetic Compatibility, SendaiConference Proceedings CD-ROM, 2004.

Hanson, G. W. and Yakovlev, A. B.: Operator Theory for Electro-magnetics, Springer, New York 2002.

Radio Science Advances in Shielding properties of a ... · Professur fur Theoretische Elektrotechnik und Numerische Feldberechnung, Helmut-Schmidt-Universit¨ at, Universit¨ ¨at

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