Analytical and quasi-analytical solutions of direct problems in eddy current testing

RIGA TECHNICAL UNIVERSITY

VALENTINA KOLISKINA

Analytical and quasi-analytical solutions of direct problems in eddy current testing Doctoral Thesis

In Partial Fulfilment of the Requirements of the Doctor Degree in Mathematics Subdiscipline of Mathematical modelling Supervisor: Prof., Dr. math. INTA VOLODKO

RIGA, 2013      

This work has been supported by the European Social Fund within the project „Support for implementation of doctoral studies at Riga Technical University”.

© Riga Technical university, 2013 © Valentina Koliskina, 2013

 2   

ABSTRACT The thesis consists of two parts. Analytical solutions of direct problems in eddy current testing are constructed in the first part of the thesis. It is assumed that the properties of a conducting medium are functions of one spatial coordinate. Solutions are found for the following cases: (a) a planar conducting multilayer medium where the magnetic permeability and electrical conductivity are exponential functions of the vertical coordinate; (b) a cylindrical conducting multilayer medium where the magnetic peremability and electrical conductivity are power functions of the radial coordinate; (c) a double conductor line above a multilayer medium with varying properties for the cases of alternating current and step current; (d) a moving planar twolayer medium with varying properties. All solutions are found by the method of integral transforms where the corresponding system of ordinary differential equations is solved in terms of different special functions. Quazi-analytical solutions are constructed in the second part of the thesis for the case of a conducting medium of finite size. It is assumed that all properties of the conducting medium are constant. Three cases of axisymmetric problems are considered: (a) a coil with alternating current above a conducting cylinder of finite size; (b) a coil with current above a conducting plate with bottom cylindrical hole; (c) a coil with current above a conducting half-space with a flaw in the form of a cylinder of finite size. The problems are solved by the method of separation of variables. The solution procedure includes two steps where numerical methods are used: (1) calculation of complex eigenvalues without good initial guess for the root and (2) solution of a system of linear algebraic equations. Results of numerical calculations are presented. Good comparison between theoretical calculations and experimental data is found. Keywords: eddy currents, integral transform, special functions, variable conductivity and permeability, flaw detection, eigenvalues

 3   

ANOTĀCIJA Promocijas darbs sastāv no divām daļām. Darba pirmajā daļā konstruēti virpuļstrāvu nesagraujošās kontroles tiešo problēmu analītiskie atrisinājumi. Tiek pieņemts, ka vadošas vides īpašības ir funkcijas, atkarīgas no vienas telpas koordinātas. Atrisinājums atrasts šādiem gadījumiem: (a) plakanai vadošai daudzslāņu videi, kuras magnētiskā caurlaidība un elektriskā vadāmība ir eksponentfunkcijas no vertikālās koordinātas; (b) cilindriskai vadošai daudzslāņu videi, kuras magnētiskā caurlaidība un elektriskā vadāmība ir radiālās koordinātas pakāpes funkcijas; (c) divvadu līnijai virs daudzslāņu vadītāja ar mainīgām īpašībām gadījumos, kad divvadu līnijā plūst maiņstrāva un lēcienveida strāva; (d) plakanai divslāņu vadošai videi ar mainīgām īpašībām, kad vide atrodas kustībā. Visi atrisinājumi atrasti pēc integrālo transformāciju metodes, kurā atbilstošo parasto diferenciālvienādojumu atrisinājumi izteikti ar dažādām speciālajām funkcijām. Darba otrajā daļā konstruēti kvazianalītiskie atrisinājumi gadījumiem, kad vadošai videi ir galīgi izmēri. Tiek uzskatīts, ka visas vadošās vides īpašības ir konstantas. Aplūkotas trīs problēmas: (a) vijums ar maiņstrāvu virs vadoša galīga izmēra cilindra; (b) vijums ar strāvu virs vadošas plāksnes ar cilindriska veida dobumu; (c) vijums ar strāvu virs vadošas pustelpas ar defektu galīga cilindra formā. Visos trijos gadījumos vijuma ass sakrīt ar atbilstošā cilindriskā ķermeņa asi. Minētās problēmas atrisinātas, izmantojot mainīgo atdalīšanas metodi. Risināšanas procedūra satur divus soļus, kuros izmantotas skaitliskās metodes: (1) komplekso īpašvērtību noteikšana, kad nav zināmi labi sākuma sakņu tuvinājumi; (2) lineāras algebrisko vienādojumu sistēmas atrisināšana. Darbā veikti skaitliski aprēķini. Ir novērota laba teorētisko aprēķinu un eksperimentālo datu sakritība. Atslēgas vārdi: virpuļstrāvas, integrālās transformācijas, speciālās funkcijas, mainīga magnētiskā caurlaidība un elektriskā vadāmība, defekta atrašana, īpašvērtības.

 4   

CONTENTS INTRODUCTION ..........................................................................................................................9 The structure of the thesis ..........................................................................................................9 Importance of the subject .........................................................................................................11 The objectives of the thesis ......................................................................................................11 Research methodology..............................................................................................................12 Scientific novelty and main results...........................................................................................13 Applications .............................................................................................................................14 Publications .............................................................................................................................14 Presentations at international conferences................................................................................16 Presentations at local conferences ...........................................................................................17 1. BASIC EQUATIONS ............................................................................................................18 1.1 Medium with constant properties .....................................................................................18 1.2 Medium with variable properties.......................................................................................20 1.2.1. Multilayer planar medium with varying electrical conductivity and magnetic permeability ............................................................................................................... 20 1.2.2. Multilayer cylindrical problems with varying electrical conductivity and magnetic permeability. ............................................................................................................... 22 1.2.3. Multilayer medium with varying electrical conductivity and magnetic permeability in Cartesian coordinates .................................................................................................. 23

1.3 Literature survey ................................................................................................................24 2. PLANAR MULTILAYER MEDIA WITH VARYING PROPERTIES ...........................27 2.1 Introduction .......................................................................................................................27 2.2 A coil above a multilayer medium with varying propertiess.............................................28 2.3 A single-turn coil above a multilayer medium with varying properties ............................28 2.4 A single-turn coil above a conducting half-space with varying properties ....................... 35 2.5 Coil of finite dimensions  ..................................................................................................38 2.6 A single-turn coil above a two-layer medium with varying properties .............................40 3. CYLINDRICAL PROBLEMS FOR MULTILAYER MEDIA WITH VARYING PROPERTIES..........................................................................................................................46 3.1 Introduction .......................................................................................................................46 3.2 A coil inside a multilayer medium with varying properties .............................................47 3.3 A coil inside a cylindrical region ......................................................................................53 3.4 A coil inside a two-layer tube ...........................................................................................56 3.5 A coil outside a multilayer tube ........................................................................................62 3.6 Other analytical solutions .................................................................................................67 3.7 Solution in terms of confluent hypergeometric function ...................................................68 4. DOUBLE CONDUCTOR LINE ABOVE A PLANAR MEDIUM WITH VARYING PROPERTIES ..........................................................................................................................75 4.1 Double conductor line above a multilayer medium with varying electrical and magnetic  5   

properties .........................................................................................................................75 4.2 Double conductor line above a conducting half-space with varying electrical and magnetic properties ..........................................................................................................78 4.3 Double conductor line above a conducting two-layer medium with varying electrical and magnetic properties ..........................................................................................................82 4.4 Pulsed eddy currents in a conducting half-space ............................................................. 87 5. MOVING PLANAR MEDIUM WITH VARYING PROPERTIES.................................. 94 5.1 Moving halfspace with varying electrical conductivity and magnetic permeability ........94 5.2 Moving two-layer medium with varying electrical conductivity and magnetic permeability .................................................................................................................... 102 6. AXISYMMETRIC PROBLEMS FOR MEDIA OF FINITE SIZE .................................. 111 6.1 A single-turn coil above a conducting cylinder of finite size .......................................... 111 6.2 A coil of finite dimensions above a conducting cylinder of finite size .......................... 121 6.3 Computational aspects ..................................................................................................... 125 6.4 A coil above a conducting plate with a flaw in the form of a circular cylinder ..............128 6.5 A coil of finite dimensions above a conducting plate with a flaw in the form of a circular cylinder .......................................................................................................... `137 6.6 A coil above a conducting half-space with a flaw in the form of a circular cylinder ..... 140 6.7 A coil of finite dimensions above a conducting half-space with a flaw in the form of a circular cylinder ............................................................................................................ 149

CONCLUSIONS .............................................................................................. 153 REFERENCES ................................................................................................. 156 APPENDIX ....................................................................................................... 161  

   

 6   

NOMENCLATURE  

List of Latin symbols

r A

amplitude of vector potential

~ A

~ r vector potential, A = Ae jωt

Ai

nonzero component of vector potential in region Ri

A0ind

induced vector potential intensity in free space

r B

amplitude of magnetic induction vector

~ B

~ r magnetic induction vector, B = Be jωt

r D

amplitude of electric induction vector

~ D

~ r electric induction vector, D = De jωt

r E

amplitude of electric field vector

~ E

~ r electric field vector, E = Ee jωt

h

height of a single-turn coil above conducting medium

h1

height og a coil above conducting medium

h2 − h1 height of a coil r H

amplitude of magnetic field vector

~ H

~ r magnetic field vector, H = He jωt

r I

amplitude current vector

r Ie

amplitude external current vector

~ I

~ r current vector, I = I e jωt

Iν (s )

modified Bessel function of the first kind of order ν

j

imaginary unit, j = − 1

 7   

Jν ( s )

Bessel function of the first kind of order ν

Kν ( s )

modified Bessel function of the second kind of order ν

r0

radius of a single-turn coil

r1

inner radius of a coil

r2

outer radius of a coil

Yν ( s )

Bessel function of the second kind of order ν

Z

dimensionless induced change in impedance

Z ind

induced change in impedance

List of Greek symbols

ε0

electric constant

ε

relative permittivity

µ

relative magnetic permeability

µ0

magnetic constant

ρ~

charge density

σ

conductivity

ψ

scalar electric potential intensity

ψ~

electric potential, ψ~ = ψe jωt

ω

frequency

 8   

INTRODUCTION The structure of the thesis The main objective of the PhD thesis is to develop mathematical models which can be used to describe nondestructive eddy current testing methods. The thesis consists of two parts. Analytical solutions of direct problems in eddy current testing are constructed in the first part of the thesis (Chapters 2 – 5). It is assumed that the properties of a conducting medium are functions of one spatial coordinate. Solutions are found for the following cases: (a) a planar conducting multilayer medium where the magnetic permeability and electrical conductivity are exponential functions of the vertical coordinate; (b) a cylindrical conducting multilayer medium where the magnetic peremability and electrical conductivity are power functions of the radial coordinate; (c) a double conductor line above a multilayer medium with exponentially varying properties for the cases of alternating current and step current; (d) a moving planar two-layer medium with varying properties. All solutions are found by the method of integral transforms where the corresponding system of ordinary differential equations is solved in terms of different special functions. Chapter 1 (Introduction) is devoted to the derivation of the basic equations used in the study. In addition, literature survey is also presented. In Chapter 2 we construct analytical solutions of eddy current testing problems for the case of a planar conducting multilayer medium. The magnetic permeability and electrical conductivity of the layers are assumed to vary with respect to the vertical coordinate. Such a situation takes place in practice during special treatment of ferromagnetic metals (for example, surface hardening). Experimental data indicate that a relatively thin layer of reduced magnetic permeability is formed as a result of surface hardening. Analysis of experimental data shows that the magnetic permeability can be reasonably well approximated by an exponential function of the vertical coordinate. It is shown in Chapter 2 that if the magnetic permeability and electrical conductivity are exponential fucntions of the vertical coordinate then the solution of the system of the Maxwell’s equations can be found in closed from in terms of improper integrals contating Bessel functions of complex argument. The solution is obtained for the case of a multilayer medium. Two special cases (conducting half-space and two-layer medium) are considered in detail. The change in impedance is computed using software package „Mathematica”. Chapter 3 is devoted to the analysis of a similar problem in cylindrical geometry where a coil with alternating current is located inside (or outside) a multilayer tube. It is assumed that the axis of the coil coincides with the axis of the tube. The electical conductivity and magnetic permeability of one or several conducting layers are power functions of the radial coordinate. It is shown in Chapter 3 that different analytical solutions can be constructed depending on the choice of the constants characterizing the power functions. Analytical solutions are derived in terms of Bessel funnctions and confluent hypergeometric functions. Results of numerical calculations are presented (calculations are done with „Mathematica”). Chapter 4 is devoted to the solution of an eddy current problem for a planar multilayer medium with exponentially varying electrical conductivity and magnetic permeability for the case where an excitation coil is modeled by means of a double conductor line (the coil is formed  9   

by two infinitely long parallel wires). The double conductor line is a sufficiently accurate model for the case where a rectangular frame with current is located above a multilayer medium under the assumption that the ratio of the sides of the frame is 1:4 or smaller. In addition, one case where an excitation current in the coil is not alternating current but is in the form of a pulse is considered in detail. The corresponding boundary value problem is solved by the method of the Laplace and Fourier transforms. The inverse Laplase transform is found in closed form for the case where a double conductor line is located above a conducting half-space with constant electrical conductivity and magnetic permeability. In Chapter 5 the results obtained in Chapter 2 are generalized for the case where an upper conducting layer is moving with constant velocity V in a horizontal direction. The problem is solved by means of the double Fourier transform in the two horizontal directions. The corresponding system of ordinary differential equations is solved analytically in terms of Bessel functions of complex agrument. The change in impedance is found in closed form in terms of double integrals. Results of numerical calculations with „Mathematica” are presented. Solutions obtained in Chapters 2 – 5 correspond to the case where a conducting medium is assumed to be infinite in one or two spatial directions. Integral transofms such as Hankel or Fourier integral transforms are used to solve the corresponding problems. Such an approach can be used in practice if a coil is sufficiently smaller than a conducting medium. However, the finite size of a conducting medium has to be taken into account if the size of the coil is comparable to the size of the conducting medium. Chapter 6 is devoted to the analysis of axisymmetric problems where a coil with alternating current is located above a conducting medium of finite size. Three cases which are important for applications are considered in detail: (a) a coil above a conducting cylinder of finite size located in free space (this model is relevant for the analysis of eddy current problem in coin validators), (b) a coil above an infinite conducting plate with a bottom cylindical hole coaxial with the axis of the coil (this problem can be used to estimate and model the effect of corrosion) and (c) a coil above a half-space with a coaxial conducting cylinder of finite size (this model is a sufficiently accurate model of spot welding where the cylinder represents a cast core which is formed in the conducting medium during welding process). Mathematical solution is based on the TREE methodology (TREE stands for „TRuncated Eigenfuntion Expansions”, see [60]). The main assumption in the TREE method is that the vector potential is exactly zero at a sufficiently large distance from the coils’ center (usual boundary conditions at infinity imply that the vector potential approaches zero as the geometrical coordinate tends to infinity). Using the TREE method quasi-analytical solution for all three problems mentioned above is constructed. We use the term „quasi-analytical solution” since the solution is found by the method of separation of variables, but there are two stages of the solution process where numerical methods should be used: (a) calculation of complex eigenvalues and (b) solution of the system of linear equations in order to determine integration constants. A computer program is developed on the basis of the algorithm described in [43] in order to compute complex eigenvalues without good initial guesses for the roots. The formula for the change in impedance in all three cases is derived. Results of numerical computations of the change in impedance using „Mathematica” are presented. Computational values of the change in impedance are compared with experimental  10   

data for the case where a coil with alternating current is located above a conducting cylinder of finite size. Good agreement between theoretical and experimental values of the change in impedance is found.

Importance of the subject Eddy current method is widely used in practice in order to control properties of electically conducting materials. Applications include determination of electrical conducticvity or magnetic permeability of conducting materials, estimation of thickness of metal coatings and analysis of the properties of coatings, detection of defects (such as voids or cracks) in a conducting medium. The solution of an inverse problem is usually required in order to solve eddy current testing problem. At this stage the difference (in some norm) between theoretical model and experimental data is minimized in order to estimate unknown parameters of the model (such as electrical conductivity of a conducting plate or parameters of the defect). Thus, in order to solve the inverse problem one has to have a convenient and reliable mathematical model for the solution of the direct problem. Analytical solutions of eddy current testing problems for the case where the properties of the conducting medium (electrical conductivity and magnetic permeability) are constant are wellknown in the literature. In addition, the conducting medium is assumed to be infinite in one or two spatial directions. In the thesis we generalize the approach developed for media with constant properties to the case where the properties of the medium vary with respect to one spatial coordinate (vertical coordinate for the case of a planar multilayer medium and radial coordinate for the case of a cylindrical multilayer tube). Analytical solutions are constructed in the thesis for the cases where the electrical conductivity and magnetic permeability of the medium are exponential functions of the vertical coordinate and power functions of the radial coordinate for planar and cylindrical cases, respectively. The solutions are found for different types of coils (circular coils and coils in the form of a double conductor line). We also analyze axisymmetric cases where a circular coil is coaxial with cylindrical region of finite size. The solutions are found by the method of separation of variables where complex eigenvalues are determined by an efficient algorithm without prior knowledge of the initial guesses for the roots. Thus, the constructed solutions are „quasi-analytical”. The obtained solutions can be used in practice to develop algorithms for coin validators, to estimate the effect of corrosion in metal plates and to assess quality of spot welding.

The objectives of the thesis All the suggested methods for eddy current testing problems analyzed in the thesis can be divided into two parts: (a) analytical solutions for planar and cylindrical multilayer medium under the assumption that the magnetic permeability and electrical conductivity of the medium are functions of one spatial coordinate and the medium is infinite in one or two spatial directions, and (b) the magnetic permeability and electrical conductivity of the conducting medium are constant but the medium is of finite size (axisymmetrical problems where the axis of the coil coincides with the axis of a cylindrical region of finite size).

 11   

The objectives of the work are as follows: (1) Obtain analytical solutions for eddy current problems where the excitation coil is a circular coil with alternating current and the properties of planar multilayer conducting medium (electrical conductivity and magnetic peremability) are exponential functions of the vertical coordinate; (2) Construct analytical solutions for the case of a planar multilayer medium with exponentially varying properties under the assumption that the coil is modeled by a double conductor line consisting of two parallel infinite wires; (3) Obtain analytical solutions for the cases where a circular coil with alternating current is located inside (or outside) a multilayer conducting tube. The coil is coaxial with the tube. In addition, the properties of conducting layers (the electrial conductivity and magnetic permeability) are power functions of the radial coordinate; (4) Develop analytical solution for the case where a double conductor line is located above a conducting half-space with varying electrical conductivity and magnetic permeability. The excitation current in the double conductor line is in the form of a pulse. Pulsed eddy current represent a convenient alternative to traditional eddy current method based on alternating excitation current; (5) Obtain analytical solutions for eddy current problem where a coil with current is located above a moving medium. Two cases of the moving medium are considered: (a) a moving half-space and (b) a moving two-layer medium where the upper layer is moving but the lower layer is fixed. The electrical conductivity and magnetic permeability of the moving medium are exponetial functions of the vertical coordinate; (6) Construct quasi-analytical solutions for axisymmetric problems where a circular coil is coaxial with cylindrical region of finite size. Three axisymmetric problems are solved in the thesis: (a) a coil with alternating current above a conducting cylinder of finite size, (b) a coil with alternating current above a conducting plate with bottom hole in the form of a cylinder coaxial with the coil and (c) a coil with alternating current above a conducting half-space with a flaw in the form of a cylinder coaxial with the coil.

Research methodology Methods of integral transforms such as Hankel, Fourier and Laplace transforms are used in Chapters 2 -5 of the thesis in order to construct analytical solutions of eddy current testing problems in planar and cylindrical cases where the medium is assumed to be infinite in one or two spatial dimensions. The electrical conductivity and magnetic permeability of the medium are functions of one spatial coordinate (exponential functions of the vertical coordinate in the case of a planar medium and power functions of the radial coordinate in the case of a cylindrical medium). The solution of the corresponding ordinary differential equations in the transformed space is found in closed form by means of the Bessel functions and confluent hypergeometric functions. The results are presented in the form of the change in impedance of the coil (for the case of a single-turn coil and coil of finite dimensions). In addition, analytical solutions are also constructed for the case of a moving planar medium under the assumption that the magnetic

 12   

permeability and electrical conductivity are exponential functions of the vertical coordinate. Calculations are performed with “Mathematica”. Method of separation of variables is used in Chapter 6 in order to solve axisymmetric problems where a coil is located above a conducting medium which contains a cylinder of finite size coaxial with the coil. A fast and reliable algorithm for the determination of complex eigenvalues is implemented in the form of the “Mathematica” code. Note that no initial guess for a complex eigenvalue is required. The change in impedance is calculated in the form of a truncated series containing Bessel functions.

Scientific novelty and main results The following new results are obtained in the thesis. (1) Analytical solution is obtained for eddy current problem where a single-turn cilcular coil or a coil in the form of a double conductor line is located above a conducting multilayer medium with varying electrical conductivity and magnetic permeability. The solution is found by the method of Hankel integral transform. The change in impedance of the coil is expressed in terms of improper integral containing Bessel functions. The particular cases of a conducting medium in the form of a half-space and two-layer medium are considered in detail. The change in impedance is also obtained for the case where a circular coil of finite dimensions is located above a multilayer medium. Results of numerical calculations with “Mathematica” are presented. (2) Method of the Fourier integral transform is used in the thesis in order to solve the eddy current problem where a coil with alternating current is located inside (or outside) a conducting multilayer tube. Analytical solution is found in the form of an improper integral containing Bessel or confluent hypergeometric functions. The change in impedance is computed with “Mathematica”. (3) Analytical solution is also found for the case where the excitation current in a double conductor line located above a multilayer conducting medium with constant electrical conductivity and magnetic permeability is assumed to be in the form of a pulse. A double conductor line is a sufficiently accurate model of a rectangular frame with current provided the ratio of the sides of the frame is 1:4 or smaller. Solution is found by the method of the Laplace transform. (4) Method of the double Fourier integral transform is used in the thesis in order to construct an analytical solution of the eddy current problem where a coil with alternating current is located above a conducting half-space or two-layer medium. The half-space (or the upper layer of the two-layer medium) is assumed to be moving in the horizontal direction with constant velocity V . The electrical conductivity and magnetic permeability of the moving medium are exponential functions of the vertical coordinate. The solution is found in closed form in terms of a double integral. Calculations of the change in impedance are performed with “Mathematica”. (5) Method of separation of variables is used in the thesis in order to construct quasi-analytical solution for eddy current problem where a circular coil is located above a conducting medium with constant properties under the assumption that the medium contains a cylindrical region  13   

of finite size (a finite cylinder in free space, or a cylindrical hole in a conducting plate, or a cylinder of finite size in a conducting half-space). The axis of the coil coincides with the axis of the cylindrical region. Calculations show good agreement between experimental data and theoretical model.

Applications Analytical and quasi-analytical methods for the solution of direct problems in eddy current testing developed in the thesis can be used in practice in order to control properties of materials and devices with the help of eddy current method. Analytical solutions obtained in Chapters 2, 3, 4 and 5 where the electrical conductivity and magnetic permeability are functions of one spatial coordinate can be used to analyze ferromagnetic metal processing methods (such as surface hardening). Experiments show that in this case a thin layer of reduced magnetic permeability is formed in the upper layer where the magnetic permeability can be accurately approximated by an exponential function of the vertical coordinate. Another application where the electrical conductivity varies with respect to the vertical coordinate is related to diffusion of aluminium in blades of gas turbines exposed to high temperatures. Mathematical models developed in the thesis can be used to compute the change in impedance of the coil where the properties of the conducting medium (electrical conductivity and magnetic permeability) are exponential functions of the vertical coordinate. Quasi-analytical solutions constructed in Chapter 6 can be used in order to test quality of products and materials. The model developed in Section 6.1 (a coil above a finite cylinder coaxial with the axis of the coil) can be implemented in coin validators. Eddy current method is one of the methods that can be used to validate coins by analyzing the electrical conductivity of a metal sample inserted into the validator. Mathematical model analyzed in Section 6.2 (a coil above a conducting plate with a bottom hole in the form of a circular cylinder coaxial with the coil) can be used to estimate the effect of corrosion in metal plates. Finally, the model developed in Section 6.3 (a coil above a conducting half-space with a flaw in the form of a circular cylinder of finite height coaxial with the coil) can be used to analyze the quality of spot welding. During welding process a cast core is formed whose properties (electrical conductivity, for example) are very close to the properties of the surrounding medium. Such a model can be quite useful in order to determine the properties of the cast core.

Publications 1. V. Koliskina, and I. Volodko, Analytical solution of eddy current problems for multilayer medium with varying electrical conductivity and magnetic permeability, International Journal of Mathematical Models and Methods in Applied Sciences, vol. 7, no. 2, pp. 174– 181, 2013 (SCOPUS). 2. V. Koliskina, and I. Volodko, The change in impedance of a single-turn coil due to a cylindrical flaw in a conducting half-space, In: Recent advances in mathematical methods, intelligent systems and materials, pp. 74 – 79, WSEAS Press, 2013 (WSEAS E-library database).

 14   

3. V. Koliskina, and I. Volodko, Eddy current problem for a moving medium with varying properties, International Journal of Mathematical Models and Methods in Applied Sciences, vol. 6, no. 8, pp. 971–978, 2012 (SCOPUS). 4. V. Koliskina, and I. Volodko, Double conductor line above a multilayer medium with varying electrical and magnetic properties, Proceedings of the International conference on applied mathematics and sustainable development, Xian, China, May 27 – 30, pp. 104–108, Scientific Research Publishing, 2012. 5. V. Koliskina, and I. Volodko, The change in impedance of a coil located above a moving half-space, In: Mathematical models and methods in modern science, pp. 107–112, WSEAS Press, 2012 (WSEAS E-library database). 6. V. Koliskina, and I. Volodko, Transient currents in a double conductor line above a conducting half-space, World Academy of Science, Engineering and Technology. – vol. 64, pp. 1147-1151, 2012. 7. V. Koliskina, and I. Volodko, Analytical solution for the change in impedance of a coil situated above a moving conducting medium, 17th International conference on mathematical modeling and analysis, Book of abstracts, p. 69, Tallinn, June 6 – 9, 2012. 8. V. Koliskina, and I. Volodko, On the electromagnetic field of a coil located above a moving half-space with varying properties, 9th Latvian mathematical conference, Book of abstracts, p. 42, Jelgava, Latvia, March 30 – 31, 2012. 9. V. Koliskina, and I. Volodko, Analytical solution of an eddy current problem for a two-layer medium with varying electric conductivity and magnetic permeability, In: Mathematical models and methods in modern science, pp. 196–200, WSEAS Press, 2011 (WSEAS Elibrary database). 10. V. Koliskina, and I. Volodko, Solution of eddy current testing problems for multilayer tubes with varying properties, International Journal of Mathematical Models and Methods in Applied Sciences, vol. 5, pp. 781–788, 2011 (SCOPUS) 11. V. Koliskina, and I. Volodko, A single-turn coil with alternating current inside a cylindrical region with varying electric conductivity and magnetic permeability, In Recent researches in communications, automation, signal processing, nanotechnology, astronomy & nuclear physics, pp. 85–88, 2011 (SCOPUS). 12. V. Koliskina, and I. Volodko, Analytical solution of an eddy current problem for a two-layer tube with varying properties, In: Recent research in communications, electrical & computer engineering, pp. 262 – 265, WSEAS Press, 2011 ((WSEAS E-library database). 13. V. Koliskina V., and I. Volodko, Analytical solution to an eddy current testing problem for a cylindrical tube with varying properties, Scientific Journal of RTU, Computer Science, vol. 46, pp. 72-75, 2011. 14. V. Koliskina, Impedance of an encircling coil due to a cylindrical tube with varying properties, World academy of science, engineering and technology, vol. 52, pp. 745 – 748, 2011 (SCOPUS). 15. V. Koliskina, and I. Volodko, The change in impedance of a double conductor line due to a two-layer medium, In: Recent research in communications & IT, pp. 228 – 232,  2011 (SCOPUS). 16. V. Koliskina, and I. Volodko, Analytical solutions of eddy current problems in cylindrical coordinates, 16th international conference on mathematical modelling and analysis, Book of abstracts, p. 74, Sigulda, Latvia, May 25 – 28, 2011.

 15   

17. V. Koliskina. Double conductor line above a half-space with varying electric and magnetic properties, Proceedings of the 10th International conference on nonlinear systems and wavelet analysis, pp. 104 -108, 2010 (SCOPUS). 18. V. Koliskina, and I. Volodko, Double conductor line above a two-layer medium with varying electric conductivity and magnetic permeability, 15th international conference on mathematical modeling and analysis, Book of abstracts, p. 47, Druskininkai, Lithuania, May 26 – 29, 2010. 19. V. Koliskina, and I. Volodko, Double conductor line above a two-layer medium with varying electric conductivity and magnetic permeability, Scientific Journal of RTU, Computer Science, vol. 45, pp. 76-80, 2010. 20. V. Koliskina, and I. Volodko, Calculation of a coil’s impedance for the case where electric and magnetic properties of a conducting half-space depend on a vertical coordinate, 8th Latvian mathematical conference, Book of abstracts, p. 38, Valmiera, Latvia, April 9 – 10, 2010. 21. V. Koliskina, and I. Volodko, Impedance of a coil above a half-space with varying electric and magnetic properties, Scientific Journal of RTU, Computer Science, vol. 41, pp. 42-46, 2009.

Presentations at international conferences 1. V. Koliskina, I. Volodko, The change in impedance of a single-turn coil due to a cylindrical flaw in a conducting half-space, 15th International conference on mathematical methods, computational techniques and intelligent systems, Limassol, Cyprus, March 21 – 23, 2013. 2. V. Koliskina, I. Volodko, The change in impedance of a coil located above a moving halfspace, 14th WSEAS International conference on mathematical methods, computational techniques and intelligent systems, Porto, Portugal, 2012. 3. V. Koliskina, and I. Volodko, Transient currents in a double conductor line above a conducting half-space, International conference on computational and mathematical sciences, Paris, France, April 25 – 26, 2012. 4. V. Koliskina, and I. Volodko, Double conductor line above a multilayer medium with varying electrical and magnetic properties, 2012 Spring World Congress on engineering and technology, Xian, China, May 27 – 30, 2012. 5. V. Koliskina, and I. Volodko, Analytical solutions for the change in impedance of a coil situated above a moving conducting medium, 17th International conference on mathematical modeling and analysis, Tallinn, Estonia, June 6 – 9, 2012. 6. V. Koliskina, and I. Volodko, Analytical solution of an eddy current problem for a two-layer medium with varying electric conductivity and magnetic permeability, International conference on mathematical methods in modern science, Puerto de la Cruz, Spain, December 10 – 12, 2011. 7. V. Koliskina, and I. Volodko, The change in impedance of a double conductor line due to a two-layer medium, 16th WSEAS International conference on systems, Corfy, Greece, July 14 – 16, 2011. 8. V. Koliskina, and I. Volodko, Analytical solutions of eddy current problems in cylindrical coordinates, 16th International conference on mathematical modeling and analysis, Sigulda, Latvia, May 25 – 28, 2011.  16   

9. V. Koliskina, Impedance of an encircling coil due to a cylindrical tube with varying properties, International conference on electrical engineering and technology, Venice, Italy, April 27 – 29, 2011. 10. V. Koliskina, and I. Volodko, Analytical solution of an eddy current problem for a two-layer tube with varying properties, 8th WSEAS International conference on applied electromagnetics, wireless and optical communications, Playa Meloneras, Spain, March 24 – 26, 2011. 11. V. Koliskina, and I. Volodko, A single-turn coil with alternating current inside a cylindrical region with varying electric conductivity and magnetic permeability, 10th WSEAS International conference on electronics, hardware, wireless and optical communications, Cambridge, UK, February 20 – 22, 2011. 12. V.Koliskina, I. Volodko. Double conductor line above a two-layer medium with varying electric conductivity and magnetic permeability, 15th International conference on mathematical modeling and analysis, Druskininkai, Lithuania, May 26 – 29, 2010. 13. V. Koliskina. Double conductor line above a half-space with varying electric and magnetic properties, 10th International conference on nonlinear systems and wavelet analysis, Kantaoui, Sousse, Tunisia, May 3 – 6, 2010.

Presentations at local conferences 1. V. Koliškina, Volodko I. Determination of complex eigenvalues in eddy current testing problems, Rīgas Tehniskās universitātes 53. starptautiskā zinātniskā konference, apakšsekcija «Datormodelēšana un robežproblēmas», RTU, 2012. gada 10. – 12. oktobrī. 2. V. Koliškina, I. Volodko. Spoles elektromagnētiskais lauks virs kustīgas pustelpas ar mainīgām īpašībām, 9. Latvijas matemātikas konference, Latvija, Jelgava, 2012. gada 30. -31. martā. 3. V. Koliškina, I. Volodko. Virpuļstrāvas metodes analītiskie atrisinājumi vidēm ar mainīgu elektrisko vadamību un magnētisko caurlaidību, Rīgas Tehniskās universitātes 52. starptautiskā zinātniskā konference, apakšsekcija «Datormodelēšana un robežproblēmas», RTU, 2011. gada 14. oktobrī. 4. V. Koliškina, Volodko I. Vijums ar strāvu cilindriskā apgabalā ar mainīgo elektrisko vadamību un magnētisko caurlaidību, Rīgas Tehniskās universitātes 51. starptautiskā zinātniskā konference, apakšsekcija «Datormodelēšana un robežproblēmas», RTU, 2010. gada 14. oktobrī. 5. V. Koliškina, I. Volodko. Calculation of a coil’s impedance for the case where electric and magnetic properties of a conducting half-space depend on a vertical coordinate, 8th Latvian Mathematical Conference, Latvia, Valmiera, 9. -10. April, 2010.

 17   

1. BASIC EQUATIONS 1.1 Medium with constant properties Maxwell’s equations for a homogeneous isotropic medium have the form (see [40]): ~ ∂B ~ , curl E = − ∂t ~ ~ ~ ~ e ∂D , curl H = I + I + ∂t ~ div B = 0, ~ div D = ρ~,

(1.1.1)

(1.1.2) (1.1.3) (1.1.4)

~ ~ (1.1.5) I = σE , ~ ~ B = µ0 µH , (1.1.6) ~ ~ D = ε 0εE , (1.1.7) ~ ~ ~ ~ where E and D are the electric field and electric induction vectors, respectively; H and B are the ~ ~ magnetic field and magnetic induction vectors, respectively; I and I e are the current and external current vectors, respectively; ρ~ is the charge density; σ is the electrical conductivity;

µ0 and ε 0 are the magnetic and electric constants, respectively; µ and ε are the constant relative magnetic permeability and relative permittivity, respectively. System (1.1.1) – (1.1.7) can be rewritten in more convenient form by introducing vector and ~ scalar potentials of the electromagnetic field. The magnetic vector potential A is defined by the relation (see [40]): ~ ~ (1.1.8) curl A = B. Equations (1.1.1) and (1.1.8) give ~ ⎛ ~ ∂A ⎞ ⎟ = 0. curl ⎜⎜ E + (1.1.9) ∂t ⎟⎠ ⎝ It follows from (1.1.9) that there exists a scalar potential of the electromagnetic field, ψ~ , which is defined as follows ~ ~ ∂A E+ = −grad ψ~. ∂t Using (1.1.6) and (1.1.8) we rewrite the left-hand side of equation (1.1.2) in the form

(1.1.10)

(

(1.1.11)

)

⎛ 1 ~⎞ ~ ~ ~ 1 1 ~ curl H = curl ⎜⎜ B ⎟⎟ = curl A = grad div A − ∆ 3 A , µ0 µ ⎝ µ0 µ ⎠ µ0 µ ~ where ∆ 3 A is the Laplacian in three dimensions. It follows from (1.1.10) and (1.1.5) that

 18   

~ ⎛ ∂A ⎞ ~ ~ ⎜ ⎟. I = −σ ⎜ grad ψ + (1.1.12) ⎟ t ∂ ⎝ ⎠ Differentiating (1.1.7) with respect to t and using (1.1.10) we obtain ~ ~ ⎛ ∂ψ~ ∂ 2 A ⎞ ∂D (1.1.13) + 2 ⎟⎟. = −ε 0ε ⎜⎜ grad ∂t ∂t ∂t ⎠ ⎝ Using (1.1.11) – (1.1.13) equation (1.1.2) can be rewritten in the form ~ ~ ⎛ ⎛ ~ ~ ∂A ⎞ ∂ψ~ ∂ 2 A ⎞ ~ ⎟ − µ 0ε 0 µε ⎜ grad ⎟ + µ 0 µI e . (1.1.14) + grad div A − ∆ 3 A = − µ 0 µσ ⎜⎜ gradψ~ + 2 ⎟ ⎟ ⎜ ∂t ⎠ ∂t ∂t ⎠ ⎝ ⎝ Equation (1.1.14) is derived under the assumption that the properties of a conducting medium (the electrical conductivity σ and magnetic permeability µ ) are constant. It follows from (1.1.8) ~ and (1.1.10) that A and ψ~ are not uniquely defined (unless additional conditions on the vector ~ potential A and scalar potential ψ~ are imposed). These additional conditions are called gauge

conditions (see the discussion on different gauge conditions in [40]). In particular, the Lorentz gauge has the form ~ ∂ψ~ = 0. div A + µ0 µσψ~ + µ0ε 0 µε ∂t Using (1.1.14) and (1.1.15) we obtain ~ ~ ~ ∂2 A ∂A ~ ∆ 3 A = µ0ε 0 µε 2 + µ 0 µσ − µ0 µI e . ∂t ∂t

(1.1.15)

(1.1.16)

Since the constants µ0 and ε 0 are very small ( µ0 = 4π ⋅ 10 −7 H/m and ε 0 = 8.85 ⋅ 10 −12 F/m, respectively), the first term on the right-hand side of (1.1.16) (which is usually referrred to as the displacement current in the literature) is much smaller than the other terms and can be neglected for frequencies up to 100 MHz. Such high frequencies are not used in eddy current testing (the depth of penetration decreases with frequency). Thus, for most applications in eddy current testing for media with constant properties (1.1.16) can be simpified to the following form ~ ~ ∂A ~ ∆ 3 A = µ0 µσ − µ0 µI e . (1.1.17) ∂t If the excitation current is periodic with respect to time then all the functions in (1.1.1) – (1.1.10) are also periodic with respect to t : ~ r ~ r ~ r A = Ae jωt , ψ~ = ψe jωt , E = Ee jωt , D = De jωt , (1.1.18) r ~ r ~ r ~ r ~ B = Be jωt , H = He jωt , I = I e jωt , I e = I e e jωt , where ω is the frequency. In this case (1.1.17) reduces to a non-homogeneous Helmoltz equation r r r ∆ 3 A + k 2 A = − µ0 µI e ,

(1.1.19)

where k 2 = − jωµ 0 µσ .  19   

1.2 Medium with variable properties 1.2.1. Multilayer planar medium with varying electrical conductivity and magnetic permeability In this section we consider an air core coil located above a multilayer conducting medium (see Fig. 1.1 ) r1

r2

h2

h1

R 0 (µ 0 = 1 , σ

R 1 (µ 1 , σ 1 )

O

d1

0

= 0)

d2

R 2 (µ 2 , σ

d n −1

R n −1 µ n −1 , σ n −1

2

(

(

Fig. 1.1 A coil above a multilayer conducting medium.

)

Rn µ n ,σ n

)

)  

      The following parameters characterize the coil:

r1 and r2 are the inner and outer radii of the

coil, respectively, h1 is the lift-off and h2 − h1 is the height of the coil. Each conducting layer Ri , i = 1,2,..., n is characterized by the electrical conductivity σ i and relative magnetic permeability µi . We use a system of cylindrical polar coordinates ( r ,ϕ , z ) centered at O, the z r axis is directed upwards. In this case due to axial symmetry the vector potential Ai in region Ri has only one non-zero component of the form   r r (1.2.1) Ai (r ,ϕ , z ) = Ai (r , z )eϕ , i = 0,1,2,..., n r where eϕ is the unit vector in the azimuthal direction. We assume that the electrical conductivity

σ i and relative magnetic permeability µi in each layer Ri , i = 1,2,..., n are functions of the vertical coordinate: σ i = σ ( z ), µi = µ ( z ). r The subscript i in Ai is omitted in sequel. Using (1.1.10) and (1.1.18) we obtain r r E + jωA = −grad ψ .

(1.2.2)

(1.2.3)

Using (1.1.18) , (1.2.3) and (1.2.2) equations (1.1.5) , (1.1.7) can be rewritten as follows r v (1.2.4) I = −σ ( z ) grad ψ + jωA , r r D = −ε 0ε gradψ + jωA . (1.2.5)

( (

) )

Equations (1.1.2) and (1.1.18) give  20   

r r r r curl H = I + I e + jωD.

(1.2.6)

The left-hand side of equation (1.2.6) can be rewritten using (1.1.6) , (1.1.8), (1.1.18) in the form r⎞ r ⎡ r⎤ r ⎛ 1 1 1 curl H = curl ⎜⎜ curl A ⎟⎟ = curl curl A + ⎢grad , curl A⎥ = µ0 µ ( z) ⎣ ⎦ ⎠ µ0 µ ( z) ⎝ µ0 µ ( z)

1 ⎛ 1 dµ ⎞ r ⎟e z ⎜⎜ − 2 µ 0 µ ( z ) µ 0 ⎝ µ ( z ) dz ⎟⎠ r ∂Aϕ r 1 ∂ (rAϕ ) r ⋅ curl A = − er + ez ∂z r ∂r r r 1 1 dµ ∂Aϕ r = grad div A − ∆A + eϕ . 2 µ0 µ ( z) µ 0 µ ( z ) dz ∂z 1

⋅ grad

(

=

(1.2.7)

)

Using (1.2.4) , (1.2.5) and (1.2.7) we transform (1.2.6) to the form r r r r 1 dµ ∂Aϕ r grad div A − ∆A + eϕ = − µ 0 µ ( z )σ ( z ) grad ψ + jωA + µ 0 µ ( z )I e − µ (z ) dz ∂z r − jωµ 0ε 0εµ ( z ) grad ψ + jωA .

(

(

)

)

Taking the projection of (1.2.8) on the ϕ − axis, imposing the condition r grad div A + µ0 µ (z )[σ ( z ) + jωε 0ε ]gradψ = 0,

(1.2.8)

(1.2.9)

using (1.2.1) and the fact that the constants µ0 and ε 0 are very small ( µ 0 = 4π ⋅ 10 −7 H/m and

ε 0 = 8.85 ⋅ 10 −12 F/m, respectively), that is, neglecting the displacement current if the frequency is not so high, we obtain ∆A − jωσ ( z ) µ0 µ ( z ) A −

1 dµ ∂A = − µ0 µ ( z ) I e , µ ( z ) dz ∂z

(1.2.10)

r r where I e (r , z ) is the amplitude of the external current: I e = I eeϕ . The operator ∆A in (1.2.10) has the form

∆A = ∆Aϕ +

2 ∂Ar Aϕ 1 ∂ ⎛ ∂u ⎞ 1 ∂ 2 u ∂ 2 u + − = ∆ = u ⎜r ⎟ + r ∂r ⎝ ∂r ⎠ r 2 ∂ϕ 2 ∂z 2 r 2 ∂ϕ r 2 =

1 ∂Aϕ

∂ 2 Aϕ

∂ 2 Aϕ

=

Aϕ

(1.2.11)

+ + − 2 r ∂r r ∂r 2 ∂z 2 Finally, using (1.2.10) and (1.2.11) we obtain the following equation for the amplitude of the vector potential in a conducting region with varying electric conductivity σ ( z ) and relative magnetic permeability µ ( z )

∂ 2 A 1 ∂A ⎛ 1 1 dµ ∂A ∂ 2 A ⎞ + − + j ωσ ( z ) µ µ ( z ) A − + 2 = −µ0 µ ( z)I e . ⎜ 2 ⎟ 0 2 ∂r r ∂r ⎝ r µ ( z ) dz ∂z ∂z ⎠

(1.2.12)

 21   

1.2.2. Multilayer cylindrical problems with varying electrical conductivity and magnetic permeability. Consider a coil situated inside an infinitely long multilayer tube ( see Fig. 1.2 ).

(

R 0 µ 0r = 1, σ 0 = 0

(

R1 µ1r , σ 1

...

r1

r0

...

( R (µ

(

)

)

) ,σ )

R i µir , σ i n

R n +1 µ nr+1 = 1, σ n +1 = 0

)

r n

n

Fig. 1.2 A single-turn coil inside a multilayer tube.

  The coil is located inside a multilayer tube where each coaxial layer (region Ri ) is described

by the inequalities:  Ri = {ri ≤ r ≤ ri+1 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ }, i = 1,2,..., n .  Here ri and ri+1 are

the inner and outer radii of the cylindrical layer, respectively. Regions R0 and Rn+1 represent free

space. In particular,  R0 = {0 ≤ r ≤ r1 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ }and 

Rn+1 = {r ≥ rn+1 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ } . Due to axial symmetry the vector potential has only

one nonzero component in each region Ri , i = 0,1,2,..., n + 1which is the function of r and z only: r r (1.2.13) Ai (r , ϕ , z ) = Ai (r , z )eϕ , i = 0,1,2,..., n + 1. Suppose that the electrical conductivity σ i and magnetic permeability µi in each region Ri , i = 1,2,..., n are functions of the radial coordinate r :

σ i = σ (r ), µi = µ (r ). r The subscript i in Ai is omitted in sequel. It follows from (1.1.10) and (1.1.18) that r r E + jωA = −grad ψ .

(1.2.14)

(1.2.15)

Using (1.1.18) , (1.2.14) and (1.2.15) equations (1.1.5) , (1.1.7) can be rewritten as follows r v (1.2.16) I = −σ (r )(grad ψ + jωA),

(

)

r r D = −ε 0ε grad ψ + jωA .

(1.2.17)

Equations (1.1.2) and (1.1.18) give r r r r curl H = I + I e + jωD.

(1.2.18)  22   

The left-hand side of equation (1.2.18) can be rewritten using (1.1.6), (1.1.8), (1.1.18) in the form r⎞ r ⎡ r⎤ r ⎛ 1 1 1 curl H = curl ⎜⎜ curl A ⎟⎟ = curl curl A + ⎢grad , curl A⎥ µ 0 µ (r ) ⎠ µ 0 µ (r ) ⎣ ⎦ ⎝ µ 0 µ (r ) dµ r 1 dµ r 1 1 1 ⋅ grad =− er − eϕ 2 2 r µ 0 µ (r ) dr µ 0 µ (r ) µ 0 µ (r ) dr r ∂Aϕ r 1 ∂ (rAϕ ) r ⋅ curl A = − er + ez ∂z r ∂r r⎤ 1 ⎡ dµ ∂ (rAϕ ) r 1 dµ ∂ (Aϕ ) r 1 1 1 ⋅ ⎢grad eϕ − ez − , curl A⎥ = 2 2 µ 0 µ (r ) r µ 0 µ (r ) dr ∂z ⎣ ⎦ r µ 0 µ (r ) dr ∂r −

r curl H =

dµ ∂ (rAϕ ) r 1 1 er 2 2 r µ 0 µ (r ) dr ∂r

dµ ⎛ 1 1 ⎜A ( grad div A − ∆A) + r µ µ (r ) dr ⎜ µ µ (r ) r

1

r

2

ϕ

+r

∂Aϕ ⎞ r ⎟eϕ . ∂r ⎟⎠

(1.2.19)

⎝ Using (1.2.16) , (1.2.17) and (1.2.19) equation (1.2.18) can be rewritten as follows r r 1 dµ ∂Aϕ r 1 1 dµ r grad div A − ∆A + eϕ + Aϕ eϕ = µ (r ) dr ∂r r µ (r ) dr r r r − µ 0 µ (r )σ (r ) grad ψ + jωA + µ 0 µ (r )I e − jωµ 0ε 0εµ (r ) grad ψ + jωA . 0

0

(

)

(

)

Imposing the condition r grad div A + µ0 µ (r )[σ (r ) + jωε 0ε ]gradψ = 0,

(1.2.20)

(1.2.21)

using (1.2.13), neglecting the displacement current and taking the projection of (1.2.20) on the ϕ − axis we obtain

r ⎞ 1 dµ ∂A ⎛ 1 1 dµ − ⎜⎜ + jωµ 0 µ (r )σ (r )⎟⎟ A = − µ 0 µ (r )I e µ (r ) dr ∂r ⎝ r µ (r ) dr ⎠ r r e where I (r , z ) is the amplitude of the external current: I e = I eeϕ . ∆A −

(1.2.22)

Finally, using (1.2.22) and (1.2.11) we obtain the following equation for the amplitude of the vector potential in a conducting region with varying electric conductivity σ ( r ) and relative magnetic permeability µ ( r )

⎞ ∂2 A ⎛ 1 1 dµ ⎞ ∂A ⎛ 1 1 dµ ∂2 A ⎜ ⎟ ⎜ ⎟ + − − + + ωσ ( ) µ µ ( ) = − µ 0 µ (r ) I e . (1.2.23) j r r A + 0 2 2 2 ⎜ ⎟ ⎜ ⎟ rµ (r ) dr ∂r ∂z ⎝ r µ (r ) dr ⎠ ∂r ⎝ r ⎠

1.2.3. Multilayer medium with varying electrical conductivity and magnetic permeability in Cartesian coordinates A double conductor line in the form of two infinitely long parallel wires is considered in the thesis as a simple model of a rectangular frame with current located parallel to the surface (we  23   

assume that the ratio of the sides of the frame is 1:4 or smaller). The vector potential in this case has the form r r r (1.2.24) A( x, y, z ) = Ax ( x, y, z )i + Ay ( x, y, z ) j . Using (1.1.6) , (1.1.8), (1.1.18) we transform the left-hand side of equation (1.2.6) to the form r⎞ r ⎡ r⎤ r ⎛ 1 1 1 curl H = curl ⎜⎜ curl A ⎟⎟ = curl curl A + ⎢grad , curl A⎥ = µ0 µ ( z ) ⎝ µ0 µ ( z ) ⎠ µ0 µ ( z ) ⎣ ⎦ 1 1 dµ r ⋅ grad =− k µ0 µ ( z ) µ 0 µ 2 ( z ) dz

r ∂A v ∂A r ⎛ ∂A ∂A ⎞ v ⋅ curl A = − y i + x j + ⎜⎜ y − x ⎟⎟k ∂z ∂z ∂y ⎠ ⎝ ∂x r r 1 1 dµ ⎛ ∂Ay r ∂Ax ⎜ grad div A − ∆A + = j+ 2 µ0 µ ( z ) µ0 µ ( z ) dz ⎜⎝ ∂z ∂z

(

)

(1.2.25)

v⎞ i ⎟⎟ . ⎠

Using (1.2.4) , (1.2.5) and (1.2.25) equation (1.2.6) can be rewritten as follows r r r 1 dµ ⎛ ∂Ax r ∂Ay r ⎞ ⎜⎜ i+ j ⎟⎟ = − µ 0 µ ( z )σ ( z ) grad ψ + jωA + grad div A − ∆A + µ (z ) dz ⎝ ∂z ∂z ⎠ r r r + µ 0 µ ( z ) I x i + I y j − jωµ 0ε 0εµ ( z ) grad ψ + jωA .

(

(

)

(

)

)

Neglecting the dispacement current and using (1.2.9) we obtain r r r r 1 dµ ⎛ ∂Ax r ∂Ay r ⎞ ⎜⎜ i+ j ⎟⎟ − jωµ 0 µ ( z )σ ( z )A = − µ 0 µ ( z ) I x i + I y j . ∆A − µ (z ) dz ⎝ ∂z ∂z ⎠

(

)

(1.2.26)

(1.2.27)

.

1.3 Literature survey In this section we present a brief literature survey of the basic methods and models used in eddy current testing. Detailed references to each particular topic analyzed in the thesis are also given in the beginning of each chapter. Methods of nondestructive testing are widely used in practice in order to estimate quality of products and materials. Eddy current method is one of the popular methods for nondestructive testing of electrically conducting materials. The method is based on the principle of electromagnetic induction discovered by M. Faraday in 1831. The idea of the method is as follows. Suppose that a coil (or any other source of alternating current) is located near an electrically conducting medium. The electromagnetic field generated by the currents in the coil is called the primary field. In accordance with the principle of electromagnetic induction eddy currents (also known as Foucault currents by the name of the French scientist Leon Foucault who discovered this phenomenon) are induced in the conducting medium. These currents represent the secondary field. Eddy currents induced in the conducting medium interact with the currents in the coil changing the primary field. As a result of the interaction the impedance of the coil also changes. The change in impedance of the coil can be measured experimentally. The following parameters characterizing the coil and the conducting medium affect the change in impedance: (a) geometrical size of the coil (the inner and outer radii of the coil, the height of the coil, the  24   

number of turns and the distance from the bottom of the coil to the conducting medium being tested, known as the lift-off in the literature), (b) parameters of the medium (electrical conductivity and magnetic permeability) and (c) the frequency of the excitation current. In addition, if a conducting medium contains a flaw then the output signal of the coil would also depend on the properties of the flaw. Physical principles of eddy current testing are described, for example, in [7]. Mathematical models can also be used in order to compute the change in impedance of the coil. These models can be divided into two groups: analytical and numerical models. The term “analytical model” refers to the case where the system of Maxwell’s equations describing the interaction of the currents in the coil with a conducting medium is solved analytically (that is, the solution is expressed in closed form in terms of known special or elementary functions). However, the change in impedance of the coil is usually expressed in terms of improper integral so that at the last stage of the analysis the integral has to be evaluated numerically. Analytical models can be developed for the cases where the conducting medium is infinite in one or two spatial dimensions. Numerical models are based on direct numerical modeling of the corresponding boundaryvalue problems. Finite element methods are widely used to model Maxwell’s equations in a conducting medium (see, for example, [22]). Analytical or numerical methods are used in practice for the solution of a direct problem in eddy current testing. The term “direct problem” is referred to as the solution of Maxwell’s equations under the assumption that all parameters of the conducting medium (in particular, electrical conductivity and magnetic permeability) as well as geometrical parameters characterizing the medium (for example, the thickness of a conducting coating for the case of a two-layer medium) are known. In practice, however, these parameters may not be known and need to be determined from the solution of an inverse problem. Examples of inverse problems in eddy current testing include determination of electrical conductivity of metal plates or the thickness of metal coatings. Usually unknown parameters are obtained by minimizing the norm of the difference between measured and computed change in impedance. Thus, efficient and reliable methods for the solution of direct problems are necessary in order to solve inverse problems. Mathematical models for eddy current testing of conducting media with constant electrical and magnetic properties are well-developed in the literature for the case where the medium is assumed to be infinite with respect to one or two special dimensions. The main results are summarized in monographs [3], [59], [60] and in several books published in Russian [17]-[19]. The solution process usually starts with the case where a single-turn circular coil is located above a conducting medium. Assuming that the Maxwell’s equations are solved for the case of a singleturn coil, the change in impedance of a coil of finite dimensions can be calculated using the superposition principle. The system of Maxwell’s equations for the case where a single-turn coil with alternating current is located above a conducting half-space with constant electrical conductivity σ is solved in [55]. Later this solution was generalized for the case of a multilayer medium [15]. The problems in [55] and [15] are solved by the method of the Hankel integral transform. Several  25   

papers are devoted to particular cases of the problem: a single-turn coil above a two-layer medium or a plate (the analysis of the corresponding solutions can be found, for example, in [60]). Different types of eddy current coils which can be used to locate cracks or other defects are described in [13], [14]. The solutions of direct problems for a multilayer medium are used in practice to solve a series of inverse problems. Thickness and conductivity of metallic layers are determined in [46]. Similar problem for determination of thickness and conductivity of thin nonmagnetic coatings on ferromagnetic conductive substrates is solved in [51]. Other applications include hardness testing of steel [45], [54], estimation of electrical conductivity of alloys [8], [53], thickness determination of metal plates using multi-frequency eddy current coils [69], [70]. Another geometrical configuration where analytical solutions can be constructed corresponds to the case where a coil with alternating current is located inside or outside a multilayer tube. Detailed solution of the problem is given in [16] and [19]. The problem has also important practical applications. For example, such coils are used to test quality of heat exchanger tubes in nuclear reactors [10]. Analytical solutions of the corresponding direct problems (under the additional assumption that the axis of a coil coincides with the axis of the tube) are obtained by the method of Fourier integral transform with respect to the longitudinal coordinate. The change in impedance of the coil in this case is obtained in terms of improper integrals containing Bessel functions. There are also examples of analytical solutions for the case where a coil is located above a conducting multilayer sphere [17]. However, these problems have quite limited range of practical applications and will not be considered in sequel. Some industrial applications require the analysis of a moving conducting medium [1], [48]. Examples include steel processing at a metallurgical plant or movement of a coin inside a coin validator. In many cases analytical solutions described above are based on the assumption of an infinite conducting medium. From a practical point of view the medium can be considered infinite in one or two spatial dimensions if the size of the coil is much smaller than the size of an object of inspection. However, there are many applications where the size of the coil is comparable with the size of the object being tested. Solution procedure has to be modified for such cases in order to take into account finite size of a conducting medium. Recently one quasi-analytical method (TREE method) is proposed in [60]. The idea of the method is based on a physical assumption that electromagnetic field can be negligibly small far from the source of alternating current. It is assumed in a “classical” theory of eddy current testing that the vector potential and its derivatives approach zero at infinity. In the TREE method the vector potential is assumed to be exactly zero at a sufficiently large distance from the coil. The corresponding boundary value problem can be solved by the method of separation of variables. However, the solution procedure includes two steps which require numerical computations: (a) calculations of complex eigenvalues and (b) solution of a linear algebraic system. The solution is then represented in the form of a truncated eigenfunction expansion. In contrast with classical problems of mathematical physics the terms of the series cannot be represented in closed form, but can be computed provided that steps (a) and (b) described above are completed.  26   

2. PLANAR MULTILAYER MEDIA WITH VARYING PROPERTIES 2.1 Introduction Mathematical models of eddy current testing problems for planar multilayer medium with constant electrical and magnetic properties are well-developed in the literature [15], [17]-[19], [3], [59]-[60]. Analytical solutions of direct problems for multilayer medium are usually constructed by means of integral transforms (such as Hankel or Fourier integral transforms). The resulting system of ordinary differential equations in the transformed space is then solved analytically if the properties of each layer of the conducting medium are constant. Alternative approach is developed in [63] where the magnetic field is assumed to be zero at a sufficiently large distance from the axis of a coil. Such a method is known as the TREE (TRuncated Eigenfuction Expansion) method in the literature [60]. In this case analytical and quasi-analytical solutions to eddy current problems can be constructed in the form of truncated series expansions. Since the range of applications of eddy current method is quite wide it is not surprising to know that in some cases the electrical conductivity and magnetic permeability of a conducting medium can vary with respect to spatial coordinates. It is shown in [57] and [58] that special type of treatment of ferromagnetic metals (such as surface hardening) can lead to the presence of a surface layer with reduced magnetic permeability which varies exponentially with respect to the vertical coordinate. In order to optimize the performance of gas turbines it is necessary to increase firing temperatures. As a result, blades are usually protected from the exposure to high temperatures by layers containing aluminium and chrome in special proportions [41]. The depletion of aluminimum in this case leads to the variation of electrical conductivity with respect to the vertical coordinate. Hence, in order to adequately describe eddy current problems in the above mentioned cases it is necessary to develop mathematical models which take into account variability of the electrical conductivity and/or magnetic permeability with respect to one geometrical coordinate (the vertical coordinate in the case of a planar multilayer medium). Two methods are usually used in such cases. One method is based on the assumption that the variation of electrical and/or magnetic properties of the medium can be represented by piecewise constant functions. In other words, a conducting layer with varying properties is divided into a large number of relatively thin sub-layers where the electrical conductivity and/or magnetic permeability of each sub-layer are constant. This approach is used in [41], [66] for a rectangular coordinate system, in [65] for a cylindrical coordinate system and in [62] for a spherical coordinate system. Note that up to 50 layers are used in [64] and up to 20 layers in [41]. In addition, it is estimated in [41] that the use of many layers affects computational efficiency by increasing computational time. Alternative approach is based on the assumption that simple model profiles (for example, in the form of exponential or power functions) can be used in order to represent variability of the properties of the conducting medium in one spatial direction (see, for example, [3], [23], [61], [24], [30], [39]). In this chapter we construct analytical solutions for the case where a coil with alternating current is located above a multilayer conducting medium. It is assumed that the electrical conductivity and magnetic permeability of each layer can vary with respect to the vertical coordinate. In particular, the electrical conductivity and magnetic permeability in each layer are assumed to be exponential functions of the vertical coordinate. Some particular cases of the suggested solutions are considered in detail. Multilayer medium with many conducting layers is  27   

rather a rare case in practice. In many applications the medium consists of one or two layers. Thus, the following particular cases of a general model described in Section 2.3 are considered: (a) a coil above a half-space and (b) a coil above a conducting two-layer medium. Results of numerical calculations are presented.

2.2 A coil above a multilayer medium with varying properties Consider an air core coil located above a multilayer medium (see Fig. 1.1). The outer and inner radii of the coil are r2 and r1 , respectively. The height of the coil is h2 − h1 , where h1 is the lift-off. The coil is located in free space (region R0 ). The thickness of each conducting layer Ri is denoted by d i , i = 1,2,..., n − 1. The bottom layer Rn is assumed to be infinite in the vertical direction. Each conducting layer Ri , i = 1,2,..., n is characterized by the two parameters: the electrical conductivity σ i and relative magnetic permeability µir . We consider a system of cylindrical polar coordinates ( r , ϕ , z ) centered at O, the z -axis is directed upwards. The coil is carrying alternating current of the form r r (2.2.1) I e = I 0 (r , h)e jωt eϕ , r where r1 ≤ r ≤ r2 , h1 ≤ h ≤ h2 , ω is the frequency, eϕ is the unit vector in the ϕ − direction and the current density I 0 (r , h) is constant over the cross-section of the coil:

NI ⎧ , r1 ≤ r ≤ r2 , h1 ≤ h ≤ h2 ⎪ I 0 (r , h) = ⎨ (r2 − r1 )(h2 − h1 ) ⎪⎩ 0, otherwise where I is constant and N is the number of turns in the coil.

r In this case it is natural to assume that the vector potential Ai in each region Ri , i = 0,1,..., n has only one non-zero component of the form r r (2.2.2) Ai = Ai (r , z )e jωt eϕ . We assume that the electrical conductivity σ i and magnetic permeability µ ir in each region

Ri depends on the vertical coordinate. More precisely, σ i and µ ir are modeled by the following relations: µ ir ( z ) = µ im exp( β i z ), σ i ( z ) = σ im exp(α i z ), (2.2.3) i = 1,2,..., n, where α i , β i , µ im , σ im are constants.

2.3 A single-turn coil above a multilayer medium with varying properties The first step in solving the problem is to find the solution for the case where a single-turn coil of radius rc is located at distance h above a conducting multilayer medium [24] ( see Fig. 2.1 ).

 28   

z

rc

h O

d1

R 0 (µ 0r = 1, σ 0 = 0 ) x R1 (µ1r , σ 1 )

(

)

d2

R 2 µ 2r , σ 2

d n −1

R n −1 µ nr−1 , σ n −1

(

(

R n µ nr , σ n

)

)

Fig. 2.1 A single-turn coil above multilayer medium.

Using (1.2.12) with µ 0r = 1, σ 0 = 0 and I e = Iδ (r − rc )δ (z − h ) we obtain the equation for the amplitude of the vector potential in region R0 : ∂ 2 A0 1 ∂A0 A0 ∂ 2 A0 + − + = − µ 0 Iδ (r − rc )δ ( z − h), ∂r 2 r ∂r r 2 ∂z 2 where δ ( x) is the Dirac delta-function.

(2.3.1)

Using (1.2.12) with I = 0 , µ ( z ) = µ ir ( z ) and σ ( z ) = σ i ( z ) we obtain the following system of equations for the amplitudes Ai (r , z ) of the vector potential in each region Ri , i = 1,2,..., n :

∂ 2 Ai 1 ∂Ai Ai ∂ 2 Ai 1 dµ ir ( z ) ∂Ai + − + − − jωσ i ( z )µ 0 µ ir ( z )Ai = 0, ∂r 2 r ∂r r 2 ∂z 2 µ ir ( z ) dz ∂z i = 1,2,..., n. Using (2.2.3) we rewrite equations (2.3.2) in the form ∂A ∂ 2 Ai 1 ∂Ai Ai ∂ 2 Ai − 2 + 2 = β i i + jωσ im µ 0 µim e (α i + β i ) z Ai , + 2 r ∂r r ∂z ∂z ∂r i = 1,2,..., n, The boundary conditions are ∂A0 1 ∂A1 A0 | z = 0 = A1 | z = 0 , | z =0 = r | z =0 , ∂z µ1 (0) ∂z

(2.3.2)

(2.3.3)

(2.3.4)

Ai | z = − dˆ = Ai +1 | z = − dˆ , i

i

∂Ai +1 ∂Ai 1 | z = − dˆ = r | ˆ , i = 1,2,.., n − 1, r i ˆ ˆ µ i − d i ∂z µ i +1 − d i ∂z z = − d i 1

( )

where

( )

(2.3.5)

i

dˆ i = ∑ d k . k =1

 29   

The following conditions hold at infinity: ∂A Ai , i → 0 as r → ∞ , i = 0,1,..., n, ∂r A0 → 0 as z → +∞, An → 0 as z → −∞.

(2.3.6) (2.3.7)

In order to solve (2.3.1) – (2.3.8) we use the Hankel integral transform of the form ∞

~ Ai (λ , z ) = ∫ Ai (r , z )rJ 1 (λr )dr , i = 0,1,.., n.

(2.3.8)

0

Using (2.3.6) , (2.3.8) we rewrite the left-hand side of equation (2.3.3) in the form (assuming uniform convergence of the corresponding integrals) ∞

∞ ∞ 2 ⎞ ⎛ ∂ 2 A 1 ∂Ai Ai ⎞ ∂ A ⎟⎟rJ1 (λr )dr = ∫ ⎜⎜ 2 i + − 2 ⎟⎟rJ1 (λr )dr + ∫ 2 i rJ 1 (λr )dr r ∂r r ⎠ ∂r ∂z ⎠ 0⎝ 0  

⎛ ∂ 2 Ai 1 ∂Ai Ai ∂ 2 Ai ∫0 ⎜⎜⎝ ∂r 2 + r ∂r − r 2 + ∂z 2

 

~ ∂ 2 Ai d 2 Ai ⋅ ∫ 2 rJ1 (λr )dr = ∂z dz 2 0 ∞

∞

(

)

u = rJ1 (λr ) → du = J1 (λr ) + λrJ1' (λr ) dr

⎛ ∂ A 1 ∂Ai Ai ⎞ ⋅ ∫ ⎜⎜ 2i + − 2 ⎟⎟rJ1 (λr )dr = ∂ 2 Ai ∂A ∂ r r ∂ r r dv = dr → v = i ⎠ 0⎝ 2 ∂r ∂r 2

= rJ1 (λr ) =

∂Ai ∂r

∞

0

∞

− ∫ J1 (λr ) 0

(

∞

∞

dv =

∞

∂Ai ∂A ∂A 1 dr − λ ∫ rJ1' (λr ) i dr + ∫ J1 (λr ) i dr − ∫ J1 (λr )Ai dr =   ∂r ∂r ∂r r 0 0 0

)

u = rJ (λr ) → du = J1' (λr ) + λrJ1'' (λr ) dr ' 1

=

∂Ai dr → v = Ai ∂r

=

∞

1 ' 1 ⎛ ⎞ = −λ rJ (λr )Ai + λ ∫ ⎜ J1'' (λr ) + J1 (λr ) − 2 2 J1 (λr )⎟ Ai rdr = 0 λr λr ⎠ 0⎝ ' 1

∞

2

1 ' 1 J 1 (x ) − 2 J 1 (x ) x x 2 ⎛ ν ⎞ 1 y ' ' ( x ) + y ' ( x ) + ⎜⎜1 − 2 ⎟⎟ y ( x ) = 0 → y = Jν ( x ) x ⎝ x ⎠

⋅ x = λr → J 1'' ( x ) + =

⋅ν = 1 → y = J 1 ( x ) J 1'' ( x ) +

∞

~ = −λ2 Ai

 

1 ' 1 J 1 (x ) − 2 J 1 (x ) = − J 1 (x ) x x

⎛ ∂ 2 Ai 1 ∂Ai Ai ∂ 2 Ai ∫0 ⎜⎜⎝ ∂r 2 + r ∂r − r 2 + ∂z 2

~ ⎞ d 2 Ai ~ ⎟⎟rJ 1 (λr )dr = − λ2 Ai , 2 dz ⎠

i = 0,1,..., n,

(2.3.9)

The right-hand side of equation (2.3.1) can be rewritten using (2.3.8) in the form  30   

∞

− µ 0 Iδ ( z − h) ∫ δ (r − rc ) rJ1 (λr )dr = − µ 0 Irc J 1 (λrc )δ ( z − h)                                                               (2.3.10) 0

Applying the Hankel transform (2.3.8) to the right-hand side of equation (2.3.3) we obtain ~ ∞ dAi ~ ⎛ ∂Ai ⎞ m m (α i + β i ) z Ai ⎟rJ1 (λr )dr = β i + jωσ im µ 0 µim e (α i + β i ) z Ai ∫0 ⎜⎝ β i ∂z + jωσ i µ0 µi e                            (2.3.11) dz ⎠

i = 1,2,..., n, Using (2.3.9), (2.3.10) and (2.3.11) we obtain ~ d 2 A0 ~ − λ2 A0 = − µ 0 Irc J 1 (λrc )δ ( z − h),                                                                                                   (2.3.12) 2 dz ~ ~ d 2 Ai dAi ~ − β − λ2 + jωσ im µ 0 µ im e (α i + β i )z Ai = 0, i = 1,2,..., n.                                                     (2.3.13) i 2 dz dz Applying the Hankel integral transform (2.3.8) to the boundary conditions we obtain ~ ~ dA0 1 dA1 ~ ~ (2.3.14) A0 | z = 0 = A1 | z = 0 , | z =0 = r | z =0 , dz µ1 (0 ) dz ~ ~ Ai | z = − dˆ = Ai +1 | z = − dˆ , i i ~ ~ (2.3.15) dAi dAi +1 1 1 | = | , i = 1 , 2 ,.., n − 1 , ˆ ˆ µ ir − dˆi dz z = − d i µ ir+1 − dˆi dz z = − d i ~ ~ A0 → 0 as z → +∞, An → 0 as z → −∞. (2.3.16)

(

( )

)

( )

In order to solve equation (2.3.12) we consider the following two sub-regions of R0 : 0 < z < h ~ ~ and z > h . The solutions in these regions are denoted by A00 and A01 , respectively. Hence, ~ d 2 A00 ~ (2.3.17) − λ2 A00 = 0, 0 < z < h, 2 dz ~ d 2 A01 ~ (2.3.18) − λ2 A01 = 0, z > h. 2 dz The general solution to (2.3.17) can be written in the form ~ A00 = C1e λz + C2 e − λz .                                                                                                                                      (2.3.19) The bounded solution to (2.3.18) is  ~ − λz   A01 = C3e .                                                                                                                                                    (2.3.20)  ~ ~ The functions A00 (λ , z ) and A01 (λ , z ) satisfy the following condition at z = h :   ~ ~ A00 | z =h = A01 | z =h . (2.3.21)  

~ The condition (2.3.21) reflects the fact that the function A0 (λ , z ) is continuous at z = h . The second condition is obtained by integrating (2.3.12) with respect to z from z = h − ε to z = h + ε and considering the limit in the resulting expression as ε → +0 :  31   

h +ε

⋅

∫

h −ε

                            

~ h +ε h +ε d 2 A0 ~ 2 dz − λ ∫ A0 dz = − µ 0 Irc J 1 (λrc ) ∫ δ ( z − h)dz dz 2 h −ε h −ε ~ h +ε h +ε dA0 ~ − λ2 ∫ A0 dz = − µ 0 Irc J 1 (λrc ) dz h −ε h −ε h +ε

(

)

~ ~ ⋅ − λ lim ∫ A0 dz = − λ2 lim 2εA0 λ , z * = 0 2

ε →0

~ dA0 ⋅ lim ε → 0 dz

ε →0

h −ε

h +ε

h −ε

~ dA01 = dz

~ dA00 − dz z =h

 

z =h

 

~ dA01 dz

~ dA00 − dz z =h

= − µ 0 Irc J 1 (λrc ).                                                                                                          (2.3.22)  z =h

Using (2.3.19) − (2.3.22) we obtain

⎧⎪C1e λh + C 2 e − λh = C3 e − λh ⇒ ⎨ ⎪⎩− λC3 e − λh − λC1e λh + λC 2 e − λh = − µ 0 Irc J 1 (λrc )                                              ⎧ C1e 2 λh + C 2 − C3 = 0 ⎪ ⇒ ⇒ ⎨ µ0 2 λh λh ( ) C e C C Ir J λ r e − + − = − ⎪ 1 c 1 c 2 3 λ ⎩

µ0 ⎧ − λh ⎪⎪C1 = 2λ Irc J 1 (λrc )e                                                                                                                        (2.3.23) ⎨ ⎪C = C + µ 0 Ir J (λr )e λh 2 c 1 c ⎪⎩ 3 2λ Using (2.3.23) we rewrite (2.3.19) and (2.3.20) in the following form µ0 ⎧~ − λz − λ (h − z ) ⎪⎪ A00 = C 2 e + 2λ Irc J 1 (λrc )e µ ~ −λ h− z                     (2.3.24) ⇒ A0 (λ , z ) = C 2 e − λz + 0 Irc J 1 (λrc )e ⎨ µ0 2λ ~ − λz −λ ( z −h ) ⎪A = C2 e + Irc J 1 (λrc )e ⎪⎩ 01 2λ  

In order to understand the meaning of each of the two terms on the right-hand side of (2.3.24) we consider the case where a coil of radius rc centered at the point (0, h) is located in an unbounded free space. The corresponding component of the vector potential will be denoted by A0free ( r , z ) . The equation for A0free is  ∂ 2 A0free 1 ∂A0free A0free ∂ 2 A0free + − 2 + = − µ0 Iδ (r − rc )δ ( z − h),                                                            (2.3.25) ∂r 2 r ∂r r ∂z 2 where δ ( x) is the Dirac delta-function. In addition, A0free is bounded at r = 0  and A0free is bounded as r → ∞ .  32   

Applying the Hankel integral transform (2.3.8), using (2.3.12) and (2.3.16) we get ~ d 2 A0free ~ − λ2 A0free = − µ 0 Irc J 1 (λrc )δ ( z − h). 2 dz The bounded solutions in regions R00 and R01 are : ~ ~ d 2 A00free R00 : − λ2 A00free = 0, z < h, 2 dz ~ → A00free (r , λ ) = C1e λz . ~ ~ d 2 A01free − λ2 A01free = 0, R01 : z > h. 2 dz ~ → A01free (r , λ ) = C3 e − λz .

(2.3.26)

(2.3.27)

(2.3.28)

Using (2.3.21) and (2.3.22) solutions (2.3.27) and (2.3.28) can be rewritten as follows    ~ µ A00free (r , λ ) = 0 Irc e −λ (h− z )                                                                                                                             (2.3.29)  2λ                 (2.3.30) µ ~ A01free (r , λ ) = 0 Irc e −λ ( z − h ) 2λ Comparing (2.3.24) and (2.3.29), (2.3.30) we see that the second term on the right-hand side of (2.3.24) represents the vector potential in an unbounded free space. Thus, the first term of the right-hand side of (2.3.24) represents the induced vector potential due to the presence of the conducting medium. The solution to (2.3.13) can be expressed in terms of Bessel functions (see [50], formula 2.1.3.10, page 247):

y xx'' + ay x' + (beλx + c )y = 0 ⇒ y = e

ν= ~ Ai (λ , z ) = e

where

−

ax 2

⎡ ⎛ 2 b λ2x ⎞ ⎛ 2 b λ2x ⎞⎤ ⎜ ⎟ ⎜ ⎟⎥ + C J e C Y ⎢ 1 ν⎜ λ 2 ν⎜ λ e ⎟ ⎟ ⎢⎣ ⎝ ⎠ ⎝ ⎠⎥⎦

a 2 − 4c

λ βi z 2

⎡ ⎛ 2 b i (α i + β i ) z ⎞ ⎛ 2 b i (α i + β i ) z ⎞⎤ 2 ⎜ ⎟ + C5 i Yν i ⎜ α i + β i e 2 ⎟⎥,   i = 1,2,..., n − 1,               (2.3.31)  ⎢C 4 i Jν i α i + β i e ⎜ ⎟ ⎜ ⎟⎥ ⎢⎣ ⎝ ⎠ ⎝ ⎠⎦

β i2 + 4λ2 , bi = − jωµ 0 µ imσ im ,   νi = αi + βi

and Jν i , Yν i are the Bessel functions of the first and second kind, respectively. The bounded solution to (2.3.13) in region Rn is βn z ⎛ 2 b n (α n + β n ) z ⎞ ~ 2 ⎟. An (λ , z ) = C 4 n e Jν n ⎜ α n + β n e 2 ⎜ ⎟ ⎝ ⎠

(2.3.32)

Using (2.3.24), (2.3.31), (2.3.32) and the boundary conditions (2.3.14) , (2.3.15) we obtain

 33   

C2 +

1

1

1

e

β i dˆ i 2

1

1

1

1

( )

(2.3.33)

1

µ0 Irc J 1 (λrc )e −λh = 2λ                   (2.3.34) C 4 ⎛ β1 ⎞ C5 ⎛ β1 ⎞ ' ' = m ⎜ J ν z 0 + b1 J ν z 0 ⎟ + m ⎜ Yν z 0 + b1 Yν z 0 ⎟, µ1 ⎝ 2 ⎠ µ1 ⎝ 2 ⎠

− λC 2 +

−

( )

µ0 Irc J 1 (λrc )e −λh = C 4 J ν z 0 + C 5 Yν z 0 , 2λ

(C

( ( ) )+ C

J ν i z1 i dˆ i

4i

=e

−

( ) 1

5i

1

( )

1

i

1

( ) 1

1

( ( )) )=

Yν i z1 i dˆ i

β i +1 dˆ i +1 2

(C

4 i +1

( ( ) )+ C

J ν i +1 z1 i dˆ i

Y

5 i +1 ν i +1

( z (dˆ ) ) ), 1i

e

β n−1 dˆ n−1 2

(C

4 n−1

( (

e

−

β n −1 dˆ n −1 2

µ

m n −1

( ( ))

( (dˆ ) )

( ( ))

( (dˆ ) )⎟⎟

))

−

β n dˆ n−1 2

⎞ ⎟+ ⎟ ⎠

( ( ))

( (

Jν n−1 z1 n−1 − dˆn −1 + C5 n−1 Yν n−1 z1 n−1 − dˆn −1 =e

                                  (2.3.35)

( ( ))

( ( ))

−

i

i

(α i + β i )dˆ i ⎛ ⎛ β ⎞ ⎞ − ⎜ i ⎜ 2 ˆ C4 i J ν i z1 i d i + bi e J ν' i z1 i dˆi ⎟ + ⎟ β i dˆ i ⎜ ⎜ 2 ⎟ ⎟ e 2 ⎜ ⎝ ⎠ ⎟ m ⎜ ⎟= ˆ ( ) + d α β µi i i i ⎛ β ⎞ ⎜ + C ⎜ i Y z dˆ + b e − 2 Y ' z dˆ ⎟ ⎟ i i νi 5i 1i ⎜⎜ ⎜ 2 ν i 1i i ⎟ ⎟⎟ ⎝ ⎠ ⎠ ⎝ (α i +1 + β i +1 )dˆ i +1 ⎛ ⎛ β − ⎜ i + 1 ⎜ 2 ˆ C 4 i +1 J ν i +1 z1 i +1 d i + bi +1 e J ν' i +1 z1 i +1 β i +1i dˆ i +1 ⎜ ⎜ 2 e 2 ⎜ ⎝ = m ⎜ (α i +1 + β i +1 )dˆ i +1 µ i +1 ⎛ − ⎜ + C ⎜ β i +1 Y 2 ˆ z d + bi +1 e Yν' i +1 z1 i +1 5 i +1 ⎜⎜ ⎜ 2 ν i +1 1 i +1 i ⎝ ⎝

( ( ))

( )

( (

C 4 n Jν n z1 n − dˆn −1

))

i

⎞

i

⎠

( ( ))

β n dˆ n

−

2

( ( ))

⎞

⎞

⎠

⎟= ⎟ ⎟ ⎟⎟             (2.3.38)  ⎠

( ( ) ) ⎟⎟ + ⎟⎟ ⎞

( ( ) ) ⎟⎟

(α n + β n ) dˆ n −1 ⎛ β − n ⎜ 2 ˆ C4 n J z d J ν' n z1 n dˆ n −1 = + bn e ⎜ 2 ν n 1 n n −1 µ nm ⎝ (α i + β i )d) i i 2 bi − 2 where z 0i =         dˆi = ∑ d .k . z1 i dˆi = z 0 i e αi + βi k =1

e

                   (2.3.36)

) ) ) =                                                                (2.3.37)

(α n −1 + β n −1 ) dˆ n −1 ⎛ ⎛ − ⎜ C ⎜ β n −1 J 2 ˆ J ν' n −1 z1 n −1 dˆ n −1 ⎜ 4 n −1 ⎜ 2 ν n −1 z1 n −1 d n −1 + bn −1 e ⎝ ⎜ ⎜ (α n −1 + β n −1 ) dˆ n −1 ⎛ − ⎜ + C ⎜ β n −1 Y 2 z dˆ Yν' n −1 z1 n −1 dˆ n −1 + bn −1 e 5 n −1 ⎜⎜ ⎜ 2 ν n −1 1 n −1 n −1 ⎝ ⎝

( ( ))

⎞ ⎟ ⎟ ⎟, ⎟ ⎟ ⎟⎟ ⎠

⎠

⎞

( ( ) ) ⎟⎟, ⎠

( )

Solving the system (2.3.33) – (2.3.38), substituting the values of the constants into (2.3.24),

 34   

(2.3.31) and (2.3.32) we obtain the solution in each region Ri in the transformed space. The solution in each region Ri (i = 0,1,..., n) is then found by means of the inverse Hankel transform of the form ∞

~ Ai (r , z ) = ∫ Ai (λ , z )λJ 1 (λr )dλ ,

i = 0,1,..., n.

(2.3.39)

0

The method described above can be used for any finite number of conducting layers. In the next section we consider one particular case – a single-turn coil above a conducting half-space with varying electrical and magnetic properties.

2.4 A single-turn coil above a conducting half-space with varying properties Consider a single-turn coil of radius rc located at a distance h above a conducting half-space (see Fig. 2.1 in the limit as d1 → ∞ ) . The geometry of the problem is shown in Fig. 2.2, where

µ1r ( z ) = µ m e β z , σ 1 ( z ) = σ m e α z (see [39]). z

rc

h

(

R 0 µ0r = 1, σ 0 = 0 x R1 (µ1r , σ 1 )

O

) Fig. 2.2 A single-turn coil above a conducting half-space.

Using the results from Section 2.3 we obtain the solutions in regions R0 and R1 (see (2.3.24) and (2.3.32)): ~ d 2 A0 ~ − λ2 A0 = − µ0 Irc J1 (λrc )δ ( z − h)    →    R0 :    2 dz

~

                                                           →      A0 (λ , z ) = C 2 e −λz +

µ0 −λ h − z Irc J 1 (λrc )e ,                             (2.4.1)  2λ

~ ~ d 2 A1 dA1 ~ − λ2 + jωσ m µ 0 µ m e (α + β )z A1 = 0    →     R1 :     2 − β dz dz

(

)

~                                                             →      A1 (λ , z ) = C 4 e

β z 2

⎛ 2 b (α +2 β ) z ⎞ ⎟.                                      (2.4.2)  J ν ⎜⎜ α + β e ⎟ ⎝ ⎠

The boundary conditions are : ~ ~ dA0 1 dA1 ~ ~ A0 | z = 0 = A1 | z = 0 , | z = 0 ,                                                                                                    (2.4.3)  | z =0 = dz µ m dz ~ ~ (2.4.4) A0 → 0 as z → +∞, A1 → 0 as z → −∞.  35   

Using (2.4.1) − (2.4.4) we obtain

µ0 ⎧ − λh ⎪ C 2 + 2λ Irc J 1 (λrc )e = C 4 J ν (z 0 ), ⎪                                                                    (2.4.5)  ⎨ µ C β ⎞ ⎛ λ ' − h 0 4 ⎪− λ C 2 + Irc J 1 (λrc )e = ⎜ J ν (z 0 ) + b J ν ( z 0 )⎟, ⎪⎩ 2 µm ⎝ 2 ⎠ 2 b where z 0 = α +β

β 2 + 4λ2 . , b = − jωµ 0 µ mσ m , ν = α +β

It follows from (2.4.5) that ⋅ C 2 = C 4 Jν ( z 0 ) −              

C4 =

µ0 Irc J 1 (λrc )e − λh 2λ

(

⋅ − λC 4 Jν ( z 0 ) + µ 0 Irc J 1 (λrc )e

µ 0 µ m Irc J 1 (λrc )e − λh β⎞ '

⎛ ⎜ λµ m + ⎟ Jν (z 0 ) + b J ν ( z 0 ) 2⎠ ⎝

− λh

⎛β ⎞ µ m = C 4 ⎜ Jν (z 0 ) + b Jν' (z 0 )⎟ ⎝2 ⎠

)

 

.                                                                                                           (2.4.6) 

In particular, the value of C2 is ⎛⎛

β⎞

⎞ ' ⎟ J ν ( z 0 ) − b Jν (z 0 ) ⎟⎟ 2⎠ ⎝⎝ ⎠ .                                                                       (2.4.7) C2 = ⎛⎛ ⎞ β⎞ 2λ ⎜⎜ ⎜ λµ m + ⎟ J ν ( z 0 ) + b Jν' (z 0 ) ⎟⎟ 2⎠ ⎝⎝ ⎠ It has been shown in Section 2.3 that the second term in (2.4.1) represents the vector potential for the case where a single-turn coil is located in an unbounded free space. Thus, the first term in (2.4.1) is the reaction of the conducting half-space on the current in the coil. This term is usually referred to as “the induced vector potential”. The induced vector potential in region R0 is ~ A0ind (λ , z ) = C2 e − λz , (2.4.8)

µ 0 Irc J 1 (λrc )e −λh ⎜⎜ ⎜ λµ m −

where C2 is given by (2.4.7). Applying the inverse Hankel transform of the form (2.3.39) to (2.4.8) we obtain

A0ind (r , z ) =

µ 0 Irc 2

∞

∫ F (λ )J (λr ) J (λr )e 1

c

1

−λ ( z + h)

dλ ,

(2.4.9)

0

where

β⎞ ⎛ ' ⎜ λµ m − ⎟ J ν (z 0 ) − b J ν (z 0 ) 2⎠ ⎝ F (λ ) = . β⎞ ⎛ ' ⎜ λµ m + ⎟ J ν (z 0 ) + b J ν ( z 0 ) 2⎠ ⎝ The induced change in impedance of the coil is given by the formula (see [60]) jω Z ind = A0ind (r , z )dl , I ∫L

(2.4.10)

 36   

where L is the contour of the coil. Substituting (2.4.9) into (2.4.10) we obtain ∞

Z ind (r , z ) = ωπµ 0 rc2 j ∫ F (λ )J 12 (λrc )e − 2λh dλ.

(2.4.11)

0

Using the following dimensionless parameters (2.4.11) can be rewritten as follows ⎫ ,⎪ ⎬ ⇒ νˆ = αˆ = α rc , βˆ = β rc , ⎪⎭

⋅ λ rc = u → λ =

u rc

βˆ 2 + 4u 2 , αˆ + βˆ

2 − jbˆ ˆ h ⋅ bˆ = ωµ 0 µ mσ m rc2 , zˆ 0 = , h= . rc αˆ + βˆ Z ind (r , z ) = ω π µ 0 rc Z ,                                                                                                                                  (2.4.12) 

where ˆ⎞ ⎛ ⎜ uµ m − β ⎟ Jνˆ ( zˆ 0 ) − − j bˆ Jν'ˆ ( zˆ 0 ) ∞⎜ 2 ⎟⎠ ˆ J 12 (u )e − 2uh du.                                                               (2.4.13) Z = j∫ ⎝ ˆ ⎞ 0 ⎛ ⎜ uµ m + β ⎟ Jνˆ ( zˆ 0 ) + − j bˆ Jν'ˆ ( zˆ 0 ) ⎜ 2 ⎟⎠ ⎝ Fig. 2.3 plots the change in impedance Z for three different values of βˆ = 1,2,3 .The other parameters of the problem are fixed at αˆ = 0, hˆ = 0.05 and µ m = 5. The calculated points in Fig. 2.3 correspond to different values of bˆ = 1,2,..., 40 (from left to right). Computations are done with “Mathematica” ( see Appendix Fig.Ap.1 ) . Im @zD 0.45

           βˆ = 1                            βˆ = 2  

0.4 0.35

                       βˆ = 3  

0.3 0.25

0.05

0.1

0.15

0.2

Re @zD

Fig. 2.3 The change in impedance Z for three different values of βˆ .

It is seen from the graph that the modulus of Z increases as the parameter βˆ increases. All calculations in the thesis are done with software package “Mathematica”. In particular, integral (2.4.13) is computed using “Mathematica” command NIntegrate. Users can specify infinity as the upper limit of integration in NIntegrate. However, it is known that computation of some improper integrals in “Mathematica” can be time-consuming. Similar problem is found in the thesis for calculation of some improper integrals. As a result, the ∞

improper integrals in the form

∫ 0

γ

f ( x)dx are replaced by ∫ f ( x)dx , where the upper limit γ is 0

 37   

chosen in order to satisfy the desired accuracy of the calculations. Example of such computation is shown in Table 2.1. It is seen from Table 2.1 that for 1 ≤ bˆ ≤ 11 and βˆ = 2   in order to construct the graph it is more than enough to choose γ = 60. “Mathematica” code for the computation of integral in (2.4.13) is shown in Appendix Fig.Ap.2.

bˆ   

γ

20 40 60 80 100 120 140

1

3

5

7

9

11

0.01283 0.37593i

0.03767 0.37150i

0.06034 0.36348i

0.08022 0.35306i

0.09725 0.34128i

0.11170 0.32894i

0.01284 0.38579i

0.03768 0.38136i

0.06036 0.37335i

0.08025 0.36292i

0.09728 0.35114i

0.11173 0.33880i

0.01284 0.38652i

0.03768 0.38209i

0.06036 0.37408i

0.08025 0.36365i

0.09728 0.35187i

0.11173 0.33953i

0.01284 0.38658i

0.03768 0.38216i

0.06036 0.37414i

0.08025 0.36371i

0.09728 0.35193i

0.11173 0.33960i

0.01284 0.38659i

0.03768 0.38216i

0.06036 0.37414i

0.08025 0.36371i

0.09728 0.35194i

0.11173 0.33960i

0.01284 0.38659i

0.03768 0.38217i

0.06036 0.37415i

0.08025 0.36372i

0.09728 0.35195i

0.11173 0.33961i

0.01284 0.38659i

0.03768 0.38217i

0.06036 0.37415i

0.08025 0.36372i

0.09728 0.35194i

0.11173 0.33961i

Table 2.1 Calculated values of the change in impedance given by (2.4.13) for different values of γ and bˆ .

2.5 Coil of finite dimensions In this section we compute the induced vector potential of a coil of finite dimensions using solution (2.4.9) for a filamentary coil. The geometry of the problem is shown in Fig.2.4 . z r2 r1

h1

x

O

h2

R 0 (µ 0r = 1, σ 0 = 0 ) y r R1 (µ1 , σ 1 ) Fig. 2.4 A coil of finite dimensions above a conducting half-space.

Assume that N is the number of turns in the coil. The inner and outer radii of the coil are r1 and r2 , respectively. The distance from the coil to the top surface of the cylinder is denoted by h1 . The height of the coil is h2 − h1. The induced vector potential in air due to currents in the whole coil is obtained as follows ind 0 coil

A

(r , z ) =

r2 h2

∫∫A

ind 0

(r , z , rc , h)drc dh .

(2.5.1)

r1 h1

Substituting (2.4.9) into (2.5.1) and using the formulas  38   

h2

⋅ ∫ e − λ h dh = − h1

1

λ

(e

− λ h2

− e − λ h1

r2

⋅ ∫ rc J1 (λrc )drc = y = λrc = r1

) 1

λ2

(2.5.2)

λ r2

∫ yJ ( y )dy 1

λ r1

we obtain the induced vector potential in air due to the presence of the conducting half-space NI ): ( the current amplitude I in this case is replaced by the current density (r2 − r1 )(h2 − h1 ) λ r2 ⎞ ⎛ F (λ ) −λ h − λ h2 − λh 1 ⎟dλ , ⎜ ( ) (2.5.3) A (r , z ) = e e J λ r e yJ ( y ) dy − 1 1 ∫λ r ⎟ 2(r2 − r1 )(h2 − h1 ) ∫0 ⎜⎝ λ3 1 ⎠ where β⎞ ⎛ ' ⎜ λµ m − ⎟ J ν (z 0 ) − b J ν (z 0 ) 2 ⎠ . F (λ ) = ⎝ β⎞ ⎛ ' ⎜ λµ m + ⎟ J ν (z 0 ) + b J ν ( z 0 ) 2⎠ ⎝ The integral with respect to y in (2.5.2) can be computed in terms of the Bessel and Struve functions [2] as follows ind 0 coil

µ 0 NI

∞

(

)

y =λ r2

λ r2

⎧π ⎫ ∫λ r yJ1 ( y)dy = ⎨⎩ 2 y[J 0 ( y) H1 ( y) − J1 ( y) H 0 ( y)]⎬⎭ y=λ r 1 1

(2.5.4)

The induced change in impedance of a coil of finite dimensions (see Fig. 2.4) is calculated by means of the following formula [15]: r2 h2 2π rr jω jω Z ind = 2 ∫∫∫ AI dv = 2 ∫ dϕ ∫ rdr ∫ A0ind coil dz I V I 0 r1 h1 r h

Z

ind 0 coil

2 2 2πjω N (r , z ) = rA0indcoil (r , z )drdz. ∫ ∫ I (r2 − r1 )(h2 − h1 ) r1 h1

(2.5.5)

Using (2.5.3) and (2.5.5) we obtain the induced change in impedance of the coil in the form 2 ∞⎛ ⎛ λ r2 ⎞ ⎞⎟ jωπµ0 N 2 ⎜ F (λ ) −λ h1 −λ h2 2 ⎜ Z (r , z ) = e yJ ( y )dy ⎟ ⎟dλ . −e ⎜ λ∫r 1 ⎟ ⎟ (r2 − r1 ) 2 (h2 − h1 ) 2 ∫0 ⎜⎜ λ6 ⎝ 1 ⎠ ⎠ ⎝ Using the following dimensionless parameters (2.5.6) can be rewritten as follows

(

ind 0 coil

u ⎫ , r1 ⎪⎬ ⇒ νˆ = αˆ = α r1 , βˆ = β r1 , ⎪⎭

⋅ λ r1 = u → λ =

)

(2.5.6)

βˆ 2 + 4u 2 , αˆ + βˆ

2 − jbˆ ˆ h1 ˆ h2 r , h1 = , h2 = , rˆ2 = 2 ⋅ bˆ = ωµ0 µ mσ m r12 , zˆ0 = ˆ r1 r1 r1 αˆ + β Z 0ind coil ( r , z ) = ω π µ 0 r1 Z ,                                                                                                                                   (2.5.7)   39   

where Z=

(hˆ

2

− hˆ1

⎛ Fˆ (u ) ˆ ˆ j ∫ ⎜⎜ 6 e −uh1 − e −uh2 u 0 ⎝

(

∞

N2

) (rˆ − 1) 2

2

2

)

2

⎛ u rˆ2 ⎞ ⎜ yJ 1 ( y )dy ⎟ ⎜∫ ⎟ ⎝u ⎠

2

⎞ ⎟du ,                                               (2.5.8) ⎟ ⎠

and ˆ⎞ ⎛ ⎜ uµ m − β ⎟ Jνˆ (zˆ 0 ) − − j bˆ Jν'ˆ ( zˆ 0 ) ⎜ 2 ⎟⎠ .                Fˆ (u ) = ⎝ ˆ⎞ ⎛ β ' ⎜ uµ m + ⎟ Jνˆ (zˆ 0 ) + − j bˆ Jνˆ ( zˆ 0 ) ⎜ 2 ⎟⎠ ⎝

Fig. 2.5 plots the change in impedance Z for three different values of βˆ = 1,2,3 .The other parameters of the problem are fixed at αˆ = 0, hˆ = 1.6, hˆ = 0.4, rˆ = 2, N = 50 and µ = 5. The 2

1

m

2

calculated points in Fig. 2.5 correspond to different values of bˆ = 1,2,...,20 (from left to right). Computations are done with “Mathematica” ( see Appendix Fig.Ap.3 ) .

           βˆ = 1  

Im z 125

                         βˆ = 2  

100 75

                       βˆ = 3  

50 25

Fig. 2.5 The change in impedance Z for three different values of βˆ .

Re z 20

40

60

80

100

-25

                                                                                                                         

2.6 A single-turn coil above a two-layer medium with varying properties In this section we consider another example of the general theory [30]. Consider a single-turn coil of radius rc located at a distance h from a two-layer medium (see Fig. 2.6). z rc

h x d

O

(

)

R 0 µ0r = 1, σ 0 = 0 y (1 1µ(1µr ,1rσ, σ1 )1 ) RR

R 2 (µ 2r , σ 2 )

Fig. 2.6 A single-turn coil above a two-layer medium.

 40   

where µ1r ( z ) = µ1m e β z , σ 1 ( z ) = σ 1m eα z ,

               µ 2r ( z ) = µ 2m ,

σ 2 (z ) = σ 2m .  

Formulation of the problem follows from (2.3.1) – (2.3.7) and has the form R0 :   

R1 :  

∂ 2 A0 1 ∂A0 A0 ∂ 2 A0 = − µ 0 Iδ (r − rc )δ ( z − h),                                                                   (2.6.1) − + + ∂z 2 ∂r 2 r ∂r r 2

∂A ∂ 2 A1 1 ∂A1 A1 ∂ 2 A1 − 2 + 2 − β 1 − jωσ 1m µ 0 µ1m e (α + β )z A1 = 0,   + 2 r ∂r r ∂z ∂z ∂r

            (2.6.2)

∂ 2 A2 1 ∂A2 A2 ∂ 2 A2            (2.6.3) − 2 + = jωσ 2m µ 0 µ 2m A2 ,   + 2 2 r ∂r r ∂z ∂r where σ 2m and µ 2m are, respectively, the constant electrical conductivity and relative magnetic

R2 :  

permeability of region  R2 .  The boundary conditions are ∂A0 1 ∂A1 | z =0 = r | z =0 , A0 | z = 0 = A1 | z = 0 , ∂z µ1 (0) ∂z A1 | z = − d = A2 | z = − d ,

(2.6.4)

∂A1 1 1 ∂A |z =−d = m 2 |z =−d . µ (− d ) ∂z µ 2 ∂z r 1

The following conditions hold at infinity: ∂A Ai , i → 0 as r → ∞ , i = 0,1,2 , ∂r A0 → 0 as z → +∞, A2 → 0 as z → −∞.

(2.6.5)

(2.6.6) (2.6.7)

Applying the Hankel transform (2.3.8) to problem (2.6.1) − (2.6.7) we obtain the solution in regions Ri , i = 0,1 : ~ d 2 A0 ~ − λ2 A0 = − µ0 Irc J1 (λrc )δ ( z − h)    →    R0 :    2 dz µ ~ −λ h − z                                  →      A0 (λ , z ) = C 2 e − λz + 0 Irc J 1 (λrc )e ,                                                       (2.6.8) 2λ ~ ~ d2A dA ~ R1 :     21 − β1 1 − λ2 + jωσ 1m µ 0 µ1m e (α + β )z A1 = 0    →     dz dz

(

~

                                 →      A1 (λ , z ) = e

where

b = − jωµ 0 µ1mσ 1m , ν =

)

β z 2

⎡ ⎛ 2 b (α + β ) z ⎞⎤ ⎛ 2 b (α + β ) z ⎞ ⎢C 4 Jν ⎜⎜ α + β e 2 ⎟⎟ + C 5Yν ⎜⎜ α + β e 2 ⎟⎟⎥,                       (2.6.9) ⎠⎦⎥ ⎝ ⎠ ⎝ ⎣⎢

β 2 + 4λ2 . α +β

The solution to (2.6.3) in the transformed space which remains bounded as z → −∞ is ~ d 2 A2 ~ ~ q z − q22 A2 = 0          →        A2 (λ , z ) = C6 e 2 , R2 : 2 dz

(2.6.10)

 41   

where q2 = λ2 + jωσ 2m µ 0 µ 2m . The boundary conditions are : ~ ~ ~ ~ dA0 1 dA A0 | z =0 = A1 | z =0 , | z =0 = m 1 | z =0 , dz µ1 dz ~ ~ ~ ~ dA1 1 1 dA2 A1 | z = − d = A2 | z = − d , |z =−d . |z=−d = m µ1r (− d ) dz µ 2 dz ~ ~ A0 → 0 as z → +∞, A2 → 0 as z → −∞.

(2.6.11) (2.6.12) (2.6.13)

Using (2.6.8) − (2.6.13) we obtain µ C 2 + 0 Irc J 1 (λrc )e −λh = C 4 J ν ( z 0 ) + C 5Yν ( z 0 ),                                                                                      (2.6.14) 2λ − λC 2 +

µ0

Irc J 1 (λrc )e −λh =

2

C4 ⎛ β ⎞ C ⎛β ⎞ J ν ( z 0 ) + b J ν' ( z 0 )⎟ + m5 ⎜ Yν ( z 0 ) + bYν' ( z 0 )⎟,             (2.6.15) m ⎜ µ1 ⎝ 2 ⎠ µ1 ⎝ 2 ⎠

 

C6 e

− q2 d

=e

−

βd 2

(C 4 J ν (z1 ) + C5Yν (z1 )),                                                                                                        (2.6.16)

 

q2

µ 2m

C 6 e − q2 d = βd

=

e

2

µ1m

(α + β )d (α + β )d ⎛ ⎛ β − − ⎞ ⎛ ⎞⎞ ⎜ C 4 ⎜ J ν ( z1 ) + be 2 J ν' ( z1 ) ⎟ + C 5 ⎜ β Yν ( z1 ) + be 2 Yν' ( z1 ) ⎟ ⎟, ⎟ ⎜ 2 ⎟⎟ ⎜ ⎜ 2 ⎠ ⎝ ⎠⎠ ⎝ ⎝

where z1 = z 0 e

−

(α + β )d 2

, z0 =

                   (2.6.17) 

2 b , b = − jωµ 0 µ1mσ 1m .   α +β

It follows from (2.6.14) − (2.6.15) that                                                                                                  

µ0 Irc J 1 (λrc )e − λh 2λ µ µ ⋅ − λC4 Jν ( z 0 ) − λC5Yν ( z 0 ) + 0 Irc J 1 (λrc )e − λh + 0 Irc J 1 (λrc )e − λh =   ⋅ C 2 = C 4 Jν (z 0 ) + C5Yν ( z 0 ) −

2

=

2

C4 ⎛ β ⎞ ⎞ C ⎛β Jν ( z 0 ) + b Jν' ( z 0 )⎟ + m5 ⎜ Yν ( z 0 ) + bYν' ( z 0 )⎟ m ⎜ µ1 ⎝ 2 ⎠ µ1 ⎝ 2 ⎠

⎛⎛ ⎞ β⎞ C 5 ⎜⎜ ⎜ λµ1m + ⎟Yν ( z 0 ) + bYν' (z 0 )⎟⎟ − µ 0 µ1m Irc J 1 (λrc )e −λh 2⎠ ⎝ ⎠ .                                                   (2.6.18) C4 = − ⎝ β ⎛ m ⎞ ' ⎜ λµ1 + ⎟ J ν ( z 0 ) + b J ν (z 0 ) 2⎠ ⎝  

Similarly, conditions (2.6.16) − (2.6.17) can be rewritten as follows

              ⋅ C 6 = e

( 2 q 2 − β )d 2

(C 4 J ν (z1 ) + C 5Yν (z1 ))

 

 42   

( β −α )d ⎞ ⎛ ⎛ β βd m ⎞ m m ⎜ ⋅ C 4 ⎜ ⎜ e µ 2 − µ1 q 2 ⎟ J ν ( z1 ) + b µ 2 e 2 J ν' ( z1 ) ⎟⎟ = ⎠ ⎠ ⎝⎝2              ( β −α )d ⎞ ⎛ ⎛β ⎞ = − C 5 ⎜⎜ ⎜ e βd µ 2m − µ1m q 2 ⎟Yν ( z1 ) + b µ 2m e 2 Yν' ( z1 ) ⎟⎟ ⎠ ⎠ ⎝⎝2

 

( β −α )d ⎞ ⎛⎛β ⎞ C5 ⎜⎜ ⎜ e βd µ 2m − µ1m q2 ⎟Yν (z1 ) + b µ 2m e 2 Yν' (z1 ) ⎟⎟ ⎠ ⎝2 ⎠ C4 = − ⎝ ( β −α )d ⎞ ⎛ β βd m ' m m ⎜ e µ 2 − µ1 q2 ⎟ Jν ( z1 ) + b µ 2 e 2 Jν (z1 ) 2 ⎠ ⎝

(2.6.19)

Constant C5 is found from (2.6.18) − (2.6.19) and has the form C5 = −

µ 0 µ1m Irc J 1 (λrc )e − λh ( D1 J ν (z1 ) + D3 J ν' (z1 ) )

( D Y (z ) + D Y (z ) )( D J (z ) + '

1 ν

3 ν

1

1

ν

2

0

) (

)(

b J ν' ( z 0 ) − D1 J ν ( z1 ) + D3 J ν' ( z1 ) D2 Yν ( z 0 ) + bYν' ( z 0 )

),

     

                                                                                                                                                                             (2.6.20)                        

where  β D1 = e βd µ 2m − µ1m q 2 , 2 β   D2 = + λµ1m , 2 D3 = b µ 2m e

( β −α )d 2

.

Using (2.6.19) − (2.6.20) we obtain C4 =

µ 0 µ1m Irc J 1 (λrc )e − λh ( D1Yν (z1 ) + D3Yν' (z1 ) )

( D Y (z ) + D Y (z ) )( D J (z ) + 1 ν

' 3 ν

1

1

2

ν

) (

)(

b J ν' ( z 0 ) − D1 J ν ( z1 ) + D3 J ν' ( z1 ) D2Yν ( z 0 ) + bYν' ( z 0 )

0

).

         

                                                                                                                                                                             (2.6.21)                        

In particular, the value of C2 is            ⋅ C 2 = C 4 Jν ( z 0 ) + C5Yν ( z 0 ) −

µ0 Irc J 1 (λrc )e − λh 2λ

µ 0 Irc J 1 (λrc )e − λh B ⋅ , 2λ D where C2 =

(2.6.22)

(( ) (( ) D = E ( D J (z ) + b J (z ) )− F ( D Y (z ) +

)

)

B = E 2λµ1m − D2 Jν (z 0 ) − b Jν' (z 0 ) − F 2λµ1m − D2 Yν (z 0 ) − bYν' (z 0 ) , 2 ν

'

0

ν

0

2 ν

0

)

bYν' (z 0 ) ,

(2.6.23) (2.6.24)

E = D1Yν ( z1 ) + D3Yν' ( z1 ),

(2.6.25)

F = D1 J ν ( z1 ) + D3 J ν' ( z1 )

(2.6.26)

and

q2 = λ2 + jωσ 2m µ 0 µ 2m ,     b = − jωµ 0 µ1mσ 1m    43   

z1 = z 0 e

−

(α + β )d 2

, z0 =

2 b , ν = α +β

β 2 + 4λ2 . α +β

The induced vector potential in region R0 is ~ A0ind (λ , z ) = C2 e − λz ,

(2.6.27)

where C2 is given by (2.6.22). Applying the inverse Hankel transform (2.3.39) to (2.6.27) we obtain the induced vector potential in the form

A0ind (r , z ) =

µ 0 Irc 2

∞

B

∫ D ⋅J (λr ) J (λr )e 1

c

−λ ( z + h)

1

dλ ,

(2.6.28)

0

where B and D are given by (2.6.23) and (2.6.24), respectively. The change in impedance of the coil is computed using (2.4.10) and (2.6.28): ∞

Z

ind

(r , z ) = ωπµ r j ∫ 2 0 c

0

B 2 ⋅ J 1 (λrc )e − 2λh dλ. D

(2.6.29)

Using the following dimensionless parameters (2.6.29) can be rewritten as follows

⎫ ,⎪ ⎬ ⇒ νˆ = αˆ = α rc , βˆ = β rc ,⎪⎭

⋅ λrc = u → λ =

u rc

βˆ 2 + 4u 2 , αˆ + βˆ

⋅ bˆ = ωµ 0 µ1mσ 1m rc2 , βˆ ˆ ˆ ⋅ Dˆ 1 = e β d µ 2m − µ1m qˆ 2 , kur qˆ 2 = u 2 + jq , q = ωσ 2m µ 0 µ 2m rc2 , 2 (βˆ −αˆ )dˆ βˆ m m ˆ ˆ ˆ ⋅ D2 = + uµ1 , ⋅ D3 = − jb µ 2 e 2 , 2 (αˆ + βˆ )dˆ − 2 − jbˆ h ⋅ zˆ 0 = , zˆ1 = zˆ 0 e 2 , hˆ = , rc αˆ + βˆ ⋅ Eˆ = Dˆ Y ˆ ( zˆ ) + Dˆ Y ˆ' ( zˆ ), ⋅ Fˆ = Dˆ J ˆ ( zˆ ) + Dˆ J 'ˆ ( zˆ ) , 1 ν

(

1

3 ν

)

1

1 ν

1

3

ν

(

1

)

⋅ Bˆ = Eˆ ⎛⎜ 2uµ1m − Dˆ 2 Jνˆ ( zˆ 0 ) − − jbˆ Jν'ˆ ( zˆ 0 ) ⎞⎟ − Fˆ ⎛⎜ 2uµ1m − Dˆ 2 Yνˆ ( zˆ 0 ) − − jbˆYνˆ' ( zˆ 0 ) ⎞⎟ , ⎝ ⎠ ⎝ ⎠ ⋅ D = Eˆ ⎛⎜ Dˆ 2 Jνˆ ( zˆ 0 ) + − jbˆ Jν'ˆ ( zˆ 0 ) ⎞⎟ − Fˆ ⎛⎜ Dˆ 2Yνˆ ( zˆ 0 ) + − jbˆYνˆ' ( zˆ 0 ) ⎞⎟ ⎝ ⎠ ⎝ ⎠ Z ind (r , z ) = ω π µ 0 rc Z ,                                                                                                                                  (2.6.30)

where                      ∞ ˆ B ˆ Z = j ∫ ⋅ J 12 (u )e − 2uh du.   ˆ 0 D

Fig. 2.7 plots the change in impedance Z for three different values of βˆ = 1,2,3 .The other  44   

parameters of the problem are fixed at dˆ = 0.2, µ1m = 1, µ 2m = 100, hˆ = 0.05, αˆ = 0 .   The calculated points in Fig. 2.7 correspond to different values of bˆ = 1,2,...,10 (from left to right). Computations are done with “Mathematica” ( see Appendix Fig.Ap.4 ) .   Im @zD

ˆ            β = 1  

0.2

ˆ                          β = 2  

0.15

ˆ                        β = 3   

0.1 0.05 0.15

0.2

0.25

0.3

Re @zD Fig. 2.7 The change in impedance Z for

three different values of βˆ .

-0.05 -0.1

 

The solution for a single-turn coil can be easily generalized for the case of a coil of finite dimensions as it is done in Section 2.5. One can apply the general theory developed in Section 2.3 to other examples with any finite number of layers. The transformed induced vector potential in free space in any case is given by the first term in formula (2.3.24) where the constant C2 is obtained from the boundary conditions (2.3.14) – (2.3.15). The induced vector potential in free space is given by (2.3.39). The change in impedance of the single-turn coil is computed by (2.4.10). Finally, the change in impedance of a coil with finite dimensions is computed using (2.5.1) and (2.5.5).  

 45   

3. CYLINDRICAL PROBLEMS FOR MULTILAYER MEDIA WITH VARYING PROPERTIES 3.1 Introduction Eddy current methods are widely used in practice in order to control properties of conducting materials. All problems where eddy current coils are used for inspection of materials can be divided into the following two categories: (a) estimation of properties and/or other parameters of conducting media (for example, electric conductivity of a conducting layer or thickness of metal coatings) and (b) detection of defects (or flaws) in a conducting medium (for example, estimation of the effect of corrosion or presence of voids or other non-metallic inclusions in the medium). In both cases theoretical models (with some unknown parameters) are usually compared with experimental data. Optimization methods (for example, the least squares method) are then used to estimate unknown parameters of the medium (see, for example, [47], [49]). Thus, a necessary step for the solution of an inverse problem (determination of unknown parameters of a medium) is the existence of a mathematical model describing the interaction of an alternating current in a coil with the conducting medium (direct problem). Mathematical models for eddy current testing problems of conducting media with constant properties are well-known in the literature [60], [3]. Analytical solutions are presented in [15] for the case where a coil with alternating current is located above a multilayer medium. Similar problems for coils encircling multiple coaxial conductors or coils inside multiple coaxial conductors are analyzed in [16]. The properties of all media in [16] are assumed to be constant. Some industrial applications (for example, surface hardening or decarbonization) modify the properties of a conducting medium (electric conductivity and/or magnetic permeability) which depend on geometrical coordinates. It is shown in [57], [58] that a thin layer of reduced magnetic permeability can exist in a medium which undergoes special treatment. In the case of a planar medium the magnetic permeability of the layer depends on the vertical coordinate. Thus, in order to analyze such cases in practice one needs to develop mathematical models where electric conductivity and magnetic permeability of conducting layers depend on geometrical coordinates. There are at least two methods that can be used in order to take into account variability of the properties of the medium with respect to one geometrical coordinate. Regions where the properties of the medium are not constant can be divided into sufficiently large number of sub-layers where the electric conductivity and magnetic permeability are assumed to be constant. Therefore, in the whole region of interest the properties of the medium are piecewise constant functions of, say, depth. For example, up to 50 layers are used in [64] to model the change in electric conductivity with respect to a vertical coordinate. Another approach is based on an attempt to use relatively simple electric conductivity and/or magnetic permeability profiles in order to model the variation of the properties of the medium with respect to one spatial coordinate. Some analytical solutions for the case where the properties of the medium depend on the vertical or radial coordinates are presented in [3]. In this chapter we follow the second approach. Analytical solutions are constructed for the case where a coil is located inside or outside a multilayer tube with varying properties. The electric conductivity and magnetic permeability are assumed to be power functions of the radial coordinate. The solution procedure is described for an arbitrary number of conducting layers with varying electric conductivity and magnetic permeability. Three cases are analyzed in detail: (a) the case of a coil inside one infinite outer layer, (b) the case of a coil inside a two-layer tube, and  46   

(c) the case of a coil outside a two-layer tube .

3.2 A coil located inside a multilayer medium with varying properties Consider a single-turn coil of radius r0 with alternating current located inside a multilayer tube

where

each

coaxial

layer

(region

Ri )

is

described

by

the

inequalities:

Ri = { ri ≤ r ≤ ri+1 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ }, i = 1,2,..., n [31]. Here ri and ri +1 are the inner and

outer radii of the cylindrical layer, respectively. Regions R0 and Rn +1 represent free space. In particular, R0 = { 0 ≤ r ≤ r1 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ } and Rn +1 = { r ≥ rn +1 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ } . The coil is located in the plane z = z 0 perpendicular to the axis of the tubes (see Fig. 3.1).

(

R 0 µ 0r = 1, σ 0 = 0

(

R1 µ1r , σ 1

...

r0

r1

...

( R (µ

)

)

) ,σ )

R i µir , σ i n

r n

n

Fig. 3.1 A single-turn coil inside a multilayer tube.

(

R n +1 µ nr+1 = 1, σ n +1 = 0

)

Due to axial symmetry the vector potential has only one nonzero component in each region Ri , i = 0,1,..., n + 1 which is the function of r and z only. We assume that the electric conductivity

σ i ( r ) and magnetic permeability µi (r ) in region Ri (i = 1,2,.., n) are modeled by the following relations µi (r ) = µ0 µir (r ) = µ0 µim r βi , σ i (r ) = σ im r αi , i = 1,2,..., n,

(3.2.1)

where α i , β i , µ im , σ im are constants, and µir (r ) is the relative magnetic permeability of region Ri . Using (1.2.23) where µ 0r = 1, σ 0 = 0 and I e = Iδ (r − r0 )δ ( z − z 0 ) we obtain the equation for the vector potential in region R0 : ∂ 2 A0 1 ∂A0 A0 ∂ 2 A0 = − µ 0 Iδ (r − r0 )δ ( z − z 0 ), − + + ∂z 2 ∂r 2 r ∂r r 2 where δ (x) is the Dirac delta-function.

(3.2.2)

I = 0   we obtain the following system of equations for the Using (1.2.23) and amplitudes Ai (r , z ) of the vector potential in each region Ri , i = 1,2,..., n :

 47   

r 1 dµi ∂ 2 Ai ⎛ 1 ⎜ + − ∂r 2 ⎜⎝ r µi r (r ) dr i = 1,2,..., n.

r ⎞ ⎞ ∂Ai ⎛ 1 ∂ 2 Ai 1 dµ i r ⎟ ⎟ ⎜ j r r A + ωσ µ µ ( ) ( ) = 0, − + + i i i 0 ⎟ ⎟ ∂r ⎜ r 2 rµ r (r ) dr ∂z 2 i ⎠ ⎠ ⎝

It follows from (3.2.1) and (3.2.3) that ∂ 2 Ai 1 ∂Ai Ai ∂ 2 Ai β i ∂Ai ⎛ β i + ⎜ 2 + jωσ im µ 0 µ im r α i + β i − 2+ 2 = + 2 r ∂r r r ∂r ⎝ r ∂z ∂r i = 1,2,..., n .

⎞ ⎟ Ai , ⎠

(3.2.3)

(3.2.4)

Using (1.2.23) with µ nr +1 = 1, σ n +1 = 0 and I = 0 we obtain the equation for the vector potential in region Rn +1 : ∂ 2 An +1 1 ∂An +1 An +1 ∂ 2 An +1 − 2 + = 0. + r ∂r ∂z 2 r ∂r 2 The boundary conditions are ∂A0 1 ∂A1 A0 |r = r1 = A1 |r = r1 , |r = r1 = r |r = r , ∂r µ1 (r1 ) ∂r 1

(3.2.5)

(3.2.6)

Ai |r = ri+1 = Ai +1 |r = ri+1 ,

(3.2.7)

∂Ai +1 ∂Ai 1 1 |r = ri+1 = r |r = ri+1 , i = 1,2,.., n . µ (ri +1 ) ∂r µ i +1 (ri +1 ) ∂r r i

In addition, A0 is bounded at r = 0 and An +1 is bounded as r → ∞ . The following conditions hold at infinity: ∂A Ai , i → 0 as z → ±∞ , i = 0,1,..., n . ∂z Applying the Fourier transform of the form ∞

~ Ai (r , λ ) =

∫ A ( r , z )e

− jλ z

i

(3.2.8)

dz , i = 0,1,.., n + 1.

(3.2.9)

−∞

to the left-hand side of equation (3.2.4) we obtain ∞

∞ ∞ ⎛ ∂ 2 Ai 1 ∂Ai Ai ∂ 2 Ai ⎞ − jλz ⎛ ∂ 2 Ai 1 ∂Ai Ai ⎞ − jλz ∂ 2 Ai − jλz ⎜ ⎟ ⎜ ⎟ + − + = + − + e dz e dz ∫−∞ ⎜⎝ ∂r 2 r ∂r r 2 ∂z 2 ⎟⎠ ∫−∞ ⎜⎝ ∂r 2 r ∂r r 2 ⎟⎠ ∫−∞ ∂z 2 e dz   ~ ~ ~ ∞ ⎛ ∂ 2 Ai 1 ∂Ai Ai ⎞ − jλz d 2 Ai 1 dAi Ai ⎜ ⎟ ⋅ ∫⎜ 2 + − e dz = + − r ∂r r 2 ⎟⎠ dr 2 r dr r 2 − ∞⎝ ∂r

u = e − jλz → du = − jλe − jλz dz ∂ 2 Ai − jλz ∂A ⋅ ∫ 2 e dz = = e − jλ z i ∂ 2 Ai ∂Ai ∂z ∂z dv = dz → v = −∞ ∂z 2 ∂z ∞

= jλ e

− jλ z

Ai

∞ −∞

+j λ 2

∞ 2

∫ Ae i

− jλ z

∞

−∞

∞

∂Ai − jλz e dz = ∂z −∞

+ jλ ∫

~ dz = −λ2 Ai

−∞

~ ~ ~ ⎛ ∂ 2 Ai 1 ∂Ai Ai ∂ 2 Ai ⎞ − jλz d 2 Ai 1 dAi Ai 2~ ∫−∞ ⎜⎜⎝ ∂r 2 + r ∂r − r 2 + ∂z 2 ⎟⎟⎠e dz = dr 2 + r dr − r 2 − λ Ai ∞

(3.2.10)  48   

Applying (3.2.9) to the right-hand side of equation (3.2.2) we obtain ∞

− µ 0 Iδ (r − r0 ) ∫ δ ( z − z 0 ) e − jλz dz = − µ 0 Ie − jλz0 δ (r − r0 )                                                                        (3.2.11) 0

Similarly, the right-hand side of equation (3.2.4) is transformed to the form ∞

⎛ β i ∂Ai ⎛ β i ⎞ ⎞ + ⎜ 2 + jωσ im µ 0 µ im r α i + β i ⎟ Ai ⎟⎟e − jλz dz = ∂r ⎝ r ⎠ ⎠ −∞                                                                (3.2.12) ~ β i dAi ⎛ β i ~ ⎞ = + ⎜ + jωσ im µ 0 µ im r α i + β i ⎟ Ai , i = 1,2,..., n . r dr ⎝ r 2 ⎠ It follows from (3.2.10), (3.2.11) and (3.2.12) that ~ ~ d 2 A0 1 dA0 ⎛ 1 ⎞~ + − ⎜ 2 + λ2 ⎟ A0 = − µ0 Ie− jλz 0 δ (r − r0 ),                                                                             (3.2.13) 2 dr r dr ⎝ r ⎠ ~ ~ d 2 Ai (1 − β i ) dAi ⎛ 1 + β i ⎞~ + − ⎜ 2 + pi2 r α i + β i + λ2 ⎟ Ai = 0, i = 1,2,..., n,                                                (3.2.14) 2 dr r dr ⎝ r ⎠

∫ ⎜⎜⎝ r

where pi =

jωµ 0 µ imσ im , j is the imaginary unit and ω is the frequency of the current in the coil.

The transformed equation in region Rn is ~ ~ d 2 An +1 1 dAn +1 ⎛ 1 ⎞~ + − ⎜ 2 + λ2 ⎟ An +1 = 0.                                                                                                      (3.2.15) 2 dr r dr ⎠ ⎝r Applying the Fourier transform to the boundary conditions we obtain ~ ~ ~ ~ 1 dA1 dA0 (3.2.16) |r =r1 = r A0 |r =r1 = A1 |r =r1 , | , dr µ1 (r1 ) dr r =r1 ~ ~ ~ ~ dAi+1 1 dAi 1 Ai |r =ri +1 = Ai+1 |r =ri +1 , | , i = 1,2,.., n . (3.2.17) | = µir (ri+1 ) dr r =ri +1 µir+1 (ri+1 ) dr r =ri+1 ~ ~ In addition, A0 is bounded at r = 0 and An +1 is bounded as r → ∞ . (3.2.18) In order to find the solution to (3.2.13) we consider two sub-regions, R00 and R01 , of

region R0 , namely,  R00 = { 0 ≤ r < r0 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ } and

R01 = { r0 < r ≤ r1 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ } . The solutions in regions R00 and R01 are denoted by ~ ~ A00 and A01 , respectively. Hence, ~ ~ d 2 A00 1 dA00 ⎛ 1 ⎞~ (3.2.19) + − ⎜ 2 + λ2 ⎟ A00 = 0 , 0 ≤ r < r0 2 dr r dr ⎝ r ⎠ ~ ~ d 2 A01 1 dA01 ⎛ 1 ⎞~ (3.2.20) + − ⎜ 2 + λ2 ⎟ A01 = 0 , r > r0 2 dr r dr ⎝ r ⎠

The solution to (3.2.19) can be expressed in terms of modified Bessel functions:

 49   

y xx'' +

1 ' ⎛ v2 ⎞ y x − ⎜⎜1 + 2 ⎟⎟ y = 0 ⇒ y = Iν ( x ) x ⎝ x ⎠

⎧⎪ y ' = λy r' (λr ) ⋅ y = y (λr ) → ⎨ r'' ⎪⎩ y r = λ2 y r'' (λr ) λ ⎛1 ⎞ ⋅ λ2 y rr'' (λr ) + y r' (λr ) − ⎜ 2 + λ2 ⎟ y (λr ) = 0 r ⎝r ⎠ → y rr'' (λr ) +

1 ' 1 ⎞ ⎛ y r (λr ) − ⎜1 + 2 2 ⎟ y (λr ) = 0 → y = I1 (λr ) λr ⎝ λr ⎠

The general solution to (3.2.19) can be written in the form ~   A00 = B1I1 (λr ) + B2 K1 (λr ),                                                                                                                       where I1 (λr ) , K1 (λr ) are the modified Bessel functions of the first and second kind, respectively. The bounded solution to (3.2.19) in region 0 ≤ r < r0 is  ~ A00 (r , λ ) = B1 I1 (λr ).                                                                                                                                        (3.2.21)  The general solution to (3.2.20) can be written in the form ~ A01 (r , λ ) = B2 I1 (λr ) + B3 K1 (λr ).                                                                                                                  (3.2.22) ~ ~ The functions A00 (r , λ ) and A01 (r , λ ) satisfy the following conditions at r = r0 :  ~ ~ (3.2.23) A00 |r = r0 = A01 |r = r0 , ~ The condition (3.2.23) reflects the fact that the function A0 (r , λ ) is continuous at r = r0 . The second condition is obtained integrating (3.2.13) with respect to r from r = r0 − ε to r = r0 + ε and considering the limit in the resulting expression as ε → +0 :

~ ~ r0 +ε ⎛ d 2 A0 1 dA0 ⎛ 1 ⎞ − jλ z 0 2 ⎞~ ⎜ ⎟ ⋅ ∫ ⎜ + − ⎜ + λ ⎟ A0 ⎟dr = − µ 0 Ie 2 ∫ δ (r − r0 )dr r dr ⎝ r 2 ⎠ ⎠ r0 −ε r0 −ε ⎝ dr ~ r0 +ε r0 +ε ~ ⎛ 1 dA0 ⎛ 1 dA0 ⎞~ ⎞ + ∫ ⎜⎜ − ⎜ 2 + λ2 ⎟ A0 ⎟⎟dr = − µ 0 Ie − jλz0 dr r −ε r0 −ε ⎝ r dr ⎝ r ⎠ ⎠ 0 r0 +ε

1 1 → du = − 2 dr r r ~ dA ~ dv = 0 dr → v = A0 dr r +ε r0 + ε ~ r0 +ε ~ r0 +ε A A 1~ 0 ~ + A0 + ∫ 20 dr − ∫ 20 dr − ∫ λ2 A0 dr = − µ 0 Ie − jλz0 r r0 −ε r0 −ε r r0 −ε r r0 −ε u=

~ dA0 dr

r0 +ε

r0 −ε

 

 

 50   

~ dA0 ⋅ lim ε →0 dr ~ A0 ⋅ lim ε →0 r

r0 −ε

r0 +ε

r0 −ε

⋅ − λ2 lim ε →0

~ dA01 = dr

r0 + ε

1 ~ = ⎛⎜ A01 r⎝

r = r0

r = r0

r0 +ε

~

∫εA dz = − λ 0

r0 −

2

~ dA00 − dr ~ − A00

r = r0

r = r0

⎞⎟ = 0 ⎠

 

~ lim 2εA0 (r * , λ ) = 0 ε →0

 

~ dA01 dr

− r = r0

~ dA00 dr

= − µ 0 Ie − jλz0 .                                                                                                              (3.2.24) r = r0

Using (3.2.21) − (3.2.24) we obtain ⎧ B1 I1 (λr0 ) = B2 I1 (λr0 ) + B3 K1 (λr0 ) ⎨ − jλ z 0 ' ' ' ⎩λB2 I1 (λr0 ) + λB3 K1 (λr0 ) − λB1 I1 (λr0 ) = − µ 0 Ie

⇒

                                             

K1 (λr0 ) ⎧ B B B = + 1 2 3 ⎪ I1 (λr0 ) ⎪ ⎨ ' ⎪ B I ' (λr ) + B K ' (λr ) − B I ' (λr ) − B I1 (λr0 )K1 (λr0 ) = − µ 0 Ie − jλz0 3 1 0 2 1 0 3 ⎪⎩ 2 1 0 I1 (λr0 ) λ ⋅ B3 (I1 (λr0 )K1' (λr0 ) − I1' (λr0 )K1 (λr0 )) = −             

W {I v ( z ), K v ( z )} = −

⎧⎪ B3 = µ0 r0 Ie − jλz 0 I1 (λr0 ) ⎨ ⎪⎩ B1 = B2 + µ0 r0 Ie − jλz 0 K1 (λr0 ),

1 z

⇒

 

µ 0 − jλ z Ie I1 (λr0 ) λ   0

                                                                                                                (3.2.25)

where W {Iν ( z ), Kν ( z )} is the Wronskian. It follows from (3.2.21), (3.2.22) and (3.2.25) that ~ A00 (r , λ ) = B2 I1 (λr ) + µ 0 r0 Ie − jλz0 K1 (λr0 )I1 (λr ) ,                                                                                    (3.2.26) ~ A01 (r , λ ) = B2 I1 (λr ) + µ 0 r0 Ie − jλz0 I1 (λr0 )K1 (λr ).                                                                                      (3.2.27) In order to understand the meaning of each of the two terms on the right-hand sides of (3.2.26) and (3.2.27) we consider the case where a coil of radius r0 centered at the point (0, z 0 ) is located in an unbounded free space. The corresponding component of the vector potential will be denoted by A0free (r , z ) . The equation for A0free is ∂ 2 A0free 1 ∂A0free A0free ∂ 2 A0free = − µ 0 Iδ (r − r0 )δ ( z − z 0 ),                                                         (3.2.28) − 2 + + r ∂r r ∂z 2 ∂r 2 where δ (x) is the Dirac delta-function.  51   

In addition, A0free is bounded at r = 0 and A0free is bounded as r → ∞ . Applying the Fourier transform (3.2.9) and using (3.2.8) and (3.2.13) we get ~ ~ d 2 A0free 1 dA0free ⎛ 1 ⎞~ (3.2.29) + − ⎜ 2 + λ2 ⎟ A0free = − µ 0 Ie − jλz0 δ (r − r0 ) 2 dr r dr ⎠ ⎝r The bounded solutions in regions R00 and R01 are : ~ ~ d 2 A00free 1 dA00free ⎛ 1 ⎞~ + − ⎜ 2 + λ2 ⎟ A00free = 0 , R00 : 0 ≤ r < r0 2 dr r dr ⎝r ⎠ ~ free                                                   → A00 (r , λ ) = B4 I1 (λr ).                                                                            (3.2.30) ~ ~ d 2 A01free 1 dA01free ⎛ 1 ⎞~ + − ⎜ 2 + λ2 ⎟ A01free = 0 , R01 :   r > r0 2 dr r dr ⎝r ⎠ ~ → A01free (r , λ ) = B5 K1 (λr ). (3.2.31) Using (3.2.23) and (3.2.24) solutions (3.2.30) and (3.2.31) can be rewritten as follows    ~ A00free (r , λ ) = µ 0 r0 Ie − jλz0 K1 (λr0 )I1 (λr ) ,                                                                                                      (3.2.32) ~ A01free (r , λ ) = µ 0 r0 Ie − jλz0 I1 (λr0 )K1 (λr ).                                                                                                       (3.2.33) Comparing (3.2.26), (3.2.27) and (3.2.32), (3.2.33) we see that the second terms on the right-hand sides of (3.2.26) and (3.2.27) represent the vector potential in an unbounded free space. Thus, the first term of the right-hand sides of (3.2.26) and (3.2.27) represents the induced vector potential due to the presence of the conducting medium. The solution to (3.2.14) can be expressed in terms of different special functions. For example, the solution to (3.2.14) for the case  α i = −1, β i = −1 (see [50]) is ~ ~ d 2 Ai 2 dAi ⎛ 2 pi2 ⎞ ~ + − ⎜ λ + 2 ⎟⎟ Ai = 0, i = 1,2,..., n   dr 2 r dr ⎜⎝ r ⎠

1 − 2a ' ⎛ 2 a 2 − v 2 ⎞ ⎟⎟ y = 0 → y ( z ) = z a Z v (bz ) y z + ⎜⎜ b + 2 z z ⎝ ⎠ 1 ⋅ 1 − 2a = 2 → a = −   2 2 2 b = λ → b = jλ ⋅ y zz'' +

1 1 pi2 = −a 2 + vi2 = − + vi2 → vi = + pi2 4 4 where pi =

jωµ 0 µ imσ im .  

(

)

1 ~ Ai (r , λ ) = B4i I vi (λr ) + B5i K vi (λr ) . (3.2.34) r The bounded solution to (3.2.5) is ~   An +1 (r , λ ) = B6 K1 (λr ).                                                                                                                                   (3.2.35)  Unknown constants in (3.2.26), (3.2.27), (3.2.34) and (3.2.35) can be found from the boundary conditions (3.2.16) – (3.2.18). Then the solution in each region Ri , i = 0,1,..., n + 1 can be found  52   

by means of the inverse Fourier transform of the form Ai (r , z ) =

1 2π

+∞

~

∫ A (r , λ )e

jλ z

i

dλ .

(3.2.36)

−∞

It can be shown that the induced vector potential in free space due to multilayer conducting tubes has the form (see [3]) A0ind (r , z ) =

1 2π

+∞

∫ B I (λr )e 2 1

jλ z

dλ .

(3.2.37)

−∞

3.3 A coil inside a cylindrical region In this section we consider one particular case of the problem analyzed in Section 3.2 where the tube consists of one unbounded conducting layer [32]. Suppose that a single-turn coil with alternating current is located inside a cylindrical region R1 = { r ≥ r1 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ } . The geometry of the problem is shown in Fig. 3.2,

(

R 0 µ 0r = 1, σ 0 = 0

r0

r1

(

R1 µ1r , σ 1

)

)

Fig. 3.2 A single-turn coil inside a cylindrical region.

where µ 0r (r ) = 1, σ 0 (r ) = 0,               µ1r (r ) = µ1m r β , σ 1 (r ) = σ 1m r α ,             Using the results from Section 3.2 we obtain the following problem :  ∂ 2 A 1 ∂A0 A0 ∂ 2 A0 − + = − µ 0 Iδ (r − r0 )δ ( z − z 0 ),                                                                 (3.3.1) R0 :    20 + ∂r ∂z 2 r ∂r r 2

R1 :  

∂ 2 A1 ∂ 2 A1 1 − β ∂A1 ⎛ 1 + β m m α +β ⎞ j r A ωµ µ σ + = 0.   + − + ⎟ ⎜ 0 1 1 1 r ∂r ⎝ r 2 ∂z 2 ∂r 2 ⎠

 

The boundary conditions are ∂A0 1 ∂A1 |r = r1 = r |r = r , A0 |r = r1 = A1 |r = r1 , ∂r µ1 (r1 ) ∂r 1

            (3.3.2)  

(3.3.3)

In addition, A0 is bounded at r = 0 and A1 is bounded as r → ∞ . The following conditions hold at infinity: ∂A Ai , i → 0 as z → ±∞ , i = 0,1. ∂z

(3.3.4)  53   

Applying the Fourier transform (3.2.9) to problem (3.3.1) – (3.3.4) and using (3.2.26) – (3.2.27), we obtain the solution in regions Ri , i = 0,1 : ~ ~ d 2 A0 1 dA0 ⎛ 1 ⎞~ + − ⎜ 2 + λ2 ⎟ A0 = − µ0 Ie − jλz 0 δ (r − r0 ),        R0 :     2 dr r dr ⎝ r ⎠ ~ ⎧⎪ A00 (r , λ ) = B2 I1 (λr ) + µ0 r0 Ie− jλz 0 K1 (λr0 )I1 (λr ),                                 →      ⎨ ~                                        (3.3.5) ⎪⎩ A01 (r , λ ) = B2 I1 (λr ) + µ0 r0 Ie− jλz 0 I1 (λr0 )K1 (λr ), ~ ~ d 2 A1 (1 − β ) dA1 ⎛ 1 + β ⎞~ R1 : + − ⎜ 2 + p12 r α + β + λ2 ⎟ A1 = 0.   2 dr r dr ⎝ r ⎠ This equation has an analytical solution, for example, for the case α = −1, β = −1 . The bounded general solution of the corresponding equation has the form ~ 1 A1 (r , λ ) = B5 K v (λr ), (3.3.6) r 1 + p12 . 4

jωµ 0 µ1mσ 1m , v =

where p1 =

The boundary conditions are ~ ~ dA0 1 dA1 ~ ~ |r = r1 = r A0 |r = r1 = A1 |r = r1 , |r = r . dr µ1 (r1 ) dr 1

(3.3.7)

Using (3.3.5) − (3.3.7) we obtain

1 ⎧ − jλz 0 I1 (λr0 )K1 (λr1 ) = B5 K v (λr1 ), ⎪ B2 I1 (λr1 ) + µ0 r0 Ie r1 ⎪ ⎨ ⎪λB I ' (λr ) + λµ r Ie− jλz 0 I (λr )K ' (λr ) = r1 ⎛⎜ − 1 B K (λr ) + λ B K ' (λr )⎞⎟. 0 0 1 0 1 1 5 v 1 ⎟ ⎪ 21 1 µ1m ⎜⎝ 2r1 r1 5 v 1 r1 ⎠ ⎩

1 K v (λr1 ) K (λr ) − µ 0 r0 Ie − jλz0 I1 (λr0 ) 1 1 I1 (λr1 ) r1 I1 (λr1 )

⋅ B2 = B5 ⋅ B5

λ r1

=−

(3.3.8)

I1' (λr1 ) 1

2µ

m 1

1 2µ

m 1

r1

r1

K v (λr1 ) K (λr ) − λµ 0 r0 Ie − jλz0 I1 (λr0 )I1' (λr1 ) 1 1 + λµ 0 r0 Ie − jλz0 I1 (λr0 )K1' (λr1 ) = I1 (λr1 ) I1 (λr1 )

(

)

B5 K v (λr1 ) − 2r1λK v' (λr1 )

(

(

)

)

B5 K v (λr1 ) 2µ1mλI1' (λr1 ) + I1 (λr1 ) − 2r1λK v' (λr1 )I1 (λr1 ) =

(

)

= λµ0 r0 Ie− jλz 0 I1 (λr0 ) K1 (λr1 )I1' (λr1 ) − I1 (λr1 )K1' (λr1 )

B5 =

2µ1m µ0 r0 Ie− jλz 0 I1 (λr0 ) ,                                                             (3.3.9) m ' ' r1 2µ1 λI1 (λr1 ) + I1 (λr1 ) K v (λr1 ) − 2r1λK v (λr1 )I1 (λr1 )

(

)

where the formula for the Wronskian W { K v (λr1 ), I v (λr1 ) } =

1 can be used. λr1  54   

In particular, the value of B2 is

B2 = µ0 r0 Ie

− jλ z 0

⎞ I1 (λr0 ) ⎛ 2µ1m K v (λr1 ) ⎜⎜ − K1 (λr1 )⎟⎟,                                       (3.3.10) ' I1 (λr1 ) ⎝ r1 D1K v (λr1 ) − 2r1λK v (λr1 )I1 (λr1 ) ⎠

1 + p12 , p1 = 4 The induced vector potential in region R0 is   ~ A0ind (r , λ ) = B2 I1 (λr ),

where D1 = 2µ1m λI1' (λr1 ) + I1 (λr1 ) , v =

jωµ 0 µ1mσ 1m .

(3.3.11)

where B2 is given by (3.3.10). Applying the inverse Fourier transform (3.2.36) to (3.3.11) we obtain the induced vector potential in free space due to the presence of a cylindrical region in the form A0ind (r , z ) =

µ 0 r0 I 2π

∞

~

I1 (λr0 )

∫ B I (λr ) I (λr ) e 2 1

−∞

1

jλ ( z − z 0 )

dλ ,

(3.3.12)

1

K v (λr1 ) ~ 2µ m where B2 = 1 − K1 (λr1 ) r1 D1 K v (λr1 ) − 2r1λK v' (λr1 )I1 (λr1 ) The induced change in impedance of the coil is given by the formula jω Z ind = A0ind (r , z )dl , I ∫L

(3.3.13)

where L is the contour of the coil. Substituting (3.3.12) into (3.3.13) we obtain Z ind = jωµ 0 r02

~ I12 (λr0 ) B ∫ 2 I1 (λr1 ) dλ. −∞ ∞

(3.3.14)

Using the following dimensionless parameters (3.3.14) can be rewritten as follows

⎫ ⎪ 1 ⎪ + jpˆ 12 , ⋅ pˆ 1 = ωµ 0 µˆ1mσˆ1m r12 ⎬ ⇒ νˆ = r 4 rˆ0 = 0 , µ1m = µˆ1m r1 , σ 1m = σˆ1m r1 ,⎪ ⎪⎭ r1 ⋅ Dˆ 1 = 2µˆ1m uI1' (u ) + I1 (u )

⋅ λr1 = u → λ =

Z ind (r , z ) = ωµ 0

u , r1

r02 Z, r1

∞ ~ˆ I 2 (urˆ ) where Z = j ∫ B2 1 0 du I1 (u ) −∞

and ~ˆ             B2 =

(3.3.15) (3.3.16)

2µˆ1m K vˆ (u ) − K1 (u ).                                                                                                                                 Dˆ 1K vˆ (u ) − 2uK vˆ' (u )I1 (u )

Formula (3.3.16) is used to compute the change in impedance of the coil. Calculations are performed with “Mathematica” since it has built-in functions to evaluate modified Bessel functions of complex order. “Mathematica” program which is used to compute the change in impedance (3.3.16) is shown in Appendix Fig.Ap.5. The results of computations are shown in  55   

Fig. 3.3 . The calculated points (from top to bottom) for each curve correspond to the following

values of pˆ 1 = 1,2,...,15 .The parameter pˆ 1 is defined by pˆ 1 = ωµ 0 µˆ1mσˆ1m r12 . The three curves in Fig. 3.3 (from right to left) correspond to the cases rˆ0 = 0.9, 0.8, and 0.7, respectively. Im @zD 0.05

0.075 -0.1

0.125

0.15

0.175

0.2

          rˆ0 = 0.9  

Re @zD

                         rˆ0 = 0.8  

-0.2

                       rˆ0 = 0.7  

-0.3 -0.4 -0.5

Fig. 3.3 The change in impedance of a coil for different values of rˆ0 .

-0.6

It is seen that the modulus of the change in impedance increases as the parameter pˆ 1 increases (that is, if the frequency increases). In addition, the change in impedance is larger when the coil is closer to the cylindrical region.

3.4 A coil inside a two-layer tube Another example of important problem in applications is considered in this section. Consider a single-turn coil of radius r0 located inside a two-layer tube [33].

(

R 0 µ 0r = 1, σ 0 = 0

r2

r0

(

)

(

)

R1 µ1r , σ 1 r1

R 2 µ 2r , σ 2

)

Fig. 3.4 A coil inside a two-layer tube.

Suppose that µ 0r (r ) = 1, σ 0 (r ) = 0,  

µ1r (r ) = µ1m r β , σ 1 (r ) = σ 1m r α ,  

                       µ 2r (r ) = µ 2m , σ 2 (r ) = σ 2m .    The electric conductivity and magnetic permeability of the inner layer are given by formula (3.2.1) while the electric conductivity σ 2m and magnetic permeability µ 2m of the second layer ( R2 ) are constants. The outer layer is unbounded in the radial direction (see Fig. 3.4).           Using the results from Section 3.2 we obtain the following equations:  56   

R0 :   

R1 :  

∂ 2 A0 1 ∂A0 A0 ∂ 2 A0 + − + = − µ 0 Iδ (r − r0 )δ ( z − z 0 ),                                                                 (3.4.1) ∂r 2 r ∂r r 2 ∂z 2

∂ 2 A1 1 − β ∂A1 ⎛ 1 + β ∂ 2 A1 2 α +β ⎞ p r A + − + + = 0,   ⎜ ⎟ 1 1 r ∂r ⎝ r 2 ∂r 2 ∂z 2 ⎠

where p1 =

            (3.4.2)

jωµ 0 µ1mσ 1m . 

The equation in region R2 is

R2 :  

∂ 2 A2 1 ∂A2 ⎛ 1 ∂ 2 A2 2⎞ + − + p A + = 0,   ⎜ 2 ⎟ 2 ∂r 2 r ∂r ⎝ r 2 ∂z 2 ⎠

where p2 =

            (3.4.3)

jωµ 0 µ 2mσ 2m .

The boundary conditions are ∂A0 1 ∂A1 |r = r1 = r |r = r , A0 |r = r1 = A1 |r = r1 , ∂r µ1 (r1 ) ∂r 1

A1 |r = r2 = A2 |r = r2 ,

1 ∂A1 1 ∂A |r = r2 = m 2 |r = r2 . µ (r2 ) ∂r µ 2 ∂r r 1

(3.4.4) (3.4.5)

In addition, A0 is bounded at r = 0 and A2 is bounded as r → ∞ . The following conditions hold at infinity: ∂A Ai , i → 0 as z → ±∞ , i = 0,1,2. (3.4.6) ∂z Applying the Fourier transform (3.2.9) to problem (3.4.1) - (3.4.6) we obtain the solution in region R0 : ~ ~ d 2 A0 1 dA0 ⎛ 1 ⎞~ + − ⎜ 2 + λ2 ⎟ A0 = − µ0 Ie − jλz 0 δ (r − r0 ),        R0 :     2 dr r dr ⎝ r ⎠ ~ ⎧⎪ A00 (r , λ ) = B2 I1 (λr ) + µ0 r0 Ie− jλz 0 K1 (λr0 )I1 (λr ),                                 →      ⎨ ~                                        (3.4.7) ⎪⎩ A01 (r , λ ) = B2 I1 (λr ) + µ0 r0 Ie− jλz 0 I1 (λr0 )K1 (λr ). The transformed equation in region R1 is ~ ~ d 2 A1 (1 − β ) dA1 ⎛ 1 + β ⎞~ R1 : + − ⎜ 2 + p12 r α + β + λ2 ⎟ A1 = 0. 2 dr r dr ⎝ r ⎠ The general solution for the case α = −1, β = −1 has the form

~ 1 (B4 I v (λr ) + B5 Kv (λr )), → A1 (r , λ ) = r where p1 =

jωµ 0 µ1mσ 1m , v =

(3.4.8)

1 + p12 .  4

The transformed equation in region R2 is ~ ~ d 2 A 1 dA2 ⎛ 1 ⎞~ R2 :   22 + − ⎜ 2 + q 2 ⎟ A2 = 0 ,   dr r dr ⎝ r ⎠  57   

so that the bounded solution of the above equation as r → ∞ is given by ~ → A2 (r , λ ) = B6 K1 (qr ), p 22 + λ2 , p 2 =

where q =

(3.4.9)

jωµ 0 µ 2mσ 2m .

The boundary conditions are ~ ~ dA0 1 dA1 ~ ~ A0 |r = r1 = A1 |r = r1 , |r = r1 = r dr µ1 (r1 ) dr ~ 1 dA1 1 ~ ~ |r = r2 = m A1 |r = r2 = A2 |r = r2 , r µ1 (r2 ) dr µ2

|r = r1 ,

(3.4.10)

~ dA2 |r = r2 . dr

(3.4.11)

Using (3.4.7) − (3.4.11) we obtain

B4 B I v (λr1 ) + 5 K v (λr1 ), r1 r1

(3.4.12)

r1 ⎛⎜ 1 ⎞⎟ − ⋅ µ1m ⎜⎝ 2r1 r1 ⎟⎠ ⋅ B4 I v (λr1 ) − 2r1λI v' (λr1 ) + B5 K v (λr1 ) − 2r1λK v' (λr1 ) ,

(3.4.13)

B2 I1 (λr1 ) + µ0 r0 Ie− jλz 0 I1 (λr0 )K1 (λr1 ) =

λB2 I1' (λr1 ) + λµ0 r0 Ie− jλz I1 (λr0 )K1' (λr1 ) = 0

( (

µ

m 2

(

))

B B4 I v (λr2 ) + 5 K v (λr2 ), r2 r2

B6 K1 (qr2 ) =

B6

)

(3.4.14)

r2 ⎛⎜ 1 ⎞⎟ ⋅ − m ⎜ µ1 ⎝ 2r2 r2 ⎟⎠ ⋅ B4 I v (λr2 ) − 2r2λI v' (λr2 ) + B5 K v (λr2 ) − 2r2λK v' (λr2 ) .

qK1' (qr2 ) =

( (

)

(

(3.4.15)

))

It follows from (3.4.12) and (3.4.13) that

⋅ B2 = ⋅

B4 I v (λr1 ) B5 K v (λr1 ) K (λr ) + − µ0 r0 Ie − jλz0 I1 (λr0 ) 1 1 I1 (λr1 ) r1 I1 (λr1 ) r1 I1 (λr1 )

B λ K v (λr1 ) ' B4 λ I v (λr1 ) ' K (λr ) I1 (λr1 ) + 5 I1 (λr1 ) − λµ 0 r0 Ie − jλz0 I1 (λr0 ) 1 1 I1' (λr1 ) + I1 (λr1 ) r1 I1 (λr1 ) r1 I1 (λr1 ) + λµ0 r0 Ie

⋅

B4 2µ

m 1

r1

=−

− jλz 0

( I (λr )(2µ B5

2µ

m 1

m 1

1

v

r1

(

)

' r1 ⎛⎜ 1 ⎞⎟⎛⎜ B4 I v (λr1 ) − 2r1λI v (λr1 ) I1 (λr0 )K (λr1 ) = m − µ1 ⎜⎝ 2r1 r1 ⎟⎠⎜⎝ + B5 K v (λr1 ) − 2r1λK v' (λr1 ) ' 1

)

λI1' (λr1 ) + I1 (λr1 )) − 2r1λI v' (λr1 )I1 (λr1 ) ) =

( K (λr )(2µ v

(

⎞ ⎟ ⎟ ⎠

m 1

1

(

λI1' (λr1 ) + I1 (λr1 )) − 2r1λK v' (λr1 )I1 (λr1 ) ) +

)

+ λµ 0 r0 Ie − jλz0 I1 (λr0 ) K1 (λr1 )I1' (λr1 ) − I1 (λr1 )K1' (λr1 )

 58   

( (

) )

K v (λr1 ) 2µ1mλI1' (λr1 ) + I1 (λr1 ) − 2r1λK v' (λr1 )I1 (λr1 ) + I v (λr1 ) 2µ1mλI1' (λr1 ) + I1 (λr1 ) − 2r1λI v' (λr1 )I1 (λr1 )

B4 = − B5

(3.4.16)

2µ1m µ0r0 Ie− jλz 0 I1 (λr0 ) , + m ' ' r1 I v (λr1 ) 2µ1 λI1 (λr1 ) + I1 (λr1 ) − 2r1λI v (λr1 )I1 (λr1 )

(

where W { K1 (λr1 ), I1 (λr1 ) } =

)

1 . λr1

Using (3.4.14) and (3.4.15) we obtain

⋅ B6 = ⋅

B4 I v (λr2 ) B5 K v (λr2 ) + r2 K1 (qr2 ) r2 K1 (qr2 )

B q K v (λr2 ) ' B4 q I v (λr2 ) ' B I v (λr2 ) − 2r2 λI v' (λr2 ) − K1 (qr2 ) + m 5 K1 (qr2 ) = − m 4 m µ 2 r2 K1 (qr2 ) µ 2 r2 K1 (qr2 ) 2µ1 r2

(

− ⋅ B4

B5 2µ

m 1

r2

)

(K (λr ) − 2r λK (λr )) v

2

2

' v

2

⎛ m K1' (qr2 ) ⎞ m ' m ⎜ ⎟= ( ) ( ) ( ) I r r I r qI r µ λ 2 λµ λ 2 µ λ − + v 2 v 2 2 2 v 2 1 2 2µ1m µ 2m ⎜⎝ K1 (qr2 ) ⎟⎠ 1 ⎛ m K1' (qr2 ) ⎞ m m ' ⎜ ⎟ = − B5 m m ⎜ µ 2 K v (λr2 ) − 2r2 λµ 2 K v (λr2 ) + 2µ1 qK v (λr2 ) 2µ1 µ 2 ⎝ K1 (qr2 ) ⎟⎠ 1

B4 = − B5

µ2m K1 (qr2 )(K v (λr2 ) − 2r2λK v' (λr2 )) + 2µ1m qKv (λr2 )K1' (qr2 ) . µ2m K1 (qr2 )(I v (λr2 ) − 2r2λI v' (λr2 )) + 2µ1m qI v (λr2 )K1' (qr2 )

(3.4.17)

Solving equations (3.4.16) and (3.4.17) we obtain

2µ1m C µ0r0 Ie− jλz0 I1 (λr0 ) 1 , B5 = D r1 where

(

(3.4.18)

)

C1 = µ2m K1 (qr2 ) I v (λr2 ) − 2r2λI v' (λr2 ) + 2µ1m qI v (λr2 )K1' (qr2 ), D = µ K1 (qr2 )D5 ( D1 − 2r2λD2 ) + 2µ1m qK1' (qr2 )D1D5 − m 2

− 2r1λµ2m K1 (qr2 )I1 (λr1 )( D3 − 2r2λD4 ) − 4r1µ1mλqK1' (qr2 )I1 (λr1 )D3.

D1 = I v (λr2 )K v (λr1 ) − I v (λr1 )K v (λr2 )

D2 = I v' (λr2 )K v (λr1 ) − I v (λr1 )K v' (λr2 ) D3 = I v (λr2 )K v' (λr1 ) − I v' (λr1 )K v (λr2 )

D4 = I v' (λr2 )K v' (λr1 ) − I v' (λr1 )K v' (λr2 ) D5 = 2µ1m λI1' (λr1 ) + I1 (λr1 )

The constant B4 is found from (3.4.17) and (3.4.18):

 59   

B4 = −

2µ1m C µ0r0 Ie− jλz0 I1 (λr0 ) 2 , D r1

where

(3.4.19)

(

)

C2 = µ2m K1 (qr2 ) Kv (λr2 ) − 2r2λKv' (λr2 ) + 2µ1m qKv (λr2 )K1' (qr2 ). The value of B2 is            ⋅ B2 =

B4 I v (λr1 ) B5 K v (λr1 ) K (λr ) + − µ 0 r0 Ie − jλz0 I1 (λr0 ) 1 1 I1 (λr1 ) r1 I1 (λr1 ) r1 I1 (λr1 )

B2 = µ0r0 Ie− jλz0

 

⎞ I1 (λr0 ) ⎛ 2µ1m µ2m K1 (qr2 )( D1 − 2r2λD2 ) + 2µ1m qK1' (qr2 )D1 ⎜⎜ − K1 (λr1 ) ⎟⎟. I1 (λr1 ) ⎝ r1 D ⎠

The induced vector potential in region R0 is ~ A0ind (r , λ ) = B2 I1 (λr ),

(3.4.20)

(3.4.21)

where B2 is given by (3.4.20). Applying the inverse Fourier transform (3.2.36) to (3.4.21) we obtain the induced vector potential in free space due to the presence of a cylindrical region in the form A0ind (r , z ) =

where

µ 0 r0 I 2π

∞

~

I1 (λr0 )

∫ B I (λr ) I (λr ) e 2 1

−∞

1

jλ ( z − z 0 )

dλ ,

(3.4.22)

1

~ 2µ1m µ2m K1 (qr2 )( D1 − 2r2λD2 ) + 2µ1m qK1' (qr2 )D1 B2 = − K1 (λr1 ). r1 D

The induced change in impedance of the coil is given by the formula jω A0ind (r , z )dl , Z ind = I ∫L

(3.4.23)

where L is the contour of the coil. Substituting (3.4.22) into (3.4.23) we obtain Z ind = jωµ 0 r02

~ I12 (λr0 ) B ∫ 2 I1 (λr1 ) dλ. −∞ ∞

(3.4.24)

Formula (3.4.24) can be rewritten as follows

⎫ ⎪ 1 ⎪ + jpˆ12 , ⋅ pˆ1 = ωµ0 µˆ1mσˆ1m r12 ⎬ ⇒ νˆ = r 4 rˆ0 = 0 , µ1m = µˆ1m r1 , σ 1m = σˆ1m r1 ,⎪ ⎪⎭ r1

⋅ λr1 = u → λ =

⋅ rˆ2 =

u , r1

µ 2mσ 2m r2 2 2 2 m m 2 m m ˆ ˆ ˆ ˆ , q = u + jp2 , p2 = ωµ0 µ 2 σ 2 r1 = ωµ0 r1 µ2 σ 2 = p1 µˆ1mσˆ1m r1

⋅ Dˆ 1 = I vˆ (urˆ2 )K vˆ (u ) − I vˆ (u )K vˆ (urˆ2 ), Dˆ 2 = I vˆ' (urˆ2 )K vˆ (u ) − I vˆ (u )K vˆ' (urˆ2 ) ⋅ Dˆ 3 = I vˆ (urˆ2 )K vˆ' (u ) − I vˆ' (u )K vˆ (urˆ2 ), Dˆ 4 = I vˆ' (urˆ2 )K vˆ' (u ) − I vˆ' (u )K vˆ' (urˆ2 ) . Dˆ 5 = 2µˆ1m uI1' (u ) + I 1 (u ), Dˆ 6 = µ 2m K1 (qˆrˆ2 ) Dˆ 1 − 2urˆ2 Dˆ 2 . Dˆ = 2µˆ m qˆK ' (qˆrˆ )Dˆ

(

7

1

1

2

)

1

 60   

Z ind (r , z ) = ωµ 0

r02 Z                                                                                                                                      (3.4.25) r1

~ˆ I12 (urˆ0 ) where          Z = j ∫ B2 du                                                                                                     (3.4.26) I1 (u ) −∞ ∞

and

~ˆ B2 =

( (

)

2µˆ1m Dˆ 6 + Dˆ 7 − K1 (u ). Dˆ 5 Dˆ 6 + Dˆ 7 − 2uµ2m K1 (qˆrˆ2 )I1 (u ) Dˆ 3 − 2urˆ2 Dˆ 4 − 4µˆ1muqˆK1' (qˆrˆ2 )I1 (u )Dˆ 3

(

)

)

Formula (3.4.26) is used to compute the change in impedance of the coil for different values of the parameters of the problem. Fig. 3.5 plots the real and imaginary parts of the change in impedance for three different values of rˆ2 : 1.1, 1.2 and 1.3. The curves correspond to the following values of pˆ 1 = 3,4,...,10 (from top to bottom). The other parameters are µ1 = 1, µˆ 2 = 6, rˆ0 = 0.9,

µ 2mσ 2m = 1.5. Computations are done with µˆ1mσˆ1m

Mathematica ( see Appendix Fig.Ap.6 ) . Im @zD

0.25

0.3

0.35

0.4

          rˆ2 = 1.1  

Re @zD

                         rˆ2 = 1.2  

-0.1

                       rˆ2 = 1.3  

-0.2 -0.3 -0.4

Fig. 3.5 The change in impedance computed by formula (3.4.26) for three different values of rˆ2 .

-0.5 -0.6

It is seen from Fig. 3.5 that for higher frequencies (larger values of pˆ 1 ) the modulus of the change in impedance decreases. The decrease is related to smaller values of the real part of the change in impedance. The values of Z for three different values of µˆ 2 , namely, µˆ 2 = 2, 4, 6 are plotted in Fig. 3.6. The points on each curve correspond to the following values of pˆ 1 = 3, 4,...,10 (from top to bottom). The other parameters are set at µ1 = 1, rˆ2 = 1.1, rˆ0 = 0.9, Im @zD

0.25 -0.1 -0.2 -0.3

0.3

0.35

0.4

Re @zD

µ 2mσ 2m = 1.5. µˆ1mσˆ1m           µˆ 2 = 2                            µˆ 2 = 4                          µˆ 2 = 6  

-0.4 -0.5 -0.6

Fig. 3.6 The change in impedance computed by formula (3.4.26) for three different values of µˆ 2 .  61   

It follows from Fig. 3.6 that for high frequencies (large values of pˆ 1 ) the calculated points are very close to one another as the parameter µˆ 2 increases.

3.5 A coil outside a multilayer tube There are many different types of eddy current probes that are used to control the properties of objects with cylindrical symmetry. Encircling coils are widely used for inspecting cylinders, rods or tubes. Mathematical models described in the previous sections can also be applied for the case of encircling coils. Consider a multilayer tube described in Section 3.2. We assume that a single-turn coil of radius r0 is located outside the tube ( r0 > r1 ). The axis of the coil coincides with the axis of the tube [35] (see Fig. 3.7).   R n+1 (µ nr+1 , σ n +1 )

r1

...

      

R i (µ ir , σ i )

r0 rn +1

...

i = 1,2,..., n  

R 0 (µ 0r = 1, σ 0 = 0 )

Fig. 3.7 A single-turn coil outside a multilayer tube.    

The system of equations for the amplitudes of the vector potential in each region Ri = { ri ≤ r ≤ ri +1 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ }, i = 1,2,..., n is given by (3.2.2), (3.2.4) and (3.2.5) where r0 (the radius of the coil) is larger than r1 (the radius of the inner cylinder of the tube). The solution procedure is essentially the same as in Section 3.2 with minor modifications. Applying the Fourier transform to (3.2.2), (3.2.4) and (3.2.5) we obtain the system of equations in the form ~ (3.2.13) − (3.2.15). The boundary conditions (3.2.16), (3.2.17) are the same. In addition, A0 is ~ bounded at r → ∞ and An +1 is bounded as r = 0 . As a result, the solution in regions R0 and Rn +1 has to be modified. We consider two sub-regions of region R0 , namely, R00 and R01 , which

are defined as follows: R00 = { r1 ≤ r < r0 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ },

R01 = { r > r0 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ }.The solutions in R00 and R01 are denoted, as before, by ~ ~ A00 and A01 , respectively. Solving (3.2.13) in regions R00 and R01 we obtain ~ ~ d 2 A0 1 dA0 ⎛ 1 ⎞~ + − ⎜ 2 + λ2 ⎟ A0 = − µ0 Ie− jλz 0 δ (r − r0 ), (3.5.1) R0 : 2 dr r dr ⎝ r ⎠ ~ ⎧⎪ A00 (r , λ ) = B1I1 (λr ) + B2 K1 (λr ),                                 →      ⎨ ~                                                                   (3.5.2) ⎪⎩ A01 (r , λ ) = B3 K1 (λr ). ~ ~ The functions A00 (r , λ ) and A01 (r , λ ) satisfy the following conditions at r = r0 :   ~ ~ (3.5.3) A00 |r =r0 = A01 |r =r0 . ~ The condition (3.5.3) reflects the fact that the function A0 (r , λ ) is continuous at r = r0 .

 62   

The second condition is obtained by integrating (3.5.1) with respect to r from r = r0 − ε to r = r0 + ε and considering the limit in the resulting expression as ε → +0 : ~ ~ dA01 dA00 − = − µ 0 Ie − jλz0 .                                                                                                                (3.5.4) dr r = r dr r = r 0

0

Using (3.5.2) − (3.5.4) we obtain ⎧ B1I1 (λr0 ) + B2 K1 (λr0 ) = B3 K1 (λr0 ), ⇒ ⎨ − jλ z 0 ' ' ' , ⎩λB3 K1 (λr0 ) − λB1I1 (λr0 ) − λB2 K1 (λr0 ) = − µ0 Ie

                                            

I1 (λr0 ) ⎧ ⎪ B3 = B2 + B1 K (λr ) , ⎪ 1 0 ⇒ ⎨ ' ⎪ B K ' (λr ) + B I1 (λr0 )K1 (λr0 ) − B I ' (λr ) − B K ' (λr ) = − µ0 Ie − jλz 0 , 1 1 1 0 2 1 0 ⎪⎩ 2 1 0 λ K1 (λr0 )

(

)

⋅ B1 I1 (λr0 )K1' (λr0 ) − I1' (λr0 )K1 (λr0 ) = −                     

µ 0 − jλ z Ie K1 (λr0 ) λ

 

0

1 W {I v (λr0 ), K v (λr0 )} = − λr0

 

⎧⎪ B1 = µ0 r0 Ie − jλz 0 K1 (λr0 ),                                                                                                                     (3.5.5) ⎨ ⎪⎩ B3 = B2 + µ0 r0 Ie − jλz 0 I1 (λr0 ). It follows from (3.5.2) and (3.5.5) that ~ A00 (r , λ ) = B2 K1 (λr ) + µ 0 r0 Ie − jλz0 K1 (λr0 )I1 (λr ) ,                                                                                     (3.5.6) ~ A01 (r , λ ) = B2 K1 (λr ) + µ 0 r0 Ie − jλz0 I1 (λr0 )K1 (λr ).                                                                                       (3.5.7)

The solution in other conducting layers can be constructed as in Section 3.2. For example, if α = −1, β = −1 the solution is ~ ~ d 2 Ai (1 − β i ) dAi ⎛ 1 + β i ⎞~ + − ⎜ 2 + pi2 r α i + β i + λ2 ⎟ Ai = 0, Ri : 2 dr r dr ⎝ r ⎠ 1 ~ B4 i I vi (λr ) + B5i K vi (λr ) → Ai (r , λ ) = r

(

1 + pi2 , pi = jωµ 0 µ imσ im , i = 1,2,..., n   4 In region Rn +1 we have ~ ~ d 2 An +1 1 dAn +1 ⎛ 1 ⎞~ + − ⎜ 2 + q 2 ⎟ An +1 = 0 ,   Rn +1 :   2 dr r dr ⎝r ⎠ ~ → An+1 (r , λ ) = B6 I1 (qr ),

)

(3.5.8)

where v =

where q =

p n2+1 + λ2 , p 2 =

(3.5.9)

jωµ 0 µ nm+1σ nm+1 .

 63   

Then the solution in each region Ri , i = 0,1,..., n + 1 can be found by means of the inverse Fourier transform of the form Ai (r , z ) =

1 2π

+∞

~

∫ A (r , λ )e

jλ z

i

dλ .

(3.5.10)

−∞

It can be shown that the induced vector potential in free space due to multilayer conducting tubes has the form A0ind (r , z ) =

1 2π

+∞

∫ B K (λr )e 2

1

jλ z

dλ .

(3.5.11)

−∞

As an example we consider the case where a single-turn coil of radius r0 is located outside a two-layer coaxial conducting tube. The conducting layer (region R1 ) is defined by the inequalities: R1 = { r1 ≤ r ≤ r2 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ } . The electric conductivity and magnetic R1 are given by (3.2.1). The properties of region permeability of region R2 = { 0 ≤ r ≤ r2 , 0 ≤ ϕ ≤ 2π , − ∞ < z < +∞ } , σ 2 and µ 2 , are assumed to be constant. (see Fig. 3.8 )

(

R 2 µ 2r , σ 2

r1

)

(

R1 µ1r , σ 1

r0

r2

)

(

R 0 µ 0r = 1, σ 0 = 0

) Fig. 3.8 A single-turn coil outside a two-layer tube.

The solutions in regions R0 , R1 and R2 can be written as follows ~ ⎧⎪ A00 (r , λ ) = B2 K1 (λr ) + µ0 r0 Ie− jλz 0 K1 (λr0 )I1 (λr ),                                                                 (3.5.12)                R0 : → ⎨ ~ ⎪⎩ A01 (r , λ ) = B2 K1 (λr ) + µ0 r0 Ie− jλz 0 I1 (λr0 )K1 (λr ),

~ 1 (B4 I v (λr ) + B5 K v (λr )), R1 : → A1 (r , λ ) = r ~ R2 : → A2 (r , λ ) = B6 I1 (qr ).

(3.5.13) (3.5.14)

Using (3.5.12) - (3.5.14) and the boundary conditions we obtain B B B2 K1 (λr1 ) + µ0r0 Ie− jλz 0 K1 (λr0 )I1 (λr1 ) = 4 I v (λr1 ) + 5 Kv (λr1 ), r1 r1

r1 ⎛⎜ 1 ⎞⎟ ⋅ − µ1m ⎜⎝ 2r1 r1 ⎟⎠ ⋅ B4 I v (λr1 ) − 2r1λI v' (λr1 ) + B5 K v (λr1 ) − 2r1λK v' (λr1 ) ,

λB2 K1' (λr1 ) + λµ0 r0 Ie− jλz K1 (λr0 )I1' (λr1 ) =

(3.5.15)

0

( (

B6 I1 (qr2 ) =

)

B B4 I v (λr2 ) + 5 Kv (λr2 ), r2 r2

(

))

(3.5.16)

(3.5.17)

 64   

r2 ⎛⎜ 1 ⎞⎟ ⋅ − µ1m ⎜⎝ 2r2 r2 ⎟⎠ ⋅ B4 I v (λr2 ) − 2r2λI v' (λr2 ) + B5 K v (λr2 ) − 2r2λK v' (λr2 ) .

B6

qI1' (qr2 ) = m

µ2

( (

)

(

(3.5.18)

))

Eliminating B2 from (3.5.15) and (3.5.16) we get

Kv (λr1 )(2µ1mλK1' (λr1 ) + K1 (λr1 )) − 2r1λKv' (λr1 )K1 (λr1 ) − I v (λr1 )(2µ1mλK1' (λr1 ) + K1 (λr1 )) − 2r1λI v' (λr1 )K1 (λr1 )

B4 = − B5

(3.5.19)

2µ m µ0r0 Ie− jλz 0 K1 (λr0 ) , − 1 m ' ' r1 I v (λr1 )(2µ1 λK1 (λr1 ) + K1 (λr1 )) − 2r1λI v (λr1 )K1 (λr1 ) where W { I1 (λr1 ), K1 (λr1 ) } = −

1 . λr1

Eliminating B6 from (3.5.17) and (3.5.18) we obtain

⋅ B6 = ⋅

B4 I v (λr2 ) B5 K v (λr2 ) , + r2 I1 (qr2 ) r2 I1 (qr2 )

B4 q I v (λr2 ) ' B q K v (λr2 ) ' B I v (λr2 ) − 2r2λI v' (λr2 ) − I1 (qr2 ) + m 5 I1 (qr2 ) = − m 4 m ( ) ( ) I qr I qr 2µ1 r2 µ2 r2 1 2 µ2 r2 1 2

(

− ⋅ B4

B5 2µ

m 1

r2

)

(K (λr ) − 2r λK (λr )), v

2

2

' v

2

⎞ ⎛ m I (λr ) ⎜⎜ µ2 I v (λr2 ) − 2r2λµ2m I v' (λr2 ) + 2µ1m q v 2 I1' (qr2 )⎟⎟ = 2µ µ ⎝ I1 (qr2 ) ⎠ 1

m 1

m 2

= − B5 B4 = − B5

⎞ ⎛ m K (λr ) ⎜⎜ µ2 Kv (λr2 ) − 2r2λµ2m K v' (λr2 ) + 2µ1m q v 2 I1' (qr2 )⎟⎟, 2µ µ ⎝ I1 (qr2 ) ⎠ 1

m 1

m 2

µ2m I1 (qr2 )(Kv (λr2 ) − 2r2λK v' (λr2 )) + 2µ1m qKv (λr2 )I1' (qr2 ) . µ2m I1 (qr2 )(I v (λr2 ) − 2r2λI v' (λr2 )) + 2µ1m qI v (λr2 )I1' (qr2 )

Solving (3.5.19) and (3.5.20) we obtain 2r C B5 = − 0 µ0 µ1m Ie− jλz 0 K1 (λr0 ) 1 , D r1

(3.5.20)

(3.5.21)

where

C1 = µ2m I1 (qr2 )(I v (λr2 ) − 2r2λI v' (λr2 )) + 2µ1m qIv (λr2 )I1' (qr2 ), D = µ2m I1 (qr2 )D5 ( D1 − 2r2λD2 ) + 2µ1m qI1' (qr2 )D1D5 −

− 2r1λµ2m I1 (qr2 )K1 (λr1 )( D3 − 2r2λD4 ) − 4r1µ1mλqI1' (qr2 )K1 (λr1 )D3.

⋅ D1 = I v (λr2 )K v (λr1 ) − I v (λr1 )K v (λr2 ) ⋅ D2 = I v' (λr2 )K v (λr1 ) − I v (λr1 )K v' (λr2 )

⋅ D3 = I v (λr2 )K v' (λr1 ) − I v' (λr1 )K v (λr2 ) ⋅ D4 = I v' (λr2 )K v' (λr1 ) − I v' (λr1 )K v' (λr2 ) ⋅ D5 = 2µ1m λK1' (λr1 ) + K1 (λr1 )

 65   

The constant B2 is given by the formula

B4 =

2r0 C µ0 µ1m Ie− jλz 0 K1 (λr0 ) 2 , D r1

where

(3.5.22)

(

)

C2 = µ2m I1 (qr2 ) K v (λr2 ) − 2r2λK v' (λr2 ) + 2µ1m qKv (λr2 )I1' (qr2 ).

Finally, the value of B2 is            ⋅ B2 =

B4 I v (λr1 ) B5 K v (λr1 ) I (λr ) + − µ 0 r0 Ie − jλz0 K1 (λr0 ) 1 1 K1 (λr1 ) r1 K1 (λr1 ) r1 K1 (λr1 )

 

⎞ K1 (λr0 ) ⎛ 2µ1m µ2m I1 (qr2 )( D1 − 2r2λD2 ) + 2µ1m qI1' (qr2 )D1 ⎜⎜ + I1 (λr1 ) ⎟⎟.              (3.5.23)  K1 (λr1 ) ⎝ r1 D ⎠

B2 = −µ0r0 Ie− jλz 0

The induced vector potential in region R0 is given by ~ A0ind (r , λ ) = B2 K1 (λr ),

(3.5.24)

where B2 is obtained from (3.5.23). Applying the inverse Fourier transform (3.5.10) to (3.5.24) we obtain the induced vector potential in free space due to the presence of a cylindrical region in the form

µrI A (r , z ) = 0 0 2π

∞

ind 0

~

K1 (λr0 )

∫ B K (λr ) K (λr ) e 2

−∞

1

1

jλ ( z − z 0 )

dλ ,

(3.5.25)

1

2µ m µ m I (qr )( D1 − 2r2 λD2 ) + 2µ1m qI1' (qr2 )D1 ~ − I1 (λr1 ) where B2 = − 1 2 1 2 r1 D The induced change in impedance of the coil is given by the formula jω Z ind = A0ind (r , z )dl , ∫ I L

(3.5.26)

where L is the contour of the coil. Substituting (3.5.25) into (3.5.26) we obtain ~ K12 (λr0 ) = jωµ r ∫ B2 dλ . K1 (λr1 ) −∞ ∞

Z

ind

2 0 0

(3.5.27)

Introducing dimensionless parameters and simplifying the resulting expression we rewrite (3.5.27) as follows

⎫ ⎪ 1 ⎪ + jpˆ12 , ⋅ pˆ1 = ωµ0 µˆ1mσˆ1m r12 ⎬ ⇒ νˆ = r 4 rˆ0 = 0 , µ1m = µˆ1m r1 , σ 1m = σˆ1m r1 ,⎪ ⎪ r1 ⎭

⋅ λr1 = u → λ =

⋅ rˆ2 =

u , r1

µ2mσ 2m r2 , qˆ = u 2 + jpˆ 22 , pˆ 2 = ωµ0 µ 2mσ 2m r12 = ωµ0 r12 µ 2mσ 2m = pˆ1 µˆ1mσˆ1m r1

 66   

⋅ Dˆ1 = I vˆ (urˆ2 )Kvˆ (u ) − I vˆ (u )Kvˆ (urˆ2 ), Dˆ 2 = I vˆ' (urˆ2 )Kvˆ (u ) − I vˆ (u )Kvˆ' (urˆ2 ) ⋅ Dˆ 3 = I vˆ (urˆ2 )Kvˆ' (u ) − I vˆ' (u )Kvˆ (urˆ2 ), Dˆ 4 = I vˆ' (urˆ2 )Kvˆ' (u ) − I vˆ' (u )Kvˆ' (urˆ2 ) Dˆ = µ m I (qˆrˆ ) Dˆ − 2urˆ Dˆ . Dˆ = 2µˆ muK ' (u ) + K (u ), 5

1

1

1

6

2

1

2

(

1

2

2

)

. Dˆ 7 = 2µˆ1m qˆI1' (qˆrˆ2 )Dˆ1 Z ind (r , z ) = ωµ 0

r02 Z ,                                                                                                                                     (3.5.28) r1

∞ ~ˆ K 2 (urˆ ) where          Z = j ∫ B2 1 0 du                                                                                                   (3.5.29) K1 (u ) −∞

and

~ˆ B2 = −

( (

)

2µˆ1m Dˆ 6 + Dˆ 7 − I1 (u ).          Dˆ 5 Dˆ 6 + Dˆ 7 − 2uµ2m I1 (qˆrˆ2 )K1 (u ) Dˆ 3 − 2urˆ2 Dˆ 4 − 4µˆ1muqˆI1' (qˆrˆ2 )K1 (u )Dˆ 3

(

)

)

Results of numerical calculations using formula (3.5.29) are shown in Fig. 3.9. The following values of the parameters are used for calculations: µ1 = 1, µˆ 2 = 5, rˆ2 = 0.9,

µ 2mσ 2m = 1.5. The µˆ1mσˆ1m

points on each curve correspond to the values of pˆ 1 = 1,2,...,10 (from top to bottom). The graphs in Fig. 3.9 are shown for the following three values of rˆ0 : 1.1, 1.3 and 1.5. It is seen from Fig. 3.9 that for larger values of rˆ0 the induced change in impedance is weaker: the modulus of Z decreases as the distance from the coil to the tube increases. Computations are done with “Mathematica” ( see Appendix Fig.Ap.7 ) . Im @zD

          rˆ0 = 1.1  

0.2

0.05

0.1 0.15

0.2

0.25

0.3

0.35

Re @zD

                         rˆ0 = 1.3                          rˆ0 = 1.5  

-0.2 -0.4 -0.6

Fig. 3.9 The change in impedance computed by formula (3.5.29) for three different values of rˆ0 .

3.6 Other analytical solutions Equation (3.2.14) is the second order ordinary differential equation with variable coefficients that depend on two parameters α i and β i . As it is shown in the previous sections the solution to (3.2.14) for the case α i = −1, β i = −1 is expressed in terms of the modified Bessel functions (see (3.2.34)). It is possible to construct closed-form solutions to (3.2.14) for other combinations of the parameters α i and β i . If α i = 0, β i = −2 equation (3.2.14) becomes  67   

~ ~ ⎞~ d 2 Ai 3 dAi ⎛ 1 − pi2 + + ⎜⎜ 2 − λ2 ⎟⎟ Ai = 0, i = 1,2,..., n                                                                                  (3.6.1) 2 dr r dr ⎝ r ⎠ Using formula 132 in [50], page 228, the general solution to (3.6.1) can be written in the form  ~ 1 Ai (r , λ ) = D4 i I pi (λr ) + D5i K pi (λr ) ,                                                                                                       (3.6.2)  r

(

)

y xx'' +

a ' ⎛ n−2 c ⎞ y x + ⎜ bx + 2 ⎟ y = 0 ⇒ x x ⎠ ⎝

⋅y=x ⋅ ν =

1− a 2

1 n

n n ⎛ ⎛ ⎞ ⎛ ⎞⎞ ⎜ C1 Jν ⎜ 2 b x 2 ⎟ + C2Yν ⎜ 2 b x 2 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟⎟ ⎜ ⎝n ⎠ ⎝n ⎠⎠ ⎝

(1 − a )2 − 4c

where n = 2, b = −λ2 , c = − pi2 + 1, a = 3 n

2 v = pi , b x 2 = − jλ r .   n If α i + β i = −2  and β i ≠ 1 equation (3.2.14) is written as follows  ~ ~ ⎞~ d 2 Ai 1 − β i dAi ⎛ 1 + β i + pi2 + − ⎜⎜ + λ2 ⎟⎟ Ai = 0, i = 1,2,..., n.                                                                (3.6.3)  2 2 dr r dr ⎝ r ⎠ The general solution to (3.6.3) is (see [50], page 228, formula 132): β

(

)

i ~ Ai (r , λ ) = r 2 C4 i Iν i (ξi r ) + C5i Kν i (ξi r ) ,                                                                                                    (3.6.4)

1 β i 2 + 4λ2 , ξ i = 1 + β i + pi2 . 2 In addition, if β i ≠ 1 , α i + β i = −1 equation (3.2.14) reduces to ~ ~ ⎞~ d 2 Ai 1 − β i dAi ⎛ 1 + β i pi2 + − ⎜⎜ 2 + + λ2 ⎟⎟ Ai = 0, i = 1,2,..., n                                                                (3.6.5)  2 dr r dr ⎝ r r ⎠

where ν i =

which is a particular form of the confluent hypergeometric equation

d2y ⎛ b ⎞ dy ⎛ d e ⎞ + ⎜a + ⎟ + ⎜c + + 2 ⎟y = 0 2 dx x ⎠ dx ⎝ x x ⎠ ⎝ with a = 0, b = 1 − β i , c = −λ2 , d = − pi2 , e = −1 − β i (see [4], page 239). In this case the solution to (3.2.14) can be expressed in terms of Whittaker functions. This case is considered in detail in Section 3.7. Probably, other analytical solutions of equation (3.2.14) can be constructed for other combinations of the parameters α i and β i .

3.7 Solution in terms of confluent hypergeometric function One-parameter family of analytical solutions of the problem described in Section 3.4 (see Fig. 3.4) for the case where α = −1 − β is obtained in the present section. Solution to (3.4.1) (3.4.6) is sought by the method of the Fourier transform in the form (3.2.9) so that the  68   

 

transformed equations are : ~ ~ d 2 A0 1 dA0 ⎛ 1 ⎞~ + − ⎜ 2 + λ2 ⎟ A0 = − µ0 Ie − jλz 0 δ (r − r0 ),                                                                     (3.7.1)                         R0 :     2 dr r dr ⎝ r ⎠ ~ ⎧⎪ A00 (r , λ ) = B2 I1 (λr ) + µ0 r0 Ie− jλz 0 K1 (λr0 )I1 (λr ),                                 →      ⎨ ~                                        (3.7.2) ⎪⎩ A01 (r , λ ) = B2 I1 (λr ) + µ0 r0 Ie− jλz 0 I1 (λr0 )K1 (λr ), ~ ~ ⎞~ d 2 A1 (1 − β ) dA1 ⎛ 1 + β p12 R1 : + − ⎜⎜ 2 + + λ2 ⎟⎟ A1 = 0, (3.7.3) 2 dr r dr ⎝ r r ⎠ where p1 = jωµ 0 µ1mσ 1m ,   ~ ~ d 2 A 1 dA2 ⎛ 1 ⎞~ R2 :   22 + − ⎜ 2 + q 2 ⎟ A2 = 0 ,                                                                                                      (3.7.4)  dr r dr ⎝ r ⎠ ~ → A2 (r , λ ) = B6 K1 (qr ), (3.7.5) p 22 + λ2 , p 2 =

and q =

jωµ 0 µ 2mσ 2m .

The boundary conditions are ~ ~ dA0 1 dA1 ~ ~ A0 |r = r1 = A1 |r = r1 , |r = r1 = r dr µ1 (r1 ) dr ~ 1 dA1 1 ~ ~ |r = r2 = m A1 |r = r2 = A2 |r = r2 , r µ1 (r2 ) dr µ2

|r = r1 ,

(3.7.6)

~ dA2 |r = r2 . dr

(3.7.7)

Using the substitution

~ A1 = r

β −1 2

~ B

(3.7.8)

(the details are shown below) ~ ⋅ A1 = r

β −1 2

~ B

~ β −1 r ⋅ A1r = 2 2

~ ⋅ Brr

~ ⋅ Brr

2

~ B+r

β −1 2

~ Br

~ ⋅ A1rr = r

β −1 2

~ Brr + (β − 1)r

β −3 2

~ β −1 β − 3 Br + r 2 2

~ β − 1 β − 3 2 ~ (β − 1) 2 ~ Br + r B− r B − (β − 1)r 2 2 2 ⎛ 1 + β p2 ⎞ β −1 ~ − ⎜⎜ 2 + 1 + λ2 ⎟⎟r 2 B = 0 → r ⎝ r ⎠ ~ 2 ~ ⎞~ β − 1 β − 3 B (β − 1) B ⎛ 1 + β p12 + − − ⎜⎜ 2 + + λ2 ⎟⎟ B = 0 → 2 2 2 2 r 2 r ⎝ r r ⎠ 2 ⎛ 1 ⎛β ⎞ ⎞ ⎜ − ⎜ + 1⎟ ⎟ 2 p ~ ⎜ 4 ⎝2 ⎠ ⎟B =0 + λ2 ⎜ − 1 − 12 + 2 2 ⎟ rλ r λ ⎟⎟ ⎜⎜ ⎠ ⎝

β −1

⋅r

β −3

~ Brr + (β − 1)r

β −3 2

β −5

2

β −5

β −3 2

β −5 2

~ B

~ Br −

 69   

we transform (3.7.3) to the Whittaker’s equation (see [4], page 237): 1 ⎛ ⎞ −ν 2 ⎟ ⎜ 1 k ~ ~ Bxx + ⎜ − + + 4 2 ⎟ B = 0, x ⎟ ⎜⎜ 4 x ⎟ ⎝ ⎠ where x = 2λr , k = −

(3.7.9)

p12 β , ν = + 1. 2λ 2

Using the substitution x

c

− ~ B = e 2 x2 y x

(3.7.10) c

− ~ ⋅ B = e 2 x2 y x c 2 ⎛ − c⎞ c ⎞ ⎞ ~ ⎛ c ⎞ ⎛⎜ 1 1 ⎛ c 2 2⎜ ⎜ ⋅ Bxx = e x y ' '+⎜ − 1⎟ y '+⎜ + 2 ⎜ − ⎟⎟ − ⎟⎟ y ⎟ ⎜ ⎝ x ⎠ ⎝ 4 x ⎝ 4 2 ⎠ 2 x ⎠ ⎟⎠ ⎝ equation (3.7.9) can be written in the form a ⎛c ⎞ (3.7.11) y ' '+⎜ − 1⎟ y '− y = 0, x ⎝x ⎠ 1 where a = ν − k + and c = 2ν + 1 . 2 General solution to (3.7.11) can be expressed in the form: (3.7.12) y = B4ψ ( a, c; x) + B5e xψ (c − a, c;− x), x

c

− ⎛ ~ ⎛ c 1⎞ ⎞ ⋅ Bx = e 2 x 2 ⎜⎜ y '+⎜ − ⎟ y ⎟⎟ ⎝ 2x 2 ⎠ ⎠ ⎝

where ψ (a, c; x ) is the confluent hypergeometric function (see [4], page 241) and B4 , B5 are arbitrary constants. Thus, a general solution to (3.7.3) can be written in the form ~ A1 = B4 r β +1e −λrψ (a, c;2λr ) + B5r β +1eλrψ (c − a, c;−2λr ). (3.7.13) In order to simplify the notations we define the following functions ϕ1 (r ) = r β +1e − λrψ (a, c;2λr ) and ϕ 2 (r ) = r β +1eλrψ (c − a, c;−2λr ). Hence, ~ A1 = B4ϕ1 (r ) + B5ϕ 2 (r ). Using (3.7.2), (3.7.5), (3.7.16) and the boundary conditions (3.7.6), (3.7.7) we obtain B2 I1 (λr1 ) + µ0r0 Ie− jλz0 I1 (λr0 )K1 (λr1 ) = B4ϕ1 (r1 ) + B5ϕ2 (r1 ),

λB2 I1' (λr1 ) + λµ0 r0 Ie− jλz I1 (λr0 )K1' (λr1 ) = 0

1

µ (r1 ) r 1

⋅ ( B4ϕ1 ' (r1 ) + B5ϕ2 ' (r1 ) ),

(3.7.14) (3.7.15) (3.7.16)

(3.7.17) (3.7.18)

B6 K1 (qr2 ) = B4ϕ1 (r2 ) + B5ϕ 2 (r2 ),

(3.7.19)

B6

(3.7.20)

µ

m 2

qK1' (qr2 ) =

1 ⋅ ( B4ϕ1 ' (r2 ) + B5ϕ2 ' (r2 ) ). µ (r2 ) r 1

 70   

Using (3.7.17) and (3.7.18) we obtain

⋅ B2 =

B4 B5 K (λr ) ϕ1 (r1 ) + ϕ2 (r1 ) − µ0 r0 Ie− jλz 0 I1 (λr0 ) 1 1 I1 (λr1 ) I1 (λr1 ) I1 (λr1 )

⎛ I ' (λr ) ⎛ I ' (λr ) ϕ ' (r ) ⎞ ϕ ' (r ) ⎞ ⋅ B4 ⎜⎜ 1 1 ϕ1 (r1 ) − 1 r 1 ⎟⎟ = − B5 ⎜⎜ 1 1 ϕ2 (r1 ) − 2 r 1 ⎟⎟ + λµ1 (r1 ) ⎠ λµ1 (r1 ) ⎠ ⎝ I1 (λr1 ) ⎝ I1 (λr1 ) I ' (λr )K (λr ) − I1 (λr0 )K1' (λr1 ) + µ0 r0 Ie− jλz 0 I1 (λr0 ) 1 1 1 1 I1 (λr1 )

⋅ µ1r (r ) = µ1m r β → µ1r (r1 ) = µ1m r1 , where µ1m = µˆ1m r1 β +1 ⋅ σ m = σ~ m r β

1

−β

⇒ µ1r (r1 ) = µˆ1m

1 1

⋅ p = jωµo µˆ1m σ~1m r1 2 1

r0 − jλz 0 Ie I1 (λr0 ) λµˆ I (λr1 )ϕ2 (r1 ) − I1 (λr1 )ϕ2 ' (r1 ) r1 + , B4 = − B5 λµˆ I (λr1 )ϕ1 (r1 ) − I1 (λr1 )ϕ1 ' (r1 ) λµˆ1m I1' (λr1 )ϕ1 (r1 ) − I1 (λr1 )ϕ1 ' (r1 ) m 1 m 1

µˆ1m µ0

' 1 ' 1

where W { K1 (λr1 ), I1 (λr1 ) } =

(3.7.21)

1 . λr1

Eliminating B6 from (3.7.19) and (3.7.20) we get

⋅ B6 = B4

ϕ1 (r2 )

K1 (qr2 )

+ B5

ϕ2 (r2 )

K1 (qr2 )

⎛ ϕ ' (r ) q ϕ (r ) ⎞ ⎛ ϕ ' (r ) q ϕ (r ) ⎞ ⋅ B4 ⎜⎜ 1r 2 − m 1 2 K1' (qr2 )⎟⎟ = − B5 ⎜⎜ 2r 2 − m 2 2 K1' (qr2 )⎟⎟ ⎝ µ1 (r2 ) µ2 K1 (qr2 ) ⎠ ⎝ µ1 (r2 ) µ2 K1 (qr2 ) ⎠ −β β β ⋅ µ1r (r ) = µ1m r β → µ1r (r2 ) = µ1m r2 , where µ1m = µˆ1m r1 ⇒ µ1r (r2 ) = µˆ1m rˆ2 → ⎛ ⎞ ⎛ ⎞ ϕ (r ) ϕ (r ) β β ⋅ B4 ⎜⎜ µ2mϕ1 ' (r2 ) − µˆ1m rˆ2 q 1 2 K1' (qr2 )⎟⎟ = − B5 ⎜⎜ µ2mϕ2 ' (r2 ) − µˆ1m rˆ2 q 2 2 K1' (qr2 )⎟⎟ K1 (qr2 ) K1 (qr2 ) ⎝ ⎠ ⎝ ⎠ B4 = − B5

µ2mϕ2 ' (r2 )K1 (qr2 ) − qµˆ1m rˆ2β ϕ2 (r2 )K1' (qr2 ) , µ2mϕ1 ' (r2 )K1 (qr2 ) − qµˆ1m rˆ2β ϕ1 (r2 )K1' (qr2 )

(3.7.22)

and

B5 = µˆ1m µ0

r0 − jλz0 C Ie I1 (λr0 ) 1 , r1 D

(3.7.23)

where

C1 = µ2mϕ1 ' (r2 )K1 (qr2 ) − qµˆ1m rˆ2β ϕ1 (r2 )K1' (qr2 ),

( )

D = λµˆ1m µ2m K1 (qr2 )I1' (λr1 )D1 + λq µˆ1m rˆ2β K1' (qr2 )I1' (λr1 )D2 + 2

+ µ2m K1 (qr2 )I1 (λr1 )D3 + qµˆ1m rˆ2β K1' (qr2 )I1 (λr1 )D4 .

⋅ D1 = ϕ1 ' (r2 )ϕ 2 (r1 ) − ϕ1 (r1 )ϕ 2 ' (r2 )

⋅ D2 = ϕ1 (r1 )ϕ 2 (r2 ) − ϕ1 (r2 )ϕ 2 (r1 )

⋅ D3 = ϕ1 ' (r1 )ϕ 2 ' (r2 ) − ϕ1 ' (r2 )ϕ 2 ' (r1 ) ⋅ D4 = ϕ1 (r2 )ϕ 2 ' (r1 ) − ϕ1 ' (r1 )ϕ 2 (r2 )  71   

The constant B4 is

B4 = − µˆ1m µ0 where

r0 − jλz0 C Ie I1 (λr0 ) 2 , r1 D

(3.7.24)

C2 = µ2mϕ2 ' (r2 )K1 (qr2 ) − qµˆ1mrˆ2βϕ2 (r2 )K1' (qr2 ).

In particular, the value of B2 is

B2 = µ0 r0 Ie− jλz 0

⎞ I1 (λr0 ) ⎛ µˆ1m µ2m K1 (qr2 )D1 + qµˆ1m rˆ2β K1' (qr2 )D2 ⎜⎜ − K1 (λr1 )⎟⎟. I1 (λr1 ) ⎝ r1 D ⎠

(3.7.25)

The induced change in impedance of the coil due to the presence of a conducting tube is given by the formula ~ A0ind (r , λ ) = B2 I1 (λr ), (3.7.26) where B2 is given by (3.7.25). Applying the inverse Fourier transform (3.2.36) to (3.7.26) we obtain the induced vector potential in free space due to the presence of a cylindrical region in the form µ 0 r0 I ∞ ~ I (λr ) ind (3.7.27) A0 ( r , z ) = B2 I1 (λr ) 1 0 e jλ ( z − z0 ) dλ , ∫ 2π − ∞ I1 (λr1 )

~ µˆ m µ m K (qr )D + qµˆ1m rˆ2β K1' (qr2 )D2 − K1 (λr1 ) where B2 = 1 2 1 2 1 r1 D

The change in impedance of the coil is given by jω A0ind (r , z )dl , Z ind (r , z ) = I ∫L where L is the contour of the coil. Substituting (3.7.27) into (3.7.28) we obtain Z ind (r , z ) = jωµ 0 r02

~ I12 (λr0 ) B ∫ 2 I1 (λr1 ) dλ. −∞

(3.7.28)

∞

(3.7.29)

Using following dimensionless parameters (3.7.29) can be rewritten as follows u ⋅ λr1 = u → λ = , r1

⋅ rˆ0 =

r0 r r , rˆ2 = 2 , rˆ1 = 1 = 1, r1 r1 r1

⋅ pˆ 1 = ωµ o µˆ 1m σ~1m r12 ,

⋅ pˆ 2 = ωµ 0 µ 2mσ 2m r12 = ωµ 0 r12 µ 2mσ 2m = pˆ 1 ⋅ qˆ = u 2 + jpˆ 22 ,

⋅ c = β + 3, a =

c pˆ 12 + 2 2u

β +1 ⋅ ϕˆ1 (rˆ2 ) = rˆ2 e −urˆ2ψ (a, c;2urˆ2 ) β +1

⋅ ϕˆ 2 (rˆ2 ) = rˆ2 e urˆ2ψ (c − a, c;−2urˆ2 ) ⋅ Dˆ = ϕˆ ' (rˆ )ϕˆ (1) − ϕˆ (1)ϕˆ ' (rˆ ) 1

1

2

2

1

2

2

⋅ Dˆ 3 = ϕˆ1 ' (1)ϕˆ 2 ' (rˆ2 ) − ϕˆ1 ' (rˆ2 )ϕˆ 2 ' (1)

µ 2mσ 2m , µˆ 1m σ~1m

⋅ ϕˆ1 (1) = e −uψ (a, c;2u ) ⋅ ϕˆ 2 (1) = e uψ (c − a, c;−2u ) ⋅ Dˆ 2 = ϕˆ1 (1)ϕˆ 2 (rˆ2 ) − ϕˆ1 (rˆ2 )ϕˆ 2 (1) ⋅ Dˆ = ϕˆ (rˆ )ϕˆ ' (1) − ϕˆ ' (1)ϕˆ (rˆ ) 4

1

2

2

1

2

2

 72   

Z ind (r , z ) = ωµ 0

r02 Z ,                                                                                                                                     (3.7.30) r1

~ˆ I12 (urˆ0 ) where          Z = j ∫ B2 du                                                                                                     (3.7.31) I1 (u ) −∞ ∞

and

(

)

~ˆ µˆ1m µ2m K1 (qˆrˆ2 )Dˆ1 + qˆµˆ1m rˆ2β K1' (qˆrˆ2 )Dˆ 2 − K1 (u ),         B2 = Dˆ 2 Dˆ = uµˆ 1m µ 2m K1 (qˆrˆ2 )I 1' (u )Dˆ 1 + uqˆ (µˆ 1m ) rˆ2β K1' (qˆrˆ2 )I 1' (u )Dˆ 2 + µ 2m K1 (qˆrˆ2 )I 1 (u )Dˆ 3 + + qˆµˆ m rˆ β K ' (qˆrˆ )I (u )Dˆ . 1

2

1

2

1

4

The results of numerical computations of the change in impedance Z using formula (3.7.31) are shown in Fig.3.10. The three curves in Fig. 3.10 correspond to three different values of β , namely, β = 0,−1 and − 2 (from bottom to top). The following values of the other parameters are used for calculations: µˆ1m = 1, rˆ0 = 0.8, rˆ2 = 1.2 . The points on each curve in Fig.3.10 correspond to different values of the parameter pˆ1 = r1 ωµ0σ~1µˆ1m ( pˆ1 increases from 1 to 3 from left to right in Fig.3.10). Calculations are done with “Mathematica” since confluent hypergeometric functions and Bessel functions are the built-in functions in “Mathematica”. ( see Appendix Fig.Ap.8 ) . Im @zD

         β = 0  

-0.06

-0.08 Re @zD 0.07

0.075

0.08

0.085

0.09

0.095

0.105 -0.12

-0.14

-0.16

                         β = −1                          β = −2  

Fig. 3.10 The change in impedance computed by means of (3.7.31) for three different values of β .

In summary, the idea of using relatively simple model one-parameter electric conductivity and magnetic permeability profiles allows one to obtain different analytical solutions that can be used in eddy current testing of objects of cylindrical shapes with varying electric conductivity and magnetic permeability. The change in impedance of a single-turn coil with alternating current located inside or outside a multilayer tube with arbitrary number of conducting layers is obtained in the present section. The electric conductivity and/or magnetic permeability of each conducting layer are assumed to be power functions of the radial coordinate. The closed-form solution is expressed in terms of improper integral containing Bessel functions. It is shown that for some combinations of the parameters the solution in a conducting layer with variable properties can be expressed in terms of different special functions (Bessel functions and Whittaker functions). Theoretical model is developed for an arbitrary number of concentric conducting layers. Three examples are considered in detail. The first two examples correspond to the case where a coil is located inside a multilayer tube: (a) the case of an infinite outer conducting layer with varying properties and  73   

(b) the case of a two-layer tube where the electric conductivity and magnetic permeability depend on the radial coordinate. In addition, the case of a coil located outside a two-layer tube with varying properties considered as well. Results of numerical calculations for all three examples are presented. Calculations are performed with “Mathematica”. There are at least two cases where analytical solutions for eddy current testing problems can be helpful. First, analytical solutions suggested in the present chapter can be used to solve inverse problems in cylindrical geometry where the electric conductivity and magnetic permeability of each conducting layer depend on the radial coordinate. Second, analytical solutions are often used to test numerical algorithms developed for more complicated cases (examples include equations with variable coefficients where the coefficients depend on more than one variable or nonlinear equations). If necessary, one can obtain formulas for the change in impedance for a coil of finite dimensions, as it is done in Chapter 2.

 74   

4. DOUBLE CONDUCTOR LINE ABOVE A PLANAR MEDIUM WITH VARYING PROPERTIES 4.1 Double conductor line above a multilayer medium with varying electrical and magnetic properties Different types of eddy current coils are used in applications: circular coils, rectangular coils, planar sensors of circular and rectangular shape. It is known [67], [72] that if the ratio of the sides of a rectangular frame is 1:4 or smaller, then a double conductor line can be considered as a sufficiently accurate model of the rectangular coil. In this chapter analytical solutions are found for the case where a double conductor line formed by two infinite parallel wires is located parallel to a multilayer conducting medium [27]. The electrical conductivity and magnetic permeability of each layer are exponential functions of the vertical coordinate. The solution is found by the method of Fourier integral transform. Corresponding system of ordinary differential equations in the transformed space is solved in closed form by means of the modified Bessel functions. Suppose that two infinitely long parallel wires are located above a multilayer conducting medium shown in Fig. 4.1. z

( y0 , h )

( y1 , h )

d2

( ( R (µ

d n −1

R n −1 µ nr−1 , σ n −1

R 0 µ 0r = 1, σ 0 = 0 y R1 µ1r , σ 1

x

O

d1

2

r 2

)

) ,σ ) 2

(

(

R n µ nr , σ n

)

Fig. 4.1 A double conductor line above a multilayer medium.

)  

 

The coordinates of the wires are ( y0 , h) and ( y1, h) , respectively. The current in the wires is ± Ie jωt where I is the amplitude of the current and ω is the frequency. Each conducting layer Ri , i = 1,2,..., n in Fig. 4.1 is characterized by the two parameters: electrical conductivity σ i and relative magnetic permeability µir . It is assumed that σ i and µir are exponential functions of the vertical coordinate of the form

µir (z ) = µime β z , σ i ( z ) = σ imeα z , i = 1,2,..., n, i

i

(4.1.1)

 75   

r where α i , β i , µ im , σ im are constants. The amplitude of the vector potential Ai in each region Ri , i = 0,1,..., n has only one non-zero component in the x - direction which is a function of y and r z only. In this case, Ai is parallel to the x - asis so that r r Ai = Ai ( y, z )ex . (4.1.2) Using

(1.2.10)

in

and µ 0r = 1, σ 0 = 0 we obtain the equation for the amplitude of the vector

Cartesian

I e = Iδ ( z − h )( δ ( y − y1 ) − δ ( y − y0 ) ) potential in region R0 :

coordinates

with

∂ 2 A0 ∂ 2 A0 + 2 = µ0 Iδ ( z − h)( δ ( y − y1 ) − δ ( y − y0 ) ), (4.1.3) ∂y 2 ∂z where δ ( x ) is the Dirac delta-function. Using (1.2.10) in Cartesian coordinates with I = 0   we obtain the following system of equations for the amplitudes Ai (r , z ) of the vector potential in each region Ri , i = 1,2,..., n : ∂ 2 Ai ∂ 2 Ai 1 dµir ( z ) ∂Ai + − − jωσ i ( z )µ0 µir ( z )Ai = 0, ∂y 2 ∂z 2 µir ( z ) dz ∂z i = 1,2,..., n. Substituting (4.1.1) into (4.1.4) we obtain ∂ 2 Ai ∂ 2 Ai ∂A + 2 = β i i + jωσ im µ0 µime(α i + β i ) z Ai , 2 ∂y ∂z ∂z i = 1,2,..., n. The boundary conditions are 1 ∂A1 ∂A0 |z = 0 , A0 |z = 0 = A1 |z = 0 , |z = 0 = r µ1 (0 ) ∂z ∂z

(4.1.4)

(4.1.5)

(4.1.6)

Ai |z = − dˆ = Ai +1 |z = − dˆ , i

i

∂Ai 1 ∂Ai +1 |z = − dˆ = r | ˆ , i = 1,2,.., n − 1, r i µi − dˆi ∂z µi +1 − dˆi ∂z z = − d i 1

( )

( )

(4.1.7)

i

where dˆi = ∑ d i . k =1

The following conditions are satisfied at infinity: ∂Ai → 0 as y 2 + z 2 → ∞, i = 0,1,..., n Ai , ∂y

(4.1.8)

Problem (4.1.3), (4.1.5) - (4.1.7) is solved by the method of integral transforms. It is convenient to represent the solution Ai in each region Ri as the sum of even and odd components of the form Ai ( y, z ) = Aieven ( y, z ) + Aiodd ( y, z ). For this purpose, we represent the right-hand side of (4.1.3) in a similar form f ( y, z ) =

(4.1.9)

1 [ f ( y, z ) + f (− y, z )] + 1 [ f ( y, z ) − f (− y, z )] = f even ( y, z ) + f odd ( y, z ) 2 2

 76   

f even ( y, z ) =

µ0 I

δ ( z − h)( δ ( y − y1 ) − δ ( y − y0 ) + δ ( y + y1 ) − δ ( y + y0 ) ) 2 µI f odd ( y, z ) = 0 δ ( z − h)( δ ( y − y1 ) − δ ( y − y0 ) − δ ( y + y1 ) + δ ( y + y0 ) ) 2 µI (4.1.10) f even ( y, z ) = f odd ( y, z ) = 0 δ ( z − h)( δ ( y − y1 ) − δ ( y − y0 ) ). 2 The even component of the solution is found by applying the Fourier cosine transform of the form ∞

~ Aieven (λ , z ) = ∫ Aieven ( y, z ) cos(λy )dy, i = 0,1,..., n.

(4.1.11)

0

The left-hand side of equation (4.1.3) can be rewritten as follows ∞

∞ 2 even ∞ 2 even ⎛ ∂ 2 Aieven ∂ 2 Aieven ⎞ ∂ Ai ∂ Ai ⎜ ⎟ ( ) ( ) + cos λ y dy cos λ y dy cos(λy )dy   = + 2 2 ∫0 ⎜⎝ ∂y 2 ∫ ∫ ⎟ ∂z ⎠ ∂y ∂z 2 0 0 ~ ∞ 2 even ∂ Ai d 2 Aieven ( ) cos ⋅∫ λ y dy = ∂z 2 dz 2 0

u = cos(λy ) → du = −λ sin (λy )dy ∂ 2 Aieven ⋅∫ = cos(λy )dy = ∂ 2 Aieven ∂Aieven 2 dv = dy → v = ∂ y 0 2 ∂y ∂y ∞

∞

=

∞

 

u = sin (λy ) → du = λ cos(λy )dy

∂Aieven ∂Aeven cos(λy ) + λ ∫ i sin (λy )dy = ∂Aieven dv = dy → v = Aieven ∂y ∂ y 0 0 ∂y

=

∞

∞ ~ = λ Aieven sin (λy ) − λ2 ∫ Aieven cos(λy )dy = −λ2 Aieven 0

0

 

~ ⎛ ∂ 2 Aieven ∂ 2 Aieven ⎞ d 2 Aieven ~ ⎟ ⎜ ( ) cos λ y dy = − λ2 Aieven , i = 0,1,..., n, + 2 2 ∫0 ⎜⎝ ∂y 2 ⎟ dz ∂z ⎠ Applying the transform (4.1.11) to (4.1.3) we obtain ∞

µ0 I 2

∞

δ ( z − h) ∫ ( δ ( y − y1 ) − δ ( y − y 0 ) ) cos(λy )dy = 0

µ0 I 2

(4.1.12)

δ ( z − h)( cos(λy1 ) − cos(λy 0 ) )         (4.1.13)

Similarly, the right-hand side of equation (4.1.5) is transformed to the form ∞

⎛ ∂Aieven m m (α i + β i ) z even ⎞ ⎜ β ωσ µ µ + j e Ai ⎟⎟ cos(λy )dy = 0 i i i ∫0 ⎜⎝ ∂z ⎠                   (4.1.14) ~ even dAi ~ ( ) α + β z = βi + jωσ im µ0 µime i i Aieven , i = 1,2,..., n. dz Using (4.1.12), (4.1.13) and (4.1.14) we obtain ~ d 2 A0even µI ~ − λ2 A0even = 0 δ ( z − h)( cos(λy1 ) − cos(λy0 ) ),                                                                       (4.1.15)  2 dz 2

 77   

~ ~ d 2 Aieven dAieven ~ − βi − λ2 + jωσ im µ0 µime(α i + β i )z Aieven = 0, i = 1,2,..., n.                                           (4.1.16)  2 dz dz Applying the Fourier cosine transform to the boundary conditions we get ~ ~ ~ even ~ even dA0even 1 dA1even (4.1.17) A0 |z = 0 = A1 |z = 0 , |z = 0 = r | , dz µ1 (0) dz z = 0 ~ ~ Aieven |z = − dˆ = Aieven +1 | z = − dˆ i , i ~ ~ (4.1.18) 1 dAieven 1 dAieven +1 | = | , i = 1 , 2 ,.., n − 1 , ˆ ˆ µir − dˆi dz z = − d i µir+1 − dˆi dz z = − d i

(

( )

)

( )

In order to solve equation (4.1.15) we consider the following two sub-regions of R0 : 0 < z < h ~ ~ and z > h . The solutions in these regions are denoted by A00even and A01even , respectively. Hence, ~ ~ d 2 A00even − λ2 A00even = 0, 0 < z < h, (4.1.19) 2 dz ~ d 2 A01even ~ (4.1.20) − λ2 A01even = 0, z > h. 2 dz The general solution to (4.1.19) is ~ A00even = C1eλz + C2e− λz .                                                                                                                                    (4.1.21) The bounded solution to (4.1.20) is given by  ~ A01even = C3e − λz .                                                                                                                                                  (4.1.22)  ~ ~ The functions A00even (λ , z ) and A01even (λ , z ) satisfy the following conditions at z = h :  ~ ~ A00even |z = h = A01even |z = h , (4.1.23) ~ The condition (4.1.23) represents continuity of the function A0even (λ , z ) at z = h . One more boundary condition is obtained by integrating (4.1.15) with respect to z from z = h − ε to z = h + ε and considering the limit in the resulting expression as ε → +0 . h +ε

⋅

∫

h −ε

           

~ h +ε h +ε µ0 I d 2 A0even ~ even 2 ( cos(λy1 ) − cos(λy0 ) ) ∫ δ ( z − h)dz dz − λ ∫ A0 dz = 2 dz 2 h −ε h −ε ~ even h +ε h +ε µI dA0 ~ − λ2 ∫ A0evendz = 0 ( cos(λy1 ) − cos(λy0 ) ) 2 dz h −ε h −ε h +ε

(

)

~ ~ ⋅ − λ lim ∫ A0evendz = − λ2 lim 2εA0even λ , z * = 0 2

ε →0

ε →0

h −ε

~ dA0even ⋅ lim ε → 0 dz

h +ε

h −ε

~ dA01even = dz

~ dA00even − dz z =h

 

z =h

 

 78   

~ dA01even dz

~ dA00even − dz z =h

= z =h

µ0 I 2

( cos(λy1 ) − cos(λy0 ) ).                                                                             (4.1.24)

Using (4.1.21) − (4.1.24) we obtain ⎧C1eλh + C2e − λh = C3e − λh ⎪ ⎨ µ0 I − λh λh − λh ( cos(λy1 ) − cos(λy0 ) ) ⎪− λC3e − λC1e + λC2e = 2 ⎩

⇒                

C1e 2 λh + C2 − C3 = 0                     

− C1e 2λh + C2 − C3 =

µ0 I ( cos(λy1 ) − cos(λy0 ) )eλh 2λ

 

µ0 I ( cos(λy0 ) − cos(λy1 ) )e − λh                                                                                                            (4.1.25) 4λ µI C3 = C2 + 0 ( cos(λy0 ) − cos(λy1 ) )e λh                                                                                                    (4.1.26) 4λ Using (4.1.21), (4.1.22), (4.1.25) and (4.1.26) we get µ0 I ⎧ ~ even − λz − λ (h − z ) ⎪⎪ A00 = C2e + 4λ ( cos(λy0 ) − cos(λy1 ) )e ⇒   ⎨ ~ even µ0 I − λz −λ (z −h ) ⎪A ( ( ) ( )) ⎪⎩ 01 = C2e + 4λ cos λy0 − cos λy1 e C1 =

~ µI −λ h − z                                                                                      (4.1.27)  A0even (λ , z ) = C2e −λz + 0 ( cos(λy0 ) − cos(λy1 ) )e 4λ

The solution to (4.1.16) can be expressed in terms of modified Bessel functions (see [50], formula 2.1.3.10, page 247):

(

λx

)

y + ay + be + c y = 0 ⇒ y = e '' xx

' x

~ Aieven (λ , z ) = e

βi z 2

−

ax 2

⎡ ⎛ 2 b λ2x ⎞ ⎛ 2 b λ2x ⎞⎤ a 2 − 4c ⎜ ⎜ ⎟ ⎟ ⎢C1Iν ⎜ λ e ⎟ + C2 Kν ⎜ λ e ⎟⎥, ν = λ ⎝ ⎠ ⎝ ⎠⎦⎥ ⎣⎢

⎡ ⎛ 2 b i (α i + β i ) z ⎞ ⎛ 2 b i (α i + β i ) z ⎞⎤ 2 ⎜ ⎟ + C5 i Kν i ⎜ α i + β i e 2 ⎟⎥,                                             (4.1.28) ⎢C4 i Iν i α i + β i e ⎜ ⎟ ⎜ ⎟⎥ ⎢⎣ ⎝ ⎠ ⎝ ⎠⎦

i = 1,2,..., n − 1,  where

νi =

β i2 + 4λ2 , bi = − jωµ 0 µ imσ im ,   αi + βi

and Iν i , Kν i are the modified Bessel functions of the first and second kind, respectively. The bounded solution to (4.1.16) in region Rn can be written as follows βn z ⎛ 2 b n (α n + β n ) z ⎞ ~ even 2 ⎟ (4.1.29) An (λ , z ) = C4 n e Iν n ⎜ α n + β n e 2 ⎟ ⎜ ⎠ ⎝ The even component in each region Ri , i = 0,1,..., n can be determined by applying the inverse Fourier cosine transform of the form

Ai

even

∞

( y, z ) =

2 ~ even ∫ Ai (λ , z ) cos(λy )dλ.

π

(4.1.30)

0

 79   

The odd components of the solution, Ai

odd

( y, z ) can be determined by the Fourier sine transform

∞

~ Aiodd (λ , z ) = ∫ Aiodd ( y, z ) sin(λy )dy

(4.1.31)

0

and the inverse Fourier sine transform

Ai

odd

( y, z ) =

2

∞

~ odd

A π∫

i

(λ , z ) sin (λy )dλ.

(4.1.32)

0

The solution to (4.1.3), (4.1.5) - (4.1.8) is then found by formula (4.1.9).

4.2 Double conductor line above a conducting half-space with varying electrical and magnetic properties In this section we consider (as an example) the case, where  d1 → ∞ (see Fig. 4.1). In other words, using the technique described above we solve the problem for the case of a conducting half-space with varying electrical and magnetic properties [37], where r βz αz µ1 (z ) = µ m e , σ 1 (z ) = σ m e (see Fig. 4.2 ). z

( y0 , h )

( y1 , h )

( (

R 0 µ 0r = 1, σ 0 = 0 y R1 µ1r , σ 1

x

O

)

)

Fig. 4.2 A double conductor line above a conducting half-space.

Using the results from Section 4.1 we obtain: ~ ~ d 2 A0even µI − λ2 A0even = 0 δ ( z − h)( cos(λy1 ) − cos(λy0 ) )    →    R0 :    2 dz 2 ~ even                                            →      A (λ , z ) = C2e−λz + µ0 I ( cos(λy0 ) − cos(λy1 ) )e −λ h− z                                    (4.2.1) 0 4λ

~ ~ ~ d 2 A1even dA1even −β − λ2 + jωσ m µ0 µme(α + β )z A1even = 0    →     R1 :     2 dz dz

(

)

~ even

                                            →      A1

β z

(λ , z ) = C4e 2

(α + β ) z ⎞ ⎛ Iν ⎜⎜ α2 + bβ e 2 ⎟⎟                                                    (4.2.2) ⎠ ⎝

The boundary conditions are ~ ~ dA0even 1 dA1even ~ ~ |z = 0 , A0even |z = 0 = A1even |z = 0 , |z = 0 = r dz µ1 (0 ) dz

(4.2.3)

 80   

~ ~ Aieven |z = − dˆ = Aieven +1 | z = − dˆ i , i ~ ~ dAieven 1 dAieven 1 +1 = | ˆ , i = 1,2,.., n − 1, | z = − dˆ i r r ˆ ˆ µi − di dz µi +1 − di dz z = − d i

( )

( )

(4.2.4)

Using (4.2.1) − (4.2.4) we obtain

µ0 I ⎧ − λh ⎪⎪ C2 + 4λ ( cos(λy0 ) − cos(λy1 ) )e = C4 Iν ( z0 ),                                                   (4.2.5)  ⎨ µ β I C ⎛ ⎞ λ ' − h 0 4 ⎪− λ C 2 + ( cos(λy0 ) − cos(λy1 ) )e = ⎜ Iν (z0 ) + b Iν (z0 )⎟, ⎪⎩ µm ⎝ 2 4 ⎠ where z 0 =

2 b α +β

, b = − jωµ 0 µ mσ m , ν =

β 2 + 4λ2 . α +β

Solving (4.2.5) we get C4 =

1 µ0 µm I ( cos(λy0 ) − cos(λy1 ) )e − λh                                                                                                       (4.2.6)  β⎞ 2 ⎛ ' ⎜ λµ m + ⎟ Iν ( z0 ) + b Iν ( z0 ) 2⎠ ⎝

The constant C2 is

β⎞ ⎛ ' ⎜ λµ m − ⎟ Iν ( z0 ) − b Iν ( z0 ) µ0 I 2⎠ even − λh ⎝ C2 = ( cos(λy0 ) − cos(λy1 ) )e .                                                     (4.2.7) β⎞ 4λ ⎛ ' ⎜ λµ m + ⎟ Iν ( z0 ) + b Iν ( z0 ) 2⎠ ⎝ Applying the Fourier sine transform we obtain β⎞ ⎛ ' ⎜ λµ m − ⎟ Iν ( z0 ) − b Iν ( z0 ) µI 2⎠ C2odd = 0 ( sin (λy0 ) − sin (λy1 ) )e − λh ⎝ .                                                       (4.2.8) β⎞ 4λ ⎛ ' ⎜ λµ m + ⎟ Iν ( z0 ) + b Iν ( z0 ) 2⎠ ⎝ The induced vector potential in region R0 is ~ A0ind (λ , z ) = C2even + C2odd e − λz . (4.2.9)

(

)

Applying the inverse Fourier cosine transform (4.1.30) to the even component in (4.2.9) and inverse Fourier sine transform (4.1.32) to the odd component in (4.2.9) we obtain

µ0 I ∞ 1 A ( y, z ) = F (λ ) ( cos(λ ( y − y0 )) − cos(λ ( y − y1 )) )e − λ ( z + h ) dλ , ∫ 2π 0 λ ind 0

(4.2.10)

where

β⎞ ⎛ ' ⎜ λµ m − ⎟ Iν ( z0 ) − b Iν ( z0 ) 2⎠ F (λ ) = ⎝ .                                                                                               (4.2.11) β⎞ ⎛ ' ⎜ λµ m + ⎟ Iν ( z0 ) + b Iν ( z0 ) 2⎠ ⎝ The induced change in impedance (per unit length) of the double conductor line is calculated by the formula

 81   

jω A0ind ( y, z )dx. I C∫ Substituting (4.2.10) into (4.2.12) we obtain the change in impedance in the form ind Z per unit length ( y , z ) =

Z

ind per unit length

µ0ω ∞ F (λ ) ( y, z ) = (1 − cos(λ ( y1 − y0 )) )e− 2λh dλ , j∫ π 0 λ

(4.2.12)

(4.2.13)

where F (λ ) is given by (4.2.11). Formula (4.2.13) can be also rewritten as follows ind Z per unit length ( y , z ) =

µ0ω Z ,                                                                                                                            (4.2.14)  π

where ⎛ βˆ ⎞ ⎜ uµ m − ⎟ Iνˆ ( zˆ0 ) − − j bˆ Iν'ˆ (zˆ0 ) ⎜ 2 ⎟⎠ ˆ du ⎝                                                               (4.2.15)  ( Z = j∫ 1 − cos u )e − 2uh ˆ⎞ ⎛ u β 0 ' ˆ ⎜ uµ m + ⎟ Iνˆ ( zˆ0 ) + − j b Iνˆ (zˆ0 ) ⎜ 2 ⎟⎠ ⎝ and the dimensionless parameters are defined as follows ∞

βˆ 2 + 4λ2 2 − j bˆ h , αˆ = α rc , βˆ = β rc , hˆ = . u = λrc , bˆ = ωµ0 µmσ m rc2 , zˆ0 = , νˆ = rc αˆ + βˆ αˆ + βˆ Here rc = y1 − y0 is the distance between the wires. Fig. 4.3 plots the change in impedance Z for three different values of βˆ = 1,2,3 .The other parameters of the problem are fixed at αˆ = 0, hˆ = 0.05 and µ = 5. The calculated points in Fig. m

4.3 correspond to different values of bˆ = 1,2,...,80 (from left to right). Computations are done with Mathematica ( see Appendix Fig.Ap.9 ) . Im @zD

           βˆ = 1  

1.3

                         βˆ = 2  

1.2

                       βˆ = 3  

1.1

0.25 0.9

0.5

0.75

1

1.25

1.5

1.75

Re @zD

Fig. 4.3 The change in impedance Z for three different values of βˆ .

It is seen from the graph that the modulus of Z increases as the parameter βˆ increases.

4.3 Double conductor line above a conducting two-layer medium with varying electrical and magnetic properties  82   

Another example is discussed in this section (see [38]). Consider two parallel infinitely long wires with current located above a two-layer conducting medium (see Fig. 4.4). z

( y0 , h )

( ((

)

))

(

)

R 0 µ 0r = 1, σ 0 = 0 y r r RR µ µ , σ , σ 1 1 1 1 1 1

x

O

d

( y1 , h )

R 2 µ 2r , σ 2

Fig. 4.4 A double conductor line above a two-layer medium.

 

We assume that the electrical conductivity and relative magnetic permeability of the upper layer are exponential functions of the vertical coordinate of the form

µ1r (z ) = µ1me β z , σ 1 (z ) = σ 1meα z .                                                                                             (4.3.1) In addition,                µ2r ( z ) = µ2m , σ 2 ( z ) = σ 2m ,                                                                                                (4.3.2)

where σ 1m , µ1m , α , β are constants. The properties of region R2 , that is, the electrical conductivity

σ 2m and the relative magnetic permeability µ 2m are assumed to be constant. The functions A0 , A1 and A2 satisfy the following system of equations: ∂ 2 A0 ∂ 2 A0 R0 :    2 + 2 = µ0 Iδ ( z − h)( δ ( y − y1 ) − δ ( y − y0 ) ),                                                                    (4.3.3) ∂y ∂z

R1 :  

∂ 2 A1 ∂ 2 A1 ∂A + 2 − β 1 − jωσ 1m µ0 µ1me(α + β ) z A1 = 0,   2 ∂y ∂z ∂z

R2 :  

∂ 2 A2 ∂ 2 A2 + 2 − jωσ 2m µ0 µ2m A2 = 0.   2 ∂y ∂z

            (4.3.4) 

             (4.3.5)

where δ ( z ) is the Dirac delta-function.                                                                                                                             The boundary conditions are 1 ∂A1 ∂A0 |z = 0 , A0 |z = 0 = A1 |z = 0 , |z = 0 = r µ1 (0 ) ∂z ∂z

(4.3.6)

∂A1 1 1 ∂A (4.3.7) |z=−d = m 2 |z =−d . µ (− d ) ∂z µ 2 ∂z The functions A0 , A1 , A2 and their partial derivatives with respect to y tend to zero at infinity: ∂Ai (4.3.8) Ai , → 0 as y 2 + z 2 → ∞, i = 0,1,2 ∂y

A1 | z = − d = A2 | z = − d ,

r 1

Problem (4.3.3) - (4.3.8) is solved by means of the Fourier cosine and sine integral transforms. Representing the solution in the form (4.1.9) and applying the Fourier cosine  83   

transform (4.1.11) to the even part of the solution (using the results from Section 4.1 ) we obtain the system of ordinary differential equations and the solutions in regions R0 , R1 and R2 in the form ~ ~ d 2 A0even µI − λ2 A0even = 0 δ ( z − h)( cos(λy1 ) − cos(λy0 ) )    →    R0 :    2 dz 2 ~ even                                            →      A (λ , z ) = C2e −λz + µ0 I ( cos(λy0 ) − cos(λy1 ) )e −λ h− z ,                                   (4.3.9) 0 4λ

~ ~ ~ d 2 A1even dA1even −β − λ2 + jωσ m µ0 µme(α + β )z A1even = 0    →     R1 :     2 dz dz

(

~ even

          →      A1

β z

(λ , z ) = e 2

)

⎡ ⎛ 2 b (α +2 β ) z ⎞⎤ ⎛ 2 b (α +2 β ) z ⎞ ⎟⎥,                                          (4.3.10) ⎜ ⎟ ⎜ C I e C K + ⎢ 4 ν ⎜ α +β 5 ν ⎜ α +β e ⎟ ⎟ ⎢⎣ ⎠⎥⎦ ⎝ ⎠ ⎝

~ d 2 A2even ~ ~ − q22 A2even = 0    →    A2even (λ , z ) = C6eq2 z , R2 :     2 dz

(4.3.11)

β 2 + 4λ2 , b = − jωµ0 µ1mσ 1m , q2 = λ2 + jωσ 2m µ 0 µ 2m . α +β

where ν =

The boundary conditions are : ~ ~ ~ ~ dA0even 1 dA1even | , A0even |z = 0 = A1even |z = 0 , |z = 0 = r dz µ1 (0) dz z = 0 ~ ~ ~ even ~ even 1 1 dA2even dA1even A1 |z = − d = A2 |z = − d , r |z = − d . |z = − d = m µ1 (− d ) dz µ 2 dz

(4.3.12)

Using (4.3.9) − (4.3.12) we obtain µI C2 + 0 ( cos(λy0 ) − cos(λy1 ) )e − λh = C4 Iν ( z0 ) + C5 Kν ( z0 ),                                                                    (4.3.13) 4λ − λ C2 +

µ0 I 4

( cos(λy0 ) − cos(λy1 ) )e− λh = C ⎛β ⎞ C ⎛β ⎞ = m4 ⎜ Iν ( z0 ) + b Iν' ( z0 )⎟ + m5 ⎜ Kν ( z0 ) + b Kν' ( z0 )⎟, µ1 ⎝ 2 ⎠ µ1 ⎝ 2 ⎠

               (4.3.14)

 

C6e − q 2 d = e

−

βd 2

(C4 Iν (z1 ) + C5 Kν (z1 )),                                                                                                            (4.3.15)

 

q2

µ 2m

C6e − q 2 d = βd

(α + β )d (α + β )d ⎛ ⎛β − − ⎞ ⎛β ⎞⎞ ' 2 ⎜ ⎜ ⎟ ⎜ Iν ( z1 ) ⎟ + C5 ⎜ Kν ( z1 ) + be 2 Kν' ( z1 ) ⎟⎟ ⎟, = m C4 ⎜ Iν ( z1 ) + be ⎟ µ1 ⎜⎝ ⎝ 2 ⎠ ⎝ 2 ⎠⎠

e

                         (4.3.16)

2

where z1 = z 0 e

−

(α + β )d 2

, µi b = − jωµ 0 µ1mσ 1m .  

 

Eliminating C2 from (4.3.13) – (4.3.14) we obtain

 84   

C4 = −

((

)

)

C5 2λµ1m + β Kν ( z0 ) + 2 b Kν' ( z0 ) − µ0 µ1m I ( cos(λy0 ) − cos(λy1 ) )e − λh .                               (4.3.17) 2λµ1m + β Iν ( z0 ) + 2 b Iν' (z0 )

(

)

Similarly, using (4.3.15) − (4.3.16) we get ( β − α )d ⎛ ⎛ β βd m ⎞ ⎞ m m ⎜ C5 ⎜ ⎜ e µ 2 − µ1 q2 ⎟ Kν ( z1 ) + b µ 2 e 2 Kν' ( z1 ) ⎟⎟ ⎝2 ⎠ ⎠. C4 = − ⎝ ( β − α )d ⎛ β βd m ⎞ ' m m ⎜ e µ 2 − µ1 q2 ⎟ Iν ( z1 ) + b µ 2 e 2 Iν ( z1 ) ⎝2 ⎠

                                                       (4.3.18)

The solution to (4.3.17) − (4.3.18) is C5 = −

µ0 µ1m I ( cos(λy0 ) − cos(λy1 ) )e− λh ( D1Iν (z1 ) + D2 Iν' ( z1 ) ) D1D3 D4 + 2 b D1D5 + D2 D3 D6 + 2 b D2 D7

where        D1 =

β

e βd µ2m − µ1m q2 , D2 = b µ2me

2 D4 = Iν ( z0 )Kν ( z1 ) − Iν ( z1 )Kν ( z0 ), D5 = Iν' ( z0 )Kν ( z1 ) − Iν ( z1 )Kν' ( z0 ),

and C4 =

               

( β −α ) d 2

,                                                              (4.3.19) 

, D3 = 2λµ1m + β ,  

D6 = Iν ( z0 )Kν' ( z1 ) − Iν' ( z1 )Kν ( z0 ),

D7 = Iν' ( z0 )Kν' ( z1 ) − Iν' ( z1 )Kν' ( z0 ),

µ0 µ1m I ( cos(λy0 ) − cos(λy1 ) )e − λh ( D1Kν (z1 ) + D2 Kν' ( z1 ) ) D1D3 D4 + 2 b D1D5 + D2 D3 D6 + 2 b D2 D7

 

.                                                              (4.3.20) 

The value of C2 is given by                C2 = C4 Iν ( z0 ) + C5 Kν ( z0 ) −

µ0 I ( cos(λy0 ) − cos(λy1 ) )e− λh   4λ

⎛ 1 ⎞ µ1m ( D1D4 + D2 D6 ) ⎟.            (4.3.21) C2even = µ0 I ( cos(λy0 ) − cos(λy1 ) )e − λh ⎜⎜ − ⎟ 4 λ 2 2 D D D b D D D D D b D D + + + 1 5 2 3 6 2 7 ⎠ ⎝ 1 3 4

Similarly, we obtain

⎛ 1 ⎞ µ1m ( D1 D4 + D2 D6 ) ⎟.              (4.3.22)                       C2odd = µ0 I ( sin (λy0 ) − sin (λy1 ) )e − λh ⎜⎜ − ⎟ 4 λ 2 2 D D D b D D D D D b D D + + + 1 5 2 3 6 2 7 ⎠ ⎝ 1 3 4

The induced vector potential in region R0 is ~ A0ind (λ , z ) = C2even + C2odd e− λz .

(

)

(4.3.23)

The solution in region R 0 can be written in the form

A0 ( y, z ) = A0free ( y, z ) + A0ind ( y, z ).

(4.3.24) Applying the inverse Fourier cosine and sine transforms (4.1.30) and (4.1.32) to (4.3.23) we obtain

A ( y, z ) = ind 0

2 µ0 I

π

∞

∫ F (λ )( cos(λ ( y − y )) − cos(λ ( y − y )) )e 0

1

−λ ( z + h)

dλ ,

(4.3.25)

0

where

 85   

F (λ ) =

µ1m ( D1D4 + D2 D6 ) D1D3 D4 + 2 b D1D5 + D2 D3 D6 + 2 b D2 D7

−

1 .                                                                     (4.3.26) 4λ

The induced change in impedance (per unit length) of the double conductor line is computed by means of the formula jω ind Z per A0ind ( y, z )dx. (4.3.27) unit length ( y , z ) = I C∫ Substituting (4.3.25) into (4.3.27) we obtain the change in impedance in the form

Z

ind per unit length

( y, z ) =

4µ0ω

π

∞

j ∫ F (λ )( 1 − cos(λ ( y1 − y0 )) )e − 2 λh dλ ,

(4.3.28)

0

where F (λ ) is given by (4.3.26). Formula (4.3.28) can be rewritten in the form 4 µ0ω ind Z per Z ,                                                                                                                         (4.3.29)  unit length ( y , z ) = π where

(

)

∞⎛ µ1m Dˆ1Dˆ 4 + Dˆ 2 Dˆ 6 1 ⎞⎟ ˆ ( Z = j∫ ⎜ 1 − cos u )e − 2uh du ,               (4.3.30) − ⎜ ˆ ˆ ˆ ˆˆ ˆ ˆˆ ˆ 4u ⎟ ˆ ˆ ˆ 0 ⎝ D1D3 D4 + 2 − jb D1D5 + D2 D3 D6 + 2 − jb D2 D7 ⎠ ˆ ˆ (β −αˆ )d βˆ ˆ ˆ and        Dˆ1 = e β d µ2m − µ1m qˆ2 , Dˆ 2 = − jbˆ µ 2me 2 , Dˆ 3 = 2uµ1m + βˆ ,   2 Dˆ 4 = Iνˆ ( zˆ0 )Kνˆ ( zˆ1 ) − Iνˆ ( zˆ1 )Kνˆ ( zˆ0 ), Dˆ = I ˆ ( zˆ )K 'ˆ ( zˆ ) − I 'ˆ ( zˆ )K ˆ ( zˆ ),

Dˆ 5 = Iν'ˆ ( zˆ0 )Kνˆ ( zˆ1 ) − Iνˆ ( zˆ1 )Kν'ˆ ( zˆ0 ),

               

6

ν

0

ν

1

ν

1

ν

0

D7 = I ( zˆ0 )Kν'ˆ ( zˆ1 ) − Iν'ˆ ( zˆ1 )Kν'ˆ ( zˆ0 ). ' νˆ

 

The dimensionless parameters are defined as follows

βˆ 2 + 4λ2 , u = λrc , bˆ = ωµ0 µ1mσ 1m rc2 , qˆ2 = u 2 + jωµ0 µ2mσ 2m rc2 , νˆ = αˆ + βˆ (αˆ + βˆ )dˆ − 2 − j bˆ , zˆ0 = zˆ1 = zˆ0e 2 , αˆ + βˆ h ˆ d αˆ = α rc , βˆ = β rc , hˆ = d= . rc

rc

Here rc = y1 − y0 is the distance between the wires. The change in impedance given by formula (4.3.30) is computed for different values of the parameters of the problem using package “Mathematica” since it allows one to calculate improper integrals and evaluate modified Bessel functions of variable order and complex argument. The results of calculations are presented in Fig. 4.5. Fig. 4.5 plots the change in impedance Z for three different values of βˆ = 1,2,3 .The other parameters of the problem are fixed at αˆ = 0, hˆ = 0.05 , dˆ = 0.05 and µ m = 1. The calculated points 1

in Fig. 4.5 correspond to different values of bˆ = 1,2,...,10 (from left to right). Computations are done with “Mathematica” ( see Appendix Fig.Ap.10 ) .

 86   

Im @zD

           βˆ = 1                            βˆ = 2  

0.38 0.36

                       βˆ = 3  

0.34 0.32 -0.08

-0.06

-0.04

-0.02

Re @zD

0.28

Fig. 4.5 The change in impedance Z for three different values of βˆ .

0.26

4.4 Pulsed eddy currents a conducting half-space Theory of an eddy current method for the case where a coil is excited by an alternating current is well-developed in the literature [15], [60]. Current excitation in the form of a pulse represents an alternative to traditional eddy current methods. There are two basic methods that are usually used to analyze non-periodic time-dependent signals in eddy current testing: fast Fourier transform and Laplace transform. The difficulty in using the Laplace transform is that the inverse transform is not always available in closed form. However, several authors [68], [9], [11], [42] have obtained analytical solutions for problems where a single-turn coil or a coil with finite dimensions is located above a conducing half-space assuming that the excitation current is in the form of a step current or exponential source current. Analytical solutions are used in practice in order to estimate thickness and conductivity of metallic layers [56]. In this section we consider transient eddy current problem for the case where an excitation coil is assumed to be of the form of a double conductor line formed by two infinitely long wires located parallel to a conducting half-space [29]. The inverse Laplace transform is found in closed form for the case of an excitation current in the form of a unit step function. Different types of eddy current coils are used in applications. One example is a coil in the form of a rectangular frame. If the ratio of the sides of the frame is 4:1 or larger then such a coil can be modeled by a double conductor line [67], [72] with a relatively small error. Suppose that two horizontal infinitely long parallel wires are situated above a conducting half-space with electrical conductivity σ and relative magnetic permeability µ (see Fig. 4.6) z

( y0 , h ) x

O

( y1 , h ) R0 y

R1

Fig. 4.6 A double conductor line above a

conducting half-space.

 87   

Assume that the current in the wires is I (t ) = ± I 0ϕ (t )

(4.4.1)

at the points ( y 0 , h) and ( y1 , h) , respectively, where I 0 is a constant, and ϕ (t ) is the Heaviside function of the form ⎧1, t > 0 ϕ (t ) = ⎨ ⎩0, t < 0

(4.4.2)

r r We assume that the electric field E0 and E1 in regions R0 = {z > 0} and R1 = {z < 0} , respectively, has only one non-zero component in the x − direction: r Ei = Ei ( y, z, t )e x , i = 0,1, (4.4.3) r where e x is the unit vector in the positive x − direction. The system of Maxwell’s equations in this case can be transformed to one equation for the electric field which has the following form in regions R0 and R1 : R0 :

∂ 2 E0 ∂ 2 E0 ∂I + = µ0 δ ( z − h)(δ ( y − y1 ) − δ ( y − y0 ) ), 2 2 ∂y ∂z ∂t

∂ 2 E1 ∂ 2 E1 ∂E + 2 − µ0 µ1mσ 1m 1 = 0, 2 ∂y ∂z ∂t where δ (ξ ) is the Dirac delta-function.

(4.4.4)

R1 :

(4.4.5)

The boundary conditions are 1 ∂E ∂E0 |z = 0 = m 1 |z = 0 . E0 |z = 0 = E1 |z = 0 , ∂z µ1 ∂z

(4.4.6)

In addition, the following conditions hold at infinity: ∂E ∂E E 0 , E1 , 0 , 1 → 0 as y 2 + z 2 → ∞. ∂y ∂y The initial conditions are E 0 |t =0 = 0, E1 |t =0 = 0.

(4.4.7) (4.4.8)

Applying the Laplace transform to (4.4.4) - (4.4.8) we obtain R0 :

∂ 2 E0 ∂ 2 E0 + = µ0 sI (s )δ ( z − h)(δ ( y − y1 ) − δ ( y − y0 ) ), ∂y 2 ∂z 2

R1 :

∂ 2 E1 ∂ 2 E1 + 2 − µ0 µ1mσ 1m sE1 = 0, ∂y 2 ∂z

E0 |z = 0 = E1 |z = 0 ,

E0 , E1 ,

∂E0 1 ∂E |z = 0 = m 1 |z = 0 , ∂z µ1 ∂z

∂E 0 ∂E1 , → 0 as ∂y ∂y

y 2 + z 2 → ∞,

(4.4.9) (4.4.10) (4.4.11) (4.4.12)

where E 0 , E1 and I are the Laplace transforms of the functions E 0 , E1 and I , respectively, s is the parameter of the Laplace transform.

 88   

Problem (4.4.9) - (4.4.12) is solved by the method of integral transforms. It is convenient to represent the solution Ei in each region Ri as the sum of even and odd components of the form Ei ( y , z , s ) = Eieven ( y , z , s ) + Eiodd ( y, z , s ), i = 0,1. For this purpose, we represent the right-hand side of (4.4.9) in a similar form 1 1 f ( y, z , s ) = f ( y, z, s ) + f (− y, z, s ) + f ( y, z , s ) − f (− y, z , s ) = 2 2 even = f ( y , z , s ) + f odd ( y, z , s )

[

f even ( y, z , s ) =

] [

(4.4.13)

]

µ0 sI (s )

δ ( z − h)( δ ( y − y1 ) − δ ( y − y0 ) + δ ( y + y1 ) − δ ( y + y0 ) ) 2 (4.4.14) µ0 sI (s ) odd f ( y, z, s ) = δ ( z − h)( δ ( y − y1 ) − δ ( y − y0 ) − δ ( y + y1 ) + δ ( y + y0 ) ) 2 The following properties of the delta-function are used in order to derive (4.4.14)( see[40] ) ⋅ δ (− y ) = δ ( y ) ⎧δ ( y + y0 ) = 0 ⋅⎨ if y ≥ 0, y0 ≥ 0, y1 ≥ 0 ⎩δ ( y + y1 ) = 0 µ sI (s ) f even ( y, z , s ) = f odd ( y, z , s ) = 0 δ ( z − h)( δ ( y − y1 ) − δ ( y − y0 ) ) (4.4.15) 2 The even component of the solution is found by applying the Fourier cosine transform of the form ∞ ~ Eieven (λ , z, s ) = ∫ Eieven ( y, z, s) cos(λy )dy, i = 0,1. (4.4.16) 0

The odd component of the solution is found by applying the Fourier sine transform of the form ∞ ~ odd Ei (λ , z, s) = ∫ Eiodd ( y, z, s) sin (λy )dy, i = 0,1. (4.4.17) 0

Applying the Fourier cosine transform (4.4.16) to (4.4.9) - (4.4.12) we obtain ~ ~ d 2 E0even µ sI ( s ) R0 : − λ2 E0even = 0 δ ( z − h)(cos λy1 − cos λy0 ), 2 dz 2 ~ ~ d 2 E1even R1 : − q 2 E1even = 0, where q = λ2 + µ0 µ1mσ1m s . 2 dz ~ ~ ~ even ~ even 1 dE1even dE0even |z = 0 = m |z = 0 , E0 |z = 0 = E1 |z = 0 , dz µ1 dz

(4.4.18) (4.4.19) (4.4.20)

In order to solve (4.4.18) we consider two sub-regions, R00 = {0 < z < h} and R01 = {z. > h} , of region R0 . The solutions to (4.4.18) in R00 and R01 are denoted by ~ ~ E00even and E01even , respectively. The general solution to (4.4.18) in R00 is ~ E00even (λ , z , s ) = C1eλz + C2e − λz .

(4.4.21)

The general solution to (4.4.18) in R01 which is bounded as z → +∞ has the form ~ E01even (λ , z , s ) = C3e − λz .

(4.4.22)  89   

Finally, the general solution to (4.4.19) which remains bounded as z → −∞ can be written as follows ~ E1even (λ , z , s ) = C4e qz . (4.4.23) Since there are four unknown constants in (4.4.21) - (4.4.23) and only two boundary conditions (4.4.20), we need two additional conditions at z = h. One additional condition represents continuity of the electric field at z = h and has the form ~ ~ (4.4.24) E00even |z = h = E01even |z = h . Integrating (4.4.18) with respect to z from h − ε to h + ε and taking the limit as ε → +0 we obtain the second additional condition in the form ~ ~ µ sI ( s) dE01even dE00even (cos(λy1 ) − cos(λy0 )). |z = h − |z = h = 0 (4.4.25) dz dz 2 Using (4.4.21), (4.4.22), (4.4.24), and (4.4.25) we obtain

⎧ C1eλh + C2e− λh = C3e − λh , ⎪ ⎨ µ0 sI (s ) − λh λh − λh ( cos(λy1 ) − cos(λy0 ) ), ⎪− λC3e − λC1e + λC2e = 2 ⎩

⇒                

C1e 2 λh + C2 − C3 = 0               

− C1e 2 λh + C2 − C3 =

  µ 0 sI (s ) ( cos(λy1 ) − cos(λy0 ) )eλh 2λ

µ 0 sI (s ) ( cos(λy0 ) − cos(λy1 ) )e− λh ,                                                                                                   (4.4.26) 4λ µ sI (s ) ( cos(λy0 ) − cos(λy1 ) )eλh .                                                                                           (4.4.27) C3 = C2 + 0 4λ C1 =

Thus, solutions (4.4.21) and (4.4.22) can be rewritten as follows µ 0 s I (s ) ⎧ ~ even − λz − λ (h − z ) ⎪⎪ E00 = C2e + 4λ ( cos(λy0 ) − cos(λy1 ) )e ⇒   ⎨ ⎪ E~ even = C e − λz + µ0 sI (s ) ( cos(λy ) − cos(λy ) )e − λ ( z − h ) 01 2 0 1 4λ ⎩⎪ ~ µ s I (s ) ( cos(λy0 ) − cos(λy1 ) )e−λ h − z .                                                                         (4.4.28) E0even = C2e − λz + 0 4λ Using (4.4.20), (4.4.21) and (4.4.23) we obtain

⎧ C1 + C2 = C4 ⎪ q ⎨ ⎪ C1 − C2 = λµ m C4 1 ⎩

C4 =

⇒      

2λµ1m C1 , λµ1m + q

λµ m − q C2 = 1 m C 4 , 2λµ1

 

µ0 µ1m sI (s ) ( cos(λy0 ) − cos(λy1 ) )e− λh ,                                                                                            (4.4.29) C4 = m 2(λµ1 + q )  90   

C2 =

λµ1m − q µ0 sI (s ) ( cos(λy0 ) − cos(λy1 ) )e− λh .                                                                                  (4.4.30) m λµ1 + q 4λ

Using the inverse Fourier cosine transform of the form ∞ 2 ~ even even Ei ( y, z , s ) = ∫ Ei (λ , z , s ) cos(λy )dλ , i = 0,1

π

(4.4.31)

0

we obtain the solution Eieven ( y , z , s ) in regions R0 , R1 : even R0 : E0even ( y, z, s) = E0even ( free ) ( y, z , s ) + E0 ( ind ) ( y, z , s ),

(4.4.32)

where even 0 ( free )

E

µ 0 sI ( s ) ∞ e − λ | z − h | ( cos(λy0 ) − cos(λy1 ) ) cos(λy )dλ , ( y, z, s) = 2π ∫0 λ

E0even ( ind ) ( y , z , s ) =

(4.4.33)

µ0 sI ( s ) ∞ λµ1m − q e − λ ( z + h ) ( cos(λy0 ) − cos(λy1 ) )cos(λy )dλ. 2π ∫0 λµ1m + q λ

R1 : E1even ( y, z, s) =

µ0 µ1m sI ( s) ∞ e qz − λh ∫0 λµ1m + q ( cos(λy0 ) − cos(λy1 ) )cos(λy )dλ . π

(4.4.34) (4.4.35)

Here E0even ( free ) ( y, z , s ) is the even component of the electric field in free space while

E0even ( ind ) ( y, z , s ) is the even component of the induced electric field due to the presence of the conducting half-space. The Fourier sine transform of the form (4.4.17) and the inverse Fourier sine transform ∞ 2 ~ Eiodd ( y, z , s) = ∫ Eiodd (λ , z , s ) sin (λy )dλ , i = 0,1

π

(4.4.36)

0

can be used in order to find the odd components of the solution. It can easily be shown that the odd components of the solution, E0odd ( y, z , s ) and E1odd ( y, z , s ) are given by (4.4.32) – (4.4.35) where cosine is replaced by sine. Finally, the solution to (4.4.9) - (4.4.12) is obtained as the sum of the even and odd components and can be written as follows E0 ( y, z, s) = E0 ( free) ( y, z, s) + E0 ( ind ) ( y, z , s), (4.4.37)

µ0 µ1m sI ( s) ∞ eqz − λh E1 ( y, z, s) = ∫0 λµ1m + q ( cos(λ ( y − y0 )) − cos(λ ( y − y1 )) )dλ . π

(4.4.38)

where

E0 ( free ) ( y, z, s) =

µ0 sI ( s) ∞ e− λ | z − h| ( cos(λ ( y − y0 )) − cos(λ ( y − y1 )) )dλ , 2π ∫0 λ

(4.4.39)

is the electric field of the double conductor line in free space and

E0

( ind ) ( y , z , s ) =

µ0 sI ( s) ∞ λµ1m − q e− λ ( z + h ) ( cos(λ ( y − y0 )) − cos(λ ( y − y1 )) )dλ 2π ∫0 λµ1m + q λ

(4.4.40)

is the induced electric field in free space due to the presence of the conducting half-space.

 91   

The Laplace transform of the induced electromotive force (EMF) in a double conductor line per unit length due to the presence of the conducting half-space is given by (see [9]): 1 v( s) = E0 ( ind ) ( y, z , s )dl , (4.4.41) I ( s ) C∫ where C is the contour of the double conductor line of length one unit in the x − direction. In order to determine the inverse Laplace transform of (4.4.40) we rewrite the expression µ0 µ1mσ 1m 2 µ1m λµ1m − q λµ1m − q in the form where . τ = = − 1 , λµ1m + q λ2 λµ1m + q µ1m + 1 + sτ

2λµ1m 2µ1m λµ1m − q q + λµ1m − 2λµ1m ⋅ =− = −1 + = m −1 λµ1m + q λµ1m + q λµ1m + λ2 + µ0 µ1mσ 1m s µ1 + 1 + sτ Consider the case of a step current in the form (4.4.1). I The Laplace transform of (4.4.1) is I ( s) = 0 so that s ⎛ ⎞ λµ m − q 2µ1m Φ (λ , s ) = sI ( s ) 1m = I 0 ⎜⎜ m − 1⎟⎟. (4.4.42) λµ1 + q ⎝ µ1 + 1 + sτ ⎠ The inverse Laplace transform of (4.4.42) can be obtained either by means of “Mathematica” command InverseLaplaceTransform or using tables of the Laplace transform (see [2], page 810): (µ1m )2 t t ⎛ 2 µ1m I 0 − τ ⎛⎜ τ t ⎞ ⎞⎟ m ⎟ − I 0δ (t ). (4.4.43) Φ (λ , t ) = − µ1 e τ erfc⎜⎜ µ1m e ⎟⎟ ⎜ πt τ τ ⎝ ⎠ ⎝ ⎠ Using (4.4.40) - (4.4.42) we obtain the inverse Laplace transform of the induced electric field in region R0 in the form E0

( ind ) ( y , z , t ) =

µ0 I 0 ∞ − λ ( z + h ) ( cos(λ ( y − y0 )) − cos(λ ( y − y1 )) ) ⋅ e 2π ∫0

(4.4.44)

(µ1m )2 t ⎞ ⎛ t⎛ m ⎛ t ⎞ ⎞⎟ dλ ⎜ 2µ1 − τ ⎜ τ ⎟ − δ (t ) ⎟ e ⋅⎜ − µ1me τ erfc⎜⎜ µ1m ⎟ ⎟ ⎜ ⎟ πt τ⎠ ⎟ λ ⎜ τ ⎝ ⎝ ⎠ ⎠ ⎝ The induced IMF is obtained from (4.4.41) and (4.4.44) in the form

(µ ) t t⎛ ⎛ ⎞ m ⎛ m t ⎞ ⎞⎟ µ 0 ∞ − 2 λh ⎜ 2µ1 − τ ⎜ τ ⎟ dλ m τ ⎟ ⎜ v(t ) = e ( 1 − cos(λ ( y1 − y0 )) )⎜ e − δ (t ) ⎟ . erfc⎜ µ1 − µ1 e ∫ ⎟ ⎜ ⎟ π 0 πt τ⎠ ⎜ τ ⎟ λ ⎝ m 2 1

⎝

⎝

⎠

⎠

(4.4.45) Using (4.4.45) we compute the integral of the EMF with respect to time: T

V (T ) = ∫ν (t )dt.

(4.4.46)

0

The integral with respect to time can be computed, for example, by Mathematica command Integrate. The result is

V (T ) =

µ 0 ∞ − 2 λh dλ e [cos λ ( y1 − y0 ) − 1]ψ (T ,τ , µ ) , ∫ π 0 λ

(4.4.47)

 92   

where

⎧⎪ ⎛ T⎞ ⎡ T ⎤ ⎫⎪ ⎡ T ( µ 2 − 1) ⎤ ⎟ + exp ⎢ 2µ 2 ⎨− 1 + µerf ⎜⎜ erfc ⎢µ ⎥⎬ ⎥ τ ⎟⎠ τ τ ⎦ ⎪⎭ ⎛ T⎞ ⎪⎩ 1 ⎣ ⎦ ⎝ ⎣ ⎟− ψ (T ,τ , µ ) = − + 2 µerf ⎜⎜ . 2 ⎟ τ µ − 2 1 ⎝ ⎠ Integral (4.4.47) is computed by using “Mathematica” (see Appendix Fig.Ap.11) for the following values of the parameters of the problem: y1 − y0 = 1, µ = 2, h = 0.1 and three values of

σ , namely, σ = 107 S / m, σ = 5 ⋅ 107 S / m and σ = 108 S / m. .The results are shown in Fig. 4.7. V@tD

           σ = 10 7 S / m  

0.8

                         σ = 5 ⋅ 10 7 S / m   0.6

                       σ = 108 S / m   0.4

0.2

t 0.02

0.04

0.06

0.08

0.1

Fig. 4.7 The integrated EMF for three different values of σ .

 93   

5. MOVING PLANAR MEDIUM WITH VARYING PROPERTIES 5.1 Moving halfspace with varying electrical conductivity and magnetic permeability In some cases (for example, during steel processing at metallurgical plants) a conducting medium is moving with constant velocity in a horizontal direction. Analytical solutions for moving media with constant properties are presented in [1], [18]. Such solutions can serve as a basis for the solution of the inverse problems. It is shown, for example, in [48] how to determine resistivity, permeability and thickness of a moving metal sheet simultaneously using a double coil method. In the present section we calculate the change in impedance of a coil located above a conducting half-space [28]. The half-space is moving with constant velocity in a horizontal direction. The properties of a conducting half-space (the electrical conductivity and magnetic permeability) are assumed to vary with depth. In particular, analytical solution is constructed for the case where both electrical conductivity and relative magnetic permeability are exponential functions of the vertical coordinate. The problem is solved by the method of a double Fourier integral transform with respect to the x and y coordinates. The change in impedance of a singleturn coil is expressed in terms of a double integral containing the parameters of the problem. Results of numerical calculations are presented. Consider a single-turn coil of radius rc located above a conducting half-space moving with constant velocity V in the x direction (see Fig. 5.1).

z

rc

h x

O V

(

R 0 µ 0r = 1, σ 0 = 0 y R1 (µ1r , σ 1 )

) Fig. 5.1 A single-turn coil located above a conducting half-space.  

In this case the vector potential has only two nonzero components of the form r r r (5.1.1) A( x, y, z ) = Ax ( x, y, z )ex + Ay ( x, y, z )e y , r r where e x and e y are the unit vectors in the x and y directions, respectively. The external current in the coil has the form r r I e = Iϕ eϕ .

(5.1.2)

r The components of the external current I e can be written in the form  94   

r r r I e = I x ex + I y e y =

r r r ex = er cos ϕ r − eϕ sin ϕ r r r r e y = er sin ϕ r + eϕ cos ϕ r I ϕ = Iδ (z − h )δ (r − rc )

r r = − I ϕ sin ϕ r eϕ + I ϕ cos ϕ r eϕ .

(5.1.3)

It can be shown that the system of the Maxwell’s equations in this case can be transformed to the form r r r 1 dµ r ⎛ ∂Ax r ∂Ay r ⎞ ∂A r r ⎜ e y ⎟⎟ − jωµ 0 µ ( z )σ ( z )A − µ 0 µ ( z )σ ( z )V ex + = ∆A − r x ∂ ∂z µ ( z ) dz ⎜⎝ ∂z (5.1.4) ⎠ r r r = − µ 0 µ ( z )(I x e x + I y e y ), where the term containing V represents the effect of a moving halfspace. Projecting (5.1.4) on the x and y axes we obtain ⎧ 1 dµ r ∆ − A ⎪ x µ r ( z ) dz ⎪ ⎨ r ⎪∆A − 1 dµ y ⎪⎩ µ r ( z ) dz

∂Ax ∂A − jωµ0 µ r ( z )σ ( z )Ax − µ0 µ r ( z )σ ( z )V x = − µ0 µ r (z )I x , ∂z ∂x ∂Ay ∂A − jωµ0 µ r ( z )σ ( z )Ay − µ0 µ r ( z )σ ( z )V y = − µ0 µ r ( z )I y . ∂z ∂x

(5.1.5)

The x component of the vector potential in region R0 (see Fig. 5.1) satisfies the following equation: R0 : µ r ( z ) = 1, σ ( z ) = 0.

∆Ax0 = µ0 Iδ ( z − h )δ (r − rc )sin ϕ r ,

(5.1.6)

where δ (x) is the Dirac delta-function. The electrical conductivity σ (z ) and magnetic permeability µ (z ) in region R1 are assumed to be of the form R1 : µ1r ( z ) = µ m e βz , σ ( z ) = σ 1 ( z ) = σ m eαz .

(5.1.7)

The equation for the x component of the vector potential in region R1 is r ( ∂Ax1 dµ1 ∂Ax1 − k12 Ax1 + V , r ∂x µ1 ( z ) dz ∂z ( r r k12 = − jωµ 0 µ1 ( z )σ 1 ( z ) , V = µ 0 µ1 ( z )σ 1 ( z )V . where

∆Ax1 =

1

The boundary conditions are ∂Ax0 1 ∂Ax1 | z =0 = r | z =0 . Ax0 | z =0 = Ax1 | z =0 , ∂z µ1 (0 ) ∂z

The following conditions hold at infinity: ∂Axi ∂Axi , → 0 as x → ±∞ , i = 0,1. Axi , ∂x ∂y

(5.1.8)

(5.1.9)

(5.1.10)

Problem (5.1.6) - (5.1.10) is solved by the method of double Fourier transform with respect to the variables x and y of the form  95   

~ Axi (λx , y, z ) =

∞

∫A

xi

( x, y, z ) exp(− jλx x )dx, i = 0,1,

(5.1.11)

−∞

~ ~ Axi (λ x , λ y , z ) =

∞

~

∫A

xi

(λ x , y, z ) exp(− jλ y y )dy, i = 0,1.

(5.1.12)

−∞

Applying the transform (5.1.11) to ∆Axi we obtain ~ ∆Axi (λx , y, z ) =

⎛ ∂ 2 Axi ∂ 2 Axi ∂ 2 Axi ∫− ∞ ⎜⎜ ∂x 2 + ∂y 2 + ∂z 2 ⎝ ∞

u = e − jλ x x → du = − jλx e − jλ x x dx =

∂ Axi

∂Axi

2

dv = u=e

=

dv =

∂x 2 − jλ x x

∂Axi ∂x

dx → v =

⎞ − jλ x x ⎟e dx = ⎟ ⎠ = jλ x

∞

∫

−∞

∂x

→ du = − jλx e − jλ x x dx dx → v = Axi

∂Axi ∂x

e

− jλ x x

dx +

~ ∂ 2 Axi ∂y 2

+

~ ∂ 2 Axi ∂z 2

=

(5.1.13)

⎛ ∂2 ⎞~ ∂2 = ⎜⎜ 2 + 2 − λ2x ⎟⎟ Axi ∂z ⎝ ∂y ⎠

Using (5.1.10), (5.1.12) and (5.1.13) we obtain ∞ ~ ⎛ ∂2 ⎞ ~ − jλ y ⎛ d2 ⎞~ ~ ~ ∂2 ∆Axi (λx , λ y , z ) = ∫ ⎜⎜ 2 + 2 − λ2x ⎟⎟ Axi e y dy = ⎜⎜ 2 − λ2 ⎟⎟ Axi                                                    (5.1.14) ∂y ∂z ⎠ ⎝ dz ⎠ −∞ ⎝

λ2 = λ2x + λ2y .

where

Applying the Fourier transforms (5.1.11), (5.1.12) to the right-hand side of equation (5.1.6) we obtain ∞ ∞

∫ ∫ µ Iδ (z − h)δ (r − r )sin ϕ 0

c

r

e

(

− j λx x + λ y y

)

dxdy =

−∞ −∞

= λx x + λ y y = λr cos ϕ λ cos ϕ r + λr sin ϕλ sin ϕ r = λr cos(ϕλ − ϕ r ) = 2π

∞

= µ0 Iδ (z − h )∫ δ (r − rc ) rdr ∫ sin ϕ r e − jλr cos (ϕλ −ϕr ) dϕ r = 0

      

0

∞

= e jz cos ϕ = J 0 ( z ) + 2∑ j k J k ( z ) cos kϕ = k =1

=

2π

∞

0

k =1

∫ sin ϕr J 0 (− λr )dϕr + 2∑ = −2 jπJ1 (λr )sin ϕλ

2π ⎛ ⎞ ⎜ cos kϕλ ∫ cos kϕ r sin ϕ r dϕ r + ⎟ ⎜ ⎟ 0 j k J k (− λr )⎜ ⎟= 2π ⎜ + sin kϕ sin kϕ sin ϕ dϕ ⎟ r r r ⎟ λ ∫ ⎜ 0 ⎝ ⎠

 

∞

= −2 jµ0π sin ϕλ Iδ ( z − h )∫ J1 (λr )δ (r − rc ) rdr = −2 jµ0π sin ϕλ I J1 (λrc )rcδ ( z − h ) ⇒ 0

 96   

∞ ∞

∫ ∫ µ Iδ (z − h)δ (r − r )sin ϕ 0

c

r

e

(

− j λx x + λ y y

)

dxdy = −2 jµ0π sin ϕλ I J1 (λrc )rcδ ( z − h )                      (5.1.15)                  

−∞ −∞

Using (5.1.14) and (5.1.15) we rewrite (5.1.6) in the form  ~ ~ ~ d 2 Ax0 ~ − λ2 Ax 0 = γδ ( z − h ),                                                                                                                          (5.1.16) 2 dz where            γ = −2 jµ 0π sin ϕλ I J1 (λrc )rc . The right-hand side of equation (5.1.8) can be rewritten in the form ∞ ∞ ( ∂Ax1 ⎞ − jλ x x ⎛ ∂Ax1 ∂ − jλ x x − jλ x 2 2 ∫−∞⎜⎜⎝ β ∂z − k1 Ax1 + V ∂x ⎟⎟⎠e dx =β ∂z −∫∞ Ax1 e dx − k1 −∫∞ Ax1 e x dx +                          ~ ( ∞ ∂Ax1 − jλ x x (~ ∂Ax1 2~ +V ∫ − k1 Ax1 + jλxV Ax1 e dx = β ∂ ∂ z x −∞ ~ ~ ~ ∞⎛ ( ~ ⎞ − jλ y y ( ~ dAx1 ∂Ax1 ~ ~ 2 2 ⎜β ⎟ (5.1.17) − k A + j λ V A e dy = β − k − j λ ϕ V Ax1 cos λ 1 x x x 1 ∫−∞⎜ ∂z 1 1 ⎟ dz ⎝ ⎠ Using (5.1.11) and (5.1.14) we transform (5.1.5) to the form  ~ ~ ~ ~ ~ d 2 Ax1 dAx1 ~ −β − q12 Ax1 = 0,                                                                                                                        (5.1.18) 2 dz dz ( ( where q12 = λ2 − k12 + jλ cos ϕλ V , k12 = − jωµ0 µ1 (z )σ 1 ( z ) , V = µ 0 µ1 ( z )σ 1 ( z )V .   ∞

(

)

Applying the Fourier transform with respect to x and y to the boundary conditions we obtain ~ ~ ~ ~ ~ ~ dAx0 ~ ~ 1 dAx1 | . (5.1.19) |z = 0 = Ax0 |z = 0 = Ax1 |z = 0 , dz µ m dz z = 0 The following conditions hold at infinity: ~ ~ Axi → 0 as x → ±∞ , i = 0,1.

(5.1.20)

In order to solve equation (5.1.16) we consider the following two sub-regions of R0 : 0 < z < h ~ ~ ~ ~ and z > h . The solutions in these regions are denoted by Ax00 and Ax01 , respectively. Hence, ~ ~ d 2 Ax00 2

dz ~ ~ d 2 Ax01

~ ~ − λ2 Ax00 = 0,

0 < z < h ,                                                                                                              (5.1.21)

~ ~ − λ2 Ax01 = 0,

z > h ,                                                                                                                      (5.1.22) dz The general solution to (5.1.21) can be written in the form ~ ~ Ax00 = C1eλz + C2e − λz .                                                                                                                                     (5.1.23) 2

The bounded solution to (5.1.22) is  97   

~ ~ Ax01 = C3e − λz .                                                                                                                                                    (5.1.24) ~ ~ ~ ~ The functions Ax00 (λx , λ y , z ) and Ax01 (λx , λ y , z ) satisfy the following conditions at z = h :  ~ ~ ~ ~ ~ ~ dAx01 dAx00 ~ ~ (5.1.25) Ax00 | z =h = Ax01 | z =h , | z =h − | z =h = γ . dz dz ~ ~ The first condition in (5.1.25) reflects the fact that the function Ax0 (λx , λ y , z ) is continuous at z = h while the second condition in (5.1.25) is obtained integrating (5.1.16) with respect to z from z = h − ε to z = h + ε and considering the limit in the resulting expression as ε → +0 .

Using (5.1.23), (5.1.24) and (5.1.25) we obtain

⎧⎪C1e λh + C2 e −λh = C3e −λh ⎨ ⎪⎩− λC3e −λh − λC1e λh + λC2 e −λh = γ

C1e 2 λh + C2 − C3 = 0

⇒

− C1e 2 λh + C2 − C3 =

γ λh e λ

⇒        

γ −λ h ⎧ ⎪⎪C1 = − 2λ e ⎨ ⎪C = C − γ e λh 2 ⎪⎩ 3 2λ

(5.1.26)

Hence, solutions (5.1.23) and (5.1.24) can be rewritten as follows ~ γ − λ (h − z ) ⎧~ A = C2 e − λz − e ~ ~ γ −λ h − z ⎪⎪ x00 2λ ⇒ Ax0 = C2e − λz − e .                                                                 (5.1.27) ⎨~ ~ γ −λ (z −h) 2λ − λz ⎪A e − x = C2 e 2λ ⎩⎪ 01 The solution to (5.1.18) can be expressed in terms of modified Bessel functions (see [50], formula 2.1.3.10, page 247): y xx'' + ay x' + (be λx + c )y = 0 ⇒ y = e

ν=

−

ax 2

⎡ ⎛ 2 b λ2x ⎞ ⎛ 2 b λ2x ⎞⎤ ⎜ ⎟ ⎜ ⎟⎥ C J e C Y + ⎢ 1 ν⎜ λ 2 ν⎜ λ e ⎟ ⎟ ⎢⎣ ⎝ ⎠ ⎝ ⎠⎥⎦

a 2 − 4c

λ

βz (α + β ) z (α + β ) z ⎤ ~ ⎡ ⎛ ⎞ ⎛ ⎞ ~ Ax1 = e 2 ⎢C4 Jν ⎜⎜ α2 + bβ e 2 ⎟⎟ + C5Yν ⎜⎜ α2 + bβ e 2 ⎟⎟⎥.                           ⎢⎣ ⎝ ⎠⎥⎦ ⎝ ⎠ The bounded solution to (5.1.18) is

βz (α + β ) z ~ ⎞ ⎛ ~ Ax1 = e 2 C4 Jν ⎜⎜ α2 + bβ e 2 ⎟⎟,                                                                                                                        (5.1.28)  ⎠ ⎝

where

( β 2 + 4λ2 , b = − jµ0 µ mσ m (ω + λ cos ϕλ V ), ν= α +β

( V = µ 0 µ1 (z )σ 1 ( z )V .

 98   

Using (5.1.19) , (5.1.27) and (5.1.28) we obtain

γ − λh e = C4 Jν ( z0 ), 2λ                                                                                         (5.1.29)  γ − λh C4 ⎛ β ⎞ ' − λ C2 − e = ⎜ J ( z ) + b Jν ( z0 )⎟, 2 µm ⎝ 2 ν 0 ⎠

C2 −

where

z0 =

2 b α +β

⋅ C2 = C4 Jν (z0 ) +

(

γ − λh e 2λ

⋅ − λC4 Jν ( z0 ) − γe

− λh

⎞ ⎛β µm = C4 ⎜ Jν ( z0 ) + b Jν' ( z0 )⎟ ⎝2 ⎠

)

                               

and C4 = −

γµ me − λh

β⎞ ⎛ ' ⎜ λµ m + ⎟ Jν ( z0 ) + b Jν ( z0 ) 2⎠ ⎝

,                                                                                                      (5.1.30)

⎛β ⎞ ' ⎜ − λµ m ⎟ Jν ( z0 ) + b Jν (z0 ) γ − λh ⎝ 2 ⎠ ,                                                                                           (5.1.31) C2 = e 2λ ⎛β ⎞ ' ( ) ( ) + λµ J z + b J z ⎜ m⎟ ν 0 ν 0 ⎝2 ⎠

where            γ = −2 jµ0π sin ϕλ I J1 (λrc )rc .

The induced vector potential in region R0 is ~ ~ Axind (λx , λ y , z ) = C2e − λz ,                                                                                                                      (5.1.32) 0 where C2 is given by (5.1.31). The inverse Fourier transform is defined as follows ∞ ∞ ~ ~ 1 Axi ( x, y, z ) = A (λx , λ y , z ) exp( j (λx x + λ y y ))dλx dλ y . 2 ∫ ∫ xi 4π −∞ −∞ ~ ~ Applying (5.1.33) to Axind we obtain 0 Axind ( x, y , z ) = 0

~ 1 ~ ind j (λ x + λ y ) A (λ x , λ y , z )e x y dλ x dλ y = 2 ∫ ∫ x0 4π −∞ −∞ 4π 2 1

∞ ∞

(5.1.33)

∞ ∞

∫ ∫C e

−λ z

2

e

(

j λx x + λ y y

)

dλ x dλ y =   

− ∞− ∞

λx x + λ y y = λr cos ϕ λ cos ϕ r + λr sin ϕ λ sin ϕ r = λr cos(ϕ λ − ϕ r )

γ = −2 jµ 0π sin ϕ λ I J1 (λrc )rc    

=

⎛β ⎞ ' ⎜ − λµ m ⎟ Jν ( z0 ) + b Jν ( z0 ) 2 ⎠ D=⎝ ⎞ ⎛β ' ⎜ + λµ m ⎟ Jν ( z0 ) + b Jν ( z0 ) ⎠ ⎝2

= 

 99   

2π

∞

=−

1 jIrc µ0 ∫ J1 (λrc )e − λ ( z + h )dλ ∫ D sin ϕλ e jλ r cos (ϕ λ −ϕ r )dϕλ 4π 0 0 e

jz cos ϕ

∞

     

= J 0 ( z ) + 2∑ j J k ( z )cos kϕ k

k =1

  Consequently,

Axind ( x, y , z ) = 0 2π

∞

=−

∞ ⎞ ⎛ 1 jIrc µ0 ∫ J1 (λrc )e − λ ( z + h )dλ ∫ D sin ϕλ ⎜ J 0 (λr ) + 2∑ j k J k (λr )cos k (ϕλ − ϕ r )⎟dϕλ . 4π k =1 ⎠ ⎝ 0 0

(5.1.34) Similarly, using (5.1.3) and (5.1.5) we obtain the equations for the y − components of the vector potential in regions R0 and R1. Equation in region R1 has the same form as (5.1.8), where Ax1 is replaced by Ay1 . Equation for the y − component of the vector potential in region R0 has the form (5.1.6) where sin ϕ r is replaced by − cos ϕ r . The solution of the corresponding problem for the y − component of the vector potential is obtained by means of the double Fourier transforms (5.1.11) and (5.1.12). Without repeating the calculations we present only the final result here: Ayind0 ( x, y, z ) =

=

2π ∞ ∞      ⎞ ⎛ 1 jIrc µ0 ∫ J1 (λrc )e − λ ( z + h )dλ ∫ D cos ϕλ ⎜ J 0 (λr ) + 2∑ j k J k (λr )cos k (ϕλ − ϕ r )⎟dϕλ . 4π k =1 ⎠ ⎝ 0 0

                                                                                                                                                                             (5.1.35) 

The induced change in impedance of the coil is given by the formula r jω r ind Z ind = A0 ( x, y, z )dl , ∫ I L where L is the contour of the coil and r r r ( x, y, z )ex + Ayind0 ( x, y, z )e y . A0ind ( x, y, z ) = Axind 0

(5.1.36)

(5.1.37)

Substituting (5.1.34) and (5.1.35) into (5.1.36) we obtain Z

ind

jωrc = I

2π

∫ (A

ind y0

0

)

cos ϕ r − Axind sin ϕ r dϕ r 0

= r = rc z =h

2π ∞ ⎡ ⎤ ⎛ ⎞ k ( ) ( ) ( ) + − + cos ϕ cos ϕ J λ r 2 j J λ r cos k ϕ ϕ d ϕ ⎜ ⎟ ⎢ ⎥ ∑ λ λ r k r r 0 ∫0 2π ∞ k =1 ⎝ ⎠ jωrc2 −λ ( z + h ) ⎢ ⎥dϕ = = jIµ 0 ∫ J1 (λrc )e dλ ∫ D 2π ⎢ ⎥ λ ∞ 4π I ⎛ ⎞ 0 0 k ⎢+ sin ϕ λ ∫ sin ϕ r ⎜ J 0 (λr ) + 2∑ j J k (λr ) cos k (ϕ λ − ϕ r )⎟dϕ r ⎥ ⎢⎣ ⎥⎦ k =1 ⎝ ⎠ 0 2π ωrc2 ∞ −λ ( z + h ) =− µ 0 ∫ J1 (λrc )e dλ ∫ 2 DjπJ1 (λr )dϕ λ 4π 0 0

r = rc z =h

  where   100   

⎞ ⎛β ' ⎜ − λµ m ⎟ Jν ( z0 ) + b Jν ( z0 ) 2 ⎠ . D=⎝ ⎛β ⎞ ' ⎜ + λµ m ⎟ Jν ( z0 ) + b Jν ( z0 ) ⎝2 ⎠ 2π

⋅ J 0 (λr ) cos ϕλ ∫ cos ϕ r dϕ r + 0

2π 2π ∞ ⎡ ⎤ + 2 cos ϕλ ∑ j k J k (λr )⎢cos kϕλ ∫ cos kϕ r cos ϕ r dϕ r + sin kϕ λ ∫ sin kϕ r cos ϕ r dϕ r ⎥ =   k =1 0 0 ⎣ ⎦ 2 = 2 jπ cos ϕ λ J1 (λr )

2π

⋅ J 0 (λr )sin ϕλ ∫ sin ϕ r dϕ r + 0

2π 2π ⎡ ⎤ + 2 sin ϕλ ∑ j J k (λr )⎢cos kϕλ ∫ cos kϕ r sin ϕ r dϕ r + sin kϕλ ∫ sin kϕ r sin ϕ r dϕ r ⎥ =   k =1 0 0 ⎣ ⎦ 2 = 2 jπ sin ϕ λ J1 (λr ) ∞

k

Hence, ⎛β ⎞ ' ⎜ − λµ m ⎟ Jν ( z0 ) + b Jν ( z0 ) j 2 2 ⎠ = − ωrc µ0 ∫ J12 (λrc )e − 2 λh dλ ∫ ⎝ dϕλ .                                          (5.1.38) β 2 ⎛ ⎞ ' 0 0 ⎜ + λµ m ⎟ Jν ( z0 ) + b Jν ( z0 ) ⎝2 ⎠ 2π

∞

Z ind

 

Using following dimensionless parameters (5.1.38) can be rewritten as follows u ⎫ , rc ⎪⎪ ⎬ ⇒ νˆ = αˆ βˆ ⎪ ⋅ α= , β= , rc rc ⎪⎭ ⋅ λrc = u → λ =

⋅

⋅

b=

β 2

)

1 − j δˆ12 + ρˆ u cos ϕλ = rc

+ λµ m =

⋅ zˆ0 =

(

βˆ 2 + 4u 2 αˆ + βˆ ⎧⎪ δˆ = rc ωµ0 µ mσ m −j ˆ b , where ⎨ 1 rc ⎪⎩ ρˆ = rc µ0 µ mσ mV

⎞ ⎛ ˆ ⎞ 1 ⎛⎜ βˆ ⎟ , ⋅ β − λµ m = 1 ⎜ β − uµ m ⎟ µ u + m ⎟ ⎟ rc ⎜⎝ 2 rc ⎜⎝ 2 2 ⎠ ⎠

2 bˆ αˆ + βˆ

 101   

Using the previous formula we obtain Z ind = ω rc µ0 Z ,                                                                                                                                                 (5.1.39)  where

⎛ βˆ ⎞ ⎜ − uµm ⎟ Jνˆ ( zˆ0 ) + − jbˆ Jν'ˆ ( zˆ0 ) ⎜2 ⎟ j ˆ ⎠ dϕλ Z = − ∫ J12 (u )e− 2u h du ∫ ⎝ ⎛ ˆ ⎞ 20 0 β ' ˆ ⎜ + uµm ⎟ Jνˆ ( zˆ0 ) + − jb Jνˆ ( zˆ0 ) ⎜2 ⎟ ⎝ ⎠ and the dimensionless parameters are defined as follows 2π

∞

(5.1.40)

βˆ 2 + 4u 2 ˆ ˆ 2 , b = δ 1 + ρˆ u cos ϕ λ , δˆ1 = rc ωµ 0 µ mσ m , ρˆ = rc µ 0 µ mσ mV .  ˆ αˆ + β

νˆ =

Formula (5.1.40) is used to compute the change in impedance of a single-turn coil for the following values of the parameters of the problem: αˆ = 0, βˆ = 2, hˆ = 0.05, d = 0.1, r = 5, µ = 5. 0

m

The results of the calculations are shown in Fig. 5.2. The points on each graph correspond to different values of δˆ = 1,2,...,20 (from left to right). 1

Computations are done with “Mathematica” ( see Appendix Fig.Ap.12 ) . Im @zD

          ρˆ = 5  

1

                         ρˆ = 10  

0.75

                       ρˆ = 15  

0.5 0.25 0.1 -0.25 -0.5

0.2

0.3

0.4

0.5

0.6

0.7

Re @zD

Fig. 5.2 The change in impedance computed by formula (5.1.40) for three values of ρˆ  

   

5.2 Moving two-layer medium with varying electrical conductivity and magnetic permeability Consider the case where an upper layer of a two-layer medium is moving in the horizontal direction with constant velocity V while the lower half-space is fixed [26] (see Fig. 5.3)).

 102   

z

rc

h x

O

d1

V

(

)

R 0 µ 0r = 1, σ 0 = 0 y r r R1R(1µ(1µ,1 σ, σ 1 )1 )

(

R 2 µ 2r , σ 2

) Fig. 5.3 A single-turn coil above a moving two-layer medium.

The electrial conductivity and magnetic permeability of the upper layer are exponential functions of the vertical coordinate:

µ ( z ) = µ1 (z ) = µme βz , σ ( z ) = σ 1 ( z ) = σ meαz ,

(5.2.1)

while the conductivity σ 2 and relative magnetic permeability µ 2 of the lower half-space are assumed to be constant. The components Axi ( x, y, z ), i = 0,1,2 of the vector potential in regions R0 , R1 and R2 (see Fig. 5.3) satisfy the following system of partial differential equations: R0 : ∆Ax0 = µ0 Iδ ( z − h )δ (r − rc )sin ϕ r ,

R1 : ∆Ax1 = where

( ∂A dµ1 ∂Ax1 − k12 Ax1 + V x1 , ∂x µ1 ( z ) dz ∂z ( k12 = − jωµ0 µ1 ( z )σ 1 ( z ) , V = µ0 µ1 ( z )σ 1 (z )V ,

1

R2 : ∆Ax 2 = − k22 Ax 2 , where

(5.2.2)

(5.2.3)

(5.2.4)

k22 = − jωµ0 µ2σ 2 .

The boundary conditions are ∂Ax0 1 ∂Ax1 Ax0 |z = 0 = Ax1 |z = 0 , |z = 0 = | , µ1 (0) ∂z z = 0 ∂z Ax1 |z = − d = Ax 2 |z = − d ,

1 ∂Ax1 1 ∂Ax 2 |z = − d = | , µ1 (− d ) ∂z µ2 ∂z z = − d

The following conditions hold at infinity: ∂Axi ∂Axi Axi , , → 0 as x → ±∞ , i = 0,1,2. ∂x ∂y Applying the Fourier transforms (5.1.11) and (5.1.12) to (5.2.2)-(5.2.7) we obtain ~ ~ ~ d 2 Ax0 2 ~ − λ A x 0 = γδ ( z − h ), dz 2

(5.2.5) (5.2.6)

(5.2.7)

(5.2.8)

 103   

~ ~ d 2 Ax1

~ ~ dAx1

~ ~ − q12 Ax1 = 0,

2

−β

2

~ ~ − q22 Ax2 = 0,

dz ~ ~ d 2 Ax 2

dz where

dz

(5.2.9) (5.2.10)

( γ = −2 jµ0π sin ϕλ I J1 (λrc )rc , q12 = λ2 − k12 + jλ cos ϕλ V , k12 = − jωµ0 µ1 ( z )σ 1 ( z ) , ( V = µ0 µ1 ( z )σ 1 (z )V , q22 = λ2 − k22 , k22 = − jωµ0 µ 2σ 2 .

The boundary conditions in the transformed space can be written as follows: ~ ~ ~ ~ ~ ~ ∂Ax0 ~ ~ 1 ∂Ax1 Ax0 |z = 0 = Ax1 |z = 0 , |z = 0 = | , µ m ∂z z = 0 ∂z ~ ~ ~ ~ ~ ~ 1 ∂Ax2 e βd ∂Ax1 ~ ~ |z = − d . |z = − d = Ax1 |z = − d = Ax2 |z = − d , µm ∂z µ2 ∂z

(5.2.11)

(5.2.12)

The following conditions hold at infinity: ~ ~ ~ ~ ~ ~ ∂Axi ∂Axi Axi , , → 0 as x → ±∞ , i = 0,1,2 ∂x ∂y As in Section  5.1, the solution of (5.2.8) in free space is given by (5.1.27) where the term containing C2 represents the induced vector potential in free space due to the presence of a twolayer conducting medium. General solution to (5.2.9) can be expressed in terms of modified Bessel functions (see [50], formula 2.1.3.10, page 247): y + ay + (be + c )y = 0 ⇒ y = e '' xx

λx

' x

ν=

−

ax 2

⎡ ⎛ 2 b λ2x ⎞ ⎛ 2 b λ2x ⎞⎤ ⎢C1 Jν ⎜⎜ λ e ⎟⎟ + C2Yν ⎜⎜ λ e ⎟⎟⎥ ⎝ ⎠ ⎝ ⎠⎦⎥ ⎣⎢

a 2 − 4c

λ

βz ~ ⎡ ⎛ 2 b (α +2 β ) z ⎞ ⎛ 2 b (α +2 β ) z ⎞⎤ ~ 2 ⎜ ⎟ ⎜ ⎟⎥, Ax1 = e ⎢C4 Jν ⎜ α + β e ⎟ + C5Yν ⎜ α + β e ⎟ ⎝ ⎠ ⎝ ⎠⎦⎥ ⎣⎢

(5.2.13)

( ( β 2 + 4λ2 , b = − jµ0 µmσ m (ω + λ cos ϕλ V ), V = µ0 µ1 ( z )σ 1 (z )V . α +β The bounded solution to (5.2.10) is ~ ~ Ax = C6e q z .                                                                                                                                                     (5.2.14)

where

ν =

2

2

Using the boundary conditions (5.1.25) and (5.2.11) −(5.2.12) we obtain C2 −

γ − λh e = C4 Jν ( z0 ) + C5Yν (z0 ),                                                                                                          (5.2.15) 2λ

 104   

− λ C2 −

γ 2

C6 e − q 2 d = C 4 e where

C4 ⎛ β ⎞ C ⎛β ⎞ ' ' ⎜ Jν ( z0 ) + b Jν ( z0 )⎟ + 5 ⎜ Yν ( z0 ) + bYν ( z0 )⎟,                                    (5.2.16)  µm ⎝ 2 ⎠ µm ⎝ 2 ⎠

e − λh =

z0 =

−

βd − ⎛ − (α + β )d ⎞ ⎛ − (α + β )d ⎞ Jν ⎜⎜ z0e 2 ⎟⎟ + C5e 2 Yν ⎜⎜ z0e 2 ⎟⎟,                                                                    (5.2.17) ⎝ ⎠ ⎝ ⎠

βd 2

2 b . α +β

It follows from (5.2.15) and (5.2.16) that ⋅ C2 = C4 Jν (z0 ) + C5Yν (z0 ) +

γ − λh e 2λ

⋅ (− λC4 Jν ( z0 ) − λC5Yν ( z0 ) − γe

− λh

     

)µm = C4 ⎛⎜ β2 Jν (z0 ) + b Jν' (z0 )⎞⎟ + C5 ⎛⎜ β2 Yν (z0 ) + bYν' (z0 )⎞⎟ ⎝ ⎠ ⎝ ⎠

⎞ ⎛⎛ β⎞ C5 ⎜⎜ ⎜ λµ m + ⎟Yν ( z0 ) + bYν' ( z0 )⎟⎟ + γµ m e − λh 2⎠ ⎝ ⎠ .                                                                           (5.2.18) C4 = − ⎝ β⎞ ⎛ ' ⎜ λµ m + ⎟ Jν ( z0 ) + b Jν ( z0 ) 2⎠ ⎝ Using (5.2.17) and (5.2.18) we obtain ⋅ C6 e − q 2 d = C 4 e

−

βd 2

βd

⋅ C6 e − q 2 d =

µ 2e 2 µ m q2

Jν ( z1 ) + C5e

−

βd 2

Yν ( z1 )

(α + β )d (α + β )d ⎛ ⎛β − − ⎛ ⎞⎞ ⎞ ⎜ C4 ⎜ Jν ( z1 ) + be 2 Jν' ( z1 )⎟ + C5 ⎜ β Yν (z1 ) + be 2 Yν' ( z1 )⎟ ⎟ ⎜2 ⎟⎟ ⎟ ⎜ ⎜2 ⎝ ⎠⎠ ⎠ ⎝ ⎝

βd βd − − ⎛ ⎞ ⋅ µ m q2 ⎜⎜ C4e 2 Jν ( z1 ) + C5e 2 Yν ( z1 )⎟⎟ = ⎝ ⎠

 

βd (α + β )d (α + β )d ⎛ ⎛β − − ⎞ ⎛β ⎞⎞ = µ 2e 2 ⎜ C4 ⎜⎜ Jν (z1 ) + be 2 Jν' ( z1 )⎟⎟ + C5 ⎜⎜ Yν ( z1 ) + be 2 Yν' (z1 )⎟⎟ ⎟ ⎜ ⎟ ⎠ ⎝2 ⎠⎠ ⎝ ⎝2

( β − α )d

⎛β ⎞ ' βd ⎜ µ 2e − µ m q2 ⎟Yν ( z1 ) + µ 2 b e 2 Yν ( z1 ) 2 ⎝ ⎠ ,                                                                 (5.2.19) C4 = −C5 ( β − α )d ⎛β ⎞ ' βd 2 Jν ( z1 ) ⎜ µ 2e − µ m q2 ⎟ Jν ( z1 ) + µ 2 b e ⎝2 ⎠ where z1 = z0e

−

(α + β )d 2

,

z0 =

2 b . α +β

Eliminating C4 from (5.2.18) and (5.2.19) we get

 105   

( β −α )d ⎛ ⎛β ⎞ ⎜ ⎜ µ 2 e βd − µ m q 2 ⎞⎟Yν ( z1 ) + µ 2 b e 2 Yν' ( z1 ) ⎛⎜ λµ m + β ⎞⎟Yν ( z 0 ) + bYν' ( z 0 ) ⎟ 2⎠ ⎜ 2 ⎟ ⎠ ⋅ C5 ⎜ ⎝ −⎝ ( β −α )d ⎟= β ⎛ ⎞ β ' ⎛ ⎞ d β ' ⎜⎜ ⎜ µ 2 e − µ m q 2 ⎟ Jν ( z1 ) + µ 2 b e 2 Jν ( z1 ) ⎜ λµ m + ⎟ Jν ( z 0 ) + b Jν (z 0 ) ⎟⎟ 2⎠   ⎝ ⎠ ⎝⎝ 2 ⎠

=

γµ m e −λh

β⎞ ⎛ ' ⎜ λµ m + ⎟ Jν ( z 0 ) + b Jν ( z 0 ) 2⎠ ⎝

( β − α )d ⎞ ⎛⎛ ⎛⎛ β ⎞ β⎞ ⎞ βd ⎜ ⋅ Cskait = ⎜ ⎜ µ 2e − µ m q2 ⎟Yν (z1 ) + µ 2 b e 2 Yν' ( z1 )⎟⎟ ⋅ ⎜⎜ ⎜ λµ m + ⎟ Jν ( z0 ) + b Jν' ( z0 )⎟⎟ − 2⎠ ⎠ ⎠ ⎠ ⎝⎝ ⎝⎝ 2 ( β − α )d ⎛⎛ β ⎞⎛ ⎛ ⎞ β⎞ ⎞ − ⎜⎜ ⎜ µ 2e βd − µ m q2 ⎟ Jν ( z1 ) + µ 2 b e 2 Jν' ( z1 )⎟⎟⎜⎜ ⎜ λµ m + ⎟Yν ( z0 ) + bYν' ( z0 )⎟⎟ = 2⎠ ⎠ ⎠ ⎝⎝ 2 ⎠⎝ ⎝ β⎞ ⎛β ⎞⎛ = ⎜ µ 2e βd − µ m q2 ⎟⎜ λµ m + ⎟( Jν ( z0 )Yν ( z1 ) − Jν ( z1 )Yν ( z0 ) ) + 2⎠ ⎝2 ⎠⎝

⎛β ⎞ + ⎜ µ 2e βd − µ m q2 ⎟ b Jν' ( z0 )Yν ( z1 ) − Jν ( z1 )Yν' ( z0 ) + 2 ⎝ ⎠

(

( β − α )d

)

( β − α )d

β⎞ ⎛ Jν ( z0 )Yν' ( z1 ) − Jν' ( z1 )Yν ( z0 ) + µ 2b e 2 Jν' ( z0 )Yν' ( z1 ) − Jν' ( z1 )Yν' ( z1 ) + ⎜ λµ m + ⎟ µ 2 b e 2 2 ⎝ ⎠ ( β − α )d ⎞⎛ ⎛ ⎛⎛ β ⎞ β⎞ ⎞ ⋅ Csauc = ⎜⎜ ⎜ µ 2e βd − µ m q2 ⎟ Jν ( z1 ) + µ 2 b e 2 Jν' ( z1 )⎟⎟⎜⎜ ⎜ λµ m + ⎟ Jν ( z0 ) + b Jν' ( z0 )⎟⎟ 2⎠ ⎠ ⎠ ⎠⎝ ⎝ ⎝⎝ 2

(

)

(

 

C5 = γµme− λh

S1 , D

(5.2.20)

where

⎛β ⎞ S1 = ⎜ µ2eβd − µm q2 ⎟ Jν ( z1 ) + µ2 b e ⎝2 ⎠

(

)

D = D5 D6 D1 + b D2 + µ2 b e

( β −α )d 2

( β −α )d 2

Jν' ( z1 ) ,

(D D + 6

3

)

b D4 ,

β

⋅ D1 = Jν ( z0 )Yν ( z1 ) − Jν ( z1 )Yν ( z0 ),

⋅ D5 =

⋅ D2 = Jν' ( z0 )Yν ( z1 ) − Jν ( z1 )Yν' ( z0 ),

⋅ D6 = λµm +

⋅ D3 = Jν ( z0 )Yν' ( z1 ) − Jν' ( z1 )Yν ( z0 ),

2

µ2e βd − µm q2 , β 2

.

⋅ D4 = Jν' ( z0 )Yν' ( z1 ) − Jν' ( z1 )Yν' ( z1 ), The constant C4 is given by

C4 = −γµme− λh

S2 , D

(5.2.21)

 106   

)

where

⎛β ⎞ S2 = ⎜ µ2e βd − µm q2 ⎟Yν ( z1 ) + µ2 b e ⎝2 ⎠ Finally, the value of C2 is ⋅ C2 = C4 Jν ( z0 ) + C5Yν ( z0 ) +

( β −α )d 2

Yν' ( z1 ) .

γ − λh e   2λ

( β −α )d ⎛ ⎛ ⎞ ⎜ ⎜ 2λµ m ⎜ D5 D1 − µ2 b e 2 D3 ⎟⎟ γ − λh ⎜ ⎝ ⎠ C2 = e ⎜ 1− ( β −α )d 2λ ⎜ D5 D6 D1 + b D2 + µ2 b e 2 D6 D3 + b D4 ⎜ ⎝

(

)

(

)

where γ = −2 jµ0π sin ϕλ I J1 (λrc )rc .

⎞ ⎟ ⎟ ⎟, ⎟ ⎟ ⎠

The induced vector potential in region R0 is ~ ~ Axind ( λ x , λ y , z ) = C2 e − λ z , 0

(5.2.22)

(5.2.23)

where C2 is given by (5.2.22). Applying the inverse Fourier transform in the form ∞ ∞ ~ ~ 1 Axi ( x, y, z ) = A (λx , λ y , z ) exp( j (λx x + λ y y ))dλx dλ y 2 ∫ ∫ x 4π − ∞ − ∞

(5.2.24)

to (5.2.23) we obtain ( x, y , z ) = Axind 0 =

1

∞

1

4π 2 −∫∞

∞ ∞

C2 e − λ z e 4π 2 −∫∞ −∫∞

⋅ C = 1−

(

~ ~ ind j (λ x x + λ y y ) A dλx dλ y = ∫ x 0 (λ x , λ y , z )e

∞

−∞

(

j λx x +λ y y

)

dλx dλ y =

( β −α )d ⎞ ⎛ ⎜ 2λµ m ⎜ D5 D1 − µ 2 b e 2 D3 ⎟⎟ ⎠ ⎝

)

D5 D6 D1 + b D2 + µ 2 b e

( β −α )d 2

(D D 6

3

+ b D4

)

= ⋅ λ x x + λ y y = λr cos ϕ λ cos ϕ r + λr sin ϕ λ sin ϕ r = λr cos(ϕ λ − ϕ r ) = ⋅ γ = −2 jµ 0π sin ϕ λ I J 1 (λrc )rc

2π

∞

=−

1 jIrc µ 0 ∫ J 1 (λrc )e −λ ( z + h ) dλ ∫ C sin ϕ λ e jλ r cos (ϕλ −ϕ r ) dϕ λ = 4π 0 0 ∞

= e jz cos ϕ = J 0 (z ) + 2∑ j k J k ( z ) cos kϕ = k =1

 107   

2π

∞

∞ 1 ⎛ ⎞ jIrc µ 0 ∫ J 1 (λrc )e −λ ( z + h ) dλ ∫ C sin ϕ λ ⎜ J 0 (λr ) + 2∑ j k J k (λr ) cos k (ϕ λ − ϕ r )⎟dϕ λ =− 4π k =1 ⎝ ⎠ 0 0

Thus, the x − component of the induced vector potential is Axind ( x, y , z ) = 0 2π

∞

=−

∞ ⎛ ⎞ 1 jIrc µ0 ∫ J1 (λrc )e − λ ( z + h )dλ ∫ C sin ϕλ ⎜ J 0 (λr ) + 2∑ j k J k (λr )cos k (ϕλ − ϕ r )⎟dϕλ . 4π k =1 ⎝ ⎠ 0 0

(5.2.25)

The y − component of the induced vector potential can be found similarly and has the form Ayind0 ( x, y, z ) = 2π

∞

∞ ⎛ ⎞ 1 jIrc µ0 ∫ J1 (λrc )e − λ ( z + h )dλ ∫ C cos ϕλ ⎜ J 0 (λr ) + 2∑ j k J k (λr )cos k (ϕλ − ϕ r )⎟dϕλ . = 4π k =1 ⎝ ⎠ 0 0

The induced change in impedance of the coil is given by the formula r jω r ind Z ind = A ( x , y , z ) d l , 0 I ∫L where L is the contour of the coil, and r r r ( x, y, z )ex + Ayind0 ( x, y, z )e y . A0ind ( x, y, z ) = Axind 0

(5.2.26)

(5.2.27)

(5.2.28)

Substituting (5.2.25) and (5.2.26) into (5.2.27) we obtain Z ind =

jωrc I

2π

∫ (A

ind y0

)

cos ϕ r − Axind sin ϕ r dϕ r 0

0

= r =rc z =h

2π ∞ ⎡ ⎤ ⎛ ⎞ ( ) + cos ϕ cos ϕ J λ r 2 j k J k (λr )cos k (ϕλ − ϕ r )⎟dϕ r + ⎥ ⎜ ⎢ ∑ 0 λ r ∫ 2 π ∞ k =1 jωrc2 ⎝ ⎠ 0 ⎥dϕ = = jIµ0 ∫ J1 (λrc )e −λ ( z +h )dλ ∫ C ⎢ 2π ⎢ ⎥ λ ∞ 4π I ⎛ ⎞ 0 0 k ⎢+ sin ϕλ ∫ sin ϕ r ⎜ J 0 (λr ) + 2∑ j J k (λr )cos k (ϕλ − ϕ r )⎟dϕ r ⎥ k =1 ⎝ ⎠ 0 ⎣⎢ ⎦⎥ 2π

⋅ J 0 (λr )cos ϕλ ∫ cos ϕ r dϕ r + 0

2π 2π ∞ ⎡ ⎤ + 2 cos ϕλ ∑ j k J k (λr )⎢cos kϕλ ∫ cos kϕ r cos ϕ r dϕ r + sin kϕλ ∫ sin kϕ r cos ϕ r dϕ r ⎥ = k =1 0 0 ⎣ ⎦ 2 = 2 jπ cos ϕλ J1 (λr ) 2π

⋅ J 0 (λr )sin ϕλ ∫ sin ϕ r dϕ r + 0

2π 2π ∞ ⎡ ⎤ + 2 sin ϕλ ∑ j k J k (λr )⎢cos kϕλ ∫ cos kϕ r sin ϕ r dϕ r + sin kϕλ ∫ sin kϕ r sin ϕ r dϕ r ⎥ = k =1 0 0 ⎣ ⎦ 2 = 2 jπ sin ϕλ J1 (λr )

 108   

2π ωrc2 ∞ −λ ( z + h ) =− µ0 ∫ J1 (λrc )e dλ ∫ C 2 jπJ1 (λr )dϕλ   4π r =r 0 0 c

z =h

Hence, the induced change in impedance of the coil is given by 2π

∞

j Z ind = − ωrc2 µ0 ∫ J12 (λrc )e − 2λh dλ ∫ Cdϕλ , 2 0 0

(5.2.29)

where

C =1−

(

( β − α )d ⎛ ⎞ 2λµ m ⎜⎜ D5 D1 − µ 2 b e 2 D3 ⎟⎟ ⎝ ⎠

)

D5 D6 D1 + b D2 + µ 2 b e

( β − α )d 2

(D D + 6

3

b D4

)

,

β

⋅ D1 = Jν ( z0 )Yν ( z1 ) − Jν ( z1 )Yν ( z0 ),

⋅ D5 =

⋅ D2 = Jν' ( z0 )Yν ( z1 ) − Jν ( z1 )Yν' ( z0 ),

⋅ D6 = λµm +

⋅ D3 = Jν ( z0 )Yν' ( z1 ) − Jν' ( z1 )Yν ( z0 ),

2

µ2e βd − µm q2 , β 2

.

⋅ D4 = Jν' ( z0 )Yν' ( z1 ) − Jν' ( z1 )Yν' ( z1 ),

Using the following dimensionless parameters (5.2.29) can be rewritten as follows u ⎫ , rc ⎪⎪ ⎬ ⇒ νˆ = αˆ βˆ ⎪ ⋅ α= , β= , rc rc ⎪⎭

⋅ λrc = u → λ =

βˆ 2 + 4u 2 αˆ + βˆ

⎧⎪ δˆ = rc ωµ0 µ mσ m −j ˆ b , where ⎨ 1 rc ⎪⎩ ρˆ = rc µ0 µ mσ mV (αˆ + βˆ )dˆ − 2 − jbˆ 1 1 2 2 ˆ u + jδ 2 = qˆ2 , zˆ1 = zˆ0e 2 , zˆ0 = ⋅ q2 = rc rc αˆ + βˆ ⋅

b=

(

)

1 − j δˆ12 + ρˆ u cos ϕλ = rc

where δˆ2 = rc ωµ0 µ 2σ 2 = rc ωµ0 µ 2σ 2 = δˆ1

µ2σ 2 , µmσ m

⋅ Dˆ1 = Jνˆ ( zˆ0 )Yνˆ ( zˆ1 ) − Jνˆ ( zˆ1 )Yνˆ ( zˆ0 ), Dˆ 2 = J vˆ' ( zˆ0 )Yνˆ ( zˆ1 ) − Jνˆ ( zˆ1 )Yvˆ' ( zˆ0 ), Dˆ 3 = Jνˆ ( zˆ0 )Yvˆ' ( zˆ1 ) − J vˆ' ( zˆ1 )Yνˆ ( zˆ0 ), ⎞ 1 1⎛ βˆ ⎞ 1 1 ⎛ βˆ ⋅ Dˆ 4 = J vˆ' ( zˆ0 )Yvˆ' ( zˆ1 ) − J vˆ' ( zˆ1 )Yvˆ' ( zˆ0 ), D5 = Dˆ 5 = ⎜⎜ µ2e βd − µm qˆ2 ⎟⎟ , D6 = Dˆ 6 = ⎜⎜ uµm + ⎟⎟, 2⎠ rc rc ⎝ 2 rc rc ⎝ ⎠ (βˆ −αˆ )dˆ 1 ˆ 1 ˆ ⋅ D7 = D7 = µ2 − jb e 2 rc rc

 109   

Finally, the induced change in impedance of the coil can be rewritten as follows Z ind = ω rc µ0 Z ,

(5.2.30)

where 2π

∞

Z =−

j 2 ˆ J1 (u )e − 2u h du ∫ Cˆ dϕλ , ∫ 20 0

Cˆ = 1 −

(5.2.31)

(

)

2uµ m Dˆ 5 Dˆ 1 − Dˆ 7 Dˆ 3   ⎞ ⎛ ⎞ ⎛ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ D5 ⎜ D6 D1 + − jb D2 ⎟ + D7 ⎜ D6 D3 + − jb D4 ⎟ ⎠ ⎝ ⎠ ⎝

Formula (5.2.31) is used to compute the change in impedance of a single-turn coil for the following values of the parameters of the problem: αˆ = 0, βˆ = 2, hˆ = 0.05, d = 0.1, r0 = 5, µ m = 5. The results of the calculations are shown in Fig. 5.4. The points on each graph correspond to different values of δˆ = 1,2,...,10 (from left to right). 1

Computations are done with “Mathematica” ( see Appendix Fig.Ap.13 ) . Im @zD

          ρˆ = 1  

-3

                         ρˆ = 3  

-3.5

                       ρˆ = 5  

-4 -4.5

-0.8

-0.7

-0.6

-0.5

-0.4

-0.3

Re @zD

Fig. 5.4 The change in impedance computed by formula (5.2.31) for three values of ρˆ

 110   

6. AXISYMMETRIC PROBLEMS FOR MEDIA OF FINITE SIZE 6.1 A single-turn coil above a conducting cylinder of finite size Mathematical models of eddy current testing problems developed in the previous chapters are based on the assumption that a conducting medium is infinite in one or two spatial dimensions. Analytical solutions of the corresponding equations for the vector potential can be obtained in such cases by the method of integral transforms (for example, Fourier or Hankel integral transforms). Recently a quasi-analytical approach for the solution of eddy current testing problems is suggested in [63]. The authors use the abbreviation TREE (TRuncated Eigenfuction Expansion) method. The main idea of the TREE method is that the vector potential is assumed to be exactly zero at a sufficiently large radial distance r = b from eddy current coil (provided that there are no other sources of alternating current). Note that in the problem formulation described in the previous chapters (for the case of an unbounded medium) the vector potential approaches zero at infinity. From a physical point of view the assumption of the TREE method (the vector potential is equal to zero at a large distance from the coil) is quite reasonable. Recommendations on the selection of the value of b are given in [60]. Thus, a solution of an eddy current problem with the TREE method is expressed in terms of a series (rather than integrals). This is the reason the term “TRuncated Eigenfuction Expansion” is used in order to describe the method. The main advantage of the TREE method in comparison with other analytical methods used for infinite domains is that with the TREE method one can also construct quasi-analytical solutions for the cases where a conducting medium has a finite size. Such models are quite important for applications since one can also model the presence of inhomogeneities (flaws) of finite size in a conducting medium. In the present section we construct a quasi-analytical solution for the problem shown in Fig 6.1. Suppose that a conducting cylinder of a finite radius c is located below a single-turn coil carrying alternating current. The axis of the coil coincides with the axis of the cylinder. The radius of the coil is r0 and the distance between the coil and the cylinder is equal to h (see Fig. 6.1). Such a problem has an important practical application for coin validators. Many coin validators use eddy current method in order to compare the conductivity of an object inserted in the validator with the conductivity of a “typical” coin. The cylinder shown in Fig. 6.1 can represent a coin.

r0 R0  

h  O

c d

R1   R2  

b 

Fig. 6.1 A single-turn coil with alternating current above a conducting cylinder.

 111   

Consider the following three regions in a three-dimensional space (see Fig. 6.1): R0 = {z > 0}, R1 = {−d < z < 0} and R2 = {z < − d } , where d is the thickness of the cylinder. The vector potential (due to azimuthal symmetry) has only one nonzero component in the ϕ -direction in each of the regions R0 , R1 and R2 (the nonzero components of the vector potential in regions R0 , R1 and R2 are denoted by A0 , A1 and A2 , respectively). Since region R1 is not homogeneous in the radial direction we use the notations A1con and A1air in order to determine the solutions in regions 0 ≤ r ≤ c and c ≤ r ≤ b , respectively. The system of equations for the functions A0 , A1 and A2 is obtained from (1.2.12) for the case where the conductivity σ is constant (note that σ = 0 in region R1 where c ≤ r ≤ b ) and the relative magnetic permeability µ is equal to one and has the form R0 :      µ ( z ) = µ 0r = 1, σ ( z ) = 0, I e = Iδ (r − r0 )δ ( z − h ),   ∂ 2 A0 1 ∂A0 A0 ∂ 2 A0 + − + = − µ 0 Iδ (r − r0 )δ ( z − h), ∂r 2 r ∂r r 2 ∂z 2 ⎧⎪ ⋅ 0 ≤ r ≤ c µ ( z ) = µ1r = 1, σ ( z ) = σ , I = 0 R1 :     ⎨   ⎪⎩ ⋅ c ≤ r ≤ b µ ( z ) = µ1r = 1, σ ( z ) = 0, I = 0 ∂ 2 A1 1 ∂A1 A1 ∂ 2 A1 ωσµ = 0, + − − j A + 0 1 ∂z 2 ∂r 2 r ∂r r 2 R2 :    µ ( z ) = µ 2r = 1, σ (z ) = 0, I = 0   ∂ 2 A2 1 ∂A2 A2 ∂ 2 A2 + − + = 0, ∂r 2 r ∂r r 2 ∂z 2 where the current in the coil is assumed to be of the form r r I e = Ie jωt eϕ .

(6.1.1)

(6.1.2)

(6.1.3) (6.1.4)

Following the basic assumption in the TREE method we assume that the functions A0 , A1 and A2 are equal to zero at r = b :

Ai |r = b = 0, i = 0,2, A1air |r = b = 0.

(6.1.5)

Continuity conditions are imposed at r = c : ∂A1con ∂A air (6.1.6) |r = c = 1 |r = c . ∂r ∂r The conditions at z = 0 and z = − d are ∂A0 ∂A (6.1.7) A0 | z = 0 = A1 | z = 0 , | z =0 = 1 | z =0 , ∂z ∂z ∂A ∂A1 (6.1.8) |z =−d = 2 |z =−d . A1 | z = − d = A2 | z = − d , ∂z ∂z Conditions (6.1.6)-(6.1.8) follow from the physical conditions of continuity of the tangent components for the intensity of electrical and magnetic field. Let us consider the solution of the corresponding homogeneous equation (6.1.1) by the method of separation of variables. Assuming that (6.1.9) A0 (r , z ) = R(r ) Z ( z ) A1con |r = c = A1air |r = c ,

 112   

we rewrite equation (6.1.1) with zero right-hand side in the form 1 1 R '' Z + R ' Z − 2 RZ + RZ '' = 0. r r Dividing (6.1.10) by RZ and separating the variables we obtain R '' 1 R ' 1 + − 2 + λ2 = 0, R r R r Z '' − λ2 = 0, Z where λ is the separation constant.

(6.1.10)

(6.1.11) (6.1.12)

Equation (6.1.11) can be rewritten in the following form 1 1⎞ ⎛ R '' + R ' + ⎜ λ2 − 2 ⎟ R = 0,                                                                                                               (6.1.13) r r ⎠ ⎝ 1 ' ⎛ 2 v2 ⎞              y + y + ⎜⎜ λ − 2 ⎟⎟ y = 0 → y = J v (λx )   x ⎠ x ⎝ ''

General solution to (6.1.13) is R (r ) = C1 J 1 (λr ) + C 2Y1 (λr ),

(6.1.14)

where J 1 (λr ) and Y1 (λr ) are the Bessel functions of order one of the first and second kind, respectively. Thus, bounded solution to (6.1.14) at r = 0 is R (r ) = C1 J 1 (λr ).

(6.1.15)

General solution to (6.1.12) can be written in the form Z ( z ) = D1e − λz + D2 e λz .

(6.1.16)

Using (6.1.9), (6.1.15), (6.1.16) and the boundary condition A0 |r =b = 0 we obtain J 1 (λi b) = 0, i = 1,2,...

λi =

αi b

(6.1.17)

where λi are eigenvalues and α i , i = 1,2,... are the roots of the equation J1 (α i ) = 0.

(6.1.18)

In order to construct the solution to (6.1.1) we consider two sub-regions of region R0 , namely, R00 = {0 < z < h} and R01 = {z > h} , respectively. The solutions in R00 and R01 are denoted by A00 and A01 , respectively. The right-hand side of (6.1.1) is equal to zero in R00 and R01 . Using the principle of superposition we represent the solutions of (6.1.1) in R00 and R01 in the form ∞

R01 : A01 (r , z ) = ∑ D1i e − λi z J 1 (λi r ),

(6.1.19)

i =1 ∞

R00 :   A00 (r , z ) = ∑ ( D2i e − λi z + D3i e λi z ) J 1 (λi r ),

(6.1.20)

i =1

where D1i , D2i and D3i are arbitrary constants.  113   

The vector potential is continuous at z = h : A00 | z = h = A01 | z = h .

(6.1.21)

Integrating (6.1.1) with respect to z from h − ε to h + ε and considering the limit as ε → +0 in the resulting equation we obtain ∂A01 ∂A (6.1.22) | z = h − 00 | z = h = − µ 0 Iδ (r − r0 ). ∂z ∂z It follows from (6.1.19)-(6.1.22) that ∞

∞

i =1

i =1

∑ D1i e −λi h J1 (λi r ) = ∑ ( D2i e −λi h + D3i e λi h ) J1 (λi r ), ∞

∑λ D e λ i =1

i

1i

− ih

∞

J1 (λi r ) + ∑ (−λi D2i e − λi h + λi D3i eλi h ) J1 (λi r ) = µ0 Iδ (r − r0 ).

(6.1.23) (6.1.24)

i =1

Multiplying (6.1.23) by rJ 1 (λ j r ) , integrating the resulting equation with respect to r from 0 to b and using the orthogonality condition 0, i≠ j b ⎧ ⎪ 2 ∫0 rJ 1 (λ j r )J1 (λi r )dr = ⎨ b J 02 (λ j b ), i = j ⎪⎩ 2 the following equation is obtained

D1 j e − λi h = D2 j e

−λ j h

D1 j = D2 j + D3 j e

(6.1.25)

λ h

+ D3 j e j .

(6.1.26)

2λ j h

Applying the same procedure to (6.1.24) we obtain b2 2 −λ h −λ h λ h D1 j e j − D2 j e j + D3 j e j λ j J 0 (λ j b ) = µ 0 Ir0 J1 (λ j r0 ). 2

(

)

Using (6.1.26) and (6.1.27) we get µ 0 Ir0 J 1 (λ j r0 ) −λ j h D3 j = e . λ j b 2 J 02 (λ j b)

(6.1.28)

Substituting (6.1.26) and (6.1.28) into (6.1.19) and (6.1.20) we obtain ∞ µ Ir ∞ J (λ r ) A00 (r , z ) = ∑ D2i e − λi z J 1 (λi r ) + 0 2 0 ∑ 1 2 i 0 e − λi ( h − z ) J 1 (λi r ), b i =1 λi J 0 (λi b) i =1 ∞

A01 (r , z ) = ∑ D2i e − λi z J 1 (λi r ) + i =1

µ 0 Ir0 b

2

(6.1.27)

∞

J 1 (λi r0 ) − λi ( z − h ) e J 1 (λi r ). 2 i 0 ( λi b )

∑λ J i =1

(6.1.29) (6.1.30)

It can be shown (see [60] that the second term in (6.1.29), (6.1.30) represents the vector potential of a single-turn coil located in an unbounded space (no conducting medium is present). The first term in (6.1.29), (6.1.30) represents the induced component of the vector potential due to the presence of a conducting cylinder (see Fig. 6.1). Let us now construct a solution to (6.1.2) in the conducting cylinder (0 ≤ r ≤ c) by the method of  114   

separation of variables. Assuming the solution in the form A1con (r , z ) = R(r ) Z ( z ) we transform (6.1.2) as follows R '' 1 R ' 1 + − 2 + q 2 = 0, R r R r

(6.1.31)

1 1⎞ ⎛ R '' + R ' + ⎜ q 2 − 2 ⎟ R = 0 r r ⎠ ⎝ Z '' − jωσµ0 − q 2 = 0, Z

(

(6.1.32)

)

Z '' − q 2 + jωσµ0 Z = 0 where q is the separation constant. Solution to (6.1.31) which is bounded at r = 0 has the form R (r ) = C3 J1 (qr ).

(6.1.33)

General solution to (6.1.32) is ~ ~ Z ( z ) = D3e pz + D4 e − pz ,

(6.1.34)

where p = q 2 + jωσµ0 . Using (6.1.33), (6.1.34) and the superposition principle the solution to (6.1.2) in region 0 ≤ r ≤ c can be written in the form ∞

A1con (r , z ) = ∑ ( D4i e pi z + D5i e − pi z ) J1 (qi r ),

(6.1.35)

i =1

where qi are unknown eigenvalues and pi = qi + jωσµ 0 . 2

In the non-conducting region (c ≤ r ≤ b) of R1 equation (6.1.2) has the form ∂ 2 A1air 1 ∂A1air A1air ∂ 2 A1air (6.1.36) + − 2 + = 0. r ∂r ∂r 2 ∂z 2 r Representing the solution to (6.1.36) in the form A1air (r , z ) = R(r ) Z ( z ) and solving it by the method of separation of variables we obtain 1 1 ⋅ R '' Z + R ' Z − 2 RZ + RZ '' = 0 : RZ r r R '' 1 R ' 1 Z '' ⋅ + − + =0 R r R r2 Z R '' 1 R ' 1 Z '' + − 2 + jωσµ0 − jωσµ0 + = 0. R r R r Z

(6.1.37)

Note that the parameter σ in (6.1.37) is equal to the conductivity of the cylinder (see Fig. 6.1).

 115   

R '' 1 R ' 1 + − 2 + jωσµ0 + q 2 = 0, R r R r ~ ~ 1 1⎞ ⎛ R '' + R ' + ⎜ p 2 − 2 ⎟ R = 0 → R(r ) = C5 J1 ( pr ) + C6Y1 ( pr )   r r ⎠ ⎝ Z '' − jωσµ0 − q 2 = 0, Z

~ ~ Z '' − p 2 Z = 0 → Z ( z ) = D5e pz + D6e − pz

where q is the separation constant and p 2 = q 2 + jωσµ0 .   Using the principle of superposition we represent the solutions of (6.1.36) in R1 (c ≤ r ≤ b) as follows ∞

[

]

A1air (r , z ) = ∑ ( D6i J 1 ( pi r ) + D7 iY1 ( pi r ) ) e pi z + ( D8i J 1 ( pi r ) + D9iY1 ( pi r ) ) e − pi z . i =1

(6.1.38)

The solution to (6.1.3) in region R2 which is bounded as z → −∞ and satisfies the zero boundary condition at r = b is ∞

A2 (r , z ) = ∑ D10i e λi z J 1 (λi r ).

(6.1.39)

i =1

Using the zero boundary condition (6.1.5) for the function A1air and linear independence of the functions e pi z and e − pi z the following system is obtained from (6.1.38): D6i J 1 ( pi b) + D7 iY1 ( pi b) = 0,

(6.1.40)

D8i J 1 ( pi b) + D9iY1 ( pi b) = 0.

(6.1.41)

Eliminating D7 i and D9i from (6.1.40) and (6.1.41) we obtain D7 i = − D6i

J 1 ( pi b ) , Y1 ( pi b)

(6.1.42)

D9i = − D8i

J 1 ( pi b ) . Y1 ( pi b)

(6.1.43)

Applying the first condition (6.1.6) to (6.1.35) and (6.1.38) gives ∞

∑[ D i =1

4i

e pi z + D5i e − pi z ]J1 (qi c) = ∞

{

= ∑ [ D6i J 1 ( pi c) + D7 iY1 ( pi c)]e i =1

pi z

+ [ D8i J1 ( pi c) + D9iY1 ( pi c)]e

− pi z

}.

(6.1.44)

Now it becomes clear why do we need the transformation in (6.1.37): add and subtract jωσµ0 . In this case we obtain the same structure of the z -dependence of the solutions in regions 0 ≤ r ≤ c and c ≤ r ≤ b . Equating coefficients in front of e pi z and e − pi z in (6.1.44) the following two equations are obtained

 116   

D4i J1 (qi c) = D6i J1 ( pi c) + D7 iY1 ( pi c),

(6.1.45)

D5i J1 (qi c) = D7 i J1 ( pi c) + D9iY1 ( pi c).

(6.1.46)

Using (6.1.42) and (6.1.43) we rewrite (6.1.45) and (6.1.46) in the form D = Dˆ [ J ( p c)Y ( p b) − J ( p b)Y ( p c)],

(6.1.47)

D5i = Dˆ 8i [ J 1 ( pi c)Y1 ( pi b) − J 1 ( pi b)Y1 ( pi c)],

(6.1.48)

4i

6i

i

1

1

i

1

i

1

i

where Dˆ 6i =

D6i D8i , Dˆ 8i = . J 1 (qi c)Y1 ( pi b) J 1 (qi c)Y1 ( pi b)

It follows from (6.1.47), (6.1.48) and (6.1.35) that ∞

A1con (r , z ) = ∑ Dˆ 6i [ J 1 ( pi c)Y ( pi b) − J 1 ( pi b)Y ( pi c)]e pi z J 1 (qi r ) + i =1 ∞

+ ∑ Dˆ 8i [ J 1 ( pi c)Y ( pi b) − J 1 ( pi b)Y ( pi c)]e − pi z J 1 (qi r ) .

(6.1.49)

i =1

Similarly, using (6.1.42) and (6.1.43) we transform (6.1.38) to the form ∞

A1air (r , z ) = ∑ Dˆ 6i J 1 (qi c)[ J 1 ( pi r )Y ( pi b) − J 1 ( pi b)Y ( pi r )]e pi z + i =1

∞

+ ∑ Dˆ 8i J 1 (qi c)[ J 1 ( pi r )Y ( pi b) − J 1 ( pi b)Y ( pi r )]e − pi z .

(6.1.50)

i =1

Introducing the notation T1 ( pi c) = J1 ( pi c)Y1 ( pib) − J1 ( pib)Y1 ( pi c),

(6.1.51)

T1 ( pi r ) = J1 ( pi r )Y1 ( pib) − J1 ( pib)Y1 ( pi r ),

(6.1.52)

we rewrite (6.1.49) and (6.1.50) in a more compact form ∞

A1con (r , z ) = ∑ J1 (qi r )T1 ( pi c)( Dˆ 6i e pi z + Dˆ 8i e − pi z ),

(6.1.53)

i =1 ∞

A1air (r , z ) = ∑ J1 (qi c)T1 ( pi r )( Dˆ 6i e pi z + Dˆ 8i e − pi z ).

(6.1.54)

i =1

Differentiating (6.1.53) and (6.1.54) with respect to r and evaluating the derivatives at r = c we obtain ∞ ∂A1con | r = c = ∑ qi J 1' (qi c)T1 ( pi c)( Dˆ 6i e pi z + Dˆ 8i e − pi z ). (6.1.55) ∂r i =1 ∞ ∂A1air |r = c = ∑ pi J 1 (qi c)T1' ( pi c)( Dˆ 6i e pi z + Dˆ 8i e − pi z ), ∂r i =1

It follows from (6.1.55) and (6.1.56) and the second boundary condition in (6.1.6) that pi J 1 (qi c)T1' ( pi c) = qi J 1' (qi c)T1 ( pi c).

(6.1.56)

(6.1.57)

Equation (6.1.57) is used to determine the eigenvalues pi and related values qi .  117   

The relationship between pi and qi is pi = qi + jωσµ 0 .    2

Thus, the solution in regions R0 , R1 and R2 is given by (6.1.29), (6.1.30), (6.1.53), (6.1.54) and (6.1.39). The four sets of constants in these formulas, namely, D , Dˆ , Dˆ and D can be 2i

6i

8i

10 i

obtained from the boundary conditions (6.1.7) and (6.1.8). Using the first condition in (6.1.7) we obtain ∞ ∞ µ 0 Ir0 ∞ J 1 (λi r0 )e − λi h D2i J 1 (λi r ) + 2 ∑ J ( λ r ) = J 1 (qi r )T1 ( pi c)( Dˆ 6i + Dˆ 8i ), 0 ≤ r ≤ c, ∑ ∑ 1 i b i =1 λi J 02 (λi b) i =1 i =1 ∞

∑ D2i J1 (λi r ) + i =1

µ 0 Ir0 b2

∞ J 1 (λi r0 )e − λi h J ( λ r ) = J 1 (qi c)T1 ( pi r )( Dˆ 6i + Dˆ 8i ), c ≤ r ≤ b. ∑ ∑ 1 i 2 λ J ( λ b ) i =1 i =1 i 0 i ∞

(6.1.58) (6.1.59)

In order to determine the coefficients Dˆ 6i and Dˆ 8i the following procedure is used. First, equations (6.1.58) and (6.1.59) are combined into one equation where the right-hand side of the resulting equation is given by different expressions on the intervals 0 ≤ r ≤ c and c ≤ r ≤ b . These expressions are defined by the right-hand sides of (6.1.58) and (6.1.59), respectively. Second, the obtained equation is multiplied by rJ 1 (λ j r ) and the resulting equation is integrated with respect to r from 0 to b . Third, we use the following orthogonality condition (6.1.25) 0, i≠ j ⎧ ⎪ 2 = λ λ ( ) ( ) rJ r J r dr b ⎨ 2 ∫0 1 j 1 i ⎪⎩ 2 J 0 (λ j b), i = j. b

The result is ∞ b2 µ Ir −λ h D2 j J 02 (λ j b) + 0 0 J1 (λ j r0 )e j = ∑ ( Dˆ 6i + Dˆ 8i )a ji , 2 2λ j i =1

(6.1.60)

where

~ a ji = T1 ( pi c)a~ ji + J1 (qi c)a~ ji .

(6.1.61)

The coefficients aij can be computed using the following formulas from [2]:

∫ (k ∫ (k

2

− l 2 )tJ 1 (kt ) J 1 (lt )dt = t[kJ 2 (kt ) J 1 (lt ) − lJ 1 (kt ) J 2 (lt )],

(6.1.62)

2

− l 2 )tJ 1 (kt )Y1 (lt )dt = t[kJ 2 (kt )Y1 (lt ) − lJ 1 (kt )Y2 (lt )].

(6.1.63)

Thus, using (6.1.62), (6.1.63) and (6.1.52) we obtain c

⋅ a~ ji = ∫ rJ 1 (λ j r )J 1 (qi r )dr = 0

c ( λ j J 2 (λ j c )J 1 (qi c ) − qi J 1 (λ j c )J 2 (qi c ) ) λ − qi2 2 j

b

b b ~ ⋅ a~ ji = ∫ rJ 1 (λ j r )T1 ( pi r )dr = Y1 ( pi b )∫ rJ 1 (λ j r )J 1 ( pi r )dr − J 1 ( pi b )∫ rJ 1 (λ j r )Y1 ( pi r )dr = c

c

c

⎧bpi J 1 (λ j b )[ J 1 ( pi b )Y2 ( pi b ) − J 2 ( pi b )Y1 ( pi b ) ] + ⎫ ⎪ ⎪ 1 = 2 + cλ j J 2 (λ j c )[ J 1 ( pi b )Y1 ( pi c ) − J 1 ( pi c )Y1 ( pi b ) ] + ⎬ 2 ⎨ λ j − pi ⎪ ⎪ ⎩+ cpi J 1 (λ j c )[ J 2 ( pi c )Y1 ( pi b ) − J 1 ( pi b )Y2 ( pi c ) ] ⎭

 118   

Applying the same procedure and using the second condition in (6.1.7) we obtain ∞ b2 µ Ir −λ h − λ j D2 j J 02 (λ j b) + 0 0 J1 (λ j r0 )e j = ∑ ( Dˆ 6i − Dˆ 8i ) pi a ji . 2 2 i =1

(6.1.64)

Multiplying (6.1.60) by λ j and adding with (6.1.64) gives ∞

∑[(λ i =1

j

−λ h + pi ) Dˆ 6i + (λ j − pi ) Dˆ 8i ]a ji = µ0 Ir0 J1 (λ j r0 )e j .

(6.1.65)

Two additional equations are obtained if the same procedure is applied to (6.1.53), (6.1.54) and (6.1.39) using boundary conditions (6.1.8). The result is shown below 2 ∞ −λ d b D10 j e j J 02 (λ j b) = ∑ ( Dˆ 6i e − pi d + Dˆ 8i e pi d )a ji , (6.1.66) 2 i =1

λ j D10 j e

−λ j d

∞ b2 2 J 0 (λ j b) = ∑ ( Dˆ 6i e − pi d − Dˆ 8i e pi d ) pi a ji . 2 i =1

(6.1.67)

Multiplying (6.1.66) by (− λ j ) and adding with (6.1.67) we obtain ∞

∑[(λ i =1

j

− pi )e − pi d Dˆ 6i + (λ j + pi )e pi d Dˆ 8i ]a ji = 0.

(6.1.68)

The four sets of equations for the unknown coefficients D2 j , Dˆ 6i , Dˆ 8i and D10 j are given by (6.1.60), (6.1.65), (6.1.68) and (6.1.66). The solution of the system is obtained as follows. First, we consider only a finite number of terms n in (6.1.60), (6.1.65), (6.1.66) and (6.1.68). Recommendations on the selection of the value of n are given in Section 6.3. Second, the coefficients Dˆ 6i , Dˆ 8i can be computed solving the system n

∑[(λ i =1

j

−λ h + pi )a ji Dˆ 6i + (λ j − pi )a ji Dˆ 8i ] = µ0 Ir0 J1 (λ j r0 )e j ,

(6.1.69)

j

− pi )a ji e − pi d Dˆ 6i + (λ j + pi )a ji e pi d Dˆ 8i ] = 0.

(6.1.70)

n

∑[(λ i =1

where n is the number of eigenvalues pi . The system (6.1.69)-(6.1.70) can be written in the matrix form r r AX = B                                                                                                                                                (6.1.71) where the coefficient matrix A is ⎛A A ⎞ A = ⎜⎜ 11 12 ⎟⎟                                                                                                                                      (6.1.72) ⎝ A21 A22 ⎠ and the block matrices A11 , A12 , A21 and A22 are

⎛ (λ1 + p1 )a11 (λ1 + p2 )a12 ... (λ1 + pn )a1n ⎞ ⎛ (λ1 − p1 )a11 (λ1 − p2 )a12 ... (λ1 − pn )a1n ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ (λ2 + p1 )a21 (λ2 + p2 )a22 ... (λ2 + pn )a2 n ⎟ ⎜ (λ2 − p1 )a21 (λ2 − p2 )a22 ... (λ2 − pn )a2 n ⎟ A11 = ⎜ ⎟ A12 = ⎜ ⎟  ....................... ....................... ⎜ ⎟ ⎜ ⎟ ⎜ (λ + p )a (λ + p )a ... (λ + p )a ⎟ ⎜ (λ − p )a (λ − p )a ... (λ − p )a ⎟ 1 n1 n 2 n2 n n nn ⎠ 1 n1 n 2 n2 n n nn ⎠ ⎝ n ⎝ n

   119   

⎛ (λ1 − p1 )a11e − p1d (λ1 − p2 )a12e − p 2 d ... (λ1 − pn )a1n e − p n d ⎞ ⎜ ⎟ ⎜ (λ2 − p1 )a21e − p1d (λ2 − p2 )a22e − p 2 d ... (λ2 − pn )a2 n e − p n d ⎟ A21 = ⎜ ⎟  .......... .......... ... ⎜ ⎟ ⎜ (λ − p )a e − p1d (λ − p )a e − p 2 d ... (λ − p )a e − p n d ⎟ n n2 n n nn 1 n1 2 ⎝ n ⎠  

⎛ (λ1 + p1 )a11e p1d (λ1 + p2 )a12e p 2 d ... (λ1 + pn )a1n e p n d ⎞ ⎟ ⎜ ⎜ (λ2 + p1 )a21e p1d (λ2 + p2 )a22e − p 2 d ... (λ2 + pn )a2 n e p n d ⎟ A22 = ⎜ ⎟  ....................... ⎟ ⎜ ⎜ (λ + p )a e p1d (λ + p )a e p 2 d ... (λ + p )a e p n d ⎟ n n2 n n nn 1 n1 2 ⎠ ⎝ n

r r The matrices X and B in (6.1.71) are r r r ⎛ X1 ⎞ r ⎛ b ⎞ X = ⎜ r ⎟  , B = ⎜⎜ ⎟⎟ , ⎜X ⎟ ⎝0 ⎠ ⎝ 2⎠ where ⎛ Dˆ 61 ⎞ ⎛ Dˆ 81 ⎞ ⎛ µ 0 Ir0 J1 (λ1r0 )e − λ1h ⎞ ⎜ ⎟ ⎜ ⎟ ⎟ ⎜ r ⎜ Dˆ 62 ⎟ r ⎜ Dˆ 82 ⎟ r ⎜ µ 0 Ir0 J1 (λ2 r0 )e −λ2h ⎟ X1 = ⎜ ⎟ ,      X 2 = ⎜ ⎟, b =⎜ ⎟ .  ... ⎜ ... ⎟ ⎜ ... ⎟ ⎟ ⎜ ⎜ ˆ ⎟ ⎜ ˆ ⎟ ⎜ µ Ir J (λ r )e −λnh ⎟ ⎠ ⎝ 0 0 1 n0 ⎝ D6 n ⎠ ⎝ D8 n ⎠

(6.1.73)

                        (6.1.74)     

Hence, (6.1.71) can be written in the form    r r r ⎧⎪ A11 X 1 + A12 X 2 = b                                                                                                                            (6.1.75) r r ⎨ ⎪⎩ A21 X 1 + A22 X 2 = 0 Solving (6.1.75) we obtain all the coefficients Dˆ and Dˆ (i = 1,2,..., n). Then D ( j = 1,2,.., n) is 6i

8i

2j

calculated using −λ jh

n µ Ir J (λ r )e 2 D2 j = 2 2 a ji ( Dˆ 6i + Dˆ 8i ) − 0 0 21 2 j 0 ∑ b J 0 (λ j b) i=1 λ j b J 0 (λ j b)

.

(6.1.76)

The induced change in impedance of the coil is given by the formula        jω Z ind = 2πr0 A0ind (r0 , h),                                                                                                                   (6.1.77)                       I where n

A0ind (r0 , h) = ∑ D2 j e j =1

−λ j h

J1 (λ j r0 ).                                                                                                       (6.1.78)

Formula (6.1.77) is used to compute the change in impedance of the coil. Calculations are performed with “Mathematica”. The program which is used to compute the change in impedance (6.1.77) is shown in Appendix Fig. Ap.14. The following parameters of the problem are selected: µ0 = 4 ⋅ 10 −7 π , σ = 9.6 Ms/m, c = 19.75 / 2 mm,  r0 = 4.5 mm, h = 0.2 mm, d = 1.93 mm,  120   

b = 60 mm. The change in impedance is computed for the following five frequencies:  f = 1125, 1598, 2270, 3224 and 4579 Hz.

 The results of calculations are shown in Fig. 6.2.The calculated points (from top to bottom) correspond to the five frequencies (from smallest to largest). The upper limit of the summation index in (6.1.78) is fixed at n = 68 . Comparison of the computational results obtained for other values of n showed that the chosen value of 68 is quite satisfactory in terms of calculation accuracy. More detailed analysis of convergence is presented in Section 6.3. Several computational steps are necessary in order to calculate the induced change in impedance. First, the set of eigenvalues λi has to be calculated. This can easily be done in “Mathematica” using a built-in routine BesselZeros. Second, a set of complex roots of (6.1.57) should be computed. Calculations are based on the method described in Section 6.3. Third, several systems of linear equations have to be solved in order to determine expansion coefficients. Finally, the change in impedance is computed using (6.1.71) - (6.1.78). Im @zD

Re @zD

0.00002 0.00004 0.00006 0.00008 -0.00001 -0.00002 -0.00003 -0.00004

Fig. 6.2 The change in impedance of the coil for five frequencies.

-0.00005 -0.00006

6.2 A coil of finite dimensions above a conducting cylinder of finite size The solution for a single-turn coil found in the previous section can be used to construct the vector potential due to a coil of finite dimensions. Consider a coil located above a conducting cylinder (see Fig. 6.3). Let N be the number of turns in the coil. The inner and outer radii of the coil are r1 and r2 , respectively. The distance from the coil to the top surface of the cylinder is denoted by z1 . The height of the coil is z 2 − z1 . r2 r1  

R0   z 2  

z1

O

c  d

R1   R2  

b 

Fig. 6.3 A coil of finite dimensions above a conducting cylinder.

It can easily be shown that the second term in (6.1.29) and (6.1.30) represents the vector potential of a single-turn coil located in an unbounded free space. In addition, the first term in (6.1.29) and (6.1.30) is the induced vector potential in air due to the presence of a conducting cylinder:  121   

n

A0ind (r , z , r0 , h) = ∑ D2 j e j =1

−λ j z

J1 (λ j r ),

(6.2.1)

where −λ j h

n µ Ir J (λ r )e 2 D2 j = 2 2 a ji ( Dˆ 6i + Dˆ 8i ) − 0 0 21 2 j 0 ∑ b J 0 (λ j b) i =1 λ j b J 0 (λ j b)

and the series are truncated at i = n . The induced vector potential in air due to currents in the whole coil is obtained as follows r2 z 2

ind 0 coil

A

(r , z ) = ∫ ∫ A0ind (r , z , r0 , h)dr0 dh .

(6.2.2)

r1 z1

There is an important difference between the calculations for a single-turn coil and coil of finite dimensions. For the case of a single-turn coil (Section 6.1) the geometrical parameters of the coil ( r0 and h ) are constants and the numerical values of r0 and h can be used to solve the system (6.1.75). If a coil has finite dimensions then one needs to integrate the solution with respect to r0 and h . Thus, numerical values to the parameters r0 and h in (6.1.75) cannot be assigned. In this case, in order to obtain the induced vector potential of a single-turn coil we need to solve (6.1.75) in matrix form. Solving the second equation in (6.1.75) we obtain r r (6.2.3) X 2 = − A22−1 A21 X 1. Substituting (6.2.3) into the first equation in (6.1.75) gives r r (A11 − A12 A22−1 A21 )X 1 = b . (6.2.4) It follows from (6.2.3) and (6.2.4) that r r ⎧⎪ X = (A − A A−1 A )−1 b 1 11 12 22 21 (6.2.5) ⎨r −1 r −1 ⎪⎩ X 2 = − A22 A21 (A11 − A12 A22−1 A21 ) b Equation (6.1.76) should be written in matrix form: r r (6.2.6) D2 = Y b , where ⎛ D21 ⎞ ⎜ ⎟ r ⎜ D22 ⎟ −1 −1 A21 A11 − A12 A22−1 A21 − Cdiag , , Y = B11 E − A22 (6.2.7) D2 = ⎜ ⎟ ... ⎜ ⎟ ⎜D ⎟ ⎝ 2n ⎠ 1 2a12 2a ⎛ 2a11 ⎛ ⎞ ⎞ 0 0 ... 0 ⎟ ... 2 2 1n ⎜ 2 2 ⎜ 2 2 ⎟ 2 2 ( ) ( ) ( ) ( ) b J b b J b b J b b J b λ λ λ λ λ 0 1 0 1 0 1 ⎜ ⎜ 1 0 1 ⎟ ⎟ ⎜ 2a ⎜ ⎟ ⎟ 1 2a22 2 a2 n 0 0 0 ... 0 ... ⎜ 2 2 21 ⎜ ⎟ ⎟ b 2 J 02 (λ2b ) ⎟ , Cdiag = ⎜ λ2b 2 J 02 (λ2b ) B11 = ⎜ b J 0 (λ2b ) b 2 J 02 (λ2b ) (6.2.8) ⎟, ⎜ ⎜ ⎟ ⎟ ....................... ....................... ⎜ ⎜ ⎟ ⎟ 2 an 2 2ann ⎟ 1 ⎜ 2an1 ⎜ ⎟ ⎜ b 2 J 2 (λ b ) b 2 J 2 (λ b ) ... b 2 J 2 (λ b ) ⎟ ⎜ 0 0 ... 0 λ b 2 J 2 (λ b ) ⎟ 0 0 0 0 n n n n n ⎝ ⎝ ⎠ ⎠ r and b is given by (6.1.74).

(

)(

)

 122   

Formula (6.2.1) for the induced vector potential of a single-turn coil can be rewritten in the form r r A0ind (r , z , r0 , h) = D2T f , (6.2.9) where ⎛ J1 (λ1r )e − λ1 z ⎞ ⎜ ⎟ r ⎜ J1 (λ2 r )e − λ 2 z ⎟ f =⎜ ⎟ .  ⎜ ... ⎟ ⎜ J ( λ r )e − λ n z ⎟ ⎝ 1 n ⎠ Substituting (6.2.9) into (6.2.2) and using the formulas z2

⋅ ∫e

−λ j h

dh = −

z1

1

λj

(e

−λ j z2

−e

− λ j z1

r2

⋅ ∫ r0 J1 (λ j r0 )dr0 = ξ = λ j r0 = r1

) (6.2.10)

λ j r2

1

λi2

∫ ξJ (ξ )dξ 1

λ j r1

we obtain the induced vector potential in air due to the presence of the conducting cylinder ( the NI ): current amplitude I in this case is replaced by the current density (r2 − r1 )( z 2 − z1 ) ind 0 coil

A

(r , z ) =

µ0 NI (r2 − r1 )( z2 − z1 )

n

∑ f ∑Y j =1

(e

n

j

i =1

− λi z1

− e − λi z 2

λ3i

ji

)

λi r2

∫ ξJ (ξ )dξ .

(6.2.11)

1

λi r1

The integral with respect to ξ in (6.2.10) can be computed in terms of the Bessel and Struve functions [2] as follows ξ = λi r2

λi r2

⎧π ⎫ ∫λ rξJ1 (ξ )dξ = ⎨⎩ 2 ξ [J 0 (ξ ) H1 (ξ ) − J1 (ξ ) H 0 (ξ )]⎬⎭ ξ = λi r1 i 1

(6.2.12)

The induced change in impedance of a coil of finite dimensions (see Fig. 6.3) is calculated by means of the following formula [15]: r2 z2 2π rr jω jω ind Z = 2 ∫∫∫ AI dv = 2 ∫ dϕ ∫ rdr ∫ A0indcoil dz I V I 0 r1 z1 r z

Z

ind

2 2 2πjω N = rA0indcoil (r , z )drdz. I (r2 − r1 )( z 2 − z1 ) r∫1 z∫1

(6.2.13)

Using (6.2.11) and (6.2.13) we obtain the induced change in impedance of the coil in the form Z

ind

(

n 2 jωπµ0 N 2 e = ∑ 2 2 (r2 − r1 ) ( z2 − z1 ) j =1

− λ j z1

−e

λ3j

−λ j z2

)

λ j r2

(e

n

∫ ξJ (ξ )dξ ∑ Y 1

λ j r1

i =1

ji

− λ i z1

− e − λi z 2

λ3i

)

λi r2

∫ ξJ (ξ )dξ 1

(6.2.14)

λi r1

Formula (6.2.14) is used to compute the change in impedance of the coil. Calculations are performed with “Mathematica”. Program which is used to compute the change in impedance (6.2.14) is shown in Appendix Fig. Ap.15. The following parameters of the problem are selected: µ0 = 4 ⋅ 10−7 π , σ = 9.6 Ms/m, c = 19.75 / 2 mm, b = 60 mm, d = 1.93 mm, r2 = 6 mm, r1 = 3 mm , z2 = 0.39 mm, z1 = 0.06 mm, N = 100 .  123   

The results of calculations are shown in Fig. 6.4.The calculated points (from top to bottom) correspond to the following values of f = 1125, 1598, 2270, 3224 and 4579 Hz. The upper limit of the summation index in (6.2.14) is fixed at n = 68 . Comparison of the computational results obtained for other values of n  showed that the chosen value of 68 is quite satisfactory in terms of calculation accuracy. Several computational steps are necessary in order to calculate the induced change in impedance. First, the set of eigenvalues λi has to be calculated. This can easily be done in “Mathematica” using a built-in routine BesselZeros. Second, a set of complex roots of (6.1.57) should be computed. Calculations are based on the method described in Section 6.3. Third, several systems of linear equations have to be solved in order to determine expansion coefficients. Finally, the change in impedance is computed using (6.2.14). Details of the numerical aspects of the procedure are given in Section 6.3. Im @zD

0.2

0.4

Re @zD

0.6

-0.1 -0.2 -0.3

Fig. 6.4 The change in impedance of a coil (calculations are done using (6.2.14)).

-0.4 -0.5 -0.6

Calculated change in impedance of the coil is compared with experimental data. Experiments are performed at Tallinn University of Technology as a part of the work for the project “Increasing EU citizen security by utilizing innovative intelligent signal processing systems for euro-coin validation and metal quality testing” in the framework of FP7 program for the period from 2010 till 2012. Measurement results are also published in [20]. The results of comparison are shown in Fig.6.5 for the same five frequencies as in Fig. 6.4. The parameters of the coil and conducting cylinder (coin) are the same as specified for Fig.6.4. The larger points on the graph are experimental points while the smaller points represent theoretical calculations. As can be seen from the graph, good agreement is found between experimental data and theory. Im @zD

0.2

0.4

0.6

Re @zD

-0.2 -0.4 -0.6 -0.8 -1

Fig. 6.5 Comparison of theoretical calculations with experimental data.                            124   

6.3 Computational aspects The procedure for calculating the change in impedance of the coil by means of the formula (6.2.14) includes several computational steps. In this section we discuss each of the steps in detail. The first decision has to be made with respect to the choice of the parameter b where the vector potential is assumed to be exactly zero (recall that b is the radial distance from the center of the coil). Recommendations on the selection of the parameter b are presented in [60] where it is recommended to choose b in the form b = sr2 , where r2 is the outer radius of the coil and the factor s should be at least five. Larger values of s can be considered as well (in general, the choice of s is directly related to the desired accuracy of the calculations). Next, we need to truncate the series in (6.2.14). In any realistic case the summation index i varies from 1 to some pre-specified value n = N s . For example, N s from 50 to 200 is used in [60] in order to compute the induced change in impedance for the case where a coil of finite dimensions is located above a conducting half-space. We have to pay the attention of the reader to the fact that the choice of the value of N s also depends on the computational accuracy. The third step includes computation of eigenvalues λi , i = 1,2,..., N s . The eigenvalues are computed by means of (6.1.17), where α i , i = 1,2,..., N s are the roots of the equation (6.1.18). The roots α i can be easily computed with “Mathematica”. The corresponding “Mathematica” script is shown in Fig. 6.6 where the first ten zeros of (6.1.18) are shown 5 then the domain G should be divided into smaller sub-domains so that the number of zeros of φ (z ) in each sub-domain be smaller than 5. The next practical issue is the shape of the contour C . Two basic shapes are recommended for calculations in [12]: circles and rectangles. The choice of the shape of the contour C is dictated by calculation of the contour integrals in (6.3.1) and (6.3.2). It is known that contour integrals can be conveniently calculated if C is a circle or rectangle. We choose rectangle as the basic shape of C since any rectangle can be easily sub-divided into smaller rectangles. Consider a rectangular contour ABCD (see Fig. 6.7)). If φ ( z ) is an analytic function inside C (where C represents the contour ABCD shown in Fig. 6.7) then x2

∫ φ ( z )dz = ∫ [φ ( x + jy ) − φ ( x + jy 1

C

x1

y2

2

)]dx + j ∫ [φ ( x2 + jy ) −φ ( x1 + jy )]dy.

(6.3.3)

y1

y

y2  

D( x1 , y 2 )  

<

∨  y1  

A( x1 , y1 ) x1

C ( x2 , y 2 )  

∧ >

B( x2 , y1 )  

x2  

Fig. 6.7 Contour of integration.

x 

In order to find eigenvalues pi and the corresponding values qi from equation (6.1.58) we define the function F ( p) by the formula (see (6.1.57)): F ( p) = pJ1 (qc)T1' ( pc) − qJ1' (qc)T1 ( pc).

(6.3.4)

 126   

Here p = x + jy is an eigenvalue and q =

p 2 − jωσµ0 . Calculating the derivative of (6.3.4) with

respect to p we obtain F ' ( p) = [ J1' ( pc)Y1 ( pb) − J1 ( pb)Y1' ( pc)][ J1 (qc) + p 2 cq −1 J1' (qc)] + pJ1 (qc)[cJ1'' ( pc)Y1 ( pb) + bJ1' ( pc)Y1' ( pb) − bJ1' ( pb)Y1' ( pc) − cY1'' ( pc) J1 ( pb)] − [ J1 ( pc)Y1 ( pb) − J1 ( pb)Y1 ( pc)][ pq −1 J1' (qc) + pcJ1'' (qc)]

(6.3.5)

− qJ1' (qc)[cJ1' ( pc)Y1 ( pb) + bJ1 ( pc)Y1' ( pb) − bJ1' ( pb)Y1 ( pc) − cY1' ( pc) J1 ( pb)] Let φ ( p ) =

F ' ( p) . Then the number of zeros of φ ( p) inside C is given by (6.3.1) where the F ( p)

contour integral is computed by means of (6.3.3). If there is only one zero, ζ 1 , of F ( p ) inside C then it is calculated as follows (see (6.3.2)) φ ' ( z) 1 (6.3.6) s1 = z dz = ζ 1. ∫ 2πi C φ ( z ) An example of numerical calculation of complex eigenlavues pi is shown in Fig. 6.8. Computations are done with “Mathematica” ( see Appendix Fig.Ap.16 ) .The parameters of the cylinder correspond to 10c euro coin (the outer radius of the coil is 10 mm). The algorithm is implemented for the case where there are at most two eigenvalues inside a rectangle. The program then subdivides the rectangle (if necessary) so that only one eigenvalue is inside a smaller rectangle. The precise value of the root is then found by means of the formula (6.3.6). Im @zD

4 3

Fig. 6.8 The first five eigenvalues pi .

2

60

80

100

120

140

160

Re @zD

The first five zeros of F ( p ) are shown in Fig. 6.8. As soon as all eigenvalues pi , i = 1,2,..., N s are calculated we can proceed to the next step of the procedure where the system of linear equations (6.1.72) - (6.1.75) is solved (note that the upper summation index in (6.1.72) - (6.1.75) is equal N s ). Finally, the change in impedance of the coil is computed by means of the formula (6.2.14). Several calculations are done in order to estimate the effect of the parameters b and n = N s on the calculation accuracy. The parameters of the problem are as follows: σ = 4Ms/m, c =8mm, r1 = 3mm, r2 = 6mm, z1 = 0.1mm, z2 = 3.1mm, d =2mm. The first table shows  127   

the change in impedance of the coil calculated by means of the formula (6.2.13) for the frequency f = 1kHz for the case where the number of eigenvalues is n = 68 and four different values of b , namely, b = 5r2 , 10r2 , 15r2 , and 20r2 are chosen. The results show that b = 10r2 is sufficient in order to calculate the induced change in impedance with the accuracy of three decimal places. This value of b is used in all other calculations in the thesis.

b n

68

5r2

10r2

15r2

20r2

0.174015 -0.0124184i

0.176349 -0.0126594i

0.176578 -0.0126837i

0.17664 -0.0126913

Table 6.1 The second table shows the calculated induced change in impedance for different values of n and five different frequencies. The value of b is fixed at b = 10r2 . As can be seen from the table, n = 68 gives quite accurate value of the change in impedance. Convergence of the method is clearly seen from the table.

f n

11 19 27 34 47 68 102 137 171 240 309 377

1000

2000

3000

4000

5000

0.171115 -0.0128815i

0.673776 -0.101349i

1.47762 -0.332801i

2.53722 -0.760574i

3.79826 -1.41964i

0.17675 -0.012734i

0.695895 -0.100212i

1.52589 -0.329288i

2.61954 -0.752716i

3.92057 -1.40577i

0.176433 -0.0126717i

0.694685 -0.0997248i

1.52338 -0.3277i

2.61556 -0.74913i

3.91522 -1.39916i

0176345 -0.0126615i

0.694348 -0.0996455i

1.52266 -0.327444i

2.61439 -0.748559i

3.91354 -1.39813i

0.17634 -0.0126593i

0.694329 -0.0996289i

1.5263 -0.327392i

2.61435 -0.748446i

3.91353 -1.39793i

0.176349 -0.0126594i

0.694366 -0.0996297i

1.52275 -0.327392i

2.6145 -0.748455i

3.91377 -1.39795i

0.176348 -0.0126592i

0.694362 -0.0996281i

1.52274 -0.327385i

2.61449 -0.748443i

3.91376 -1.39793i

0.176348 -0.0126592i

0.694363 -0.0996281i

1.52275 -0.327385i

2.6145 -0.748444i

3.91377 -1.39793i

0.176348 -0.0126592i

0.694363 -0.099628i

1.52274 -0.327385i

2.6145 -0.748443i

3.91377 -1.39793i

0.176348 -0.0126592i

0.694363 -0.099628i

1.52274 -0.327385i

2.6145 -0.748443i

3.91377 -1.39793i

0.176348 -0.0126592i

0.694363 -0.099628i

1.52274 -0.327385i

2.6145 -0.748443i

3.91377 -1.39793i

0.176348 -0.0126592i

0.694363 -0.099628i

1.52274 -0.327385i

2.6145 -0.748443i

3.91377 -1.39793i

Table 6.2

6.4 A coil above a conducting plate with a flaw in the form of a circular cylinder In this section we consider a mathematical model of eddy current problem which can be used to model corrosion. Consider a coil of radius r0 located at a distance h above a conducting plate (see Fig. 6.9). The plate has a cylindrical hole of height d 2 and radius c . The thickness of the plate is d1 + d 2 .

 128   

r0 R0  

h 

O

R1   R2  

d1 d2

c b 

R3  

Fig. 6.9 A single-turn coil above a conducting plate with a flaw in the form of a circular cylinder.

We use the TREE method to solve the problem (the solution based on the layer approximation is given in [52]). The system of equations for the components of the vector potential in regions Ri , i = 0,1,2,3 has the form R0 :    µ (z ) = µ 0r = 1, σ (z ) = 0, I e = Iδ (r − r0 )δ (z − h )   ∂ 2 A0 1 ∂A0 A0 ∂ 2 A0 + − + = − µ 0 Iδ (r − r0 )δ ( z − h), ∂r 2 r ∂r r 2 ∂z 2 R1 :    µ (z ) = µ1r = 1, σ (z ) = σ , I = 0   ∂ 2 A1 1 ∂A1 A1 ∂ 2 A1 j A + − − + ωσµ = 0, 0 1 ∂r 2 r ∂r r 2 ∂z 2 ⎧⎪ ⋅ 0 ≤ r ≤ c µ ( z ) = µ 2r = 1, σ ( z ) = 0, I = 0 R2 :     ⎨       ⎪⎩ ⋅ c ≤ r ≤ b µ ( z ) = µ 2r = 1, σ ( z ) = σ , I = 0 ∂ 2 A2 1 ∂A2 A2 ∂ 2 A2 + − − j ωσµ A + = 0, 0 2 ∂r 2 r ∂r r 2 ∂z 2 R3 :     µ ( z ) = µ3r = 1, σ ( z ) = 0, I = 0  

(6.4.1)

(6.4.2)

(6.4.3)

∂ 2 A3 1 ∂A3 A3 ∂ 2 A3 + − + = 0, (6.4.4) ∂r 2 r ∂r r 2 ∂z 2 where Ai is the solution in region Ri , i = 0,1,2,3 (note that σ = 0 in region R2 where 0 ≤ r ≤ c ). As in Section 6.1 we use the notations A2air and A2con in regions 0 ≤ r ≤ c and c ≤ r ≤ b , respectively. The boundary conditions are Ai |r =b = 0, i = 0,1,3, A2con |r =b = 0, A0 | z =0 = A1 | z =0 ,

∂A0 ∂A | z =0 = 1 | z =0 , 0 ≤ r ≤ b, ∂z ∂z

A1 | z =− d1 = A2air | z =− d1 ,

∂A1 ∂Aair | z =− d1 = 2 | z =− d1 , 0 ≤ r ≤ c, ∂z ∂z

∂A1 ∂Acon | z =− d1 = 2 | z =− d1 , c ≤ r ≤ b, ∂z ∂z air ∂A2 ∂A | z =− d1−d2 = A3 | z =− d1−d2 , | z =− d1−d2 = 2 | z =− d1 −d2 , 0 ≤ r ≤ c, ∂z ∂z

(6.4.5) (6.4.6) (6.4.7)

A1 | z =− d1 = A2con | z =− d1 ,

(6.4.8)

A2air

(6.4.9)

A2con | z =− d1−d2 = A3 | z =− d1−d2 ,

∂A2con ∂A | z =− d1−d2 = 3 | z =− d1−d2 , c ≤ r ≤ b, ∂z ∂z

(6.4.10)  129   

A2con |r =c = A2air |r =c ,

∂A2con ∂Aair | r =c = 2 | r = c . ∂r ∂r

(6.4.11)

Solution to (6.4.1) is obtained as in Section 6.1 and is given by (6.1.29) and (6.1.30) in regions R00 = {0 < z < h} and R01 = {z > h} , respectively. Using the method of separation of variables we obtain the general solution to (6.4.2) in region R1 in the form ∞

A1 (r , z ) = ∑ ( D4i e p1i z + D5i e − p1i z ) J1 (λi r ),

(6.4.12)

i =1

where p1i = λi2 + jωσµ0 . General solution to (6.4.3) can be written in the form (see the details in Section 6.1): ∞

A2air (r , z ) = ∑ [ D6i e pi z + D7 i e − pi z ]J1 ( pi r ),

(6.4.13)

i =1

∞

{

}

A2con (r , z ) = ∑ [ D8i J1 (qi r ) + D9iY1 (qi r )]e pi z + [ D10i J1 (qi r ) + D11iY1 (qi r )]e − pi z ,

(6.4.14)

i =1

where pi = qi2 + jωσµ0 . General solution to (6.4.4) which is bounded as z → −∞ is written as follows ∞

A3 (r , z ) = ∑ D12i e λi z J1 (λi r ).

(6.4.15)

i =1

Using (6.4.14) and the last boundary condition in (6.4.5) we obtain the following two equations (6.4.16) D8i J1 (qi b) + D9iY1 (qi b) = 0, D10i J1 (qi b) + D11iY1 (qi b) = 0.

(6.4.17)

Eliminating D9i and D11i from (6.4.16) and (6.4.17) we obtain D9i = − D8i

J1 (qi b) , Y1 (qi b)

D11i = − D10i

(6.4.18)

J1 (qi b) . Y1 (qi b)

(6.4.19)

Using continuity of the functions A2air and A2con at r = c (the first condition in (6.4.11)) we get ∞

∑[D i =1

6i

e pi z + D7 i e − pi z ]J 1 ( pi c) = ∞

{

= ∑ [ D8i J 1 (qi c) + D9iY1 (qi c)]e i =1

pi z

+ [ D10i J 1 (qi c) + D11iY1 (qi c)]e

− pi z

}.

(6.4.20)

The following two relationships are obtained from (6.4.20): D6i J 1 ( pi c) = D8i J 1 (qi c) + D9iY1 (qi c),

(6.4.21)

D7 i J 1 ( pi c) = D10i J 1 (qi c) + D11iY1 (qi c).

(6.4.22)

 130   

Combining equations (6.4.18) and (6.4.21) we obtain D = Dˆ [ J (q c)Y (q b) − J (q b)Y (q c)], 6i

8i

1

i

1

i

1

i

1

i

(6.4.23)

where

Dˆ 8i =

D8i . J1 ( pi c)Y1 (qi b)

(6.4.24)

It follows from (6.4.19) and (6.4.22) that D = Dˆ [ J (q c)Y (q b) − J (q b)Y (q c)], 7i

10 i

1

i

1

i

1

i

1

(6.4.25)

i

where

Dˆ 10i =

D10i . J1 ( pi c)Y1 (qi b)

(6.4.26)

Solutions (6.4.13) and (6.4.14) can be rewritten in the following form (here we use the relationships (6.4.18), (6.4.19), (6.4.23) and (6.4.25)): ∞

A2air (r , z ) = ∑ T1 (qi c)[ Dˆ 8i e pi z + Dˆ 10i e − pi z ]J1 ( pi r ),

(6.4.27)

i =1

∞

A2con (r , z ) = ∑ J1 ( pi c)[ Dˆ 8i e pi z + Dˆ 10i e − pi z ]T1 (qi r ),

(6.4.28)

where T1 (qi r ) = J1 (qi r )Y1 (qib) − J1 (qib)Y1 (qi r ),

(6.4.29)

T1 (qi c) = J1 (qi c )Y1 (qib) − J1 (qib)Y1 (qi c) .

(6.4.30)

i =1

Differentiating (6.4.27) and (6.4.28) with respect to r and evaluating the derivatives at r = c we obtain ∞ ∂A2air |r = c = ∑ pi J 1' ( pi c)T1 (qi c )( Dˆ 6i e pi z + Dˆ 8i e − pi z ), ∂r i =1 ∞ ∂A2con |r = c = ∑ qiT1' (qi c) J 1 ( pi c)( Dˆ 6i e pi z + Dˆ 8i e − pi z ). ∂r i =1

(6.4.31) (6.4.32)

It follows from (6.4.31) and (6.4.32) and the second boundary condition in (6.4.11) that pi J1' ( pi c)T1 (qi c) = qiT1' (qi c) J1 ( pi c). (6.4.33) Equation (6.4.33) is used to determine the eigenvalues pi and related values qi . The relationship between pi and qi is pi = qi + jωσµ 0 .    2

Thus, the solution in regions R0 , R1 , R2 and R3 is given by (6.1.29), (6.1.30), (6.4.12), (6.4.27), (6.4.28) and (6.4.15). The six sets of constants in these formulas, namely, D , D , D , Dˆ , Dˆ and D can be obtained from the boundary conditions (6.4.6) – 2i

4i

5i

8i

10 i

12 i

(6.4.10). Using the condition (6.4.6) we obtain

 131   

∞

∑ D2i J1 (λi r ) +

µ0 Ir0 b2

i =1

∞

− ∑ λi D2i J1 (λi r ) +

∞ J1 (λi r0 )e − λi h J ( λ r ) = ( D4i + D5i ) J1 (λi r ), ∑ ∑ 1 i 2 i =1 λi J 0 (λi b) i =1 ∞

µ0 Ir0 b2

i =1

∞ J1 (λi r0 )e − λi h J1 (λi r ) = ∑ p1i ( D4i − D5i ) J1 (λi r ) . ∑ 2 J 0 (λib) i =1 i =1 ∞

(6.4.34) (6.4.35)

The obtained equations are multiplied by rJ 1 (λ j r ) and the resulting equation is integrated with respect to r from 0 to b . Using the orthogonality condition (6.1.25) we obtain D2 j

b2 2 µ Ir b2 −λ h J 0 (λ j b) + 0 0 J1 (λ j r0 )e j = (D4 j + D5 j ) J 02 (λ j b) , 2 2λ j 2

− λ j D2 j

b2 2 µ Ir b2 −λ h J 0 (λ j b) + 0 0 J1 (λ j r0 )e j = p1 j (D4 j − D5 j ) J 02 (λ j b) . 2 2 2

(6.4.36) (6.4.37)

From (6.4.36) we find D2 j = D4 j + D5 j − µ0 Ir0

J1 (λ j r0 )e

−λ j h

λ j b 2 J 02 (λ j b)

(6.4.38)

.

We multiply (6.4.36) by λ j and add the resulting equation to (6.4.37): (λ j + p1 j ) D4 j + (λ j − p1 j ) D5 j =

2 µ0 Ir0 J1 (λ j r0 )e b 2 J 02 (λ j b)

−λ j h

(6.4.39)

.

Using the first conditions in (6.4.7) and in (6.4.8) we obtain ∞

∑ (D i =1

4i

e − p1i d1 + D5i e p1i d1 ) J1 (λi r ) = ∞

= ∑ T1 (qi c) J1 ( pi r )( Dˆ 8i e − pi d1 + Dˆ10i e pi d1 ), 0 ≤ r ≤ c,

(6.4.40)

i =1

∞

∑ (D i =1

4i

e − p1i d1 + D5i e p1i d1 ) J1 (λi r ) = ∞

= ∑ T1 (qi r ) J1 ( pi c)( Dˆ 8i e i =1

(6.4.41) − pi d1

+ Dˆ 10i e

pi d1

), c ≤ r ≤ b.

In order to determine the coefficients Dˆ 8i and Dˆ 10i the following procedure is used. First, equations (6.4.40) and (6.4.41) are combined into one equation where the right-hand side of the resulting equation is given by different expressions on the intervals 0 ≤ r ≤ c and c ≤ r ≤ b . These expressions are defined by the right-hand sides of (6.4.40) and (6.4.41), respectively. Second, the obtained equation is multiplied by rJ 1 (λ j r ) and the resulting equation is integrated with respect to r from 0 to b . Third, we use the orthogonality condition (6.1.25) and formulas (6.4.29), (6.4.30), (6.1.62) and (6.1.63). The result is D4 j e

− p1 j d1

+ D5 j e

p1 j d1

=

∞ 2 ( Dˆ 8i e − pi d1 + Dˆ 10i e pi d1 )a ji , ∑ 2 2 b J 0 (λ j b) i =1

(6.4.42)

 132   

~ a ji = T1 (qi c)a~ ji + J1 ( pi c )a~ ji .

where

(6.4.43)

Thus, using (6.1.62), (6.1.63) and (6.4.29) we obtain c

⋅ a~ ji = ∫ rJ1 (λ j r )J1 ( pi r )dr = 0

c ( λ J (λ c )J ( p c ) − pi J1 (λ j c )J 2 ( pic ) ) λ − pi2 j 2 j 1 i 2 j

b

b b ~ ⋅ a~ ji = ∫ rJ1 (λ j r )T1 (qi r )dr = Y1 (qib )∫ rJ1 (λ j r )J1 (qi r )dr − J1 (qib )∫ rJ1 (λ j r )Y1 (qi r )dr = c

c

c

⎧bqi J1 (λ j b )[ J1 (qib )Y2 (qib ) − J 2 (qib )Y1 (qib ) ] + ⎫ ⎪ 1 ⎪ ( ) ( ) ( ) ( ) ( ) c λ J λ c J q b Y q c J q c Y q b [ ] = 2 + − + ⎨ ⎬ j j i i i i 2 1 1 1 1 λ j − qi2 ⎪ ⎪ ⎩+ cqi J1 (λ j c )[ J 2 (qi c )Y1 (qib ) − J1 (qib )Y2 (qi c ) ] ⎭ Using the same procedure and applying the second condition in (6.4.7) and in (6.4.8) we obtain ∞ 2 p d −p d p1 j D4 j e 1 j 1 − D5 j e 1 j 1 = 2 2 (6.4.44) ∑ pi ( Dˆ 8ie− pi d1 − Dˆ10ie pi d1 )a ji . b J 0 (λ j b) i =1

(

)

We multiply (6.4.42) by p1 j and add the resulting equation to (6.4.44) p d

∞ e 1j 1 D4 j = ∑ p1 j b 2 J 02 (λ j b) i =1

{( ( p

1j

) }

+ pi )e − pi d1 Dˆ 8i + ( p1 j − pi )e pi d1 Dˆ 10i a ji .

(6.4.45)

We multiply (6.4.42) by (− p1 j ) and add the resulting equation to (6.4.44) −p d

∞ e 1j 1 ∑ p1 j b 2 J 02 (λ j b) i =1

D5 j =

{( ( p

1j

) }

− pi )e − pi d1 Dˆ 8i + ( p1 j + pi )e pi d1 Dˆ 10i a ji .

(6.4.46)

Two additional equations are obtained if the same procedure is applied to (6.4.27), (6.4.28) and (6.4.15) using boundary conditions (6.4.9) and (6.4.10). The result is shown below D12 j e

−λ j d 3

λ j D12 j e

∞ b2 2 J 0 (λ j b) = ∑ ( Dˆ 8i e − pi d 3 + Dˆ10i e pi d 3 )a ji , 2 i =1

−λ j d3

(6.4.47)

∞ b2 2 J 0 (λ j b) = ∑ pi ( Dˆ 8i e − pi d 3 − Dˆ10i e pi d 3 )a ji , 2 i =1

(6.4.48)

where d 3 = d1 + d 2

Similarly, multiplying (6.4.47) by (− λ j ) and adding the resulting equation to (6.4.48) we obtain ∞

∑[(λ i =1

j

− pi )e − pi d 3 Dˆ 8i + (λ j + pi )e pi d 3 Dˆ 10i ]a ji = 0.

(6.4.49)

Thus, using (6.4.45), (6.4.46) and (6.4.39) we obtain ( p − pi )d1 − ( p + p )d ∞ ⎛ (λ + p )( p + p )e 1 j + (λ j − p1 j )( p1 j − pi )e 1 j i 1 a ji Dˆ 8i + ⎞⎟ j i 1j 1j ⎜ = ∑ ⎜ + (λ + p )( p − p )e ( p1 j + pi )d1 + (λ − p )( p + p )e − ( p1 j − pi )d1 a Dˆ ⎟ i =1 j i j i ji 10 i ⎠ 1j 1j 1j 1j ⎝

(

)

(

= 2 p j µ0 Ir0 J1 (λ j r0 )e

)

(6.4.50)

−λ j h

 133   

The six sets of equations for the unknown coefficients D2 j , D4 j , D5 j , Dˆ 8i , Dˆ 10i and D12 j are given by (6.4.38), (6.4.45), (6.4.46), (6.4.49), (6.4.50) and (6.4.47). The solution of the system is obtained as follows. First, we consider only a finite number of terms n in (6.4.38), (6.4.45), (6.4.46), (6.4.47), (6.4.49) and (6.4.50). Recommendations on the selection of the value of n are given in Section 6.3. Second, the coefficients Dˆ 8i , Dˆ 10i can be computed solving the system

(

)

⎛ (λ j + p1 j )( p1 j + pi )e ( p1 j − pi )d1 + (λ j − p1 j )( p1 j − pi )e − ( p1 j + pi )d1 a ji Dˆ 8i + ⎞ ⎟= ⎜ ∑ ( p1 j + pi )d1 − ( p1 j − p i )d1 ⎜ i =1 a ji Dˆ10i ⎟⎠ + (λ j − p1 j )( p1 j + pi )e ⎝ + (λ j + p1 j )( p1 j − pi )e n

(

= 2 p1 j µ0 Ir0 J1 (λ j r0 )e n

∑[(λ i =1

j

)

(6.4.51)

−λ j h

− pi )e − pi d 3 Dˆ 8i + (λ j + pi )e pi d 3 Dˆ10i ]a ji = 0.

(6.4.52)

where n is the number of eigenvalues pi . The system (6.4.51)-(6.4.52) can be written in the matrix form r r AX = B                                                                                                                                                (6.4.53) where the coefficient matrix A is ⎛A A ⎞ A = ⎜⎜ 11 12 ⎟⎟                                                                                                                                      (6.4.54) ⎝ A21 A22 ⎠ and the block matrices A11 , A12 , A21 and A22 are

 

((

)(

)

(

)(

)

)

⎛ λ1 + p1 p1 + p1 e ( p11 − p1 )d1 + λ1 − p1 p1 − p1 e − ( p11 + p1 )d1 a11 ............................. ⎞ 1 1 1 1 ⎟ ⎜ ⎜ ... λ + p p + p e ( p12 − p 2 )d1 + λ − p p − p e − ( p12 + p 2 )d1 a ........................ ⎟ 2 12 12 2 2 12 12 2 22 ⎟  A11 = ⎜ ⎟ ⎜ ............................................................................................ ⎟ ⎜ ⎜ .......................... λn + p1 p1 + pn e ( p1n − p n )d1 + λn − p1 p1 − pn e − ( p1n + p n )d1 ann ⎟ n n n n ⎠ ⎝

((

)(

)

((

((

)(

(

)(

)

)(

)

(

)(

)

(

)(

)

)

)

)

)

⎛ λ1 + p1 p1 − p1 e( p11 + p1 )d1 + λ1 − p1 p1 + p1 e − ( p11 − p1 )d1 a11 ............................. ⎞ 1 1 1 1 ⎜ ⎟ ⎜ ... λ + p p − p e( p12 + p 2 )d1 + λ − p p + p e − ( p12 − p 2 )d1 a ........................⎟ 2 12 12 2 2 12 12 2 22 ⎟  A12 = ⎜ ⎜ ⎟ ............................................................................................ ⎜ ⎟ ⎜ .......................... λn + p1 p1 − pn e( p1n + p n )d1 + λn − p1 p1 + pn e − ( p1n − p n )d1 ann ⎟ n n n n ⎝ ⎠

((

)(

((

)

)(

(

)

)(

(

)

)(

)

)

)

⎛ (λ1 − p1 )e − p1d3 a11 (λ1 − p2 )e − p2d3 a12 ......... (λ1 − pn )e − pnd3 a1n ⎞ ⎟ ⎜ ⎜ (λ2 − p1 )e − p1d3 a21 (λ2 − p2 )e − p2d3 a22 ........ (λ2 − pn )e − pnd3 a2 n ⎟ A21 = ⎜ ⎟  .......... .......... .......... .......... ........ ⎟ ⎜ ⎜ (λ − p )e − p1d3 a (λ − p )e − p2d3 a ........ (λ − p )e − pnd3 a ⎟ 1 2 n1 n n2 n n nn ⎠ ⎝ n  

 134   

⎛ (λ1 + p1 )e p1d3 a11 (λ1 + p2 )e p2d3 a12 ............. (λ1 + pn )e pnd3 a1n ⎞ ⎟ ⎜ ⎜ (λ2 + p1 )e p1d3 a21 (λ2 + p2 )e p2d3 a22 ............ (λ2 + pn )e pnd3 a2 n ⎟ A22 = ⎜ ⎟  .......... .......... .......... .......... ........ ⎟ ⎜ p2 d 3 pn d 3 ⎜ (λ + p )e p1d3 a (λn + p2 )e an 2 ............ (λn + pn )e ann ⎟⎠ 1 n1 ⎝ n r r The matrices X and B in (6.4.53) are r r r ⎛ X1 ⎞ r ⎛ b ⎞ ⎜ ⎟ X = r  , B = ⎜⎜ ⎟⎟ , ⎜X ⎟ ⎝0 ⎠ ⎝ 2⎠ where ⎛ Dˆ 10 ⎞ ⎛ Dˆ 8 ⎞ ⎛ 2 p11 µ0 Ir0 J1 (λ1r0 )e − λ1h ⎞ ⎜ 1⎟ ⎜ 1⎟ ⎟ ⎜ r ⎜ Dˆ 8 ⎟ r ⎜ Dˆ 10 ⎟ r ⎜ 2 p12 µ0 Ir0 J1 (λ2 r0 )e − λ2 h ⎟ ⎟. X1 = ⎜ 2 ⎟ , X 2 = ⎜ 2 ⎟ , b = ⎜ ⎜ ... ⎟ ⎜ ... ⎟ ⎜ ⎟ ............... ⎟ ⎜ ⎟ ⎜ ⎜⎜ − λn h ⎟ ⎟ ⎜ Dˆ 8 ⎟ ⎜ Dˆ 10 ⎟ 2 p µ Ir J ( λ r ) e 1 0 0 1 0 n n ⎝ ⎠ ⎝ n⎠ ⎝ n⎠

(6.4.55)

(6.4.56)

Hence, (6.4.53) can be written in the form    r r r ⎧⎪ A11 X 1 + A12 X 2 = b                                                                                                                            (6.4.57) r r ⎨ ⎪⎩ A21 X 1 + A22 X 2 = 0 Solving (6.4.57) we obtain all the coefficients Dˆ 8i and Dˆ10i (i = 1,2,..., n). Then D4 j and D5 j ( j = 1,2,.., n) are calculated using ( p + p )d ⎧⎪ ⎛ ( p1 j + pi )e ( p1 j − pi )d1 a ji ( p1 j − pi )e 1 j i 1 a ji ˆ ⎞⎟⎫⎪ ˆ ⎜ D4 j = ∑ ⎨ D8i + D10i ⎬ , ⎜ ⎟⎪ p1 j b 2 J 02 (λ j b) p1 j b 2 J 02 (λ j b) i =1 ⎪ ⎝ ⎠⎭ ⎩ − ( p − p )d n ⎧ ⎛ ( p − p )e − ( p1 j + p i )d 1 a ( p1 j + pi )e 1 j i 1 a ji ˆ ⎞⎟⎫⎪ ⎪ ⎜ 1j i ji ˆ D5 j = ∑ ⎨ D8i + D10i ⎬. ⎜ ⎟⎪ p1 j b 2 J 02 (λ j b) p1 j b 2 J 02 (λ j b) i =1 ⎪ ⎝ ⎠⎭ ⎩ n

(6.4.58)

(6.4.59)

The system (6.4.58)-(6.4.59) can be written in the matrix form r r r ⎧⎪ D4 = B11 X 1 + B12 X 2 r r                                                                                                                          (6.4.60) ⎨r ⎪⎩ D5 = B21 X 1 + B22 X 2 The matrices B11 , B12 , B21 and B22 are ( p − p )d ( p − p )d ⎛ p1 + p1 e( p11 − p1 )d1 a11 p11 + p2 e 11 2 1 a12 p11 + pn e 11 n 1 a1n ⎞⎟ ⎜ 1 ... ⎜ ⎟ p11 b 2 J 02 (λ1b ) p11 b 2 J 02 (λ1b ) p11 b 2 J 02 (λ1b ) ⎜ ⎟ ( p − p )d ( p − p )d ⎜ p12 + p1 e( p12 − p1 )d1 a21 p12 + p2 e 12 2 1 a22 p12 + pn e 12 n 1 a2 n ⎟ ... ⎜ ⎟ B11 = ⎜ p12 b 2 J 02 (λ2b ) p12 b 2 J 02 (λ2b ) p12 b 2 J 02 (λ2b ) ⎟                (6.4.61)  ⎜ ⎟ ................................................ ⎜ ⎟ ( p1n − p 2 )d1 ( p1n − p n )d1 ⎟ ⎜ p1 + p1 e( p1n − p1 )d1 an1 p1n + p2 e an 2 p1n + pn e ann ⎜ n ⎟ ... 2 2 2 2 2 2 ⎜ ⎟ ( ) ( ) ( ) p b J λ b p b J λ b p b J λ b 1 0 1 0 1 0 n n n n n n ⎝ ⎠

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

 135   

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

(

)

(

)

(

)

(

)

(

)

(

)

( p + p )d ( p + p )d ⎛ p1 − p1 e ( p11 + p1 )d1 a11 p11 − p2 e 11 2 1 a12 p11 − pn e 11 n 1 a1n ⎞⎟ ⎜ 1 ... ⎟ ⎜ p11 b 2 J 02 (λ1b ) p11 b 2 J 02 (λ1b ) p11 b 2 J 02 (λ1b ) ⎟ ⎜ ( p + p )d ( p + p )d ⎜ p12 − p1 e( p12 + p1 )d1 a21 p12 − p2 e 12 2 1 a22 p12 − pn e 12 n 1 a2 n ⎟ ... ⎟ ⎜ B12 = ⎜ p12 b 2 J 02 (λ2b ) p12 b 2 J 02 (λ2b ) p12 b 2 J 02 (λ2b ) ⎟                 (6.4.62)  ⎟ ⎜ ................................................ ⎟ ⎜ ( p1n + p 2 )d1 ( p1n + p n )d1 ⎟ ⎜ p1 − p1 e ( p1n + p1 )d1 an1 p p e a p p e a − − n2 n nn 1n 2 1n ⎟ ⎜ n ... 2 2 2 2 2 2 ⎟ ⎜ ( ) ( ) ( ) p b J λ b p b J λ b p b J λ b n n n 1 0 1 0 1 0 n n n ⎠ ⎝ − ( p + p )d − ( p + p )d ⎛ p1 − p1 e − ( p11 + p1 )d1 a11 p11 − p2 e 11 2 1 a12 p11 − pn e 11 n 1 a1n ⎞⎟ ⎜ 1 ... ⎟ ⎜ p11 b 2 J 02 (λ1b ) p11 b 2 J 02 (λ1b ) p11 b 2 J 02 (λ1b ) ⎟ ⎜ − ( p + p )d − ( p + p )d ⎜ p12 − p1 e − ( p12 + p1 )d1 a21 p12 − p2 e 12 2 1 a22 p12 − pn e 12 n 1 a2 n ⎟ ... ⎟ ⎜ B21 = ⎜ p12 b 2 J 02 (λ2b ) p12 b 2 J 02 (λ2b ) p12 b 2 J 02 (λ2b ) ⎟             (6.4.63)  ⎟ ⎜ ................................................ ⎟ ⎜ − ( p1n + p 2 )d1 − ( p1n + p n )d1 ⎟ ⎜ p1 − p1 e − ( p1n + p1 )d1 an1 p p e a p p e a − − n2 n nn 1n 1n 2 ⎟ ⎜ n ... 2 2 2 2 2 2 ⎟ ⎜ ( ) ( ) ( ) p b J λ b p b J λ b p b J λ b n n 1 0 n 1 0 1 0 n n n ⎠ ⎝  

)

− ( p − p )d − ( p − p )d ⎛ p1 + p1 e − ( p11 − p1 )d1 a11 p11 + p2 e 11 2 1 a12 p11 + pn e 11 n 1 a1n ⎞⎟ ⎜ 1 ... ⎟ ⎜ p11 b 2 J 02 (λ1b ) p11 b 2 J 02 (λ1b ) p11 b 2 J 02 (λ1b ) ⎟ ⎜ − ( p − p )d − ( p − p )d ⎜ p12 + p1 e − ( p12 − p1 )d1 a21 p12 + p2 e 12 2 1 a22 p12 + pn e 12 n 1 a2 n ⎟ ... ⎟ ⎜ B22 = ⎜ p12 b 2 J 02 (λ2b ) p12 b 2 J 02 (λ2b ) p12 b 2 J 02 (λ2b ) ⎟            (6.4.64) ⎟ ⎜ ................................................ ⎟ ⎜ − ( p1n − p 2 )d 1 − ( p1n − p n )d1 ⎟ ⎜ p1 + p1 e − ( p1n − p1 )d1 an1 p p e a p p e a + + n2 n nn 1n 2 1n ⎟ ⎜ n ... 2 2 2 2 2 2 ⎟ ⎜ ( ) ( ) ( ) p b J λ b p b J λ b p b J λ b n n n 1 0 1 0 1 0 n n n ⎠ ⎝

Solving (6.4.60) we obtain all the coefficients Dˆ 4 j and Dˆ 5 j ( j = 1,2,..., n). Then D2 j is calculated using D2 j = D4 j + D5 j − µ0 Ir0

J1 (λ j r0 )e

−λ j h

λ j b 2 J 02 (λ j b)

.

(6.4.65)

The induced change in impedance of the coil is given by the formula        jω 2πr0 A0ind (r0 , h)                                                                                                                    (6.4.66)                      Z ind = I n

where A0ind (r0 , h) = ∑ D2 j e j =1

−λ j h

J1 (λ j r0 )                                                                                            (6.4.67)

Formula (6.4.66) is used to compute the change in impedance of the coil. Calculations are performed with “Mathematica”. The program which is used to compute the change in impedance  136   

(6.4.66) is shown in Appendix Fig. Ap.17. The following parameters of the problem are selected: µ0 = 4 ⋅ 10−7 π , σ = 3 Ms/m, c = 2.2 mm,  r0 = 4.5 mm, h = 1.4 mm, d1 = 0.7 mm, d 2 = 0.3 mm, b = 55 mm. The change in impedance is computed for the following seven frequencies:  f = 1000, 2000, 3000, 4000, 5000, 6000, and 7000 Hz.The results of calculations are shown in Fig. 6.10.The calculated points (from top to bottom) correspond to the seven frequencies (from smallest to largest). The upper limit of the summation index in (6.4.67) is fixed at n = 62 . Comparison of the computational results obtained for other values of n showed that the chosen value of 62 is quite satisfactory in terms of calculation accuracy. More detailed analysis of convergence is presented in Section 6.3. Several computational steps are necessary in order to calculate the induced change in impedance. First, the set of eigenvalues λi has to be calculated. This can easily be done in “Mathematica” using a built-in routine BesselZeros. Second, a set of complex roots of (6.4.33) should be computed. Calculations are based on the method described in Section 6.3. Third, several systems of linear equations have to be solved in order to determine expansion coefficients. Finally, the change in impedance is computed using (6.4.53) -(6.4.67) . Im @zD

Re @zD

-2 ×10 - 6 -4 ×10 - 6 -6 ×10 - 6 -8 ×10 - 6

Fig. 6.10 The change in impedance of the coil for seven frequencies.

-0.00001 -0.000012

6.5 A coil of finite dimensions above a conducting plate with a flaw in the form of a circular cylinder Consider the model shown in Fig. 6.11. The inner and outer radii of the coil are r1 and r2 , respectively. The bottom of the coil is located at the distance z1 from the conducting plate. The height of the coil is z 2 − z1 and the number of turns is N .

r2 r1   R0  

R1   R2   R3  

z2  

z1

O d1 d2

c  b 

Fig. 6.11 A coil of finite dimensions above a conducting plate with a cylindrical hole.  137   

It can easily be shown that the second term in (6.1.29) and (6.1.30) represents the vector potential of a single-turn coil located in an unbounded free space. In addition, the first term in (6.1.29) and (6.1.30) is the induced vector potential in air due to the presence of a conducting cylinder: n

A0ind (r , z , r0 , h) = ∑ D2 j e

−λ j z

j =1

J1 (λ j r ),

(6.5.1)

where D2 j = D4 j + D5 j − µ0 Ir0

J1 (λ j r0 )e

−λ j h

λ j b 2 J 02 (λ j b)

.

The induced vector potential in air due to currents in the whole coil is obtained as follows r2 z 2

ind 0 coil

A

(r , z ) = ∫ ∫ A0ind (r , z , r0 , h)dr0 dh .

(6.5.2)

r1 z1

There is an important difference between the calculations for a single-turn coil and coil of finite dimensions. For the case of a single-turn coil (Section 6.4) the geometrical parameters of the coil ( r0 and h ) are constants and the numerical values of r0 and h can be used to solve the system (6.4.57). If a coil has finite dimensions then one needs to integrate the solution with respect to r0 and h . Thus, numerical values to the parameters r0 and h in (6.4.57) cannot be assigned. In this case, in order to obtain the induced vector potential of a single-turn coil we need to solve (6.4.57) in matrix form. Solving the second equation in (6.4.57) we obtain r r (6.5.3) X 2 = − A22−1 A21 X 1. Substituting (6.5.3) into the first equation in (6.4.57) gives r r (A11 − A12 A22−1 A21 )X 1 = b .

(6.5.4)

It follows from (6.5.3) and (6.5.4) that r r ⎧⎪ X = (A − A A−1 A )−1 b 1 11 12 22 21 ⎨r −1 r −1 ⎪⎩ X 2 = − A22 A21 (A11 − A12 A22−1 A21 ) b

(6.5.5)

Using (6.4.60) - (6.4.64) equation (6.4.65) should be written in matrix form: r r D2 = Y b ,

(

)(

−1 A21 A11 − A12 A22−1 A21 where Y = B11 + B21 − (B12 + B22 )A22

⎛ D21 ⎞ ⎜ ⎟ r ⎜ D2 2 ⎟ D2 = ⎜ ⎟, ... ⎟ ⎜ ⎜D ⎟ ⎝ 2n ⎠

Cdiag

1 ⎞ ⎛ 0 0 0 ... 0 ⎟ ⎜ 2 2 ⎟ ⎜ 2 p11 λ1b J 0 (λ1b ) ⎟ ⎜ 1 ⎟ ⎜0 0 0 ... 0 2 2 ⎟ = ⎜ 2 p12 λ2b J 0 (λ2b ) ⎟ ⎜ ....................... ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜⎜ 0 0 ... 0 0 2 2 2 p1n λnb J 0 (λnb ) ⎟⎠ ⎝

)

−1

(6.5.6)

− Cdiag  ,                                          (6.5.7)

(6.5.8)

r and b is given by (6.4.56).

 138   

Formula (6.5.1) for the induced vector potential of a single-turn coil can be rewritten in the form r r A0ind (r , z , r0 , h) = D2T f , (6.5.9) where ⎛ J1 (λ1r )e − λ1 z ⎞ ⎟ ⎜ r ⎜ J1 (λ2 r )e − λ2 z ⎟ f =⎜ ⎟ .  ⎟ ⎜ ... ⎜ J ( λ r )e − λ n z ⎟ ⎠ ⎝ 1 n Substituting (6.5.9) into (6.5.2) and using the formulas z2

⋅ ∫e

−λ j h

dh = −

z1

1

λj

(e

−λ j z2

−e

− λ j z1

r2

⋅ ∫ r0 J1 (λ j r0 )dr0 = ξ = λ j r0 = r1

) (6.5.10)

λ j r2

1

λi2

∫ ξJ (ξ )dξ 1

λ j r1

we obtain the induced vector potential in air due to the presence of the conducting cylinder ( the current amplitude I in this case is replaced by the current NI density ): (r2 − r1 )( z 2 − z1 ) ind 0 coil

A

(r , z ) =

µ0 NI (r2 − r1 )( z2 − z1 )

n

∑ f ∑Y j =1

(e

n

j

i =1

− λi z1

− e − λi z 2

λ3i

ji

)

λi r2

∫ ξJ (ξ )dξ .

(6.5.11)

1

λi r1

The integral with respect to ξ in (6.5.10) can be computed in terms of the Bessel and Struve functions [2] as follows ξ = λi r2

λi r2

⎧π ⎫ ∫λ rξJ1 (ξ )dξ = ⎨⎩ 2 ξ [J 0 (ξ ) H1 (ξ ) − J1 (ξ ) H 0 (ξ )]⎬⎭ ξ = λi r1 i 1

(6.5.12)

The induced change in impedance of a coil of finite dimensions (see Fig. 6.11) is calculated by means of the following formula [15]: r2 z2 2π rr jω jω ind Z = 2 ∫∫∫ AI dv = 2 ∫ dϕ ∫ rdr ∫ A0indcoil dz I V I 0 r1 z1 r z

Z

ind

2 2 2πjω N = rA0indcoil (r , z )drdz. I (r2 − r1 )( z 2 − z1 ) ∫r1 z∫1

(6.5.13)

Using (6.5.11) and (6.5.13) we obtain the induced change in impedance of the coil in the form Z

ind

(

n 2 jωπµ0 N 2 e = ∑ 2 2 (r2 − r1 ) ( z2 − z1 ) j =1

− λ j z1

−e

λ3j

−λ j z2

)

λ j r2

∫ ξJ (ξ )dξ ∑ Y 1

λ j r1

(e

n

i =1

ji

− λ i z1

− e − λi z 2

λ3i

)

λi r2

∫ ξJ (ξ )dξ 1

(6.5.14)

λi r1

 139   

Formula (6.5.14) is used to compute the change in impedance of the coil. Calculations are performed with “Mathematica”. Program which is used to compute the change in impedance (6.5.14) is shown in Appendix Fig. Ap.18. The following parameters of the problem are selected: µ0 = 4 ⋅ 10−7 π , σ = 3 Ms/m, c = 2.2 mm,  r2 = 5.5 mm, r1 = 3.5 mm , z1 = 0.3 mm, z2 = 2.6 mm, d1 = 0.7 mm, d 2 = 0.3 mm, b = 55 mm, N = 400 . The change in impedance is

computed for the following seven frequencies:  f = 1000, 2000, 3000, 4000, 5000, 6000, and 7000 Hz.The results of calculations are shown in Fig. 6.12.The calculated points (from top to bottom) correspond to the seven frequencies (from smallest to largest). The upper limit of the summation index in (6.5.14) is fixed at n = 62 . Comparison of the computational results obtained for other values of n  showed that the chosen value of 62 is quite satisfactory in terms of calculation accuracy. Several computational steps are necessary in order to calculate the induced change in impedance. First, the set of eigenvalues λi has to be calculated. This can easily be done

in “Mathematica” using a built-in routine BesselZeros. Second, a set of complex roots of (6.4.33)  should be computed. Calculations are based on the method described in Section 6.3. Third, several systems of linear equations have to be solved in order to determine expansion coefficients. Finally, the change in impedance is computed using (6.5.14). Details of the numerical aspects of the procedure are given in Section 6.3. Im @zD

1

2

3

4

5

Re @zD

-0.5 -1

-1.5

Fig. 6.12 The change in impedance of a coil (calculations are done using formula (6.5.14)).

-2

6.6 A coil above a conducting half-space with a flaw in the form of a circular cylinder In this section we consider a mathematical model of eddy current problem which can be used for quality control of spot welding [5], [6], [21], [71]. In this case metal parts are welded only at separate points (not along the whole surface of contact). Spot welding of two metal pieces produces cast core whose electrical conductivity is close to the conductivity of the surrounding medium.

 140   

r0   R0  

h 

R1   R2  

b 

O 

d1

c 

d2

R3  

Fig. 6.13. A single-turn coil above a conducting plate.

Consider a coil of radius r0 located at a distance h above a conducting plate (see Fig. 6.13). The plate has a cylindrical hole of height d 2 and radius c . The thickness of the plate is d1 + d 2 .We use the TREE method to solve the problem [25]. The system of equations for the components of the vector potential in regions Ri , i = 0,1,2,3 has the form R0 :    µ (z ) = µ 0r = 1, σ (z ) = 0, I e = Iδ (r − r0 )δ (z − h ) ∂ 2 A0 1 ∂A0 A0 ∂ 2 A0 = − µ 0 Iδ (r − r0 )δ ( z − h), − + + ∂z 2 ∂r 2 r ∂r r 2 R1 :    µ ( z ) = µ1r = 1, σ ( z ) = σ 1 , I = 0   ∂ 2 A1 1 ∂A1 A1 ∂ 2 A1 + − − jωσ 1 µ 0 A1 + 2 = 0, ∂r 2 r ∂r r 2 ∂z r ⎧⎪ ⋅ 0 ≤ r ≤ c µ ( z ) = µ 2 = 1, σ ( z ) = σ 1 − σ 2 , I = 0 R2 : ⎨ ⎪⎩ ⋅ c ≤ r ≤ b µ ( z ) = µ 2r = 1, σ ( z ) = σ 1 , I = 0 ∂ 2 A2 1 ∂A2 A2 ∂ 2 A2 ωσµ = 0, + − − j A + 0 2 ∂z 2 ∂r 2 r ∂r r 2 R3 :     µ ( z ) = µ3r = 1, σ ( z ) = σ 1 , I = 0  

(6.6.1)

(6.6.2)

(6.6.3)

∂ 2 A3 1 ∂A3 A3 ∂ 2 A3 = 0, (6.6.4) + − − jωσ 1 µ 0 A3 + ∂z 2 ∂r 2 r ∂r r 2 where Ai is the solution in region Ri , i = 0,1,2,3 (note that σ = σ 1 − σ 2 and σ = σ 1 in region R2 where 0 ≤ r ≤ c and c ≤ r ≤ b , respectively).

The boundary conditions are Ai |r = b = 0, i = 0,1,3, A2con |r =b = 0, ∂A0 ∂A | z =0 = 1 | z =0 , 0 ≤ r ≤ b, ∂z ∂z ∂A1 ∂Acc A1 | z = − d1 = A2cc | z = − d1 , | z = − d1 = 2 | z = − d 1 , 0 ≤ r ≤ c , ∂z ∂z

A0 | z =0 = A1 | z =0 ,

(6.6.5) (6.6.6) (6.6.7)

 141   

A1 | z =− d1 = A2con | z =− d1 ,

∂A1 ∂Acon | z =− d1 = 2 | z =− d1 , c ≤ r ≤ b, ∂z ∂z

A2cc |z = − d1 − d 2 = A3 |z = − d1 − d 2 ,

(6.6.8)

∂A ∂A2cc | z = − d 1 − d 2 = 2 | z = − d 1 − d 2 , 0 ≤ r ≤ c, ∂z ∂z

(6.6.9)

∂A2con ∂A | z =− d1−d2 = 3 | z =− d1−d2 , c ≤ r ≤ b, ∂z ∂z con cc ∂A ∂A2 A2con |r = c = A2cc |r = c , |r = c = 2 |r = c . ∂r ∂r A2con | z =− d1−d2 = A3 | z =− d1−d2 ,

(6.6.10) (6.6.11)

Here we used the abbreviations “cc” and “con” in region R2 with the reference to conducting cylinder (cast core) and homogeneous conducting region, respectively. Solution to (6.6.1) is obtained as in Section 6.1 and is given by (6.1.29) and (6.1.30) in regions R00 = {0 < z < h} and R01 = {z > h} , respectively. Using the method of separation of variables we obtain the general solution to (6.6.2) in region R1 in the form ∞

A1 (r , z ) = ∑ ( D4i e pi z + D5i e − pi z ) J1 (λi r ),

(6.6.12)

i =1

where pi = λi2 + jωσ 1µ0 . General solution to (6.6.3) can be written in the form (see the details in Section 6.1): ∞

A2cc (r , z ) = ∑ [ D10i e p1i z + D11i e − p1i z ]J 1 ( p2i r ),

(6.6.13)

i =1

∞

{

}

A2con (r , z ) = ∑ [ D6i J1 (qi r ) + D7 iY1 (qi r )]e p1i z + [ D8i J1 (qi r ) + D9iY1 (qi r )]e − p1i z .

(6.6.14)

i =1

where p1i = qi2 + jωσ 1µ0 ,

p2i = qi2 + jωσ 2 µ0 .

General solution to (6.6.4) which is bounded as z → −∞ is written as follows ∞

A3 (r , z ) = ∑ D12i e pi z J1 (λi r ).

(6.6.15)

Using (6.6.14) and the last boundary condition in (6.6.5) we obtain the following two equations D6i J 1 (qi b) + D7 iY1 (qi b) = 0,

(6.6.16)

D8i J1 (qi b) + D9iY1 (qi b) = 0.

(6.6.17)

i =1

Eliminating D7 i and D9i from (6.6.16) and (6.6.17) we obtain D7 i = − D6i

J 1 (qi b) , Y1 (qi b)

(6.6.18)

D9i = − D8i

J1 (qi b) . Y1 (qi b)

(6.6.19)

 142   

Using continuity of the functions A2cc and A2con at r = c (the first condition in (6.6.11)) we get ∞

∑[ D i =1

10 i

e p1i z + D11i e − p1i z ]J 1 ( p2i c) =

{

∞

= ∑ [ D6i J1 (qi c) + D7 iY1 (qi c)]e

p1i z

i =1

+ [ D8i J 1 (qi c) + D9iY1 (qi c)]e

− p1i z

}.

(6.6.20)

The following two relationships are obtained from (6.6.20): D10i J 1 ( p2i c) = D6i J 1 (qi c) + D7 iY1 (qi c),

(6.6.21)

D11i J 1 ( p2i c) = D8i J 1 (qi c) + D9iY1 (qi c).

(6.6.22)

Combining equations (6.6.18) and (6.6.21) we obtain D = Dˆ [ J (q c)Y (q b) − J (q b)Y (q c)], 10 i

6i

1

i

i

1

1

i

1

i

(6.6.23)

where Dˆ 6i =

D6i . J 1 ( p2i c)Y1 (qi b)

(6.6.24)

It follows from (6.6.19) and (6.6.22) that D = Dˆ [ J (q c)Y (q b) − J (q b)Y (q c)], 11i

8i

1

i

1

i

1

i

1

i

(6.6.25)

where Dˆ 8i =

D8i . J 1 ( p2i c)Y1 (qi b)

(6.6.26)

Solutions (6.6.13) and (6.6.14) can be rewritten in the following form (here we use the relationships (6.6.18), (6.6.19), (6.6.23) and (6.6.25)): ∞

A2cc (r , z ) = ∑ T1 (qi c) J 1 ( p2i r )[ Dˆ 6i e p1i z + Dˆ 8i e − p1i z ],

(6.6.27)

i =1

∞

A2con (r , z ) = ∑ T1 (qi r ) J 1 ( p2i c)[ Dˆ 6i e p1i z + Dˆ 8i e − p1i z ],

(6.6.28)

where T1 (qi r ) = J 1 (qi r )Y1 (qi b) − J 1 (qi b)Y1 (qi r ).

(6.6.29)

T1 (qi c) = J 1 (qi c)Y1 (qi b) − J 1 (qi b)Y1 (qi c)

(6.6.30)

i =1

Differentiating (6.6.27) and (6.6.28) with respect to r and evaluating the derivatives at r = c we obtain ∞ ∂A2cc |r = c = ∑ p2i J1' ( p2i c)T1 (qi c)( Dˆ 6i e p1i z + Dˆ 8i e − p1i z ), ∂r i =1

(6.6.31)

∞ ∂A2con | r = c = ∑ qiT1' (qi c) J 1 ( p2i c)( Dˆ 6i e p1i z + Dˆ 8i e − p1i z ). ∂r i =1

(6.6.32)

It follows from (6.6.31) and (6.6.32) and the second boundary condition in (6.6.11) that (6.6.33) p2i J 1' ( p2i c)T1 (qi c) = qiT1' (qi c) J 1 ( p2i c).  143   

Equation (6.6.33) is used to determine the eigenvalues p2i and related values qi . The relationship between p2i and qi is p2i = qi + jωσ 2 µ 0 .     2

Thus, the solution in regions R0 , R1 , R2 and R3 is given by (6.1.29), (6.1.30), (6.6.12), (6.6.27), (6.6.28) and (6.6.15). The six sets of constants in these formulas, namely, D , D , D , Dˆ , Dˆ 2i

4i

5i

6i

8i

and D12i can be obtained from the boundary conditions (6.6.6) – (6.6.10). Using the condition (6.6.6) (see the details in Section 6.4) we obtain µ Ir J (λ r ) −λ h D2 j = D4 j + D5 j − 02 0 12 j 0 e j , b λ j J 0 (λ j b) (λ j + p j ) D4 j + (λ j − p j ) D5 j =

2µ0 Ir0 J1 (λ j r0 )e b 2 J 02 (λ j b)

(6.6.34)

−λ j h

(6.6.35)

.

where p j = λ2j + jωσ 1µ0 .

(6.6.36)

Using the first conditions in (6.6.7) and in (6.6.8) we obtain ∞

∑ (D i =1

4i

e − pi d1 + D5i e pi d1 ) J1 (λi r ) = ∞

= ∑ T1 (qi c) J1 ( p2i r )( Dˆ 6i e − p1i d1 + Dˆ 8i e p1i d1 ), 0 ≤ r ≤ c,

(6.6.37)

i =1

∞

∑ (D i =1

4i

e − pi d1 + D5i e pi d1 ) J1 (λi r ) = ∞

= ∑ T1 (qi r ) J1 ( p2i c)( Dˆ 6i e i =1

(6.6.38) − p1i d1

+ Dˆ 8i e

p1i d1

), c ≤ r ≤ b,

In order to determine the coefficients Dˆ 6i and Dˆ 8i the following procedure is used. First, equations (6.6.37) and (6.6.38) are combined into one equation where the right-hand side of the resulting equation is given by different expressions on the intervals 0 ≤ r ≤ c and c ≤ r ≤ b . These expressions are defined by the right-hand sides of (6.6.37) and (6.6.38), respectively. Second, the obtained equation is multiplied by rJ 1 (λ j r ) and the resulting equation is integrated with respect to r from 0 to b . Third, we use the orthogonality condition (6.1.25) and formulas (6.6.29), (6.6.30), (6.1.62) and (6.1.63). The result is ∞ 2 ∑ ( Dˆ 6ie− p1i d1 + Dˆ 8ie p1i d1 )a ji , b 2 J 02 (λ j b) i =1 ~ where a ji = T1 (qi c)a~ ji + J1 ( p2i c)a~ ji .

D4 j e

− p j d1

+ D5 j e

p j d1

=

(6.6.39) (6.6.40)

Thus, using (6.1.64), (6.1.65) and (6.6.29) we obtain

 144   

c

⋅ a~ ji = ∫ rJ1 (λ j r )J1 ( p2i r )dr = 0

c ( λ J (λ c )J ( p c ) − p2i J1 (λ j c )J 2 ( p2ic ) ) λ − p22i j 2 j 1 2i 2 j

b b ~ ⋅ a~ ji = ∫ rJ1 (λ j r )T1 (qi r )dr = Y1 (qib )∫ rJ1 (λ j r )J1 (qi r )dr − J1 (qib )∫ rJ1 (λ j r )Y1 (qi r )dr = b

c

c

c

⎧bqi J1 (λ j b )[ J1 (qib )Y2 (qib ) − J 2 (qib )Y1 (qib ) ] + ⎫ ⎪ 1 ⎪ = 2 + cλ j J 2 (λ j c )[ J1 (qib )Y1 (qi c ) − J1 (qi c )Y1 (qib ) ] + ⎬ 2 ⎨ λ j − qi ⎪ ⎪ ⎩+ cqi J1 (λ j c )[ J 2 (qi c )Y1 (qib ) − J1 (qib )Y2 (qi c ) ] ⎭ Using the same procedure and applying the second condition in (6.6.7) and in (6.6.8) we obtain ∞ 2 −p d p d D4 j e j 1 − D5 j e j 1 p j = 2 2 p1i ( Dˆ 6i e − p1i d1 − Dˆ 8i e p1i d1 )a ji , (6.6.41) ∑ b J 0 (λ j b) i =1

(

)

Two additional equations are obtained if the same procedure is applied to (6.6.27), (6.6.28) and (6.6.15) using boundary conditions (6.6.9) and (6.6.10). The result is shown below 2 2 ∞ − p d b J 0 (λ j b ) D12 j e j 3 = ∑ ( Dˆ 6i e − p1i d 3 + Dˆ 8i e p1i d 3 )a ji , (6.6.42) 2 i =1 D12 j p j e

− p j d3

b 2 J 02 (λ j b) 2

∞

= ∑ p1i ( Dˆ 6i e − p1i d 3 − Dˆ 8i e p1i d 3 )a ji ,

(6.6.43)

i =1

where d 3 = d1 + d 2 . Multiplying (6.6.39) by p j and adding with (6.6.41) we obtain p d

∞ e j1 D4 j = ∑ 2 2 p j b J 0 (λ j b) i =1

{( ( p

j

) }

+ p1i )e − p1i d1 Dˆ 6i + ( p j − p1i )e p1i d1 Dˆ 8i a ji .

(6.6.44)

Multiplying (6.6.39) by (− p j ) and adding with (6.6.41) we obtain −p d

D5 j =

∞ e j1 ∑ p j b 2 J 02 (λ j b) i =1

{( ( p

j

) }

− p1i )e − p1i d1 Dˆ 6i + ( p j + p1i )e p1i d1 Dˆ 8i a ji .

(6.6.45)

Thus, using (6.6.44), (6.6.45) and (6.6.35)we obtain

(

)

⎛ (λ j + p j )( p j + p1i )e ( p j − p1i )d1 + (λ j − p j )( p j − p1i )e − ( p j + p1i )d1 a ji Dˆ 6i + ⎞ ⎟= ⎜ ∑ ( p j + p1i )d1 − ( p j − p1i )d1 ⎜ ˆ i =1 a ji D8i ⎟⎠ + (λ j − p j )( p j + p1i )e ⎝ + (λ j + p j )( p j − p1i )e ∞

(

= 2 p j µ0 Ir0 J1 (λ j r0 )e

)

(6.6.46)

−λ j h

Multiplying (6.6.42) by (− p j ) and adding with (6.6.43) we obtain

 145   

∑ ((p

)

∞

i =1

− p1i )e − p1i d 3 a ji Dˆ 6i + ( p j + p1i )e p1i d 3 a ji Dˆ 8i = 0

j

(6.6.47)

The six sets of equations for the unknown coefficients D2 j , D4 j , D5 j , Dˆ 6i , Dˆ 8i and D12 j are given by (6.6.34), (6.6.44), (6.6.45), (6.6.46), (6.6.47) and (6.6.42). The solution of the system is obtained as follows. First, we consider only a finite number of terms n in (6.6.34), (6.6.44), (6.6.45), (6.6.46), (6.6.47) and (6.6.42). Recommendations on the selection of the value of n are given in Section 6.3. Second, the coefficients Dˆ 6i , Dˆ 8i can be computed solving the system

(

)

⎛ (λ j + p j )( p j + p1i )e ( p j − p1i )d1 + (λ j − p j )( p j − p1i )e − ( p j + p1i )d1 a ji Dˆ 6i + ⎞ ⎜ ⎟= ∑ ( p j + p1i )d1 − ( p j − p1i )d1 ⎜ i =1 a ji Dˆ 8i ⎟⎠ + (λ j − p j )( p j + p1i )e ⎝ + (λ j + p j )( p j − p1i )e n

(

= 2 p j µ0 Ir0 J1 (λ j r0 )e n

∑[( p i =1

j

)

(6.6.48)

−λ j h

− p1i )e − p1i d 3 a ji Dˆ 6i + ( p j + p1i )e p1i d 3 a ji Dˆ 8i ] = 0,

(6.6.49)

where n is the number of eigenvalues p2 i . The system (6.6.48)-(6.6.49) can be written in the matrix form r r AX = B                                                                                                                                                (6.6.50) where the coefficient matrix A is ⎛ A11 A12 ⎞ ⎟⎟                                                                                                                                      (6.6.51) A = ⎜⎜ A A ⎝ 21 22 ⎠ and the block matrices A11 , A12 , A21 and A22 are ⎛ (λ1 + p1 ) p1 + p1 e ( p1 − p11 )d1 + (λ1 − p1 ) p1 − p1 e − ( p1 + p11 )d1 a11 ............................. ⎞ 1 1 ⎟ ⎜ ⎜ ... (λ + p ) p + p e( p 2 − p12 )d1 + (λ − p ) p − p e − ( p 2 + p12 )d1 a ........................ ⎟ 2 2 2 12 2 2 2 12 22 ⎟ A11 = ⎜ ⎟ ⎜ ............................................................................................ ⎟ ⎜ ⎜ .......................... (λn + pn ) pn + p1 e( p n − p1n )d1 + (λn − pn ) pn − p1 e − ( p n + p1n )d1 ann ⎟ n n ⎠ ⎝   ⎛ (λ1 + p1 ) p1 − p1 e( p1 + p11 )d1 + (λ1 − p1 ) p1 + p1 e − ( p1 − p11 )d1 a11 ............................. ⎞ 1 1 ⎟ ⎜ ⎜ ... (λ + p ) p − p e ( p 2 + p12 )d1 + (λ − p ) p + p e − ( p 2 − p12 )d1 a ........................ ⎟ 2 2 2 12 2 2 2 12 22 ⎟  A12 = ⎜ ⎟ ⎜ ............................................................................................ ⎟ ⎜ ⎜ .......................... (λn + pn ) pn − p1 e ( p n + p1n )d1 + (λn − pn ) pn + p1 e − ( p n − p1n )d1 ann ⎟ n n ⎠ ⎝  

(

(

(

)

(

)

(

(

(

(

(

)

(

(

(

)

)

(

)

)

(

(

)

(

)

)

(

)

)

)

) )

( (

) )

( (

) )

(

)

(

)

(

)

)

)

)

)

(

( (

)

⎛ p1 − p1 e − p11 d3 a11 p1 − p1 e − p12 d3 a12 ......... p1 − p1 e − p1n d3 a1n ⎞ 1 2 n ⎜ ⎟ − p1n d3 − p12 d3 ⎜ p − p e − p11d3 a p2 − p12 e a22 ........ p2 − p1n e a2 n ⎟⎟ 2 11 21 ⎜ A21 =   ⎜ ⎟ ................................................ ⎜ ⎟ ⎜ pn − p1 e − p11d3 an1 pn − p1 e − p12 d3 an 2 ........ pn − p1 e − p1n d3 ann ⎟ 1 2 n ⎠ ⎝  146   

( (

) )

( (

) )

( (

) )

(

)

(

)

(

)

⎛ p1 + p1 e p11d3 a11 p1 + p1 e p12 d3 a12 ......... p1 + p1 e p1n d3 a1n ⎞ 1 2 n ⎜ ⎟ p1n d3 p12 d3 ⎜ p + p e p11d3 a p2 + p12 e a22 ........ p2 + p1n e a2 n ⎟⎟ 2 11 21 ⎜ A22 =    ⎜ ⎟ ................................................ ⎜ ⎟ ⎜ pn + p1 e p11d3 an1 pn + p1 e p12 d3 an 2 ........ pn + p1 e p1n d3 ann ⎟ 1 2 n ⎝ ⎠ r r The matrices X and B in (6.6.50) are r r r ⎛ X1 ⎞ r ⎛ b ⎞ X = ⎜ r ⎟  , B = ⎜⎜ ⎟⎟ , ⎜X ⎟ ⎝0 ⎠ ⎝ 2⎠ where ⎛ Dˆ 8 ⎞ ⎛ Dˆ 6 ⎞ ⎛ 2 p1µ0 Ir0 J1 (λ1r0 )e − λ1h ⎞ ⎜ 1⎟ ⎜ 1⎟ ⎜ ⎟ r ⎜ Dˆ 6 ⎟ r ⎜ Dˆ 8 ⎟ r ⎜ 2 p2 µ0 Ir0 J1 (λ2 r0 )e −λ2h ⎟ X1 = ⎜ 2 ⎟ , X 2 = ⎜ 2 ⎟ , b = ⎜ ⎟. ⎜ ... ⎟ ⎜ ... ⎟ ............... ⎜ ⎟ ⎟ ⎟ ⎜ ⎜ −λn h ⎟ ⎜ ⎜ Dˆ 8 ⎟ ⎜ Dˆ 6 ⎟ ⎝ 2 pn µ0 Ir0 J1 (λn r0 )e ⎠ ⎝ n⎠ ⎝ n⎠

(6.6.52)

(6.6.53)

Hence, (6.6.50) can be written in the form    r r r ⎧⎪ A11 X 1 + A12 X 2 = b                                                                                                                            (6.6.54) r r ⎨ ⎪⎩ A21 X 1 + A22 X 2 = 0 Solving (6.6.54) we obtain all the coefficients Dˆ 6i and Dˆ 8i (i = 1,2,..., n). Then D4 j and D5 j ( j = 1,2,.., n) is calculated using ( p + p )d ⎧⎪ ⎛ ( p j + p1i )e( p j − p1i )d1 a ji ( p j − p1i )e j 1i 1 a ji ˆ ⎞⎟⎫⎪ ˆ ⎜ D4 j = ∑ ⎨ D6i + D8i ⎬ , (6.6.55) ⎜ ⎟⎪ p j b 2 J 02 (λ j b) p j b 2 J 02 (λ j b) i =1 ⎪ ⎝ ⎠ ⎩ ⎭ ( ) ( ) − + − − p p d p p d j 1i 1 n ⎧ ⎛ ( p − p )e ( p j + p1i )e j 1i 1 a ji Dˆ ⎞⎟⎫⎪. a ji ˆ ⎪ 1i + D5 j = ∑ ⎨ ⎜ j D (6.6.56) 6i 8i ⎬ ⎜ ⎟⎪ p j b 2 J 02 (λ j b) p j b 2 J 02 (λ j b) i =1 ⎪ ⎝ ⎠⎭ ⎩ The system (6.6.55)-(6.6.57) can be written in the matrix form r r r ⎧⎪ D4 = B11 X 1 + B12 X 2 r r                                                                                                                          (6.6.57) ⎨r ⎪⎩ D5 = B21 X 1 + B22 X 2 The matrices B11 , B12 , B21 and B22 are ( p − p )d ( p − p )d ⎛ p1 + p1 e ( p1 − p11 )d1 a11 p1 + p1n e 1 1n 1 a1n ⎞⎟ p1 + p12 e 1 12 1 a12 1 ⎜ ... ⎟ ⎜ p1b 2 J 02 (λ1b ) p1b 2 J 02 (λ1b ) p1b 2 J 02 (λ1b ) ⎟ ⎜ ( p − p )d ( p − p )d ( p − p )d p2 + p1n e 2 1n 1 a2 n ⎟ p2 + p12 e 2 12 1 a22 ⎜ p2 + p11 e 2 11 1 a21 ... ⎟               (6.6.58)  2 2 2 2 B11 = ⎜ ( ) ( ) p b J b p b J b p2b 2 J 02 (λ2b ) λ λ 2 0 2 2 0 2 ⎟ ⎜ ⎟ ⎜ ................................................ ⎟ ⎜ ( p − p )d ( p − p )d ( p − p )d pn + p1n e n 1n 1 ann ⎟ pn + p12 e n 12 1 an 2 ⎜ pn + p11 e n 11 1 an1 ... ⎟ ⎜ pnb 2 J 02 (λnb ) pnb 2 J 02 (λnb ) pnb 2 J 02 (λnb ) ⎠ ⎝ n

(

)

)

(

)

(

(

)

(

)

(

)

(

)

(

)

(

)

 147   

(

)

)

(

)

(

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

( p + p )d ( p + p )d ⎛ p1 − p1 e( p1 + p11 )d1 a11 p1 − p1n e 1 1n 1 a1n ⎞⎟ p1 − p12 e 1 12 1 a12 1 ⎜ ... ⎟ ⎜ p1b 2 J 02 (λ1b ) p1b 2 J 02 (λ1b ) p1b 2 J 02 (λ1b ) ⎟ ⎜ ( p + p )d ( p + p )d ( p + p )d p2 − p1n e 2 1n 1 a2 n ⎟ p2 − p12 e 2 12 1 a22 ⎜ p2 − p11 e 2 11 1 a21 ... ⎟               (6.6.59) B12 = ⎜ p2b 2 J 02 (λ2b ) p2b 2 J 02 (λ2b ) p2b 2 J 02 (λ2b ) ⎟ ⎜ ⎟ ⎜ ................................................ ⎟ ⎜ ( p + p )d ( p + p )d ( p + p )d pn − p1n e n 1n 1 ann ⎟ pn − p12 e n 12 1 an 2 ⎜ pn − p11 e n 11 1 an1 ... ⎟ ⎜ pnb 2 J 02 (λnb ) pnb 2 J 02 (λnb ) pnb 2 J 02 (λnb ) ⎠ ⎝  

 

− ( p + p )d − ( p + p )d ⎛ p1 − p1 e − ( p1 + p11 )d1 a11 p1 − p1n e 1 1n 1 a1n ⎞⎟ p1 − p12 e 1 12 1 a12 1 ⎜ ... ⎟ ⎜ p1b 2 J 02 (λ1b ) p1b 2 J 02 (λ1b ) p1b 2 J 02 (λ1b ) ⎟ ⎜ − ( p + p )d − ( p + p )d − ( p + p )d p2 − p1n e 2 1n 1 a2 n ⎟ p2 − p12 e 2 12 1 a22 ⎜ p2 − p11 e 2 11 1 a21 ... ⎟           (6.6.60) B21 = ⎜ p2b 2 J 02 (λ2b ) p2b 2 J 02 (λ2b ) p2b 2 J 02 (λ2b ) ⎟ ⎜ ⎟ ⎜ ................................................ ⎟ ⎜ − ( p + p )d − ( p + p )d − ( p + p )d pn − p1n e n 1n 1 ann ⎟ pn − p12 e n 12 1 an 2 ⎜ pn − p11 e n 11 1 an1 ... ⎟ ⎜ pnb 2 J 02 (λnb ) pnb 2 J 02 (λnb ) pnb 2 J 02 (λnb ) ⎠ ⎝

(

)

− ( p − p )d − ( p − p )d ⎛ p1 + p1 e − ( p1 − p11 )d1 a11 p1 + p1n e 1 1n 1 a1n ⎞⎟ p1 + p12 e 1 12 1 a12 1 ⎜ ... ⎟ ⎜ p1b 2 J 02 (λ1b ) p1b 2 J 02 (λ1b ) p1b 2 J 02 (λ1b ) ⎟ ⎜ − ( p − p )d − ( p − p )d − ( p − p )d p2 + p1n e 2 1n 1 a2 n ⎟ p2 + p12 e 2 12 1 a22 ⎜ p2 + p11 e 2 11 1 a21 ... ⎟           (6.6.61) B22 = ⎜ p2b 2 J 02 (λ2b ) p2b 2 J 02 (λ2b ) p2b 2 J 02 (λ2b ) ⎟ ⎜ ⎟ ⎜ ................................................ ⎟ ⎜ − ( p − p )d − ( p − p )d − ( p − p )d pn + p1n e n 1n 1 ann ⎟ pn + p12 e n 12 1 an 2 ⎜ pn + p11 e n 11 1 an1 ... ⎟ ⎜ pnb 2 J 02 (λnb ) pnb 2 J 02 (λnb ) pnb 2 J 02 (λnb ) ⎠ ⎝ Solving (6.6.57) we obtain all the coefficients Dˆ 4 j and Dˆ 5 j ( j = 1,2,..., n).

Then D2 j is calculated using D2 j = D4 j + D5 j − µ0 Ir0

J1 (λ j r0 )e

−λ j h

λ j b 2 J 02 (λ j b)

.

(6.6.62)

The induced change in impedance of the coil is given by the formula        jω Z ind = 2πr0 A0ind ( r0 , h)                                                                                                                    (6.6.63)                      I n

where A0ind (r0 , h) = ∑ D2 j e j =1

−λ j h

J1 (λ j r0 )                                                                                            (6.6.64)

Formula (6.6.63) is used to compute the change in impedance of the coil. Calculations are performed with “Mathematica”. The program which is used to compute the change in impedance (6.6.63) is shown in Appendix Fig. Ap.19. The following parameters of the problem are selected: µ0 = 4 ⋅ 10−7 π , σ 1 = 18.5 Ms/m, σ 2 = 3 Ms/m, c = 2.2 mm,  r0 = 4.5 mm, h = 1.4 mm,  148   

d1 = 0.7 mm, d 2 = 0.3 mm, b = 55 mm. The change in impedance is computed for the following seven frequencies:  f = 1000, 2000, 3000, 4000, 5000, 6000, and 7000 Hz. The results of calculations are shown in Fig. 6.14.The calculated points (from top to bottom) correspond to the seven frequencies (from smallest to largest). The upper limit of the summation index in (6.6.64) is fixed at n = 62 . Comparison of the computational results obtained for other values of n showed that the chosen value of 62 is quite satisfactory in terms of calculation accuracy. More detailed analysis of convergence is presented in Section 6.3. Several computational steps are necessary in order to calculate the induced change in impedance. First, the set of eigenvalues λi has to be calculated. This can easily be done in Mathematica using a built-in routine BesselZeros. Second, a set of complex roots of (6.6.33) should be computed. Calculations are based on the method described in Section 6.3. Third, several systems of linear equations have to be solved in order to determine expansion coefficients. Finally, the change in impedance is computed using (6.6.50) - (6.6.64). Im @zD

Re @zD

0.00001 0.00002 0.00003 0.00004 -0.00002 -0.00004 -0.00006

Fig. 6.14 The change in impedance of the coil for seven frequencies.

-0.00008

6.7 A coil of finite dimensions above a conducting half-space with a flaw in the form of a circular cylinder Consider the model shown in Fig. 6.15. The inner and outer radii of the coil are r1 and r2 , respectively. The bottom of the coil is located at the distance z1 from the conducting plate. The height of the coil is z 2 − z1 and the number of turns is N .

r2   r1   R0  

R1   R2  

z2  

z1  

O  d1

c 

d2

R3   Fig. 6.15 A coil of finite dimensions above a conducting plate.

b   149   

It can easily be shown that the second term in (6.1.29) and (6.1.30) represents the vector potential of a single-turn coil located in an unbounded free space. In addition, the first term in (6.1.29) and (6.1.30) is the induced vector potential in air due to the presence of a conducting cylinder: n

A0ind (r , z , r0 , h) = ∑ D2 j e

−λ j z

j =1

J1 (λ j r ),

(6.7.1)

where D2 j = D4 j + D5 j − µ0 Ir0

J1 (λ j r0 )e

−λ j h

λ j b 2 J 02 (λ j b)

.

The induced vector potential in air due to currents in the whole coil is obtained as follows r2 z 2

A0indcoil (r , z ) = ∫ ∫ A0ind (r , z , r0 , h)dr0 dh .

(6.7.2)

r1 z1

There is an important difference between the calculations for a single-turn coil and coil of finite dimensions. For the case of a single-turn coil (Section 6.6) the geometrical parameters of the coil ( r0 and h ) are constants and the numerical values of r0 and h can be used to solve the system (6.6.57). If a coil has finite dimensions then one needs to integrate the solution with respect to r0 and h . Thus, numerical values to the parameters r0 and h in (6.6.57) cannot be assigned. In this case, in order to obtain the induced vector potential of a single-turn coil we need to solve (6.6.57) in matrix form. Solving the second equation in (6.6.57) we obtain r r (6.7.3) X 2 = − A22−1 A21 X 1. Substituting (6.7.3) into the first equation in (6.6.57) gives r r (A11 − A12 A22−1 A21 )X 1 = b .

(6.7.4)

It follows from (6.7.3) and (6.7.4) that r r ⎧⎪ X = (A − A A−1 A )−1 b 1 11 12 22 21 ⎨r −1 r ⎪⎩ X 2 = − A22−1 A21 (A11 − A12 A22−1 A21 ) b

(6.7.5)

Using (6.6.57) - (6.6.61) equation (6.6.62) should be written in matrix form: r r D2 = Y b , where

(

)(

Y = B11 + B21 − (B12 + B22 )A22−1 A21 A11 − A12 A22−1 A21

⎛ D21 ⎞ ⎜ ⎟ r ⎜ D2 2 ⎟ D2 = ⎜ ⎟, ... ⎜ ⎟ ⎜D ⎟ ⎝ 2n ⎠

Cdiag

1 ⎛ ⎞ 0 0 0 ... 0 ⎟ ⎜ 2 2 ⎜ 2 p1λ1b J 0 (λ1b ) ⎟ ⎜ ⎟ 1 0 0 ... 0 ⎜0 ⎟ = ⎜ 2 p2λ2b 2 J 02 (λ2b ) ⎟ ⎜ ⎟ ....................... ⎜ ⎟ 1 ⎜ ⎟ ⎜ 0 0 ... 0 0 2 p λ b 2 J 2 (λ b ) ⎟ n n n 0 ⎝ ⎠

)

−1

(6.7.6)

− Cdiag                                            (6.7.7)

(6.7.8)

r and b is given by (6.6.53).

 150   

Formula (6.7.1) for the induced vector potential of a single-turn coil can be rewritten in the form r r A0ind (r , z , r0 , h) = D2T f , (6.7.9) where ⎛ J1 (λ1r )e − λ1 z ⎞ ⎟ ⎜ r ⎜ J1 (λ2 r )e − λ2 z ⎟ f =⎜ ⎟ .  ⎟ ⎜ ... ⎜ J ( λ r )e − λ n z ⎟ ⎠ ⎝ 1 n Substituting (6.7.9) into (6.7.2) and using the formulas z2

⋅ ∫e

−λ j h

dh = −

z1

1

λj

(e

−λ j z2

−e

− λ j z1

r2

⋅ ∫ r0 J1 (λ j r0 )dr0 = ξ = λ j r0 = r1

) (6.7.10)

λ j r2

1

λi2

∫ ξJ (ξ )dξ 1

λ j r1

we obtain the induced vector potential in air due to the presence of the conducting cylinder ( the current amplitude I in this case is replaced by the current NI density ): (r2 − r1 )( z 2 − z1 ) ind 0 coil

A

(r , z ) =

µ0 NI (r2 − r1 )( z2 − z1 )

n

∑ f ∑Y j =1

(e

n

j

i =1

− λi z1

− e − λi z 2

λ3i

ji

)

λi r2

∫ ξJ (ξ )dξ .

(6.7.11)

1

λi r1

The integral with respect to ξ in (6.7.10) can be computed in terms of the Bessel and Struve functions [2] as follows ξ = λi r2

λi r2

⎧π ⎫ ∫λ rξJ1 (ξ )dξ = ⎨⎩ 2 ξ [J 0 (ξ ) H1 (ξ ) − J1 (ξ ) H 0 (ξ )]⎬⎭ ξ = λi r1 i 1

(6.7.12)

The induced change in impedance of a coil of finite dimensions (see Fig. 6.15) is calculated by means of the following formula [15]: r2 z2 2π rr jω jω ind Z = 2 ∫∫∫ AI dv = 2 ∫ dϕ ∫ rdr ∫ A0indcoil dz I V I 0 r1 z1 r z

Z

ind

2 2 2πjω N rA0indcoil (r , z )drdz. = I (r2 − r1 )( z 2 − z1 ) ∫r1 z∫1

(6.7.13)

Using (6.7.11) and (6.7.13) we obtain the induced change in impedance of the coil in the form Z

ind

(

n 2 jωπµ0 N 2 e = 2 2 ∑ (r2 − r1 ) ( z2 − z1 ) j =1

− λ j z1

−e

λ3j

−λ j z2

)

λ j r2

∫ ξJ (ξ )dξ ∑ Y 1

λ j r1

(e

n

i =1

ji

− λ i z1

− e − λi z 2

λ3i

)

λi r2

∫ ξJ (ξ )dξ 1

(6.7.14)

λi r1

Formula (6.7.14) is used to compute the change in impedance of the coil. Calculations are performed with “Mathematica”. Program which is used to compute the change in impedance (6.7.14) is shown in Appendix Fig. Ap.20. The following parameters of the problem are  151   

selected: µ0 = 4 ⋅ 10−7 π , σ 1 = 18.5 Ms/m, σ 2 = 3 Ms/m, c = 2.2 mm, r2 = 5.5 mm, r1 = 3.5 mm , z1 = 0.3 mm, z2 = 2.6 mm, d1 = 0.7 mm, d 2 = 0.3 mm, b = 55 mm, N = 200 . The change in impedance is computed for the following seven frequencies: f = 1000, 2000, 3000, 4000, 5000, 6000, and 7000 Hz.The results of calculations are shown in Fig. 6.16.The calculated points (from top to bottom) correspond to the seven frequencies (from smallest to largest). The upper limit of the summation index in (6.7.14) is fixed at n = 62 . Comparison of the computational results obtained for other values of n showed that the chosen value of 62 is quite satisfactory in terms of calculation accuracy. Several computational steps are necessary in order to calculate the induced change in impedance. First, the set of eigenvalues λi has to be calculated. This can easily be done in “Mathematica” using a built-in routine BesselZeros. Second, a set of complex roots of (6.6.33) should be computed. Calculations are based on the method described in Section 6.3. Third, several systems of linear equations have to be solved in order to determine expansion coefficients. Finally, the change in impedance is computed using (6.7.14). Details of the numerical aspects of the procedure are given in Section 6.3. Im @zD

0.25

0.5

0.75

1

1.25

1.5

1.75

Re @zD

-0.5 -1 -1.5 -2 -2.5 -3 -3.5

Fig. 6.16 The change in impedance of the coil for seven frequencies.

 

 152   

CONCLUSIONS The thesis is devoted to the development and analysis of mathematical models of eddy current testing problems. Eddy current method is widely used in practice in order to control properties of conducting materials. The idea of the method is based on the principle of electromagnetic induction and can be explain as follows. Suppose that a coil with alternating current is located above an electrically conducting medium being tested. The currents in the coil induce eddy currents (or Foucault currents) in the conducting medium. The eddy currents, in turn, interact with the currents in the coil and change the impedance of the coil. By monitoring the change in impedance of the coil one can make conclusions with respect to the quality of the conducting medium. In particular, parameters of the medium (such as the electrical conductivity and magnetic permeability) can be estimated. In addition, if the medium contains defects (or flaws) then eddy current method can be used to estimate the parameters of the flaw (geometrical size or electrical conductivity). The determination of unknown parameters of the conducting medium requires the solution of the inverse problem. Usually in such cases theoretical model is compared with experimental data and the difference between theoretical model and experimental data is minimized (in some norm). Thus, in order to solve the inverse problem one needs to have an accurate and reliable mathematical model for the solution of the direct problem. The thesis is devoted to the development and analysis of direct problems that occur in eddy current testing. Two major groups of problems are analyzed in the thesis. The first group deals with the case where a coil with alternating current is located in the vicinity of a conducting medium. It is assumed that the properties of the conducting medium are not constant but vary with respect to one spatial coordinate. Such a situation often occurs in applications. One important example is metal processing (such as surface hardening) at metallurgical plants. It is known that surface hardening can produce a thin upper layer of reduced magnetic permeability. In this case models based on the assumption of constant properties of the conducting material (for example, electrical conductivity) do not work. Another example is related to the electric industry where the thermal efficiency of gas turbines can be considerably improved by increasing firing temperatures. In this case blades of gas turbines have to be protected from hot corrosion and high temperature oxidation. This is achieved using protective coatings containing aluminium and chrome. However, operation under extreme conditions may lead to loss of aluminium in the coating and may result in blade failure. Experiments show that in this case the electrical conductivity of the alloy varies with depth. Thus, the two above mentioned problems demonstrate the necessity of developing mathematical models where the electric conductivity and magnetic permeability are functions of one spatial coordinate. Two different geometrical configurations are considered in the thesis: (a) a coil with alternating current located above a conducting multilayer medium where the electrical conductivity and magnetic permeability of the upper layer are exponential functions of the vertical coordinate and (b) a coil inside or outside a conducting multilayer tube where the conducting layer closest to the coil has varying electrical conductivity and magnetic permeability (in particular, the electrical conductivity and magnetic permeability are power functions of the radial coordinate). Two different types of excitation coils are considered in part (a): a circular coil of finite cross-section and a double conductor line consisting of two infinitely long parallel wires. Analytical solutions of the above mentioned problems are constructed in the thesis. The methods of integral transforms (such as Hankel or Fourier integral transforms) are used to reduce the problem to the system of ordinary differential equations with variable coefficients. The solutions of the corresponding ordinary differential equations are found in  153   

closed form in terms of different special functions such as Bessel functions or confluent hypergeometric functions. The change in impedance of the coil is computed in closed form in terms of improper integrals containing special functions. Thus, solution of direct eddy current testing problems for media with varying electrical conductivity and magnetic permeability is obtained in closed form. There is another advantage of using analytical solutions for media with varying properties. These solutions can be used as benchmark for more complicated problem where the solution can be found by numerical methods such as finite element method. Often analytical solutions are used to test numerical algorithms developed for more complicated cases where both electrical conductivity and magnetic permeability are functions of several spatial coordinates. In Chapter 5 eddy current problems analyzed in Chapter 2 (multilayer planar conducting medium) are generalized for the case where the conducting medium is moving in the horizontal direction with constant velocity. Analytical solutions are found for the case where the electrical conductivity and magnetic permeability of the upper layer are exponential functions of the vertical coordinate and the layer is moving in the horizontal direction with constant velocity. The change in impedance is found in closed form in terms of double integrals. The second group of problems analyzed in the thesis is related to the solution of axisymmetric problems for the case of a circular coil located above a conducting medium with constant properties containing a cylinder of finite size whose axis coincides with the axis of the coil. Three problems are considered in the thesis: (i) a circular coil above a conducting cylinder of finite radius and height, (ii) a circular coil above an infinite conducting plate with a bottom cylindrical hole and (iii) a circular coil above a conducting half-space with a flaw in the form of a circular cylinder of finite height. All three problems have important practical applications. Problem in part (i) is relevant to coin validators. One of the principles which is used in coin validators is based on the comparison of the electrical conductivity of a conducting sample inserted into the validator (“fake coin”) with the electrical conductivity of real coin (such as Euro coin, for example). The model analyzed in part (i) can be used to compute the change in impedance of a coil in cases where the size of the coil is comparable to the size of a conducting sample (such as coin). Problem described in part (ii) can be used in order to estimate the effect of corrosion of metal plates (in this case the parameter of interest is the height of the cylindrical hole). Finally, problem analyzed in part (iii) can be used to test the quality of spot welding. In spot welding, metal pieces are welded only at separate points (not along the whole surface). A cast core is formed as the result of welding process. The properties of the cast core are of great interest to engineers since the quality of welding process can be directly assessed by analyzing the size of the core and its electrical conductivity. Solution developed in the thesis can be served as the solution for the direct problem which is used to model spot welding and assess its quality. Mathematical basis for the second group of problems is so-called TREE method (“TRuncated Eigenfunction Expansions”) suggested by Prof. T. Theodoulidis. In any “classical” eddy current problem the vector potential approaches zero as the distance from the coil tends to infinity (assuming that there are no other sources of current except the current in the coil). “Classical” eddy current problems are usually solved by means of integral transform method (such as Fourier or Hankel integral transform). The idea of the TREE method is that one assumes that the vector potential is exactly zero at a sufficiently large distance from the coil. As a result, the solution is sought in finite domain and the corresponding eddy current problem can be solved by the method of separation of variables. Thus, the solution (for example, the change in impedance) is constructed in terms of series. There are several numerical aspects of the procedure which need to be taken care of in the TREE method. First, the distance b (usually the radial distance) from the  154   

coil where the vector potential is zero should be specified. The choice of b depends on the accuracy desired and can be controlled by the user. Second, a set of eigenvalues (in many cases, complex eigenvalues) should be computed. As a result, one needs a fast and reliable procedure for calculation of eigenvalues. In addition, nothing is usually known about initial guesses for the roots. Thus, Newton’s method for the calculation of eigenvalues cannot be used. An algorithm is implemented in the thesis which is based on theory of complex variable. Using well-known argument principle we can compute the number of zeros of an analytic function inside a contour of integration (usually rectangle with sides parallel to the coordinate axes). If the number of roots inside the contour is larger than one then a smaller contours of integration are considered. Finally, if there is only one root of an analytic function inside the contour, then explicit formula (in terms of contour integral) can be used to compute the root with any desired accuracy. A computer program in “Mathematica” is developed in the thesis which can be used to locate and compute the roots in case where the number of roots inside the contour is at most two. Third, constants of integration are obtained solving the corresponding linear system of equations. Finally, the change in impedance of the coil is computed in the form of a series where the number of terms in the series is equal to the number of complex eigenvalues (this number can be adjusted, that is, increased, if higher accuracy is desired). As one can see, the TREE method can be called quasianalytical since the solution is obtained in the form of a series expansion but there are some elements of the procedure which require the use of numerical methods (computation of complex eigenvalues and solution of the system of linear equations). The author believes that the TREE method has high potential for solving eddy current problems since the class of problems which can be solved by the TREE method is much wider than the class of problems which can be solved by the method of integral transforms.          

 155   

REFERENCES [1] I.G. Ablamunets, and V.E. Shaternikov, Parameters of eddy-current transducers placed above a moving plate, The Soviet Journal of Nondestructive Testing, Vol. 13, No. 1, pp. 7-12, 1977. [2] M. Abramowitz, and I.A. Stegun, Handbook of mathematical functions with formulas, graphs and mathematical tables. New York: Wiley, 1972. [3] M.Ya. Antimirov, A.A. Kolyshkin, and R. Vaillancourt, Mathematical models for eddy current testing. Montreal: CRM, 1997. [4] H. Bateman and, A. Erdelyi , Higher transcendental functions. New York: Mcgraw-Hill, 1953. [5] R. Becker, K. Betzold, K. D. Boness, R. Collins, C.C. Holt, and J. Simkin, The modeling of electrical current NDT methods. Its application to weld testing (Part 1), British Journal of NDT, Vol. 28, No. 5, pp. 286-294, 1986. [6] R. Becker, K. Betzold, K. D. Boness, R. Collins, C.C. Holt, and J. Simkin, The modeling of electrical current NDT methods. Its application to weld testing (Part 2), British Journal of NDT, Vol. 28, No. 6, pp. 361-370, 1986. [7] J. Blitz, Electrical and magnetic methods of nondestructive testing. New York: Adam Hilger, 1991. [8] M. P. Blodgett, and P. B. Nagy, Eddy current assessment of near-surface residual stress in shot-peened Nickel-based alloys, Journal of Nondestructive Evaluation, vol. 23, No. 3, pp. 107–123, 2004. [9]     J. Bowler, and M. Johnson, Pulsed eddy-current response to a conducting half-space, IEEE Transactions on Magnetics, vol. 33, pp. 2258–2264, 2006. [10] V. S. Cecco, G. van Drunen, and F. L. Sharp, Eddy current manual, vol. 1 and 2. Chalk River, Ontario: Chalk River Nuclear Laboratories, 1983-1984. [11]    A.F. Chub, N.L. Bondarenko, and A.P. Stipura, Nonstationary electromagnetic field of a coil located above a ferromagnetic half-space, Soviet Journal of Nondestructive Testing, vol. 13, pp. 328–334, 1977. [12] L.M. Delves, and J.N. Lynness, A numerical method for locating the zeros of an analytic function, Mathematics of Computation, vol. 21, pp. 543 – 560, 1967. [13] R. J. Ditchburn, S. K. Burke, and M. Posada, Eddy current nondestructive inspection with thin spiral coils: long cracks in steel, Journal of Nondestructive Evaluation, Vol. 22, No. 2, pp. 63-77, 2003. [14] R. J. Ditchburn, and S. K. Burke, Planar rectangular spiral coils in eddy-current nondestructive inspection, NDT&E International, Vol. 38, No. 8, pp. 690-700, 2005. [15] C.V. Dodd, and W.E. Deeds, Analytical solutions to eddy-current probe-coil problems, J. Appl. Phys., vol. 39, pp. 2829–2838, 1968. [16] C.V. Dodd, C.C. Cheng, and W.E. Deeds, Induction coils coaxial with an arbitrary number of cylindrical conductors, J. Appl. Phys., vol. 45, pp. 638–647, 1974. [17] A. L. Dorofeev, A. I. Nikitin, and A. L. Rubin, Induction measurements of thickness. Moscow: Energya, 1978 (In Russian). [18] V. V. Dyakin, and V. A. Sandovskii, Theory and computation of superposed eddy current transducers. Moscow: Nauka, 1981 (In Russian). [19] V. G. Gerasimov, Electromagnetic control of one-layer and multi-layer products. Moscow: Energya, 1972 (In Russian).  156   

[20] R. Gordon, O. Martens, R. Land, M. Min, M. Rist, A. Gavrijaseva, A. Pokatilov, and A. Kolyshkin, Eddy-current validation of euro coins, Lecture Notes on Impedance Spectroscopy: Measurement, Modeling and Applications, vol. 3, pp. 47 – 63, 2012. [21] T. Hayashi, Y. Kawasaki, H. Yamada, T. Kiwa, M. Tamazumi, and K. Tsukada, Magnetic image detection of the stainless-steel welding part inside a multi-layered tube structure, NDT & E International, vol. 42, pp. 308–315, 2009. [22] N. Ida, Numerical modeling for electromagnetic non-destructive evaluation. London: Chapman & Hall, 1995. [23] G.A. Kasimov, and Yu.V. Kulaev, An applied electromagnetic transducer above the object of inspection with the electrical and magnetic properties of the material changing with depth, Soviet Journal of Nondestructive Testing, vol. 14, pp. 550–552, 1978. [24] V. Koliskina, I. Volodko, Analytical solution of eddy current problems for multilayer medium with varying electrical conductivity and magnetic permeability, International Journal of Mathematical Models and Methods in Applied Sciences, vol. 7, no. 2, pp. 174–181, 2013. [25] V. Koliskina, I. Volodko, The change in impedance of a single-turn coil due to a cylindrical flaw in a conducting half-space, In: Recent advances in mathematical methods, intelligent systems and materials, pp. 74 – 79, WSEAS Press, 2013. [26] V. Koliskina, I. Volodko, Eddy current problem for a moving medium with varying properties, International Journal of Mathematical Models and Methods in Applied Sciences, vol. 6, no. 8, pp. 971–978, 2012. [27] V. Koliskina, I. Volodko, Double conductor line above a multilayer medium with varying electrical and magnetic properties, Proceedings of the International conference on applied mathematics and sustainable development, Xian, China, May 27 – 30, pp. 104–108, Scientific Research Publishing, 2012. [28] V. Koliskina, I. Volodko, The change in impedance of a coil located above a moving halfspace, In: Mathematical models and methods in modern science, pp. 107–112, WSEAS Press, 2012. [29] V. Koliskina, I. Volodko. Transient currents in a double conductor line above a conducting half-space, World Academy of Science, Engineering and Technology. – vol. 64, pp. 11471151, 2012. [30] V. Koliskina, I. Volodko, Analytical solution of an eddy current problem for a two-layer medium with varying electric conductivity and magnetic permeability, In: Mathematical models and methods in modern science, pp. 196–200, WSEAS Press, 2011. [31] V. Koliskina, I. Volodko, Solution of eddy current testing problems for multilayer tubes with varying properties, International Journal of Mathematical Models and Methods in Applied Sciences, vol. 5, pp. 781–788, 2011. [32] V. Koliskina, and I. Volodko, A single-turn coil with alternating current inside a cylindrical region with varying electric conductivity and magnetic permeability, In Recent researches in communications, automation, signal processing, nanotechnology, astronomy & nuclear physics, pp. 85–88, 2011. [33] V. Koliskina, and I. Volodko, Analytical solution of an eddy current problem for a twolayer tube with varying properties, In: Recent research in communications, electrical & computer engineering, pp. 262 – 265, WSEAS Press, 2011.

 157   

[34] V. Koliskina V., and I. Volodko, Analytical solution to an eddy current testing problem for a cylindrical tube with varying properties, Scientific Journal of RTU, Computer Science, vol. 46, pp. 72-75, 2011. [35] V. Koliskina, Impedance of an encircling coil due to a cylindrical tube with varying properties, World academy of science, engineering and technology, vol. 76, pp. 687 – 690, 2011. [36] V. Koliskina, and I. Volodko, The change in impedance of a double conductor line due to a two-layer medium, In: Recent research in communications & IT, pp. 228 – 232, 2011. [37] V. Koliskina. Double conductor line above a half-space with varying electric and magnetic properties, Proceedings of the 10th International conference on nonlinear systems and wavelet analysis, pp. 104 -108, 2010. [38] V. Koliskina, and I. Volodko, Double conductor line above a two-layer medium with varying electric conductivity and magnetic permeability, Scientific Journal of RTU, Computer Science, vol. 45, pp. 76-80, 2010. [39] V. Koliskina, and I. Volodko, Impedance of a coil above a half-space with varying electric and magnetic properties, Scientific Journal of RTU, Computer Science, vol. 41, pp. 42-46, 2009. [40] N. S. Koshlyakov, E. B. Gliner, and M. M. Smirnov, Basic differential equations of mathematical physics. Moscow: Fizmatgiz, 1962 (In Russian). [41] M. Lambert, F. Nouguier, and R. Zorgati, Eddy-current modeling of a continuous conductivity profile resulting from a diffusion process, IEEE Transactions on Magnetics, vol. 47, pp. 2093–2099, 2011. [42]    R. Ludwig, and X.W. Dai, Numerical and analytical modeling of pulsed eddy-currents in a conducting half-space, IEEE Tranactions on Magnetics, vol. 26, pp. 299–307, 1990. [43] J.N. Lyness, Numerical algorithms based on the theory of complex variable, Proc. ACM, pp. 125 – 133, 1967. [44] X. Ma, A. J. Peyton, and Y. Y. Zhao, Eddy current measurements of electrical conductivity and magnetic permeability of porous metals, NDT & E International, vol. 45, pp. 478–482, 1996. [45] D. Mercier, J. Lesage, X. Decoopman, and D. Chicot, Eddy currents and hardness testing for evaluation of steel decarburizing, NDT & E International, vol. 39, pp. 652–660, 2006. [46] J. C. Moulder, E. Uzal, and J. H. Rose, Thickness and conductivity of metallic layers from eddy current measurements, Review of Scientific Instruments, Vol. 63, No. 6, pp. 34553465, 1992. [47] S. M. Nair, and J. H. Rose, Reconstruction of three-dimensional conductivity variations from eddy current (electromagnetic induction) data, Inverse Problems, vol. 6, pp. 1007– 1030, 1990. [48] Y. Nonaka, A double coil method for simultaneously measuring the resistivity, permeability and thickness of a moving metal sheet, IEEE Transactions on Instrumentation and Measurement, vol. 45, pp. 478–482, 1996. [49] S. J. Norton, and J. R. Bowler, Theory of eddy current inversion, Journal of Applied Physics, vol. 73, pp. 501–512, 1993. [50] A.D. Polyanin, and V.F. Zaitsev, Handbook of exact solutions for ordinary differential equations .Florida: Chapman&Hall/CRC, 2003. [51] A. Ptchelintsev, and B. de Halleux, Thickness and conductivity determination of thin  158   

nonmagnetic coatings on ferromagnetic conductive substrates using surface coils, Review of Scientific Instruments, vol. 69, pp. 1488–1494, 1998. [52] R. Satveli, J. C. Moulder, B. Wang, and J. H. Rose, Impedance of a coil near an imperfectly layered metal structure: The layer approximation, Journal of Applied Physics, vol. 79, No. 6, pp. 2811–2821, 1996. [53] Y. Shen, C. Lee, C. C. H. Lo, and N. Nakagawa, Conductivity profile determination by eddy current for shot-peened superalloy surfaces toward residual stress assessment, Journal of Applied Physics, vol. 101, pp. 014907-1–014907-10, 2007. [54] H. Sun, J. R. Bowler, N. Bowler, and M. J. Johnson, Eddy current measurements on case hardened steel, AIP Conference Proceedings, vol. 615, pp. 1561–1568, 2002. [55] V. S. Sobolev, On the theory of a method of superposed transducer with eddy current testing, Defektoskopia, vol. 1, pp. 6–15, 1965 (In Russian). [56] C.- C. Tai, J. H. Rose, and J. C. Moulder, Thickness and conductivity of metallic layers from pulsed eddy-current measurements, Review of Scientific Instruments, Vol. 67, No. 11, pp. 3965-3973, 1996. [57] C.-C. Tai, Characterization of coatings on magnetic metal using the swept-currency eddy current method, Rev. Sci. Instr., vol. 71, pp. 3161–3167, 2000. [58] C.-C. Tai, H.-C. Yang, and Y.-H. Liu, Modeling the surface condition of ferromagnetic metal by the swept-frequency eddy current method, IEEE Trans. Magn., vol. 38, pp. 205– 210, 2002. [59] J. A. Tegopoulos, and E.E. Kriezis, Eddy currents in linear conducting media, in Studies in electrical and electronic engineering, vol. 16. Amsterdam: Elsevier, 1985. [60] T.P. Theodoulidis, and E.E. Kriezis, Eddy current canonical problems (with applications to nondestructive evaluation). New York: Tech Science Press, 2006. [61] T. P. Theodoulidis, T. D. Tsiboukis, and E. E. Kriezis, Analytical solutions in eddy current testing of layered metals with continuous conductivity profiles, IEEE Transactions on Magnetics, vol. 31, No. 3, pp. 2254–2260, 1995. [62] T.P. Theodoulidis, and E.E. Kriezis, Coil impedance due to a sphere of arbitrary radial conductivity and permeability profiles, IEEE Transactions on Magnetics, vol. 38, pp. 1452–1460, 2002. [63] T.P. Theodoulidis, and E.E. Kriezis, Series expansions in eddy current nondestructive evaluation models, Journal of Materials Processing Technology, vol. 161, pp. 343–347, 2005. [64] E. Uzal, J.C. Moulder, and J.H. Rose, Impedance of coils over layered metals with continuously variable conductivity and permeability: Theory and experiment, J. Appl. Phys., vol. 74, pp. 2076–2089, 1993. [65] E. Uzal, I. Ozkol, and M.O. Kaya, Impedance of a coil surrounding an infinite cylinder with an arbitrary radial conductivity profile, IEEE Transactions on Magnetics, vol. 34, pp. 213–217, 1998. [66]    E. Uzal, and J.H. Rose, The impedance of eddy current probes above layered metals whose conductivity and permeability vary continuously, IEEE Transactions on Magnetics, vol. 29, pp. 1869–1873, 1993. [67] V. V. Vlasov, and V. A. Komarov, Interaction of the magnetic field of a long single-turn loop with a conducting ferromagnetic cylinder, The Soviet Journal of Nondestructive Testing, vol. 8, pp. 443–440, 1972.  159   

[68]    J.R. Wait, Propagation of electromagnetic pulses in a homogeneous conducting earth, Appied. Scientific Research, vol. 8, pp. 213–253, 1960. [69] W. Yin, A. J. Peyton, and S. J. Dickinson, Simultaneous measurements of distance and thickness of a thin metal plate with an electromagnetic sensor using a simplified model, IEEE Transactions on Instrumentation and Measurement, vol. 53, pp. 1335–1338, 2004. [70] W. Yin, and A. J. Peyton, Thickness measurement of non-magnetic plates using multifrequency eddy current sensors, NDT & E International, vol. 40, pp. 43–48, 2007. [71] N. Yusa, L. Janousek, M. Rebican, Z. Chen, and K. Miya, Eddy current inversions of defects in rough welds using a simplified computational model, Nondestructive Testing and Evaluation, vol. 20, pp. 191–199, 2005. [72]    N.N Zatsepin, and V.M. Tatarnikov, Theory of a superposed transducer, Soviet Journal of Nondestructive Testing, vol. 18, pp. 81–87, 1982.

 160   

APPENDIX. Fig. Ap.1 ( Section 2.4 ) “Mathematica” program for numerical calculation of the change in impedance Z for three different values of βˆ .

 161   

Fig. Ap.2 ( Section 2.4 ) “Mathematica” program for the selection of the upper limit of integration in (2.4.13). Remove@"Global`∗"D α = 0.0; β = 2; h = 0.05; µm = 5; DoA

9DoA

9If@n 1, 8r1 = 1, r2 = 0, r3 = 0