Efficient calculation of the mutual inductance of arbitrarily oriented circular filaments via a generalisation of the Kalantarov-Zeitlin method
EEfficient calculation of the mutual inductance ofarbitrarily oriented circular filaments via ageneralisation of the Kalantarov-Zeitlin method
Kirill V. Poletkin a, ∗ , Jan G. Korvink a a The Institute of Microstructure Technology, Karlsruhe Institute of Technology,Hermann-von-Helmholtz-Platz 1, 76344 Eggenstein-Leopoldshafen, Germany
Abstract
In this article, we present a new analytical formulation for calculation of the mu-tual inductance between two circular filaments arbitrarily oriented with respectto each other, as an alternative to Grover [1] and Babiˇc [2] expressions reportedin 1944 and 2010, respectively. The formula is derived via a generalisation ofthe Kalantarov-Zeitlin method, which showed that the calculation of mutual in-ductance between a circular primary filament and any other secondary filamenthaving an arbitrary shape and any desired position with respect to the primaryfilament is reduced to a line integral. In particular, the obtained formula pro-vides a solution for the singularity issue arising in the Grover and Babiˇc formulasfor the case when the planes of the primary and secondary circular filaments aremutually perpendicular. The efficiency and flexibility of the Kalantarov-Zeitlinmethod allow us to extend immediately the application of the obtained resultto a case of the calculation of the mutual inductance between a primary cir-cular filament and its projection on a tilted plane. Newly developed formulashave been successfully validated through a number of examples available in theliterature, and by a direct comparison with the results of calculation performedby the
FastHenry software.
Keywords:
Inductance, circular filaments, coils, line integral, electromagnetic ∗ Corresponding author
Email address: [email protected] (Kirill V. Poletkin)
Preprint submitted to Journal of Magnetism and Magnetic Materials March 26, 2019 a r X i v : . [ phy s i c s . c l a ss - ph ] M a r ystem, electromagnetic levitation
1. Introduction
Analytical and semi-analytical methods in the calculation of inductances,and in particular the mutual inductances of filament wires and their loops, playan important role in power transfer, wireless communication, and sensing andactuation, and is applied in different fields of science, including electrical andelectronic engineering, medicine, physics, nuclear magnetic resonance, mecha-tronics and robotics, to name the most prominent. Collections of formulas forthe calculation of mutual inductance between filaments of different geometricalshapes covering a wide spectrum of practical arrangements have variously beenpresented in classical handbooks by Rosa [3], Grover [4], Dwight [5], Snow [6],Zeitlin [7], Kalantarov [8], among others.The availability of efficient numerical methods such as
FastHenry [9] (basedon the multipole expansion) currently provides an accurate and fast solutionfor the calculation of mutual and self-inductance for any circumstance, includ-ing the use of arbitrary materials, conductor cross-sections, loop shapes, andarrangements. However, analytical methods allow to obtain the result in theform of a final formula with a finite number of input parameters, which whenapplicable may significantly reduce computation effort. It will also facilitatemathematical analysis, for example when derivatives of the mutual inductancew.r.t. one or more parameters are required to evaluate electromagnetic forcesvia the stored magnetic energy, or when optimization is performed.Analytical methods applied to the calculation of the mutual inductance be-tween two circular filaments is a prime example, and has been successfully usedin an increasing number of applications, including electromagnetic levitation[10], superconducting levitation [11, 12, 13] , magnetic force interaction [14] ,wireless power transfer [15, 16, 17], electromagnetic actuation [18, 19, 20, 21],micro-machined contactless inductive suspensions [22, 23, 24, 25] and hybridsuspensions [26, 27, 28], biomedical applications [29, 30], topology optimiza-2ion [31], nuclear magnetic resonance [32, 33], indoor positioning systems [34],navigation sensors [35], and magneto-inductive wireless communications [36].The original formula of the mutual inductance between two coaxial circularfilaments was derived by Maxwell [37, page 340, Art. 701] and expressed in termsof elliptic integrals. Butterworth obtained a formula covering the case of circularfilaments with parallel axes [38]. Then, a general formal expression made forcases where the axes of the circles are parallel, and where their axes intersect,was derived by Snow [39]. However, the Butterworth and Snow expressionssuffer from a low rate of convergence. This issue was recognized and solved byGrover, who developed the most general method in the form of a single integral[1]. Using the vector potential method, as opposed to the Grover means, thegeneral case for calculating the mutual inductance between inclined circularfilaments arbitrarily positioned with respect to each other was subsequentlyobtained by Babiˇc et al. [2].Kalantarov and Zeitlin showed that the calculation of mutual inductancebetween a circular primary filament and any other secondary filament havingan arbitrary shape and any desired position with respect to the primary filamentcan be reduced to a line integral [8, Sec. 1-12, page 49]. In the present paper,we report an adaptation of this method to the case of two circular filamentsand then derive a new analytical formula for calculating the mutual inductancebetween two circular filaments having any desired position with respect to eachother as an alternative to the Grover and Babiˇc expressions.In particular, the obtained formula provides a solution for the singularityissue arising in the Grover and Babiˇc formulas for the case when the planes ofthe primary and secondary circular filaments are mutually perpendicular. Theefficiency and flexibility of the Kalantarov-Zeitlin method allow us to extendimmediately the application of the obtained result to a case of the calculationof the mutual inductance between a primary circular filament and its projectionon a tilted plane. For instance, this particular case appears in micro-machinedinductive suspensions and has a direct practical application in studying theirstability [25] and pull-in dynamics [27, 28]. The new analytical formulae were3
YZ O x yz BR p R s P Figure 1: General scheme of arbitrarily positioning two circular filaments with respect to eachother: P is an arbitrary point on the secondary filament. verified by comparison with series of reference examples covering all cases givenby Grover[4], Kalantarov and Zeitlin [8], and using direct numerical calculationsperformed by the Babiˇc Matlab function [2] and the FastHenry software [9].
2. Preliminary discussion
Two circular filaments having radii of R p and R s for the primary circularfilament (the primary circle) and the secondary circular filament (the secondarycircle), respectively are considered to be arbitrarily positioned in space, namely,they have a linear and angular misalignment, as is shown in Figure 1. Let usassign a coordinate frame (CF) denoted as XY Z to the primary circle in a suchway that the Z axis is coincident with the circle axis and the XOY plane ofthe CF lies on the circle’s plane, where the origin O corresponds to the centreof primary circle. In turn, the xyz CF is assigned to the secondary circle in asimilar way so that its origin B is coincident with the centre of the secondarycircle.The linear position of the secondary circle with respect to the primary oneis defined by the coordinates of the centre B ( x B , y B , z B ). The angular positionof the secondary circle can be defined in two ways. Firstly, the angular positionis defined by the angle θ and η corresponding to the angular rotation around anaxis passing through the diameter of the secondary circle, and then the rotation4 y,y''z B y'z' θ η θ η z''x',x'' (a) Manner I (b) Manner II x y,y''z B y'x',x'' z'z'' αα ββ β α Figure 2: Two manners for determining the angular position of the secondary circle withrespect to the primary one: x (cid:48) y (cid:48) z (cid:48) is the auxiliary CF the axes of which are parallel to theaxes of XY Z , respectively; x (cid:48)(cid:48) y (cid:48)(cid:48) z (cid:48)(cid:48) is the auxiliary CF defined in such a way that the x (cid:48) and x (cid:48)(cid:48) are coincide, but the z (cid:48)(cid:48) and y (cid:48)(cid:48) axis is rotated by the α angle with respect to the z (cid:48) and y (cid:48) axis, respectively. of this axis lying on the surface x (cid:48) By (cid:48) around the vertical z (cid:48) axis, respectively, asit is shown in Figure 2(a). These angles for determination of angular position ofthe secondary circle was proposed by Grover and used in his formula numberedby (179) in [4, page 207] addressing the general case for calculation of the mutualinductance between two circular filaments.The same angular position can be determined through the α and β angle,which corresponds to the angular rotation around the x (cid:48) axis and then aroundthe y (cid:48)(cid:48) axis, respectively, as it is shown in Figure 2(b). This additional secondmanner is more convenient in a case of study dynamics and stability issues, forinstance, applying to axially symmetric inductive levitation systems [23, 25] incompared with the Grover manner. These two pairs of angles have the followingrelationship with respect to each other such as: sin β = sin η sin θ ;cos β sin α = cos η sin θ. (1)The details of the derivation of this set presented above are shown in Ap-pendix A. 5 o sX YZO x yz B PQ ρ x B y B z B '' Figure 3: The Kalantarov-Zeitlin method: s = (cid:112) x B + y B is the distance to the centre B onthe XOY plane.
3. The Kalantarov-Zeitlin method
Using the general scheme for two circular filaments shown in Figure 1 asan illustrative one, the Kalantarov-Zeitlin method is presented. The methodreduces the calculation of mutual inductance between a circular primary filamentand any other secondary filament having an arbitrary shape and any desiredposition with respect to the primary circular filament to a line integral [8, Sec.1-12, page 49].Indeed, let us choose an arbitrary point P of the secondary filament (as it hasbeen mentioned above the filament can have any shape), as shown in Figure 1.An element of length d(cid:96) (cid:48)(cid:48) of the secondary filament at the point P is considered.Also, the point P is connected the point Q lying on the Z axis by a line, whichis perpendicular to the Z axis and has a length of ρ , as shown in Figure 3. Thenthe element d(cid:96) (cid:48)(cid:48) can be decomposed on dz along the Z axis and on dρ along the ρ line and dλ along the λ -circle having radius of ρ (see, Figure 4). It is obviousthat the mutual inductance between dz and the primary circular filament isequal to zero because dz is perpendicular to a plane of primary circle. But themutual inductance between dρ and the primary circular filament is also equal tozero because of the symmetry of the primary circle relative to the dρ direction.Thus, the mutual inductance dM between element d(cid:96) (cid:48)(cid:48) and the primary circle6 XYO ρ d λ P d ρ Figure 4: The Kalantarov-Zeitlin method: projection of the secondary filament on the ρ -planepassed through the point P and parallel to the plane of the primary circular filament; d(cid:96) isthe projection of the element d(cid:96) (cid:48)(cid:48) on the ρ -plane. is equal to the mutual inductance dM λ between element dλ and the primarycircle. Moreover, due to the fact that the primary and the λ -circle are coaxialand, consequently, symmetric then we can write: dM λ M λ = dλλ = dλ πρ , (2)where M λ is the mutual inductance of the primary coil and λ -circle.From Figure 4, it is directly seen that dλ = dy cos ϕ − dx sin ϕ = (cos ζ cos ϕ − cos ε sin ϕ ) d(cid:96), (3)where cos ε and cos ζ are the direction cosines of element d(cid:96) relative to the X and Y axis, respectively. Hence, accounting for (2) and (3), we can write: dM = dM λ = M λ cos ζ cos ϕ − cos ε sin ϕ πρ d(cid:96), (4)and as a result, a line integral for calculation mutual inductance between theprimary circle and a filament is M = 12 π (cid:90) (cid:96) M λ cos ζ cos ϕ − cos ε sin ϕρ d(cid:96), (5)where M λ is defined by the Maxwell formula for mutual inductance betweentwo coaxial circles [37, page 340, Art. 701]. Note that during integrating, the Z x' y'B PO,QX Y λ s r Figure 5: Determination of the position of the point P on the ρ -plane through the fixedparameter s and the distance r . coordinate of the element d(cid:96) is also changing and this dependency is taken intoaccount by the M λ function directly.
4. Derivation of Formulas
Due to the particular geometry of secondary filament under consideration,its projection on the ρ -plane (the ρ -plane is parallel to the primary circle planeand passed through the point P ) is an ellipse, which can be defined in a polarcoordinate by a function r = r ( ϕ ) with the origin at the point B as it is shownin Figure 5. Hence, the distance ρ can be expressed in terms of the parameter s , which is fixed, and the distance r from the origin B , which is varied with x's ρ B r y' λ γ γγ d λ dr Figure 6: The relationship between dλ and dϕ . ϕ . Introducing the angle γ as shown in Figure 6, for thedistance ρ the following equations can be written: ρ cos γ = r + s cos( ξ − ϕ ) ,ρ sin γ = s sin( ξ − ϕ ) . (6)Due to (6), we have: ρ = r + r · s cos( ξ − ϕ ) + s , (7)where the function r = r ( ϕ ) can be defined as [40]: r = R s cos θ (cid:113) sin ( ϕ − η ) + cos θ cos ( ϕ − η ) . (8)The angle θ and η defines the angular position of the secondary circle withrespect to the primary one according to manner I considered in Sec. 2. Notethat the function r can be also defined via the angles α and β of manner IIalso considered in Sec. 2 as it is shown in Appendix B. However, for thefurther derivation, the angular position of the secondary circle is defined throughmanner I, since it is convenient for the direct comparison with Grover’s andBabiˇc’ results.According to Figure 6, the relationship between the element dλ of the λ -circleand an increment of the angle ϕ is as follows: dλ = r · dϕ cos γ − dr sin γ = (cid:18) r cos γ − drdϕ sin γ (cid:19) dϕ. (9)Then, accounting for (9), (7) and (6), line integral (5) can be replaced by adefinite integral for the calculation of mutual inductance as follows: M = 12 π (cid:90) π M λ r + r · s cos( ξ − ϕ ) − drdϕ s sin( ξ − ϕ ) ρ dϕ. (10)Now, let us introduce the following dimensionless parameters such as:¯ x B = x B R s ; ¯ y B = y B R s ; ¯ z B = z B R s ; ¯ r = rR s ;¯ ρ = ρR s ; ¯ s = (cid:113) ¯ x B + ¯ y B . (11)9 Bs XY ρ d λ P x'y'O,Q η Figure 7: The special case: the two filament circles are mutually perpendicular to each other.
The ϕ -derivative of ¯ r is d ¯ rdϕ = 12 ¯ r tan θ sin(2( ϕ − η )) , (12)The mutual inductance M λ is M λ = µ k Ψ( k ) (cid:112) R p R s ¯ ρ, (13)where µ is the magnetic permeability of free space, andΨ( k ) = (cid:18) − k (cid:19) K ( k ) − E ( k ) , (14)where K ( k ) and E ( k ) are the complete elliptic functions of the first and secondkind, respectively, and k = 4 ν ¯ ρ ( ν ¯ ρ + 1) + ν ¯ z λ , (15)where ν = R s /R p and ¯ z λ = ¯ z B + ¯ r tan θ sin( ϕ − η ). Accounting for dimensionlessparameters (11) and substituting (12) and (14) into integral (10), the new for-mula to calculate the mutual inductance between two circular filaments havingany desired position with respect to each other becomes M = µ (cid:112) R p R s π (cid:90) π ¯ r + t · cos ϕ + t · sin ϕk ¯ ρ . · ¯ r · Ψ( k ) dϕ, (16)10here terms t and t are defined as t = ¯ x B + 0 . r tan θ sin(2( ϕ − η )) · ¯ y B ; t = ¯ y B − . r tan θ sin(2( ϕ − η )) · ¯ x B , (17)and ¯ ρ = (cid:112) ¯ r + 2¯ r · ¯ s cos( ξ − ϕ ) + ¯ s .Formula (16) can be applied to any possible cases, but one is excluded whenthe two filament circles are mutually perpendicular to each other. In this casethe projection of the secondary circle onto the ρ -plane becomes simply a line asit is shown in Fig. 7 and as a result to integrate with respect to ϕ is no longerpossible.For the treatment of this case, the Kalantarov-Zeitlin formula (5) is directlyused. Let us introduce the dimensionless variable ¯ (cid:96) = (cid:96)/R s and then the inte-gration of (5) is preformed with respect to this dimensionless variable ¯ (cid:96) withininterval − ≤ ¯ (cid:96) ≤
1. The direction cosines cos ζ and cos ε become as sin η andcos η , respectively (see, Fig. 7). Accounting for ρ cos ϕ = s cos ξ + (cid:96) cos η,ρ sin ϕ = s sin ξ + (cid:96) sin η, (18)and the Maxwell formula (14) and (15), where the Z -coordinate of the element d ¯ (cid:96) is defined as ¯ z λ = ¯ z B ± (cid:112) − ¯ (cid:96) , (19)then the formula to calculate the mutual inductance between two filament cir-cles, which are mutually perpendicular to each other, becomes as follows: M = µ (cid:112) R p R s π (cid:20)(cid:90) − t − t k ¯ ρ . · Ψ( k ) d ¯ (cid:96) + (cid:90) − t − t k ¯ ρ . · Ψ( k ) d ¯ (cid:96) (cid:21) , (20)where terms t and t are defined as t = sin η (¯ x B + ¯ (cid:96) cos η ); t = cos η (¯ y B + ¯ (cid:96) sin η ) , (21)11nd ¯ ρ = (cid:112) ¯ s + 2¯ (cid:96) · ¯ s cos( ξ − η ) + ¯ (cid:96) . Note that integrating (20) between − r = r ( ϕ ) is constant and defined throughthe radius of primary coil as r = R p . Since the centre of the projection iscoincide with the Z -axis, thus s = 0. Then, the formula is derived from (16) asits particular case (¯ s = 0 and ¯ r = ¯ ρ = 1) and becomes, simply, M = µ R p π (cid:90) π k · Ψ( k ) dϕ. (22)The obtained formulas can be easily programmed, they are intuitively under-standable for application. Also, the singularity arises in Grover’s and Babiˇc’sformula for the calculate of the mutual inductance between two filament circles,which are mutually perpendicular to each other, is solved in developed formula(20). The Matlab files with the implemented formulas (16), (20) and (22) areavailable from the authors as supplementary materials to this article. Also, inAppendix B the the developed formulas can be rewritten through the pair ofthe angle α and β .
5. Examples of Calculation. Numerical Verification
In this section, developed new formulas (16), (20) and (22) are verified by theexamples taken from Grover [4] and Kalantarov [8] books and Babiˇc article [2].The special attention was addressed to the singularity case arisen when the twofilament circles are mutually perpendicular to each other. Then, formula (22)was validated with the
FastHenry software [9]. All calculations for consideredcases proved the robustness and efficiency of developed formulas.Note that the notation proposed in Grover’s and Kalantarov’s books in orderto define the linear misalignment of the secondary coil is different from the12 a d
X YZOB
Figure 8: Geometrical scheme of coaxial circular filaments denoted via Grover’s notation: radii a and A of secondary and primary coils, respectively; d is the distance between the planes ofcircles. notation used in the Babiˇc article and in our article as well. Also, the angularmisalignment in the Babiˇc formula must be defined through the parameters ofthe secondary coil plane. These particularities of the notation will be discussedspecifically for each case. For all calculation, the primary coil is located on theplane XOY and its centre at the origin O (0,0,0). Let us consider the circular filaments, which are coaxial and have a distancebetween their centres, as shown in Fig. 8. Then, this case in the notationproposed in this article is defined as R p = A , R s = a , the linear misalignmentis z B = d , x B = y B = 0, the angular misalignment (manner I, Sec. 2) is θ = 0 and η = 0. For the Babiˇc formula the linear misalignment is definedin the same way, but for angular one the parameters of the secondary circleplane must be calculated and becomes a = 0, b = 0 and c = 1 (the Babiˇcnotation). These parameters have the following relationship with the angle θ and η : a = sin η sin θ ; b = − cos η sin θ and c = cos θ [2, Eq. (27), page 3597].13 xample 1 (Example 24, page 78 in Grover’s book [4]) Let us suppose that two circles of radii a =20 cm and A =25 cm with theirplanes d =10 cm apart are given. The results of calculation areGrover’s book The Babiˇc formula This work, Eq. (16) M , nH 248.79 248.7874 248.7874 Example 2 (Example 25, page 78 in Grover’s book [4])
Two circles of radii a =2 in= 5 .
08 cm and A =5 in=12 . d =4 in=10 .
16 cm apart, the results becomeGrover’s book The Babiˇc formula This work, Eq. (16) M , nH 18.38 18.3811 18.3811 Example 3 (Example 5-4, page 215 in Kalantarov’s book [8])
For two circles having the same radii of 10 .
00 cm with their planes d =4 cmapart, the calculation shows the followingKalantarov’s book The Babiˇc formula This work, Eq. (16) M , nH 135.1 135.0739 135.0739 Example 4 (Example 5-5, page 215 in Kalantarov’s book [8])
Circles having the same radii as in
Example 3 , but their planes d =50 cmapart are given. The results areKalantarov’s book The Babiˇc formula This work, Eq. (16) M , nH 1.41 1.4106 1.410614 d X YZO B
A x yzr ρ Figure 9: Geometrical scheme of circular filaments with parallel axes denoted via Grover’snotation: ρ is the distance between axes; r is the distance between the centres; ϕ is the anglebetween the Z -axis and the radius vector r . Example 5 (Example 5-6, page 224 in Kalantarov’s book [8])
Two circular coaxial filaments, radii of which are A =25 cm and a =20 cm,with their planes d =8 cm apart are given. The results are as followsKalantarov’s book The Babiˇc formula This work, Eq. (16) M , nH 289.11 289.0404 289.0404 The scheme for calculation of the mutual inductance between circular fila-ments with parallel axes is shown in Fig. 9. The linear misalignment in theGrover notation can be defined by d is the distance between the planes of circles(the same parameter as in Sec. 5.1) and ρ is the distance between axes or via r isthe distance between the centres and ϕ is the angle between the Z -axis and theradius vector r . These parameters have the following relationship to the notationdefined in this article, namely, z B = d = r cos ϕ and ρ = (cid:112) x B + y B = r sin ϕ .The angular misalignment is defined in the same way as described in Sec. 5.1). Example 6 (Example 62, page 178 in Grover’s book [4]))
Two circles of radii a = A =15 cm have a distance between their centres r =20 cm and an angle ϕ = cos − . Z -axis and the radius vector15 (please, see Fig. 9). Assuming that y B = ρ = r sin ϕ =12 cm and z B = r cos ϕ =16 cm, the results of calculation are as followsGrover’s book The Babiˇc formula This work, Eq. (16) M , nH 45.31 45.3342 45.3342 Example 7 (Example 63, page 178 in Grover’s book [4])
Two circles of the same diameter of 2 a = 2 A =48 in=121 .
92 cm are arrangedso that the distance between their planes d =15 in=38 . ρ =47 . .
158 cm (please, see Fig. 9). Thus, we have R p = R s =60 .
96 cm, y B = ρ and z B = d , the results of calculation are as followsGrover’s book The Babiˇc formula This work, Eq. (16) M , nH − . − . − . Example 8 (Example 65, page 183 in Grover’s book [4])
Two circles with radii of A =10 cm and a =8 cm have the distance betweentheir centres r =50 cm and an angle of cos ϕ = 0 . R p = A and R s = a , assuming that y B = r sin ϕ =45 .
83 cm and z B = r cos ϕ =20 . M , nH − . − . − . Example 9 (Example 66, page 184 in Grover’s book [4])
Two circles with radii of A =10 cm and a =8 cm have the distance betweentheir centres r =20 cm and an angle of cos ϕ = 0 . R p = A and16 s = a , assuming that y B = r sin ϕ =16 . z B = r cos ϕ =12 . M , nH 4 .
405 4 .
465 4 . Example 10 (Example 5-8, page 231 in Kalantarov’s book [8])
Two circular filaments of the same radius of A = a =5 cm are arrangedthat the distance between their centres is r =40 cm and an angle of cos ϕ = 0 . R p = R s =5 cm, assuming that y B = r sin ϕ =36 .
66 cm and z B = r cos ϕ =16 . M , nH − . − . − . Example 11 (Example 5-9, page 233 in Kalantarov’s book [8])
Two circular filaments of radii of A =10 cm and a =5 cm are arranged thatthe distance between their centres is r =20 cm and an angle of cos ϕ = 0 . R p =10 cm and R s =5 cm, assuming that y B = r sin ϕ =36 .
66 cmand z B = r cos ϕ =16 . M , nH 2 .
95 3 . . Example 12 (Example 5-10, page 234 in Kalantarov’s book [8])
Two circular filaments of radii of A =20 cm and a =4 cm are arranged thatthe distance between their centres is r =2 cm and an angle of cos ϕ = 0 . R p =20 cm and R s =4 cm, assuming that y B = r sin ϕ =1 . dS X YZO B A ρ θ x x Figure 10: Geometrical scheme of inclined circular filaments with intersect axes denoted viaGrover’s notation: x and x are the distances from S , θ is the angle of inclination of theaxes. and z B = r cos ϕ =1 .
32 cm, the results become as followsKalantarov’s book The Babiˇc formula This work,Eq. (16) M , nH 15 .
99 15 . . The general scheme of the arrangement of two inclined circular filamentswhose axes intersect for calculation of mutual inductance is shown in Fig. 10.In Grover’s notation, we have A and a are radii of circular filaments and S isthe point of intersection of the circles axes. The centres of circles are at thedistances x and x from S of the primary and secondary circle, respectively. θ is the angle of inclination of the axes. Also, the following relationships aretrue, namely, d = x − x cos θ and ρ = x sin θ . Thus, the linear misalignmentin the notation of this paper is defined again through z B = d , ρ = (cid:112) x B + y B and the angular misalignment is defined by the angle θ , but η is equal to zero.For the Babiˇc formula, the angular misalignment is defined by the parametersof the secondary circle plane as follows: a = 0; b = − sin θ and c = cos θ .From the general scheme shown in Fig. 10, two particular cases can berecognized. Namely, the first case is corresponded to concentric circles, when x = x = 0 and the second case is corresponded to circular filaments whose18xes intersect at the centre of one of the circle, when x = 0. For the firstcase, to calculate the mutual inductance, the angle θ and radii of circles mustbe known. For the second particular case, in addition to the distance d betweenthe centre B of the secondary circle and the plane of the primary circle must begiven. Example 13 (Example 5-7, page 227 in Kalantarov’s book [8])
Two circular filaments of radii of A =10 cm and a =2 . θ =60 ◦ . Hence,assuming that x B = y B = z B = 0, the calculation of mutual inductance showsKalantarov’s book The Babiˇc formula This work, Eq. (16) M , nH 6 .
044 6 . . Example 14 (Example 69, page 194 in Grover’s book [4])
Two concentric circles with radii of A =20 cm and a =14 cm are arrangedthat an angle of inclination of the plane of the secondary circle is cos θ = 0 . x B = y B = z B = 0 and θ =72 . ◦ , the results ofcalculation are as followsGrover’s book The Babiˇc formula This work, Eq. (16) M , nH 47 .
44 47 . . Example 15 (Example 70, page 194 in Grover’s book [4])
Two circles with radii of A =10 in=25 . a =3 in=7 .
62 cm are ar-ranged that an angle of inclination of the plane of the secondary circle iscos θ = 0 . d =3 in=7 .
62 cm. Hence, assuming that x B = y B = 0,19 B = d and θ =66 . ◦ , the results of calculation are as followsGrover’s book The Babiˇc formula This work, Eq. (16) M , nH 15 .
543 15 . . Example 16 (Example 71, page 201 in Grover’s book [4])
Two circles of radii A =20 . a =10 . d , of 20 . θ , of 30 ◦ . The results ofcalculation showGrover’s book The Babiˇc formula This work, Eq. (16) M , nH 29 .
436 29 . . Example 17 (Example 5-11, page 235 in Kalantarov’s book [8])
Two circular filaments have radii of A =10 cm and a =8 cm. The axisthe primary circle is crossed through the centre of the secondary circle at adistance d of 8 cm between their centres. The axes are to be inclined at anangle, cos θ = 0 .
7. Hence, assuming that x B = y B = 0, z B = d and an angle of45 . ◦ , the results of calculation areKalantarov’s book The Babiˇc formula This work, Eq. (16) M , nH 23 . . . Example 18 (Example 73, page 204 in Grover’s book [4])
Two circles of radii A =16 . a =10 . S in such a way that distances x and x are to be 20 . . θ = 0 .
5. Hence,20ssuming that x B = 0, y B =4 . z B =17 . . ◦ , wehave Grover’s book The Babiˇc formula This work, Eq. (16) M , nH 13 .
612 13 . . Table 1: Calculation of mutual inductance for Example 19 η The Grover formula, The Babiˇc This work,[4, Eq. (179)] formula, [2, Eq. (24)] Eq. (16) M , nH M , nH M , nH0 13.6113 13.6113 13.6113 π/ π/ π/ π/ π/ π/ π/ π π/ π/ π/ π/ π/ π/ π/ π .4. Mutual inductance of circular filaments arbitrarily positioned in the space The validation of the developed formulas (16) and (20) for the general case,when the angular misalignment is defined through the angle θ and η as shown inFig. 2(a) in an range from 0 to 360 ◦ , the examples from the Babiˇc article [2] wereused. Also, we utilized the Matlab functions with the Grover formula [4, page207, Eq. (179)] and the Babiˇc formula [2, page 3593, Eq. (24)] implemented byF. Sirois and S. Babiˇc. Example 19 (Example 12, page 3597 in the Babiˇc article [2])
Using the geometrical arrangement as in Example 18 (two circles with radii A =16 . a =10 . x B = 0, y B =4 . z B =17 . θ of 60 . ◦ ), but the angle η is varied in a range from 0 to 360 ◦ . The results of calculation are summed upin Table 1. Analysis of Table 1 shows that the developed formula (16) worksidentically to the Grover and Babiˇc formula. Example 20 (Example 11, page 3596 in the Babiˇc article [2])
Let us consider two circular filaments having radii of R p =40 cm and R s =10 cm,which are mutually perpendicular to each other that angles of η =0 and θ =90 . ◦ .The centre of the secondary circle has the following coordinates: x B = 0, y B =20 cm, and z B =10 cm. The problem illustrates the application of newformula (20). The results areThe Grover formula The Babiˇc formula This work,Eq. (20) M , nH − . − . − . Example 21
Now we again apply formula (20) to the problem considered in Example20, but in this case the centre of the secondary coil is located at origin, thus22 − M , n H η , degree − Figure 11: Distribution of the error of the Babiˇc formula in dependent on changing the η -anglewithin a range 0 < η ≤ o for Example 21 ( x B = y B = z B = 0). x B = y B = z B = 0. Hence, we haveThe Grover formula The Babiˇc formula This work, Eq. (20) M , nH N aN N aN η in a range 0 < η ≤ o , wereveal that the calculation of mutual inductance performed by developed for-mula (20) shows zero within this range of the η -angle, but the Babiˇc formulademonstrates a small error, which is not exceeded M =2 . × − nH anddistributed with the η -angle as shown in Figure 11. Example 22
Let us consider mutually perpendicular circles (angles of θ =90 . ◦ and η =0)having the same radii as in Example 20, but the centre of the secondary coiloccupies a position on the XOY -surface with the following coordinates x B = Not-a-Number , n H η , degree
50 100 150 200 250 300 350 − −
88 89 90 91 9222.533.544.5x 10 −
88 9290
268 270 2722.42.62.833.23.43.6x 10 −
268 272270
Figure 12: Chaotic distribution of the error of the Babiˇc formula in dependent on changingthe η -angle within a range 0 < η ≤ o for Example 22 ( x B = y B =10 cm and z B = 0): thescaled-up images show the interruption of continuity of the curve at η =90 ◦ and 270 ◦ . Projection of the primary circle θ X YZOx yz B R p Figure 13: Geometrical scheme for calculation of mutual inductance between a primary cir-cular filament and its projection. The angular misalignment is given by an angle of θ , whilethe linear misalignment by the coordinate z B . y B =10 cm and z B = 0. Results of calculation areThe Grover formula The Babiˇc formula This work, Eq. (20) M , nH 4 . × − . × − η in a range 0 < η ≤ o , the calculation of mutual24 able 2: Calculation of mutual inductance for Example 23 θ This work, Eq. (22)
FastHenry [9] M , nH M , nH0 135.0739 135.07610 ◦ ◦ η -angle, but the Babiˇc formula demonstrates the chaotic distribution ofthe small error of calculation as shown in Fig. 12, which is in a range from − . × − to 2 . × − nH and at the η -angle of 90 ◦ and 270 ◦ the calcu-lation of mutual inductance is indeterminate (see, the scaled-up images of Fig.12, which present the interruption of continuity at η =90 ◦ and 270 ◦ ). In this section new formula (22) for calculation of mutual inductance betweena primary circular filament and its projection on a tilted plane is validated bycomparison with the calculation performed via the
FastHenry software [9]. Theangular misalignment is given by the angle θ and η , while the linear misalign-ment is defined by the coordinate z B of the point B crossing the tilted planeand the Z -axis. Fig. 13 shows a geometrical scheme for the calculation. Theshown arrangement of the primary circle and its projection on the tilted planecorresponds to a particular case, when η = 0. Worth noting that the η -anglehas no effect on the result of the calculation of the mutual inductance. Example 23
Let us consider primary circle having a radius of 10 .
00 cm and a tilting planecrosses the Z -axis at the point z B =4 cm. When a tilting angle of zero, then thegeometry and arrangement corresponds to Example 3 (Example 5-4, page 21525n Kalantarov’s book) for the case of two coaxial circles with the same radii.We calculate the mutual inductance for three values of a tilted angle at 0 , ◦ and 15 ◦ . The results of calculation are shown in Table 2.Although, there is the small deviation between results obtained with the FastHenry software and analytical formula (22), but this deviation is not signif-icant and can be explained by the fact that the circles in the
FastHenry softwareare divided into straight segments with a finite cross section in comparing withthe analytical formula where the circles have no segments and a cross section.We are concluding the validity of developed formula (22).
Table 3: Calculation of mutual inductance for Example 24 θ This work, Eq. (22)
FastHenry [9] M , nH M , nH0 1.4106 1.37615 ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ xample 24 In this last example, we increase a distance between the centre of the primarycircle with a radius of 10 .
00 cm and a tilting plane to z B =50 cm. Hence, arange of the tilted angle becomes larger then in example 23. Note that for zerotilting angle the geometry of the considered problem corresponds to example 4(Example 5-5, page 215 in Kalantarov’s book). The results of calculation areshown in Table 3. Analysis of Table 3 shows a good agreement between thecalculations, which confirms the validity of developed formula (22).
6. Conclusion
We derived and validated new formulas (16) and (20) for calculation of themutual inductance between two circular filaments arbitrarily oriented with re-spect to each other. These analytic formulas have been developed based onthe Kalantarov-Zeitlin method, which showed that the calculation of mutual in-ductance between a circular primary filament and any other secondary filamenthaving an arbitrary shape and any desired position with respect to the primaryfilament is reduced to a line integral. In particular, the developed formula (20)provides a solution for the singularity issue arising in Grover’s and Babiˇc’ formu-las for the case when the planes of the primary and secondary circular filamentsare mutually perpendicular.Moreover, a curious reader can already recognize that formula (20) can beapplied for calculation of the mutual inductance between the circular filamentand a line, position of which with respect to the circle is defined through thelinear and angular misalignment. For this reason, in Eq. (19) we assume that¯ z λ = ¯ z B and formula (20) is integrated only from − FastHenry software shows a good agreement. Besides, the obtained formulas can be easilyprogrammed, they are intuitively understandable for application.
Acknowledgment
Kirill Poletkin acknowledges with thanks to Prof. Ulrike Wallrabe for thecontinued support of his research. Also, Kirill Poletkin acknowledges withthanks the support from German Research Foundation (Grant KO 1883/26-1). z z'z'' y,y''y' α β α β x θθ θ ηη x' o - η o o β I II
Figure .14: The relationship between the angles of two manners for determining angular mis-alignment of the secondary circle: I and II are denoted for two spherical triangles highlightedby arcs in red color. ppendix A. Determination of angular position of the secondary cir-cular filament The angular position of the secondary circle can be defined through the pairof angle θ and η corresponding to manner I and the angle α and β manner II.The relationship between two pairs of angles can be determined via two sphericaltriangles denoted in Roman number I and II as shown in Fig. .14. According tothe law of sines, for spherical triangle I we can write the following relationship:sin η sin π/ β sin θ . (A.1)For spherical triangle II, we havesin( π/ − η )sin( π/ − β ) = sin α sin θ . (A.2)Accounting for (A.1) and (A.2), the final set determining the relationship be-tween two pairs of angles becomes as sin β = sin η sin θ ;cos β sin α = cos η sin θ. (A.3) Appendix B. Presentation of developed formulas via the pair of an-gles α and β Using set (A.3), we can write the following equations: cos θ = cos β (1 + sin α );sin θ = sin β + cos β sin α ;tan θ = sin β + cos β sin α cos β (1 + sin α ) ;cos η = cos β sin α/ sin θ ;sin η = sin β/ sin θ. (B.1)Now, applying set (B.1) to (8), the square of the dimensionless function ¯ r be-comes as ¯ r = cos β (1 + sin α )(sin β + cos β sin α )(sin ϕ cos β sin α − cos ϕ sin β ) +cos β (1 + sin α )(cos ϕ cos β sin α + sin ϕ sin β ) . (B.2)29hen, for the dimensionless parameter ¯ z λ , we have¯ z λ = ¯ z B + ¯ r sin ϕ cos β sin α − cos ϕ sin β (cid:113) cos β (1 + sin α ) . (B.3)Substituting (B.2), (B.3) and t = ¯ x B + ¯ y B · ¯ r sin β + cos β sin α cos β (1 + sin α ) × (sin ϕ cos β sin α − cos ϕ sin β ) × (cos ϕ cos β sin α + sin ϕ sin β ); t = ¯ y B − ¯ x B · ¯ r sin β + cos β sin α cos β (1 + sin α ) × (sin ϕ cos β sin α − cos ϕ sin β ) × (cos ϕ cos β sin α + sin ϕ sin β ) , (B.4)into Eq. 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