MUTUAL INDUCTANCE AND FORCE EXERTED

BETWEEN THICK COILS

Romain Ravaud, Guy Lemarquand, Valerie Lemarquand, Slobodan Babic,

Cevdet Akyel

To cite this version:

Romain Ravaud, Guy Lemarquand, Valerie Lemarquand, Slobodan Babic, Cevdet Akyel. MU-TUAL INDUCTANCE AND FORCE EXERTED BETWEEN THICK COILS. PIER - ProgressIn Electromagnetics Research, 2010, 102, pp.367-380.

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MUTUAL INDUCTANCE AND FORCE EXERTEDBETWEEN THICK COILS

R. Ravaud, G. Lemarquand and V. Lemarquand

Laboratoire dAcoustique de lUniversite du Maine, UMR CNRS 6613Avenue Olivier Messiaen, 72085 Le Mans, Franceguy.lemarquand@ieee.org

S. Babic and C. Akyel

Ecole Polytechnique, Departement de Genie PhysiqueQC H3C 3A7, Montreal, Canada

AbstractWe present exact three-dimensional semi-analytical ex-pressions of the force exerted between two coaxial thick coils with rect-angular cross-sections. Then, we present a semi-analytical formulationof their mutual inductance. For this purpose, we have to calculate sixand seven integrations for calculating the force and the mutual induc-tance respectively. After mathematical manipulations, we can obtainsemi-analytical formulations based on only two integrations. It is tobe noted that such integrals can be evaluated numerically as they aresmooth and derivable. Then, we compare our results with the fila-ment and the finite element methods. All the results are in excellentagreement.

1. INTRODUCTION

Various electromagnetic applications are composed of two thick coilsthat form a loosely coupled transformer. The first coil generates amagnetic field in all points in space, and this magnetic field is partlypicked up by the secondary coil. This is an efficient way of transferingpower wirelessly. However, a decrease in power transfer efficiency canbe caused by a lower mutual inductance between two coils [1][2]. Inother words, it is very useful to know the accurate value of the mutualinductance or the force exerted between two coils. Indeed, the forceis directed linked to the mutual inductance as it is proportional to itsgradient.

2 R. Ravaud et al

The calculation of mutual inductance for circular coils has beenstudied by many authors [3]-[12]. These papers are generally based onthe application of Maxwells formula, Neumanns formula and the BiotSavart law. By using these approaches, the mutual inductance of cir-cular coils can be expressed in terms of analytical and semi-analyticalfunctions, as the elliptic integrals of the first, second and third kind,the Heummanns Lambda function or the Bessel functions [13][15].Such analytical methods are also suitable for calculating the magneticnear-field or far-field from circular cylindrical magnetic sources [16][19]or for the determination of the forces exerted between them [20]-[25].On the other hand, authors generally use the finite element methodor the boundary element method for solving such magnetostatic prob-lems. However, as stated in [26], it is interesting to obtain analyticalor semi-analytical exact expressions having a lower computational costfor optimization purposes.

In this paper, we propose to replace each coil by a toroidalconductor carrying uniform current volume density for calculating boththe force exerted between two thick coils and their mutual inductance.These toroidal conductors are assumed to be perfectly circular andradially centered. We define an equivalent current volume densityj which is linked to the number of loops and the coil dimensions.We use the Lorentz force for evaluating the exact axial force exertedbetween two thick coils carrying uniform current volume densities. Weobtain a semi-analytical expression of the force that requires only twonumerical integrations. However, its computational cost remains verylow compared to the finite element method one as we need only about0.05 s for calculating the axial force for a given configuration. Then,we compare our semi-analytical approach with the filament method.This comparison is a way for us to verify the accuracy of our semi-analytical model and to study the differences that occur between thesethree methods when two coils are close to each other. The analyticaland numerical simulations are in very good agreement.The second part of this paper deals with the analytical calculation ofthe mutual inductance of two thick coils in air. We use the potentialvector and the Stokes Theoreom for reducing the number of numericalintegrations required for evaluating this mutual inductance. We obtaina semi-analytical expression based on two numerical integrations.However, its computational cost remains very low (less than 0.5 s).Furthermore, we do not use any simplifying assumptions in theintegral formulations of the mutual inductance. Then, we compare ourapproach with both the finite element method and a method proposedby Kajikawa [27][28]. Here again, the results are in excellent agreement

Mutual Inductance and Force Exerted Between Thick Coils 3

and show that our approach is exact.

2. AXIAL FORCE FORMULATION

We present in this section our three-dimensional analytical formulationfor calculating the force exerted between two thick coils withrectangular cross sections. It is emphasized here that the two thickcoils considered are replaced by two toroidal conductors having uniformcurrent densities. However, for the rest of this paper, we will talk aboutthick coils rather than toroidal conductors.

2.1. Notation and geometry

Let us first consider Fig.1 where we have represented two thick coilscarrying uniform current volume densities:j1 =

N1I1

(r2r1)(z2z1)and j2 =

N2I2

(r4r3)(z4z3).

The parameters we use are defined as follows: for the lower coil,(respectively the upper one): r1,r2 (r3,r4) inner and outer radius[m], z1,z2 (z3,z4) lower and upper height [m], d = z3 z2: axialdistance between the two thick coils [m].

2.2. Expressions of the axial force

The first step for calculating the axial force exerted between two thickcoils is to express the magnetic induction field produced by the lowerthick coil shown in Fig.1. By using the Biot-Savart Law, the magnetic

induction field ~B1 is expressed as follows:

~B1(r, z) =0

4

V1

~j1dv1 {G(~r,~r)

}(1)

The axial force, exerted by coil 1 on coil 2, is derived with thefollowing equation:

~F =

V2

~j2dv2 ~B1(r, z) (2)

By inserting (1) in (2), we have:

~F =

V2

~j2dv2 0

4

V1

~j1dv1

{G(~r,~r)

}(3)

where the Greens function G(~r,~r) is defined as follows:

G(~r,~r) =1

r2+ r2 2rr cos() + (z z)2

(4)

4 R. Ravaud et al

r

0

r

uz

z

z

r

z

z 1

2

3

4

I

I

1

2

N

N

2

1u

r1

2

3r

4

u

Figure 1. Geometry considered: two thick coils carrying uniformcurrent volume densities. For the lower thick coil, the inner radius isr1, the outer one is r2, the lower height is z1, the upper one is z2; forthe upper thick coil, the inner radius is r3, the outer one is r4, thelower height is z3, the upper one is z4.

Therefore, (3) becomes:

~F =0

4

V1

V2

~j2dv2 {~j1dv1

{G(~r, ~r)

}}(5)

We obtain the final expression:

~F =0

4

V1

V2

~j2 {~j1

{G(~r, ~r)

}}dv1dv2 (6)

where

dv1dv2 = rrdrddzdrddz (7)

In cylindrical coordinates, G(~r,~r) is reduced to the following form:

G(~r,~r) =r r cos()(

r2+ r2 2rr cos() + (z z)2

) 32

~ur

Mutual Inductance and Force Exerted Between Thick Coils 5

+r sin()(

r2+ r2 2rr cos() + (z z)2

) 32

~u

+z z(

r2+ r2 2rr cos() + (z z)2

) 32

~uz

By projecting ~j1, ~j2 and G(~r,~r) in cartesian coordinates, we find

the following form of the axial force exerted between two thick coilsradially centered:

Fz =0j1j2

4

V1

V2

(z z

)cos()rrdrddzdrd

dz

(r2+ r2 2rr cos() + (z z)2

) 32

(8)

The previous relation can be directly reduced to the following form:

Fz =0j1j2

2

V1

r4r3

z4z3

(z z

)cos()rrdrddzdrdz

(r2+ r2 2rr cos() + (z z)2

) 32

(9)

After integrating with respect to r, z and z, we obtain the followingreduced analytical expression of the axial force exerted between twothick coils radially centered:

Fz =0j1j2

2

r2r1

21

2i=1

(1)i4

j,k=3

(1)j+k(f gr2i cos()

)cos()drd

(10)with

f = ri

2+

2ln [+ ]

g = zj + arctan

[

] arctan

[

]+ ln [+ ] + ln [ + ]

(11)

The parameters , , , , are defined in Table 1. It is emphasizedhere that Eq.(10) is an exact semi-analytical expression. This impliesthat it can be used whatever the thick coil dimensions and may be usedas a robust tool for evaluating the accuracy of any numerical methods.

6 R. Ravaud et al

Parameters

r2 + r2i 2rri cos() + (zj zk)

2

r2 + r2i 2rri cos() r ri cos() ri sin() zj zk

Table 1. Definition of the parameters used in ( 10)

3. COMPARISON OF THE AMPERIAN CURRENTMODEL AND THE FILAMENT METHOD

We present in this section a comparison between the amperiancurrent model and the filament method for calculating the axial forceexerted between two thick thick coils carrying uniform current volumedensities. We compare two configurations and we discuss the accuracyof each method.

3.1. First configuration: two thick coils having the samedimensions

The first configuration consists of two thick coil having the samedimensions and the same uniform current volume densities. We takethe following dimensions for our numerical simulations:r1 = r3= 0.0875 mr2 = r4= 0.1125 mz2 z1 = z4 z3= 0.025 mj1 = j2= 320 000 A/m

2, I= 1 A, N = 200 turnsd = axial distance between the two thick coils [m]It is useful to mention that the volume densities j1 and j2 correspondto a current I = I1 = I2 that equals 1 A. We represent in Fig.2 theaxial force exerted between the two coils versus the axial distance dbetween them. When d = 0, the two thick coils are in contact. Wehave also presented the numerical results of our analytical method andthe filament method in Table 2. Table 2 and Eq.2 show a very goodagreement between the filament method, our amperian current modeland the finite element method for calculating the axial force exertedbetween two thick coils in air. The filament method was employed with10*10*10*10 filaments and the computational cost was 25 s. However,this computational cost can be reduced by employing 5*5*5*5 filamentsand reach only 1.7 s. The computational cost of our semi-analyticalmethod is 0.18 s.

Mutual Inductance and Force Exerted Between Thick Coils 7

0 0.2 0.4 0.6 0.8 1d @mD

0255075

100125150175

F@m

ND

Figure 2. Representation of the axial force exerted between twothick coils versus the axial distance d = z3 z2. We take thefollowing dimensions: r1 = r3 = 0.0875 m, r2 = r4 = 0.1125 m,z2 z1 = z4 z3 = 0.025 m, j1 = j2 = 320 000 A/m

2. Line =our analytical method, Points = Filament method

d [m] Filament method amperian current model Flux 3D0 185.35236819 185.15890560.01 128.17153945 128.04563716 1280.02 94.85815069 94.75916322 950.03 73.13895302 73.05562878 720.04 57.93815379 57.865606359 580.05 46.77311227 46.708817766 470.1 18.79270092 18.7547813112 170.2 4.45130774 4.4378405952 40.5 0.26663880 0.26555243641

Table 2. Comparison between our analytical approach, the filamentmethod and the finite element method for calculating the axial force(mN) exerted between two thick coils having the same dimensions.

3.2. Second configuration: two thick coils having differentdimensions

The second configuration consists of two thick coils having differentdimensions. We use the following parameters:N1 = 400 and N2 = 800We take the following dimensions:r1 = 0.1 m, r2 = 0.2 m, r3 = 0.3 m, r4 = 0.4 m, z2 z1 = 0.2 m,

8 R. Ravaud et al

d [m] Filament method amperian current model Flux 3D5 filaments

0.3 0 0 00.15 70.1056457641 70.401487264358 700 76.7078644327 77.0031431689693 78+0.05 66.487894642 66.7305538188 67+0.3 22.2685946360 22.2794952389 21+1 1.678574304037 1.67154193012364+5.0 0.006895024572 0.0068086780910152

Table 3. Comparison between our analytical approach, the filamentmethod and the finite element method for calculating the axial force(mN) exerted between two thick coils having different dimensions.r1 = 0.1 m, r2 = 0.2 m, r3 = 0.3 m, r4 = 0.4 m, z2 z1 = 0.2 m,z4 z3 = 0.4 m, j1 = j2 = 20 000 A/m

2

-0.2 0 0.2 0.4 0.6 0.8 1d @mD

0

20

40

60

80

F@m

ND

Figure 3. Representation of the axial force exerted between twotoroidal conductors versus the axial distance d. We take the followingdimensions: r1 = r3=0.0875 m, r2 = r4=0.1125 m, z2 z1 =z4z3=0.025 m, j1 = j2=320 000 A/m

2. Line = our analytical method,Points = filament method.

z4 z3 = 0.4 m, j1 = j2 = 20 000 A/m2, d = axial distance between

the two thick coils [m]It is useful to mention that the volume densities j1 and j2 correspondto a current I = I1 = I2 that equals 1 A.

We represent in Fig.3 the axial force between two thick coils havingdifferent dimensions versus the axial displacement d and in Table 2the numerical results in some points. Table 3 and Fig.3 show thatthe three methods presented in this paper are in excellent agreement.

Mutual Inductance and Force Exerted Between Thick Coils 9

Nevertheless, our semi-analytical approach has a lower computationalcost than the finite element method or the filament method.

4. MUTUAL INDUCTANCE OF TWO THICK COILS

Let us first consider the expression of the axial magnetic field producedby the lower thick coil. By considering the number of loops of eachthick coil, the axial component of the magnetic field created by thelower coil is expressed as follows:

Hz(r, z) =N1I1

4S1

r2r1

21

z2z1

(r r cos())rdrddz(r2+ r2 2rr cos() + (z z)2

) 32

(12)

It is emphasized here that the calculation of the force requiresthe knowledge of the radial component created by the lower thickcoil whereas the calculation of the mutual inductance requires theknowledge of the magnetic field axial component created by this lowerthick coil.The flux across one elementary loop of the second thick coil whoseradius is r is expressed as follows:

=

2pi0

r

0Hz(r, z)rd

(13)

By using the Stokes Theorem, we can write that:

=

2pi0

r

0

{0 ~A

}

rd (14)

where ~A is the vector potential created by the lower thick coil. Theprevious relation can be transformed as follows:

(r, z) = 20{A

(r, z

)}(15)

We obtain:

(r, z) =0N1I1

2S1

r2r1

21

z2z1

cos()rrdrddzr2 + r 2rr cos() + (z z)2

(16)

10 R. Ravaud et al

The total magnetic flux across the second thick coil is given as follows:

=0N1N2I1

2S1S2

r4r3

z4z3

r2r1

21

z2z1

cos()rrdrddzr2 + r 2rr cos() + (z z)2

drdz

(17)The mutual inductance can be deducted from the previous expressionas follows:

M =

I1=

0N1N2

2S1S2

r4r3

z4z3

r2r1

21

z2z1

cos()rrdrddzr2 + r 2rr cos() + (z z)2

drdz

(18)After integrating with respect to r, z and z, we obtain a semi-analyticalexpression based on only two numerical integrations:

M =0

8

N1N2

S1S2

2pi0

r4r3

2i=1

(1)(i)4

j,k=3

(1)(j+k)rf(ri, zj , zk, cos())drd

(19)with

f(ri, zj , zk, x) =2

3

(2r2i rirx+ r

2(2 3x2) (zj zk)2)

2r2(zj zk) ln [zj zk + ]

+r2i (zj zk) + 2rirx(zj zk)

4rxr2(2)(zj zk) arctan

[ri rx

r2(1 x2)

]

+r2(2x2 1)(zj zk) ln[r2i + r

2 2rirx

]

2rx(r22 + (zj zk)

2)ln [ri rx+ ]

+r2

(2x 2x3 + 2x2

)(zj zk) ln [A1]

+r2

(2x+ 2x3 + 2x2

)(zj zk) ln [A2]

where

A1 =rir

2 + r2(+x2 1)( x) (zj zk)(zj zk + )

r2 (+2x 2x3 + 2x2) (+ri + r( x)) (zj zk)3

A2 =rir

2 r2(x2 1)( + x) (zj zk)(zj zk + )

r2 (2x 2x3 + 2x2) (ri + r(+ x)) (zj zk)3

(20)

Mutual Inductance and Force Exerted Between Thick Coils 11

0 0.1 0.2 0.3 0.4 0.5d @mD

0

0.2

0.4

0.6

M@m

HD

Figure 4. Representation of the mutual inductance between two thickcoils versus the axial distance d; r1 = r3=0.025 m, r2 = r4=0.03 m,z2 z1 = z4 z3=0.025 m, N1 = N2=200.

d [m] Kajikawa amperian current model0 0.776 0.77539040.005 0.571 0.57128720.01 0.435 0.43483170.02 0.267 0.26677990.03 0.173 0.17295280.05 0.0823 0.08232390.1 0.021 0.02096590.15 0.00803 0.00802930.2 0.00386 0.00385790.3 0.0013 0.001301990.4 0.000586 0.00058600.5 0.000312 0.0003119

Table 4. Comparison between our analytical model, the approach ofKajikawa[28] and the finite element method for calculating the mutualinductance (mH). r1=0.025 m, r2=0.03 m, r3=0.025 m, r4=0.03 m,z2 z1=0.025 m, z4 z3=0.025 m, N1=200, N2=200

We represent in Fig.4 the mutual inductance versus the axial distanced with our analytical method and in Table 4 the numerical resultsobtained with the finite element method, the approach of Kajikawaand our analytical approach. Table 4 and Fig.4 clearly show that allthe results are in excellent agreement. This confirms the accuracy ofour analytical approach.

12 R. Ravaud et al

5. CONCLUSION

We have presented exact semi-analytical expressions of the forceexerted between two thick coils carrying uniform current volumedensities and of their mutual inductance. For this purpose, wehave replaced each thick coil by a toroidal conductor having uniformcurrent volume density. Our expressions are based on two numericalintegrations of continuous and derivable functions. Consequently, theircomputational cost remain very low compared to the finite elementmethod and are also lower than the filament method. This exactexpression can be used for calculating the force exerted between twothick coils in air as our results are in excellent agreement with thefilament method and the finite element method.

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