Magneitc Force Between Inclined Filaments Placed in Any Desired PositionMagnetic Force Between Inclined Circular Filaments Placed in Any Desired Position Slobodan Babic1 and Cevdet

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IEEE TRANSACTIONS ON MAGNETICS, VOL. 48, NO. 1, JANUARY 2012

Magnetic Force Between Inclined Circular Filaments Placedin Any Desired PositionSlobodan Babic1 and Cevdet Akyel2

Département de Génie Physique,École Polytechnique de Montréal, Montréal, QC H3C3A7, CanadaDépartement de Génie Électrique, École Polytechnique de Montréal, Montréal, QC H3C3A7, Canada

This paper presents new general formulas for calculating the magnetic force between inclined circular laments placed in any de-sired position. We use two approaches to calculate the magnetic force, one based on Biot-Savart law and another based on the mutualinductance between these coils. All mathematical procedures are completely described to dene coil positions that lead to relatively easymethod for calculating the magnetic force between inclined circular laments in any desired position. Two formulas obtained by thesimple integration of different expressions in terms of complete elliptic integrals of the rst and second kind give the same numericalresults. The presented methods are understandable, numerically suitable and easy applicable for engineers and physicists. We validatedthe new formulas through a series of examples, which are presented here.

Index Terms— Biot-Savart law, circular laments, magnetic force, mutual inductance.

I. INTRODUCTION

THE magnetic force calculation between coaxial circularlaments has been thoroughly treated by a number of

authors since the time of Maxwell [1]–[16]. According to ourknowledge up until now there have been few papers or booksthat deal with the magnetic force calculation between circularcoils with parallel or inclined axes [17]–[27]. Today, with pow-erful numerical methods, such as nite element method (FEM)and boundary element method (BEM), [28], [29], it is possibleto accurately and rapidly calculate this important physicalquantity. Also, this problem can be tackled by semianalyticalmethods that can considerably reduce the computational timeand the enormous mathematical procedures. In this paper,we give the general approach to calculate the magnetic force

between inclined circular laments in any desired positions.We treated the most general case using some elementary math-ematical transformations to describe coil positions in differentplanes. The magnetic force has been obtained by a simpleintegral whose kernel function contains some combinations of the complete elliptic integrals of the rst and second kind. Weused two approaches to calculate the magnetic force betweeninclined circular loops one based on Biot-Savart law and an-other based on the mutual inductance between these loops.

The obtained new formulas for the magnetic force are verysuitable for numerical treatment. Presented formulas can beeasily used in the calculation of the magnetic force betweeninclined circular coils of rectangular cross section using thelament method.

II. BASICEXPRESSIONS

Let’s take into consideration twocircular laments as showedin Fig. 1, where the center of the larger circle (primary coil) of the radius is placed at the plane with the axis of along the axis of this circle. The smaller circle (secondary coil)

Manuscript received June 29, 2011; revised August 03, 2011; acceptedAugust 07, 2011. Date of publication August 18, 2011; date of current versionDecember 23, 2011. Corresponding author: S. Babic (e-mail: [email protected]).

Digital Object Identier 10.1109/TMAG.2011.2165077

Fig. 1. Filamentary circular coils with angular and lateral misalignment (mogeneral case).

of the radius is placed in an inclined plane whose generaequation is

(1)

Also, we dene the center of the secondary coil in the plan, and a point on this circle, .

The coordinates of the point can be given by the couple

orwhere ,

, [26].Thus, we dene the positions of two coils in 3-D space tha

will permit us to calculate all necessary parameters in the calculation of the magnetic force between them.

For coils (see Fig. 1), we dene:1) The primary coil of radius is placed in the plane

with the center at .An arbitrary point of this coil has para-metric coordinates

(2)0018-9464/$26.00 © 2011 IEEE



70 IEEE TRANSACTIONS ON MAGNETICS, VOL. 48, NO. 1, JANUARY 20

2) The differential element of the primary coil is given by

(3)

3) The unit vector at the point (center of the secondary coil)laying in the plane , is dened by

(4)

4) The unit vector between two points and that areplaced in the plane is

(5)

5) We dene the unit vector as the cross product of the unitvectors and as follows:

(6)

6) An arbitrary point of the secondary coilhas parametric coordinates

(7)

This is thewell-known parametricequation of circle in 3-Dspace.

7) The differential element of the secondary coil is given by

(8)

III. CALCULATIONMETHOD

A. General Case (Biot-Savart Approach)

The magnetic force between two circular laments can becalculated by the Biot-Savart law [4]

(9)

where

Circular laments carry corresponding currents and. —the permeability of free space

(vacuum). The magnetic force components in , and directions are given in the nal form (See Appendix I)

(10)

where



BABIC AND AKYEL: MAGNETIC FORCE BETWEEN INCLINED CIRCULAR FILAMENTS PLACED IN ANY DESIRED POSITION

complete elliptic integral of the rst kind [31],[32];complete elliptic integral of the second kind[31], [32].

It is necessaryto payattention topossiblesingular cases.Eventhough thin circular loops are of negligible cross section, themagnetic force can be singular if loops touch or overlap. In thecalculation, some singularities cannot be avoided because theyare natural singularities caused by negligible thickness of coils.This is the way all possible singular cases haveto bechecked andtreated carefully to nd the corresponding magnetic force. Wegive special cases to cover all calculation possibilities.

1) Special Cases:a) , (The secondary coil is parallel to the

plane ).This special case is not covered by the general case (10)and it is necessary to carefully solve it. For this specialcase we keep (10) with changes

b) , (The secondary coil is parallel tothe plane ). For this special case we keep (10) withchanges

c) , (The secondary coil is parallel tothe plane ). For this special case, we keep (10) withchanges

In the case of two coaxial coils (the center of the sec-ondary coil is laying on the axis, ), themagnetic force components are

(11)

B. General Case (Mutual Inductance Approach)

The mutual inductance between two inclined circular laments (see Fig. 1) in any desired position can be calculateby [26]

(12)




where

The magnetic force between two inclined circular lamentsin any desired position can be calculated by [1]

(13)

where and are currents in the coils, is their mutualinductance, and or

(14)

Finding the rst derivatives in (14) (see Appendix II), themagnetic force can be obtained by the following components:

(15)

where

1) Special Cases:

a) , (The secondary coil is parallel to theplane ).




This special case is not covered by the general case (15)and it is necessary to carefully solve it. For this specialcase we keep (15) with changes

where

b) , (The secondary coil is parallel tothe plane ). For this special case we keep (15) withchanges

c) , (The secondary coil is parallel tothe plane ). For this special case we keep (15) withchanges

In the case of two coaxial coils (the center of the sec-ondary coil is laying on the axis, ), themagnetic force components are

(16)

where

IV. EXAMPLES

To verify the validity of the new formulas, we present thfollowing set of examples.

Example 1: Thecenterof the primary coilof the radiusis and the center of the secondary coil of the

radius is. The secondary coil is located in the plane

. Calculate themagnetic force between coils. Allcurrents areunit.




The magnetic force obtained by (10) (Biot-Savart approach)is

The magnetic force obtained by (15) (mutual inductance ap-proach) is

Example 2: The center of the primary coil of of the radiusis and the center of the secondary coil of

the radius is

. The secondary coil is located in the plane. Calculate themagnetic force between coils. All currents areunit.



Example 3: The centerof the primary coil of the radiusis and the center of the secondary coil of the


. Calculate the magnetic force between coils. All currents areunit.



Example 4: The center of the primary coil of the ra-dius is and the center of the secondary coil of the radius is

. Thesecondary coil is located in the plane .Calculate the magnetic force between coils. All currents ar

unit.The magnetic force obtained by (10) (Biot-Savart approachis

The magnetic force obtained by (15) (mutual inductance approach) is



. Calculate themagnetic force between coils. All currents areunit but of the opposite sign.

The magnetic force obtained by (10) (Biot-Savart approachis

The magnetic force obtained by (15) (Mutual inductance approach) is

Example 6: In this example, we calculated the restoring (ra-dial) magnetic force and the propulsive (axial) magnetiforce between the primary circular coiland the secondary circular coil with the axial dis-placement in the function of the distance betweenplanes , [30]. All currents are equal to 1 A.

In this example we have two circular coils with parallel axeUsing the presented methods in this paper wecan use either (10or (15) for which we have: ; ;

; ,between 0 and 11 mm; ; .

Prof. J. T. Conway (private communication) kindly providethe author with magnetic force results of an independently dveloped method [30]. Applying either (10) or (15) we obtai




TABLE IRESTORING(RADIAL) FORCE AS AFUNCTION OF THEDISPLACEMENTd

OF TWO NON-COAXIALFILAMENTARYCOILS FORTHEAXIALDISPLACEMENTp = 3 m m

TABLE IIPROPULSIVEFORCE AS AFUNCTION OF THEDISPLACEMENTd

OF TWO NON-COAXIALFILAMENTARYCOILS FORTHEAXIALDISPLACEMENTp = 3 m m

the magnetic force results given in Table I and Table II for therestoring (radial) and propulsive (axial) force. All results arein an excellent agreement with those obtained by Conway’s

method. Example 7: In Table III we give comparative calculation fortwo unequal loops and with the distancebetween loops planes as a function of the axialdisplacement [30].

For these circular coils we have: ; ;; ; ;

.From this example, we can give some conclusions about pos-

sible singularities. The possible singularities could appear in thepoints where coplanar coils touch ( and ).In these points, we obtained the values of the magnetic forceconsiderably increased regarding to values for other points of the calculation. Obviously, the numeric calculations in thesepoints can be erroneous because of possible singularities. (See

TABLE IIIMAGNETICFORCE F FOR Z = d = 1 0 m

magnetic forces obtained by two comparative approaches fom and m.)

From Tables III and IV, we can see that all results are in aexcellent agreementwhen coils are in parallel planeswhose dis-tance is innitesimally small. We have practically two loops ithe same plane. To have the exact information about the magnetic force components for loops which are in the same planwhere they touchor overlap it is necessary to nd limits of thesexpressions if they exist. Inspecting presented expressions (10or (15), singularities appear if module for which theelliptic integral of the rst kind . In these cases,integrals are not absolutely convergent because of oscillationof the integrand [4].

Example 8: In the following examples, we verify the validityof each special case obtained by two different approaches.The center of the primary coil of the radius is

and the center of the secondary coil of the radiusis . The

secondary coil is located in the plane m. Calculate themagnetic force between coils. All currents are unit.

Applying the general case (10) or (15) the components of thmagnetic force are




TABLE IVMAGNETICFORCE F FOR Z = d = 1 0 m

Applying the special case (10 (c)) or (15 (c)),the components of the magnetic force are

The special cases (10 (c)) or (15 (c)) are directly included inthe general case (10) or (15).


radius is .The secondary coil is located in the plane m. Calculatethe magnetic force between coils. All currents are unit.

Applying the general case (10) or (15), the components of themagnetic force are

Applyingthe specialcase (10 (b))or (15 (b)),, the components of the magnetic force are

The special case (10 (b)) or (15 (b)) are directly included ithe general case (10) or (15) also.

Example 10: The center of the primary coil of the radius

is and the center of the secondary coil ofthe radius is .The secondary coil is located in the plane m. Calculatthe magnetic force between coils. All currents are unit.

This case is singular if the general case (10)or (15) is directly applied. To verify the validity of the speciacases (10(a)) or (15(a)), let’s put in (10) or (15) an

or and .Applying the general case (10) or (15), the components of th

magnetic force are

Applying directly the special case (10 (a)) or (15 (a)),the components of the magnetic force are

Signicant gures which agree are bolded. Thus, we con

rmed the validity of this singular case (10(a)) or (15(a)). Froprevious examples, we can conclude that all results obtained btwo approaches for calculating the magnetic force between cicular loops positioned inany desired positionare in an excellenagreement that conrm the validity of both of them. All calculations were executed using Mathematica programming. Botmethods presented in this paper give thesame numerical resultsAlso in Matlabprogramming using Romberg numerical integration both methods give the same numerical results. In all calculations of the magnetic force components, the average computational time was about 0.0075 s. A Mathematica implementatioof previous formulas is available from the authors on request

V. CONCLUSION

In this paper, we give new formulas for calculating the magnetic force between inclined circular laments placed in any dsired positions. The most general formulas are derived eithefrom the mutual inductance between inclined circular lamentplaced in any desired position or by direct application of thBiot-Savart law. In order to use new formulas, whose nal expressions are given per (10) or per (15), one needs to providthe radius of the primary and secondary coils, the position othe center of the secondary coil (the primary coil is assumeto be centered at origin), and the plane equation with all unvectors in which the secondary coil is located. With these parameters, the problem is completely dened. All possible casewere tested with new formulas, and none of them failed. W




note that our new formulas are general, very suitable, and easilyapplicable for engineers and physicists. In these formulas, ker-nels are relatively simple and expressed by elliptic integrals of the rst and second kind so that their integration using Math-ematica or Matlab programming is accurate with signicantlyreduced computational time.

APPENDIXIIn (9) the double cross product gives

so that the magnetic force components in , and directionsare,

where

The rst integration is made regarding to the variable witsubstitution [32], where

The rst integration gives

where




Solving twointegrals andusing thefollowing transformations[31], [32]:

we obtain

I x

=

4 k

( 2 p )3

2 ( M 1

s i n + N 1

c o s ) K ( k )

0

2 ( M 1

s i n + N 1

c o s ) 0 ( P 1

+ M 1

s i n + N 1

c o s ) k

2

1 0 k

2

E ( k )

=

4 k

( 2 p )

3

9 1

( ; k ) :

In the same way we obtainI

y =

2

0

( M 2

s i n t + N 2

c o s t + P 2

)

r

d t

=

4 k

( 2 p )

3

2 ( M 2

s i n + N 2

c o s ) K ( k )

2 ( M 2

s i n + N 2

c o s ) 0 ( P 2

+ M 2

s i n + N 2

c o s ) k

2

1 0 k

2

E ( k )

=

4 k

( 2 p )

3

9 2

( ; k )

I z

=

2

0

( M 3

s i n t + N 3

c o s t )

r

d t

=

4 k ( M 3

s i n + N 3

c o s )

( 2 p )3

2 K ( k ) 0

2 0 k

2

1 0 k 2

E ( k )

=

4 k

( 2 p )

3

9 3

( ; k )

where9

1( ; k ) = 2 ( M

1s i n + N

1c o s ) K ( k )

0

2 ( M 1

s i n + N 1

c o s ) 0 ( P 1

+ M 1

s i n + N 1

c o s ) k

2

1 0 k 2

E ( k )

9 2

( ; k ) = 2 ( M 2

s i n + N 2

c o s ) K ( k )

0

2 ( M 2

s i n + N 2

c o s ) 0 ( P 2

+ M 2

s i n + N 2

c o s ) k

2

1 0 k

2

E ( k )

9 3

( ; k ) = ( M 3

s i n + N 3

c o s ) 2 K ( k ) 0

2 0 k

2

1 0 k 2

E ( k ) :

The magnetic force components in , and directions aregiven in the nal form

where

For the special cases we use the same reasoning to calculatthe three components of the magnetic force as in the generacase.

APPENDIXIImagnetic force component:




where

magnetic force component:In the similar way for the component we obtain

where

magnetic force component:In the similar way for the component we obtain

For the special cases, we use the same reasoning to calculatthe three components of the magnetic force as in the generacase.

ACKNOWLEDGMENT

This work was supported by the Natural Science and En

gineering Research Council of Canada (NSERC) under GranRGPIN 4476-05 NSERC NIP 11963.The authors would like to thank Prof. J. T. Conway of the

University of Agder, Grimstad, Norway for providing very higprecision calculations for the magnetic force calculation, whichave proven invaluable in validating the methods presentedhere.

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Magneitc Force Between Inclined Filaments Placed in Any Desired PositionMagnetic Force Between Inclined Circular Filaments Placed in Any Desired Position Slobodan Babic1 and Cevdet