# Calculation of Mutual Inductance between Circular and Arbitrarily Shaped
Filaments via Segmentation Method

Kirill V. Poletkin, Slobodan Babic, Sreejith Sasi Kumar and Emil R. Mamleyev
E. R. Mamleyev and K. V. Poletkin are with Institute of Microstructure
Technology, Karlsruhe Institute of Technology, Hermann-von-Helmholtz-Platz 1,
76344 Eggenstein-Leopoldshafen, Germany.K. V. Poletkin is also with New
Uzbekistan University, Mustaqillik ave. 54, 100007 Tashkent, Uzbekistan
e-mail<EMAIL_ADDRESS>Babic is an independent researcher; 53 Berlioz
101, H3E 1N2, Montréal, Québec, Canada e-mail<EMAIL_ADDRESS>S. Kumar
is with Technische Universität Ilmenau, Ehrenbergstraße 29, 98693 Ilmenau,
Germany.

###### Abstract

In this article, two analytical formulas for the calculation of mutual
inductance between a circular filament and line segment arbitrarily
positioning in the space are derived by using Mutual Inductance Method (MIM)
and Babic’s Method (BM), respectively. Using the fact that any curve can be
interpolated by a set of line segments, a method for calculation of mutual
inductance between a circular filament and filament having an arbitrary shape
in the space is proposed based on the derived analytical formulas. The derived
two formulas and the proposed method (Segmentation Method) were numerically
validated by using FastHenry software and reference examples from the
literature. In particular, the proposed method was successfully applied to the
calculation of mutual inductance between the circular filament and the
following special curves such as circle, circular arc, elliptic arc, ellipse,
spiral, helices and conical helices. All results of the calculation are in
good agreement with the reference examples.

###### Index Terms:

mutual inductance, analytical formula, circular filament, special functions,
complete elliptic functions, segment.

##  I Introduction

Analytical and semi-analytical methods in the calculation of self- and mutual-
inductances of conducting elements of electrical circuits have played an
important role in development of power transfer, wireless communication,
sensing and actuation and have been applied in a broad fields of science,
including electrical and electronic engineering, medicine, physics, nuclear
magnetic resonance, mechatronics and robotics, to designate the most
prominent.

Although, a number of efficient numerical methodes implemented in the
commercially developed software are available, analytical and semi-analytical
methods allow to obtaining the result of calculations in the form of a final
formula with a finite number of input parameters, which when applicable may
significantly reduce computation effort. Providing the direct access to a
calculational formula for a user in such methods facilitates mathematical
analysis of obtained results of calculation and opens an opportunity for their
further optimization.

In particular, calculation of mutual inductance between a circular filament
and systems of filaments of different shapes is a prime example of application
of such analytical methods. Albeit, collections of formulas for the
calculation of mutual inductance between filaments of different geometrical
shapes covering a wide spectrum of practical arrangements have variously been
presented in classical handbooks by Rosa[1], Grover[2], Dwight[3], Snow [4],
Zeitlin[5], Kalantarov [6], among others. However, these analytical methods
have proved their efficiency and have been successfully employed in an
increasing number of applications, including electromagnetic levitation [7,
8], superconducting levitation [9], calculation of mutual inductance between
thick coils [10, 11, 12, 13, 14, 15], magnetic force and torque calculation
between circular coils [16, 17, 18, 19, 20], calculation of magnetic stiffness
[21, 22], wireless power transfer [23, 24, 25], electromagnetic actuation [26,
27, 28], micro-machined contactless inductive suspensions [29, 30, 31] and
hybrid contactless suspensions [32, 33, 34, 35], biomedical applications [36,
37], topology optimization of coils [38], nuclear magnetic resonance [39, 40],
indoor positioning systems [41], navigation sensors [42], wireless power
transfer systems [43, 44], magneto-inductive wireless communications [45] and
others.

Kalantarov and Zeitlin showed that the calculation of mutual inductance
between a circular primary filament and any other secondary filament having an
arbitrary shape and any desired position with respect to the primary filament
can be reduced to a line integral [6, Sec. 1-12, page 49]. Adapting this
result, Poletkin derived the analytical formulas for calculating the mutual
inductance between two circular filaments having any desired position with
respect to each other in work [12], as an alternative to Grover’s and Babic’s
expressions reported in works [2] and [11], respectively.

Moreover, it was mentioned in work [12] that the obtained formula for the
treatment of the singular case, when the circular filaments are mutually
perpendicular, can be applied also to the calculation of the mutual inductance
between a circle and line segment after taking a minor modification of the
original formula. Using this fact and an alternative approach developed by
Babic [11], two analytical formulas for calculation of mutual inductance
between a circular filament and line segment arbitrarily positioning in the
space are derived by means of MIM and BM, respectively.

Since, any curve can be interpolated with a desired accuracy by a finite
number of line segments, a segmentation method for calculation of the mutual
inductance between the primary circle and a filament having an arbitrary shape
is proposed and successfully developed based on the two derived formulas for
calculation of the mutual inductance between the circle and line segment. It
is shown that for calculation of mutual inductance by means of the
segmentation method, a set of points belonging to the arbitrarily shaped
filament as the input data is only needed. The developed methodology is
successfully applied to the calculation of mutual inductance between the
circular filament and the following special curves such as circle, circular
arc, elliptic arc, ellipse, spiral, helices and conical helices. All results
of calculation are in a good agreement with the reference examples and
successfully validated by the FastHenry software [46] .

Figure 1: General scheme of arbitrarily positioning the line segment with
respect to the circular filament: $R_{p}$ is the radius of the circle and
$L_{s}$ is the length of the line segment.

##  II Mutual Inductance between Circular Filament and Line Segment

Let us consider the filament system consisting of a primary circular filament
having a radius of $R_{p}$ and a secondary filament represented as a line
segment having a length of $L_{s}$, which is arbitrarily positioned in the
space with respect to the primary circle as shown in Fig. 1. A coordinate
frame (CF) denoted as $XYZ$ is assigned to the primary circle in a such way
that the $Z$ axis is coincident with the circle axis and the $XOY$ plane of
the CF lies on circle’s plane, where the origin $O$ corresponds to the centre
of primary circle. In turn, the $xyz$ CF is assigned to the secondary filament
so that its origin $C$ is attached to one of its endpoints. The axes of the
$xyz$ CF are parallel to the axes of $XYZ$, respectively.

The linear position of the secondary filament with respect to the primary one
is defined by the coordinates of the origin $C$ ($x_{C},y_{C},z_{C}$). The
angular position of the line segment is defined by the angle $\eta$ and
$\theta$ corresponding to the angular rotation around the $z$-axis of the
plane passing through the $z$-axis and line segment, and then the rotation of
the segment around an axis, which is perpendicular to this segment-$z$ plane
and passing through the origin $C$, respectively, as shown in Fig. 1.

###  II-A Mutual Inductance Method (MIM)

The mutual inductance between the circular filament and the line segment
arbitrarily orientated in the space can be calculated by the following
formula, which is derived by using Kalantarov-Zeitlin’s approach reported in
work [12]. Introducing the following dimensionless coordinates:

${\displaystyle{x}=\frac{x_{C}}{L_{s}},\;{y}=\frac{y_{C}}{L_{s}},\;{z}=\frac{z_{C}}{L_{s}},{s}=\sqrt{{x}^{2}+{y}^{2}},}\\\
$ (1)

the formula can be written as

$\begin{array}[]{l}{\displaystyle
M=\frac{\mu_{0}\sqrt{R_{p}L_{s}}}{\pi}\int_{0}^{1}U\cdot\Phi(k)d\bar{\ell}},\end{array}$
(2)

where $\mu_{0}$ is the magnetic permeability of free space,
$\bar{\ell}=\ell/L_{s}$ is the dimensionless integrating variable,

$U=U({x},{y},\eta,\theta)=\frac{t_{1}-t_{2}}{{\rho}^{1.5}}\cdot\cos\theta,$
(3)
$\begin{array}[]{l}t_{1}=t_{1}({x},\eta,\theta)=\sin\eta\cdot({x}+\bar{\ell}\cos\theta\cos\eta),\\\
t_{2}=t_{2}({y},\eta,\theta)=\cos\eta\cdot({y}+\bar{\ell}\cos\theta\sin\eta),\\\
{\rho}={\rho}({x},{y},\eta,\theta)=\\\
\sqrt{{s}^{2}+2\bar{\ell}\cos\theta\cdot\left({x}\cos(\eta)+{y}\sin(\eta)\right)+\bar{\ell}^{2}\cos^{2}\theta},\end{array}$
(4) $\Phi(k)=\frac{1}{k}\left[\left(1-\frac{k^{2}}{2}\right)K(k)-E(k)\right],$
(5)

and $K(k)$ and $E(k)$ are the complete elliptic functions of the first and
second kind, respectively, and

$\begin{array}[]{l}{\displaystyle
k^{2}=k^{2}({x},{y},{z},\theta,\eta)=\frac{4\nu{\rho}}{(\nu{\rho}+1)^{2}+\nu^{2}{z}_{\lambda}^{2}},}\\\
{\displaystyle\nu=L_{s}/R_{p},\;{z}_{\lambda}={z}+\bar{\ell}\sin\theta}.\end{array}$
(6)

###  II-B Babic’s Method (BM)

Let us consider again the calculation scheme shown in Fig. 1, but in a
difference with MIM the line segment is given by coordinates of its two
endpoints, namely, $p_{0}(x_{0},y_{0},z_{0})$ and $p_{1}(x_{1},y_{1},z_{1})$.

Then, the coordinates of an arbitrary point $S$ $(x_{s},y_{s},z_{s})$ of the
secondary line segment can be written parametrically as follows:

$x_{s}=q\bar{\ell}+x_{0},\;y_{s}=r\bar{\ell}+y_{0},\;z_{s}=s\bar{\ell}+z_{0},\;\bar{\ell}\in[0,1],$
(7)

where $q=x_{1}-x_{0}$, $r=y_{1}-y_{0}$, $s=z_{1}-z_{0}$. Hence, the
differential element of the line segment becomes

$d\vec{l}_{s}=\\{q,r,s\\}d\bar{\ell},\;\;\bar{\ell}\in[0,1].$ (8)

While, parametric coordinates of an arbitrary point $P$ $(x_{p},y_{p},z_{p})$
of the primary circular filament can be defined by the following expressions:

$x_{p}=R_{p}\cos\phi,\;y_{p}=R_{p}\sin\phi,\;z_{p}=0,\;\phi\in[0,2\pi].$ (9)

The differential of the primary circular filament is given by

$d\vec{l}_{p}=R_{p}\\{-\sin\phi,\cos\phi,0\\}d\phi,\;\phi\in[0,2\pi].$ (10)

Accounting for equations (10) and (8), the mutual inductance between circular
filament and the line segment can be calculated by

$\begin{array}[]{c}{\displaystyle
M=\frac{\mu_{0}}{4\pi}\int_{0}^{1}\int_{0}^{2\pi}\frac{d\vec{l}_{p}\cdot
d\vec{l}_{s}}{r_{ps}}=}\\\
{\displaystyle\frac{\mu_{0}R_{p}}{4\pi}\int_{0}^{1}\int_{0}^{2\pi}\frac{-q\sin\phi+r\cos\phi}{r_{ps}}d\phi
d\bar{\ell}},\end{array}$ (11)

where

$r_{ps}^{2}=\left(x_{s}-R_{p}\cos(\phi)\right)^{2}+\left(y_{s}-R_{p}\sin(\phi)\right)^{2}+z_{s}^{2}.$
(12)

The following integral

$\begin{array}[]{c}{\displaystyle
I=\int_{0}^{2\pi}\frac{-q\sin\phi+r\cos\phi}{r_{ps}}d\phi}\end{array}$ (13)

can be expressed via the complete elliptic functions of the first and second
kind $K(k)$ and $E(k)$, respectively, as follows

$\begin{array}[]{c}{\displaystyle
I=\frac{2V}{\sqrt{R_{p}}\sqrt{p^{3}}}\Psi(k),}\end{array}$ (14)

where

$\begin{array}[]{l}{\displaystyle V=r\cdot x_{s}-q\cdot
y_{s},\;p=\sqrt{x_{s}^{2}+y_{s}^{2}},}\\\
{\displaystyle\Psi(k)=\left(\frac{2}{k}-k\right)K(k)-\frac{2}{k}E(k),}\\\
{\displaystyle k^{2}=\frac{4R_{p}p}{(R_{p}+p)^{2}+{z}_{s}^{2}}.}\end{array}$
(15)

Hence, replacing the integral (13) in (11) by (14), the final formula for
calculation of mutual inductance becomes

$\begin{array}[]{c}{\displaystyle
M=\frac{\mu_{0}\sqrt{R_{p}}}{2\pi}\int_{0}^{1}\frac{V}{\sqrt{p^{3}}}\Psi(k)d\bar{\ell}}.\\\
\end{array}$ (16)

The input parameters for Babic’s and MI method are related by the following
relationships:

$\begin{array}[]{l}{\displaystyle L^{2}_{s}=q^{2}+r^{2}+s^{2}},\\\
{\displaystyle\eta=\tan^{-1}\left(\frac{r}{q}\right)},\;{\displaystyle\theta=\tan^{-1}\left(\frac{s}{\sqrt{q^{2}+r^{2}}}\right)}.\end{array}$
(17)

Worth noting that according to (2) and (16) the mutual inductance is equal to
zero, when the line segment crosses the $Z$ axis or is located on a line
crossing the $Z$ axis.

Figure 2: General scheme of arbitrary shaped filament and the circular
filament.

##  III  Segmentation Method

The secondary filament having an arbitrary shape and the primary circular
filament of a radius $R_{p}$ are considered as shown in Fig. 2. The secondary
filament is given, for instance, by a 3D parametric curve
$\sigma=\sigma(\ell)$, where $\ell$ is the parameter defined within a finite
interval $\left[\varphi_{0},\varphi_{1}\right]$, and the following inequality
$\varphi_{1}>\varphi_{0}\geq 0$ is valid.

Figure 3: Interpolation of the $\sigma$ curve by line segments.

It is assumed that the curve is sampled by $n$ points, so that we have
$\underline{p}_{0}(h_{0})$, $\underline{p}_{1}(h_{1})$,
$\underline{p}_{2}(h_{2})$, $\ldots$ , $\underline{p}_{n-1}(h_{n-1})$,
$\underline{p}_{n}(h_{n})$. The points can be defined in the following way

$\underline{p}_{i}(h_{i})=\left[x_{\sigma}(h_{i})\;y_{\sigma}(h_{i})\;z_{\sigma}(h_{i})\right]^{T},\;i=0\ldots
n,$ (18)

as ($3\times 1$)-column-matrices, where $x_{\sigma}$, $y_{\sigma}$ and
$z_{\sigma}$ are the coordinates of the $\sigma$-curve and

$h_{i}=h_{i-1}+(\varphi_{1}-\varphi_{0})/n,\;i=1\ldots n,$ (19)

and so $h_{0}=\varphi_{0}$. Using the set of points (18) and (19), the
$\sigma$-curve can be interpolated by line segments as shown in Fig. 3. The
length of each line segment can be calculated by

$L_{si}^{2}=(\underline{p}_{i}-\underline{p}_{i-1})^{T}(\underline{p}_{i}-\underline{p}_{i-1}),$
(20)

while the angles and dimensionless coordinates of $i$-th line segment become

$\begin{array}[]{c}{\displaystyle\eta_{i}=\tan^{-1}\left(\frac{\underline{p}_{yi}-\underline{p}_{yi-1}}{\underline{p}_{xi}-\underline{p}_{xi-1}}\right),}\\\
{\displaystyle\theta_{i}=\tan^{-1}\left(\frac{\underline{p}_{zi}-\underline{p}_{zi-1}}{\sqrt{\left(\underline{p}_{xi}-\underline{p}_{xi-1}\right)^{2}+\left(\underline{p}_{yi}-\underline{p}_{yi-1}\right)^{2}}}\right)};\end{array}$
(21)

and

${\displaystyle{x_{i}}=\frac{\underline{p}_{xi-1}}{L_{si}},\;{y_{i}}=\frac{\underline{p}_{yi-1}}{L_{si}},\;{z_{i}}=\frac{\underline{p}_{zi-1}}{L_{si}},{s_{i}}=\sqrt{{x_{i}}^{2}+{y_{i}}^{2}},}\\\
$ (22)

respectively. Substituting (20), (21) and (22) into (2), the mutual inductance
$M_{i}$ between the circular filament and $i$-th line segment can be
calculated. Hence, performing summation of all $n$ terms of $M_{i}$, the
formula for calculation of the mutual inductance between the arbitrary shape
and the primary circular filament can be written as follows

$\begin{array}[]{l}{\displaystyle
M_{\sigma}=\frac{\mu_{0}\sqrt{R_{p}}}{\pi}\int_{0}^{1}\sum_{i=1}^{n}\sqrt{L_{si}}\cdot
U_{i}\cdot\Phi(k_{i})d\bar{\ell}}.\end{array}$ (23)

Alternatively, the calculation can be performed by using Babic’s formula (16),
namely,

$\begin{array}[]{l}{\displaystyle
M_{\sigma}=\frac{\mu_{0}\sqrt{R_{p}}}{2\pi}\int_{0}^{1}\sum_{i=1}^{n}\frac{V_{i}}{\sqrt{p_{i}^{3}}}\Psi(k_{i})d\bar{\ell}}.\end{array}$
(24)

Thus, interpolating the parametric $\sigma$-curve by $n$ line segments, the
calculation of the mutual inductance between the arbitrary shape and the
primary circular filament can be performed by means of using developed
formulas, namely, (2) and (16), which are applied $n$ times to each
interpolating line segment followed by summation executed by (23) and (24) for
all obtained calculation results. The Matlab files with the implemented
formulas (2), (16), (23) and (24) are available from the authors as
supplementary materials to this article.

Figure 4: A regular polygon of $n$ sides arbitrarily positioned in the space
with respect to the circle of radius $R_{p}$.

##  IV  Examples of Calculation. Numerical Verification

In this section the developed formulas (2) and (16), as well as the proposed
segmentation method based on (23) and (24) are validated by using FastHenry
software and reference examples from the literature.

###  IV-A Mutual inductance of line segment and circle

####  Example 1

Let us suppose that the radius of primary circle is
$R_{p}$=$1\text{\,}\mathrm{m}$ and the line segment defined by two endpoints
having the following coordinates, namely,
$p_{0}($1\text{\,}\mathrm{m}$,$2\text{\,}\mathrm{m}$,$3\text{\,}\mathrm{m}$)$
and
$p_{1}($2\text{\,}\mathrm{m}$,$3\text{\,}\mathrm{m}$,$4\text{\,}\mathrm{m}$)$.
The results of calculation are

| FastHenry | BM, (16) | MIM, (2)
---|---|---|---
$M$, nH | $-3.53653$ | $-3.401894$ | $-3.401894$

####  Example 2

Let us suppose that the radius of primary circle is
$R_{p}$=$1\text{\,}\mathrm{m}$ and the line segment defined by two endpoints
having the following coordinates, namely,
$p_{0}($1\text{\,}\mathrm{m}$,$1\text{\,}\mathrm{m}$,$1\text{\,}\mathrm{m}$)$
and
$p_{1}($0\text{\,}\mathrm{m}$,$1\text{\,}\mathrm{m}$,$1\text{\,}\mathrm{m}$)$.
The segment is parallel to the plane of the circle. The results of calculation
are

| FastHenry | BM, (16) | MIM, (2)
---|---|---|---
$M$, nH | $69.4492$ | $69.51806$ | $69.51806$

####  Example 3

Let us suppose that the radius of primary circle is
$R_{p}$=$0.03\text{\,}\mathrm{m}$ and the line segment defined by two
endpoints having the following coordinates, namely,
$p_{0}($0.0175\text{\,}\mathrm{m}$,$-0.002\,990\,4\text{\,}\mathrm{m}$,$0.004\,019\,2\text{\,}\mathrm{m}$)$
and
$p_{1}($0.0025\text{\,}\mathrm{m}$,$0.022\,99\text{\,}\mathrm{m}$,$0.055\,981\text{\,}\mathrm{m}$)$.
The results of calculation are

| FastHenry | BM, (16) | MIM, (2)
---|---|---|---
$M$, nH | $1.82457$ | $1.83574$ | $1.83574$

Figure 5: Scheme for Example 4.

###  IV-B Mutual inductance between circle and a regular polygon of $n$ sides
arbitrarily positioned in the space

In this section, the mutual inductance between a regular polygon of $n$ sides,
which is arbitrarily positioned in the space with respect to the circle of
radius $R_{p}$ is calculated by segmentation method based on (23) and (24). It
is assumed that the polygon is inscribed in a circle of a radius $R_{\sigma}$
lying on the $xy$ plane as shown in Fig. 4. The parameter $\ell$ is defined
within the finite interval $\left[0,2\pi\right]$. This interval determines the
distribution of the polygon vertices on the $xy$ plane. In particular, $n$-th
vertex within such interval of the parameter $\ell$ is located on the $x$
axis. The linear misalignment of the polygon is defined by coordinates of
point $C$ corresponding to the centre of the secondary circle, while the
angular misalignment can be defined by Grover’s angles denoted by $\theta$ and
$\eta$ [12].

Then, input points can be calculated by

$\underline{p}_{i}(h_{i})=\left[\begin{array}[]{c}x_{C}\\\ y_{C}\\\
z_{C}\end{array}\right]+\underline{\Lambda}_{\theta}^{T}\underline{\Lambda}_{\eta}^{T}\left[\begin{array}[]{c}R_{\sigma}\cos
h_{i}\\\ R_{\sigma}\sin h_{i}\\\ 0\end{array}\right],\;i=0\ldots n,$ (25)

where

$\underline{\Lambda}_{\eta}=\left[\begin{array}[]{ccc}\cos\eta&\sin\eta&0\\\
-\sin\eta&\cos\eta&0\\\
0&0&1\end{array}\right],\;\underline{\Lambda}_{\theta}=\left[\begin{array}[]{ccc}1&0&0\\\
0&\cos\theta&\sin\theta\\\ 0&-\sin\theta&\cos\theta\end{array}\right].$ (26)

####  Example 4 (Example 1-3, page 50 in Kalantarov’s book [6])

The primary circle of radius $R_{p}$=$1\text{\,}\mathrm{m}$ and the square
filament of a side length of $1\text{\,}\mathrm{m}$ (the polygon of $n=4$
sides, $R_{\sigma}=\sqrt{2}\cdot$$0.5\text{\,}\mathrm{m}$) are considered. The
square is lying on the $XY$ plane and its origin coincides with point $O$ as
shown in Fig. 5. The results of calculation show

| Kalantarov’s | FastHenry | BM, | MIM,
---|---|---|---|---
| book |  | (24) | (23)
$M_{\sigma}$, $\mu$H | $0.880$ | $0.72949$ | $0.73075$ | $0.73075$

####  Example 5

Considering the same arrangement as in Example 4, but the centre of square
filament is located at point $C$ having the following coordinates
$x_{C}=y_{C}=$$0\text{\,}\mathrm{m}$ and
$z_{C}=$$1\text{\,}\mathrm{m}$$/\sqrt{2}$. The results of calculation are

| FastHenry | BM, (24) | MIM, (23)
---|---|---|---
$M_{\sigma}$, $\mu$H | $0.317264$ | $0.31754544$ | $0.31754544$

####  Example 6

The primary circle has a radius of $R_{p}=$$16\text{\,}\mathrm{cm}$. The
triangular filament is inscribed in a circle having a radius of
$R_{\sigma}=$$10\text{\,}\mathrm{cm}$. The centre of the secondary circle has
the following coordinates: $x_{C}=0$, $y_{C}=$$4.3301\text{\,}\mathrm{cm}$,
and $z_{C}=$$17.5\text{\,}\mathrm{cm}$. While, the angular misalignment is
defined by the following Grover’s angles, namely,
$\eta=$$45\text{\,}\mathrm{\SIUnitSymbolDegree}$ and
$\theta=$$60.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ as shown in Fig. 6. The
results of calculation are

| FastHenry | BM, (24) | MIM, (23)
---|---|---|---
$M_{\sigma}$, nH | $5.94445$ | $5.927887$ | $5.927887$

Figure 6: Scheme for Example 6.

####  Example 7

Using the same arrangement as in previous Example 6, the square polygon is
inscribed. The results of calculation are

| FastHenry | BM, (24) | MIM, (23)
---|---|---|---
$M_{\sigma}$, nH | $9.40649$ | $9.435092$ | $9.435092$

####  Example 8

The hexagon ($n=6$ sides) is inscribed in the secondary circle of the same
arrangement as in Example 6. The results of calculation are

| FastHenry | BM, (24) | MIM, (23)
---|---|---|---
$M_{\sigma}$, nH | $12.4966$ | $12.54164$ | $12.54164$

Figure 7: Scheme for Example 9.

####  Example 9

The primary circle has a radius of $R_{p}=$$40\text{\,}\mathrm{cm}$. The
triangular filament is inscribed in a circle having a radius of
$R_{\sigma}=$$10\text{\,}\mathrm{cm}$, which is perpendicular to the primary
circle that Grover’s angles of $\eta=$$0\text{\,}\mathrm{\SIUnitSymbolDegree}$
and $\theta=$$90.0\text{\,}\mathrm{\SIUnitSymbolDegree}$. The centre of the
secondary circle has the following coordinates: $x_{C}=0$,
$y_{C}=$$20\text{\,}\mathrm{cm}$, and $z_{C}=$$10\text{\,}\mathrm{cm}$ as
shown Fig. 7. The results of calculation become

| FastHenry | BM, (24) | MIM, (23)
---|---|---|---
$M_{\sigma}$, nH | $-4.65652$ | $-4.686079$ | $-4.686079$

####  Example 10

Using the same arrangement as in Example 9, the square polygon is inscribed.
The results of calculation are

| FastHenry | BM, (24) | MIM, (23)
---|---|---|---
$M_{\sigma}$, nH | $-7.07775$ | $-7.094651$ | $-7.094651$

####  Example 11

Using the same arrangement as in Example 9, the hexagon is inscribed. The
results of calculation are

| FastHenry | BM, (24) | MIM, (23)
---|---|---|---
$M_{\sigma}$, nH | $-9.0490$ | $-9.0334$ | $-9.0334$

###  IV-C Mutual inductance between two circular filaments

In this section, the method of segmentation is applied to the calculation of
mutual inductance between two circular filaments arbitrarily orientated in the
space with respect to each other. Similar to Sec.IV-B, the secondary circle is
lying on the $xy$ plane. The linear misalignment is defined by coordinates of
point $C$ corresponding to the centre of the secondary circle, and the angular
misalignment is defined by Grover’s angles denoted by $\theta$ and $\eta$
[12]. The results of the calculation are compared with results obtained by
means of analytical formulas developed by Babic in work [11] and Poletkin in
[12].

Figure 8: The required number of line segments for given relative error of
interpolation of a circle, $n=n(\varepsilon)$.

The accuracy of interpolation of a circle by line segments can be measured by
the relative error $\varepsilon$, which can be estimated as the deference
between the radius of the circle and a length of a line joining a centre of
the circle and mid point of a line segment divided by the radius:

$\varepsilon=1-\cos\frac{\pi}{n}.$ (27)

Form (27) follows the obvious conclusion that the larger the number of line
segments the smaller the error. The required number of line segments for the
given relative error can be also evaluated by (27) and presented, for
instance, in the logarithmic scale as shown in Fig. 8. As seen from Fig. 8, to
interpolate a circle with the relative error, for instance, less than
$10^{-3}$, the number of line segments must be more than 100.

For generating input points, equation (25) can be used with the only
difference that the number of line segments is being chosen in according to
the desired accuracy of interpolation of a circle, which in turn effects on
the accuracy of calculation of mutual inductance.

Figure 9: The relative error of calculation of mutual inductance versus the
relative error of interpolation of a circle by line segments.  Figure 10:
Arrangement of circles in the illustrative example.

Indeed, the accuracy of calculation of mutual inductance can be measured also
by the relative error, which is defined as follows

$\varepsilon_{\sigma}=\frac{M-M_{\sigma}}{M},$ (28)

where $M$ is the mutual inductance calculated by Babic’s or Poletkin’s formula
as mentioned above. Clearly that it is dependent on the number of line
segments. To exam this point let us consider the following geometrical
arrangement taken from Babic’s work [11, page 3597, Example 12], as an
illustrative example, in which two circles with radii
$R_{p}=$$16.0\text{\,}\mathrm{cm}$ and $R_{\sigma}=$$10.0\text{\,}\mathrm{cm}$
and the centre of the secondary $\sigma$-circle is located at $x_{C}=0$,
$y_{C}=$$4.3301\text{\,}\mathrm{cm}$, $z_{C}=$$17.5\text{\,}\mathrm{cm}$ and
the angle, $\theta$ of $60.0\text{\,}\mathrm{\SIUnitSymbolDegree}$, but the
$\eta$ angle is chosen to be 45.0$\mathrm{\SIUnitSymbolDegree}$ as shown in
Fig. 10. Performing calculation for different numbers of line segments, the
results are tabulated as follows

$n$ | $\varepsilon_{\sigma}$ | BM, (24), nH | MIM, (23), nH
---|---|---|---
22 | $0.015173$ | $15.25271$ | $15.25271$
70 | $0.001507$ | $15.46438$ | $15.46438$
222 | $0.000150$ | $15.48539$ | $15.48539$
702 | $0.000015$ | $15.48748$ | $15.48748$

Using the obtained results shown in the above table, the relative error
$\varepsilon_{\sigma}$ as a function of the number $n$ of line segments can be
defined, namely, $\varepsilon_{\sigma}=\varepsilon_{\sigma}(n)$. Moreover,
both functions $\varepsilon_{\sigma}=\varepsilon_{\sigma}(n)$ and
$n=n(\varepsilon)$ can be joined into one function exhibiting the dependency
of the relative error of calculation mutual inductance on the relative error
of interpolation of a circle as shown in Fig. 9. Although, Fig. 9 shows the
evaluation of the error for the particular arrangement. However, the result is
surprisely consistence and can be used for the preliminary estimation of the
error of calculation in other arrangements as shown below. For further
calculation, the number of line segments is set to 200 that, according to Fig.
9, the error, $\varepsilon_{\sigma}$, is estimated to be around $0.00015$.

Figure 11: Scheme for Example 12.

####  Example 12

The primary and secondary circles have radii of
$R_{p}=$$5\text{\,}\mathrm{mm}$ and $R_{s}=$$1\text{\,}\mathrm{mm}$,
respectively. The centre of the secondary circle is located at
$x_{C}=$3\text{\,}\mathrm{mm}$$, $y_{C}=$$1\text{\,}\mathrm{mm}$,
$z_{C}=$$0.5\text{\,}\mathrm{mm}$ and its angular misalignment are defined by
$\theta=$57.6885\text{\,}\mathrm{\SIUnitSymbolDegree}$$ and
$\eta=$108.4349\text{\,}\mathrm{\SIUnitSymbolDegree}$$ as shown in Fig. 11.
The result of calculation is

$n$ | $\varepsilon_{\sigma}$ | BM, (24), nH | MIM, (23), nH
---|---|---|---
200 | $0.0001566$ | $0.3577388$ | $0.3577388$

####  Example 13

The primary and secondary circles have radii $R_{p}=$$40\text{\,}\mathrm{cm}$
and $R_{s}=$$5\text{\,}\mathrm{cm}$, respectively. The centre of the secondary
circle is located at $x_{C}=$10\text{\,}\mathrm{cm}$$,
$y_{C}=$$15\text{\,}\mathrm{cm}$, $z_{C}=$$0.0\text{\,}\mathrm{cm}$. The
angular misalignment of the secondary filament is defined as follows
$\theta=$74.4986\text{\,}\mathrm{\SIUnitSymbolDegree}$$ and
$\eta=$123.6901\text{\,}\mathrm{\SIUnitSymbolDegree}$$. As it has mentioned
above, for calculation the number of line segments of $n=200$ is taken. The
result of calculation is

$n$ | $\varepsilon_{\sigma}$ | BM, (24), nH | MIM, (23), nH
---|---|---|---
200 | $0.0001615$ | $3.848737$ | $3.848737$

####  Example 14 (Example 11, page 3596 in the Babič article [11])

Let us consider two circular filaments having radii of
$R_{p}=$$40\text{\,}\mathrm{cm}$ and $R_{s}=$$10\text{\,}\mathrm{cm}$, which
are mutually perpendicular to each other that angles of $\eta=$$0\text{\,}$
and $\theta=$$90.0\text{\,}\mathrm{\SIUnitSymbolDegree}$. The centre of the
secondary circle has the following coordinates: $x_{C}=0$,
$y_{C}=$$20\text{\,}\mathrm{cm}$, and $z_{C}=$$10\text{\,}\mathrm{cm}$ as
shown in Fig. 12. The result becomes

$n$ | $\varepsilon_{\sigma}$ | BM, (24), nH | MIM, (23), nH
---|---|---|---
200 | $0.0001467$ | $-10.72715$ | $-10.72715$

Figure 12: Scheme for Example 14.  Figure 13: The chaotic distribution of the
absolute error against the $\eta$ angle in Example 15.

####  Example 15 (Example 21, Poletkin’s article [12])

Mutually perpendicular circles (angles of
$\theta=$$90.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ and $\eta=$$0\text{\,}$)
having the same radii as in Example 14 are considered, but the centre of the
secondary coil is located at the origin $O$. Results of calculation are

$n$ | BM, (24), nH | MIM, (23), nH
---|---|---
200 | $3.27233\times 10^{-15}$ | $2.91992\times 10^{-15}$

The results show the small error, which has the chaotic nature. Indeed,
changing the angle $\eta$ in a range of $0<\eta\leq 360^{o}$ the chaotic
distribution of the error is observed as shown in Fig. 13. Analysis of the
figure depicts that the error does not exceed of
$4\text{\times}{10}^{-15}\text{\,}\mathrm{nH}$. Worth noting that varying the
number of line segments, $n$, alters the error distribution, but keeps its
value below the observed maximum.

####  Example 16 (Example 22, Poletkin’s article [12])

Now we again consider mutually perpendicular circles having the same radii as
in Example 14, but in this case the centre of the secondary circle occupies a
position on the $XOY$-surface with the following coordinates
$x_{C}=y_{C}=$$10\text{\,}\mathrm{cm}$ and $z_{C}=0$. Hence, we have

$n$ | BM, (24), nH | MIM, (23), nH
---|---|---
200 | $-6.23859\times 10^{-15}$ | $-8.05774\times 10^{-15}$

Figure 14: The chaotic distribution of the absolute error against the $\eta$
angle in Example 16.

Then, again let us rotate the $\eta$ angle in a range $0<\eta\leq 360^{o}$,
the calculation of mutual inductance demonstrates the chaotic distribution of
the small error of calculation, which does not exceeded of
$15\text{\times}{10}^{-15}\text{\,}\mathrm{nH}$ as shown in Fig. 14. The error
is continually distributed without interruption of its continuity at
$\eta=$$90\text{\,}\mathrm{\SIUnitSymbolDegree}$ and
$270\text{\,}\mathrm{\SIUnitSymbolDegree}$ in comparing with the application
of Babic’s and Grover’s formula (please, see Example 22, in article [12]).
Thus, the segmentation method for the calculation of mutual inductance between
two circular filaments arbitrarily orientated in the space with respect to
each other does not suffer from the singularity case, when two circles are
mutually perpendicular.

###  IV-D Mutual inductance between circle and a filament of arbitrary shape

Using the proposed methodology based on (23) and (24), the segmentation method
is applied to the calculation of the mutual inductance between the primary
circle and filaments arbitrary positioning in the space and having different
shapes, which can be defined, for instance, by parametric equations of special
curves to generate the input set of points (18). In particular, the following
curves such as circular and elliptic arcs, ellipse, spiral, helices and
conical helices are considered in the proceeding section.

Figure 15:  Elliptic arc on the $xy$ plane is determined as a part of ellipse
having the semi-major axis $A_{\sigma}$ and semi-minor $B_{\sigma}$ within the
finite interval $\left[\varphi_{0},\varphi_{1}\right]$ (the left side figure);
The position of elliptic arc arbitrary orientated in the space with respect to
the primary circle (the right side figure).

Beginning with the calculation of mutual inductance between the circle and
elliptic arc arbitrary positioning in the space, the secondary filament can be
defined as a part of ellipse lying on the $xy$ plane. The ellipse is
characterized by its semi-major axis $A_{\sigma}$ directed along the $x$-axis
and semi-minor $B_{\sigma}$ along the $y$-axis. The parameter $\ell$ is
defined within the finite interval $\left[\varphi_{0},\varphi_{1}\right]$ as
shown in Fig. 15. The linear misalignment of the arc is defined by coordinates
of point $C$ corresponding to the centre of the ellipse, while the angular
misalignment can be determined by Euler’s angles denoted by $\alpha$, $\beta$
and $\gamma$. The following relationship between Grover’s and Euler’s angles
is true [12]:

$\left\\{\begin{array}[]{l}\sin\beta=\sin\eta\sin\theta;\\\
\cos\beta\sin\alpha=\cos\eta\sin\theta.\end{array}\right.$ (29)

According to Fig. 15, the input points can be calculated as follows

$\underline{p}_{i}(h_{i})=\left[\begin{array}[]{c}x_{C}\\\ y_{C}\\\
z_{C}\end{array}\right]+\underline{\Lambda}_{\gamma}^{T}\underline{\Lambda}_{\alpha}^{T}\underline{\Lambda}_{\beta}^{T}\left[\begin{array}[]{c}A_{\sigma}\cos
h_{i}\\\ B_{\sigma}\sin h_{i}\\\ 0\end{array}\right],\;i=0\ldots n,$ (30)

where

$\begin{array}[]{c}\underline{\Lambda}_{\gamma}=\left[\begin{array}[]{ccc}\cos\gamma&\sin\gamma&0\\\
-\sin\gamma&\cos\gamma&0\\\
0&0&1\end{array}\right],\;\underline{\Lambda}_{\alpha}=\left[\begin{array}[]{ccc}1&0&0\\\
0&\cos\alpha&\sin\alpha\\\ 0&-\sin\alpha&\cos\alpha\end{array}\right],\\\
\underline{\Lambda}_{\beta}=\left[\begin{array}[]{ccc}\cos\beta&0&-\sin\beta\\\
0&1&0\\\ \sin\beta&0&\cos\beta\end{array}\right].\end{array}$ (31)

####  Example 17

The primary circle has a radius of $R_{p}=$$1\text{\,}\mathrm{m}$, while the
elliptic arc is defined by the ellipse having the semi-major axis
$A_{\sigma}=$ $1\text{\,}\mathrm{m}$ and semi-minor one
$B_{\sigma}=$$0.5\text{\,}\mathrm{m}$ and within the finite interval
$\left[\varphi_{0}=$10\text{\,}\mathrm{\SIUnitSymbolDegree}$,\varphi_{1}=$110\text{\,}\mathrm{\SIUnitSymbolDegree}$\right]$.
The centre of the ellipse is located at $x_{C}=$10.0\text{\,}\mathrm{cm}$$,
$y_{C}=$$10.0\text{\,}\mathrm{cm}$, $z_{C}=$$10.0\text{\,}\mathrm{cm}$. The
angular misalignment of the secondary filament is defined by
$\alpha=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$,
$\beta=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$, but $\gamma$ is changed
in a range from 0 to $325.0\text{\,}\mathrm{\SIUnitSymbolDegree}$. The number
of line segments of $n=200$ is taken. The results of calculation are

$\gamma$ | FastHenry, nH | BM,(24), nH | MIM, (23), nH
---|---|---|---
$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $368.066$ | $368.191$ | $368.191$
$35.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $330.527$ | $329.896$ | $329.896$
$100.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $243.600$ | $242.784$ | $242.784$
$180.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $214.997$ | $214.6251$ | $214.6251$
$250.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $288.439$ | $289.2692$ | $289.2692$
$300.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $354.949$ | $356.677$ | $356.677$
$325.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $373.121$ | $375.319$ | $375.319$

It is clear that an elliptic arc can be transformed into the other curves such
as a circular one ($A_{\sigma}=B_{\sigma}$, $\left[\varphi_{0}\geq
0,\varphi_{1}<2\pi\right]$), an ellipse ($A_{\sigma}\neq B_{\sigma}$,
$\left[\varphi_{0}=0,\varphi_{1}=2\pi\right]$) and a circle
($A_{\sigma}=B_{\sigma}$, $\left[\varphi_{0}=0,\varphi_{1}=2\pi\right]$).

Figure 16:  The scheme for Example 18 (circular arc):
$\alpha=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$,
$\beta=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ and
$\gamma=$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ .

####  Example 18 (circular arc)

The primary circle has a radius of $R_{p}=$$1\text{\,}\mathrm{m}$, while the
circular arc is defined by $A_{\sigma}=B_{\sigma}=$$1\text{\,}\mathrm{m}$
within the finite interval
$\left[\varphi_{0}=$10\text{\,}\mathrm{\SIUnitSymbolDegree}$,\varphi_{1}=$110\text{\,}\mathrm{\SIUnitSymbolDegree}$\right]$.
The centre of the secondary circle is located at
$x_{C}=$10.0\text{\,}\mathrm{cm}$$, $y_{C}=$$10.0\text{\,}\mathrm{cm}$,
$z_{C}=$$10.0\text{\,}\mathrm{cm}$. The angular misalignment of the secondary
filament is defined as follows
$\alpha=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$,
$\beta=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ but $\gamma$ is changed in
a range from 0 to $325.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ as shown in
Fig. 16. The number of line segments of $n=200$ is taken. The results of
calculation are

$\gamma$ | FastHenry, nH | BM, (24), nH | MIM, (23), nH
---|---|---|---
$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $452.190$ | $452.632$ | $452.632$
$35.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $463.948$ | $463.5728$ | $463.5728$
$100.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $577.401$ | $577.2027$ | $577.2027$
$180.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $461.735$ | $461.3058$ | $461.3058$
$250.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $522.928$ | $523.709$ | $523.709$
$300.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $517.435$ | $519.920$ | $519.920$
$325.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $474.681$ | $477.143$ | $477.143$

####  Example 19 (ellipse)

The primary circle has a radius of $R_{p}=$$1\text{\,}\mathrm{m}$, while the
ellipse is defined by $A_{\sigma}=$ $1.0\text{\,}\mathrm{m}$
$B_{\sigma}=$$0.5\text{\,}\mathrm{m}$. The centre of the ellipse is located at
$x_{C}=$10.0\text{\,}\mathrm{cm}$$, $y_{C}=$$10.0\text{\,}\mathrm{cm}$,
$z_{C}=$$10.0\text{\,}\mathrm{cm}$. The angular misalignment of the secondary
filament is defined by $\alpha=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$,
$\beta=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ and
$\gamma=$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ as shown in Fig. 17. The
number of line segments of $n=200$ is taken. The result of calculation is

Figure 17:  The scheme for Example 19 (ellipse):
$\alpha=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$,
$\beta=$20.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ and
$\gamma=$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ .

FastHenry, nH | BM, (24), nH | MIM, (23), nH
---|---|---
$904.848$ | $905.9695$ | $905.9695$

Figure 18:  Spiral on the $xy$ plane is determined by the inner radius of
$R_{\sigma}$, the distance between turns $d$ and number of turns $t$ (the left
side figure); The position of the spiral arbitrary orientated in the space
with respect to the primary circle (the right side figure).

For the case of the calculation of mutual inductance between the circle and a
spiral, the secondary filament is lying on the $xy$ plane and its geometry can
be characterized by the inner radius $R_{\sigma}$, the number of turns $t$ and
the distance between the turns $d$ as shown on Fig. 18. The boundaries of the
finite interval of the parameter $\ell$ are estimated as follows

${\displaystyle\varphi_{0}=2\pi R_{\sigma}/d,\;\varphi_{1}=\varphi_{0}+2\pi
t.}\\\ $ (32)

Similar to the elliptic arc, the linear misalignment of the spiral is defined
by coordinates of point $C$, and the angular misalignment is defined by
Euler’s angles as shown in Fig. 18. The input points are generated by the
following equation:

$\underline{p}_{i}(h_{i})=\left[\begin{array}[]{c}x_{C}\\\ y_{C}\\\
z_{C}\end{array}\right]+\underline{\Lambda}_{\gamma}^{T}\underline{\Lambda}_{\alpha}^{T}\underline{\Lambda}_{\beta}^{T}\left[\begin{array}[]{c}{\displaystyle\frac{dh_{i}}{2\pi}}\cos
h_{i}\\\ {\displaystyle\frac{dh_{i}}{2\pi}}\sin h_{i}\\\
0\end{array}\right],\;i=0\ldots n,$ (33)

where cosine matrices are determined by Eq. (31).

####  Example 20 (spiral)

The primary circle has a radius of $R_{p}=$$0.5\text{\,}\mathrm{m}$. The
spiral is defined by the following parameters: the inner radius $R_{\sigma}$
is $0.05\text{\,}\mathrm{m}$, the distance $d$ is $0.04\text{\,}\mathrm{m}$
and the number of turns $t$ is $9$. The centre $C$ is located at
$x_{C}=$0.6\text{\,}\mathrm{m}$$, $y_{C}=$$0.1\text{\,}\mathrm{m}$, and
$z_{C}=$$0.7\text{\,}\mathrm{m}$. The angular misalignment is defined as
follows $\alpha=$45.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$,
$\beta=$-45.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$, but $\gamma$ is changed
in a range from $0$ to $350.0\text{\,}\mathrm{\SIUnitSymbolDegree}$. The
number of line segments $n$ is chosen to be $900$ (It is recommended to use at
least 100 line segments per turn to have a reasonable accuracy). The results
of calculation are

$\gamma$ | FastHenry, nH | BM, (24), nH | MIM, (23), nH
---|---|---|---
$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-72.7737$ | $-73.645$ | $-73.645$
$35.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $8.58253$ | $8.73071$ | $8.73071$
$100.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $179.826$ | $180.625$ | $180.625$
$180.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $217.839$ | $217.859$ | $217.859$
$250.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $74.2666$ | $73.3870$ | $73.3870$
$300.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-63.6821$ | $-64.2813$ | $-64.2813$
$350.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-86.1984$ | $-87.1184$ | $-87.1184$

Figure 19: The helices is determined by its radius $R_{\sigma}$, the pitch $p$
and the number of turns $t$ (the left side figure); The position of the
helices arbitrary orientated in the space with respect to the primary circle
(the right side figure).

Further, the calculation of mutual inductance between the circle and a helices
is considered. The geometry of helices can be defined on the $xyz$ CF by its
radius $R_{\sigma}$, the number of turns $t$ and the pitch $p$ as shown in
Fig. 19. The boundaries of the finite interval of the parameter $\ell$ are

${\displaystyle\varphi_{0}=0,\;\varphi_{1}=2\pi t.}\\\ $ (34)

Similar to previous cases, the linear misalignment of the helices is defined
by coordinates of origin $C$ of the $xyz$ CF, and the angular misalignment is
defined by Euler’s angles as shown in Fig. 19. The following equation can be
used to generate the input points:

$\underline{p}_{i}(h_{i})=\left[\begin{array}[]{c}x_{C}\\\ y_{C}\\\
z_{C}\end{array}\right]+\underline{\Lambda}_{\gamma}^{T}\underline{\Lambda}_{\alpha}^{T}\underline{\Lambda}_{\beta}^{T}\left[\begin{array}[]{c}R_{\sigma}\cos
h_{i}\\\ {\displaystyle-R_{\sigma}}\sin h_{i}\\\
{\displaystyle\frac{ph_{i}}{2\pi}}\end{array}\right],\;i=0\ldots n.$ (35)

####  Example 21 (helices)

The primary circle has radius of $R_{p}=$$0.9\text{\,}\mathrm{m}$. The helices
is defined by the following parameters: the radius $R_{\sigma}$ is
$0.6\text{\,}\mathrm{m}$, the pitch $p$ is $0.05\text{\,}\mathrm{m}$ and the
number of turns $t$ is $4$. The origin $C$ is located at
$x_{C}=$0.3\text{\,}\mathrm{m}$$, $y_{C}=$$0.2\text{\,}\mathrm{m}$,
$z_{C}=$$0.5\text{\,}\mathrm{m}$. The angular misalignment is characterized by
$\alpha=$54.7356\text{\,}\mathrm{\SIUnitSymbolDegree}$$,
$\beta=$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ and $\gamma$ is taken from
a range from $0$ to $335.0\text{\,}\mathrm{\SIUnitSymbolDegree}$. The number
of line segments $n$ is chosen to be $400$. The results of calculation are

$\gamma$ | FastHenry, nH | BM, (24), nH | MIM, (23), nH
---|---|---|---
$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-963.374$ | $-965.106$ | $-965.106$
$35.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-1242.69$ | $-1245.91$ | $-1245.91$
$135.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-1892.12$ | $-1893.43$ | $-1893.43$
$180.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-1537.51$ | $-1537.38$ | $-1537.38$
$235.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-1025.08$ | $-1024.81$ | $-1024.81$
$300.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-771.485$ | $-771.878$ | $-771.878$
$335.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $-835.499$ | $-836.041$ | $-836.041$

Figure 20: The conical helices is determined by its radius $R_{\sigma}$, the
pitch $p$, the number of turns $t$ and the distance between the turns $d$ (the
left side figure); The position of the conical helices arbitrary orientated in
the space with respect to the primary circle (the right side figure).

Finally, the calculation of mutual inductance between the circle and a conical
helices is carried out. The geometry of helices can be defined on the $xyz$ CF
by its radius $R_{\sigma}$, the number of turns $t$, the pitch $p$ and the
distance between turns $d$ as shown in Fig. 20. The boundaries of the finite
interval of the parameter $\ell$ are

${\displaystyle\varphi_{0}=2\pi R_{\sigma}/d,\;\varphi_{1}=\varphi_{0}+2\pi
t.}\\\ $ (36)

The linear misalignment of the conical helices is defined by coordinates of
origin $C$ of the $xyz$ CF, and the angular misalignment is defined by Euler’s
angles as shown in Fig. 20. The following equation generates the input points:

$\underline{p}_{i}(h_{i})=\left[\begin{array}[]{c}x_{C}\\\ y_{C}\\\
z_{C}\end{array}\right]+\underline{\Lambda}_{\gamma}^{T}\underline{\Lambda}_{\alpha}^{T}\underline{\Lambda}_{\beta}^{T}\left[\begin{array}[]{c}R_{\sigma}h_{i}\cos
h_{i}\\\ {\displaystyle-R_{\sigma}}h_{i}\sin h_{i}\\\
{\displaystyle\frac{p(h_{i}-\varphi_{0})}{2\pi}}\end{array}\right],\;i=0\ldots
n.$ (37)

####  Example 22 (conical helices)

The primary circle has radius of $R_{p}=$$0.9\text{\,}\mathrm{m}$. The conical
helices is defined by the following parameters: the radius $R_{\sigma}$ is
$0.0\text{\,}\mathrm{m}$, the pitch $p$ is $0.1\text{\,}\mathrm{m}$, the
number of turns $t$ is $4$ and the distance $d$ is $0.04\text{\,}\mathrm{m}$.
The origin $C$ is located at $x_{C}=$0.3\text{\,}\mathrm{m}$$,
$y_{C}=$$0.2\text{\,}\mathrm{m}$, $z_{C}=$$0.5\text{\,}\mathrm{m}$. The
angular misalignment is characterized by
$\alpha=$135\text{\,}\mathrm{\SIUnitSymbolDegree}$$,
$\beta=$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ and $\gamma$ is taken from
$0$ to $335.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ as shown in Fig. 21. The
number of line segments $n$ is chosen to be $400$. The results of calculation
are

Figure 21:  The scheme for Example 22 (conical helices):
$\alpha=$135.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$,
$\beta=$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ and
$\gamma=$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$$ .

$\gamma$ | FastHenry, nH | BM, (24), nH | MIM, (23), nH
---|---|---|---
$0.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $27.2802$ | $27.2225$ | $27.2225$
$35.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $29.1615$ | $28.8732$ | $28.8732$
$135.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $68.2821$ | $68.401$ | $68.401$
$180.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $84.7305$ | $85.15068$ | $85.15068$
$235.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $79.8669$ | $79.6694$ | $79.6694$
$300.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $47.0447$ | $46.7936$ | $46.7936$
$335.0\text{\,}\mathrm{\SIUnitSymbolDegree}$ | $32.3496$ | $32.1987$ | $32.1987$

##  V Conclusion

In this article, two novel formulas for calculation of mutual inductance
between a circular filament and line segment arbitrarily positioning in the
space were derived by means of mutual inductance method and Babic’s method,
respectively. These two formulas were expressed via the integral
relationships, whose kernels include the elliptic functions of the first and
second kinds. The two formulas are successfully validated to each other
mutually and numerically via the FastHenry software.

Using the fact that any curve can be interpolated with a desired accuracy by a
finite number of line segments, a segmentation method for calculation of the
mutual inductance between the primary circle and a filament having an
arbitrary shape is proposed and successfully developed based on the two
derived formulas. It is shown that for calculation of mutual inductance
between the primary circle and the secondary filament of an arbitrarily shape
by means of segmentation method based on (23) and (24), the set of points
belonging to the secondary filament as the input data is only needed.

The proposed segmentation method is successfully applied to the calculation of
the mutual inductance between the primary circle and filaments arbitrary
positioning in the space and having different shapes such as circular and
elliptic arcs, ellipse, spiral, helices and conical helices. All results of
calculation are in a good agreement with the reference examples and
successfully validated by the FastHenry software.

Considering a limit case, when the radius of the primary circle is trending to
zero, the developed methodology can be used for calculation of the magnetic
flux density and its gradient generated by a current-carrying arbitrarily
shaped filament. As a result, the force and stiffness between two arbitrarily
shaped current-carrying filaments can be calculated by using developed
methodology. This matter can be considered as our future work.

## Acknowledgment

S.S.K., E.R.M., and K.V.P. acknowledge with thanks the support from German
Research Foundation (Grant KO 1883/37-1) under the priority programme SPP
2206.

## References

  * [1] E. B. Rosa, _The self and mutual inductances of linear conductors_. US Department of Commerce and Labor, Bureau of Standards, 1908.
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