Calculation of Restoring and Repulsive Magnetic Forces between two Non-coaxial Coils of Rectangular Cross-Section with Parallel Axes

1. Introduction

The calculation of mutual inductance and magnetic force between coils of various shapes and positions has been the subject of numerous studies [1,2,3,4]. Most of these calculations are presented in analytical form for coils with simple configurations.

Analytical and semi-analytical methods for calculating self and mutual inductances of conducting elements in electrical circuits, as well as magnetic force interactions between these elements, have emerged as powerful mathematical tools [5,6,7,8,9,10,11,12]. These methods have been instrumental in advancing power transfer, wireless communication, sensing and actuation technologies, and have found applications across a wide range of scientific disciplines, including electrical and electronic engineering, medicine, physics, nuclear magnetic resonance, mechatronics, and robotics, among others.

While there are several efficient numerical methods available in commercially developed software, analytical and semi-analytical methods offer the advantage of providing calculation results in the form of a final formula with a finite number of input parameters. This feature can significantly reduce computational effort when applicable. In this paper, we present new formulas for calculating the restoring (radial) and repulsive (axial) magnetic forces between two non-coaxial circular loops with parallel axes. Using these formulas and the filament method, we calculate the forces between two non-coaxial coils of rectangular cross-sections with parallel axes. We emphasize the importance of accuracy and computational efficiency by selecting different numbers of subdivisions for the coils. Our analysis demonstrates that using the same number of subdivisions is not preferable due to significant computational time. Therefore, we choose varying numbers of subdivisions that considerably reduce computational time without compromising accuracy, which is crucial from an engineering perspective.

The results are compared with those obtained and presented in terms of line integrals [13]. These semi-analytical expressions may not be familiar to most engineers, who may require simpler expressions for practical use. In this paper, we provide straightforward formulas that also utilize single integrals for Maxwell coils, as employed in the filament method. This approach is highly accessible and suitable for professionals, including engineers and physicists, working in this field. Utilizing programming languages such as MATLAB, or Mathematica these simple formulas can be easily implemented.

We also employ another method [5], where the single integral is replaced by the summation of its kernel function over very small segments within the integration interval [0, π]. This method achieves considerably reduced computational time with satisfactory accuracy. All methods are validated according to the methods given in and [14].

2. Basic Expressions

In this paper we use the general formula for calculating the mutual inductance between two circular loops with angular misalignment, whose axes intersect but not at the center either of one, to calculate the restoring and the repulsive magnetic force between them (See Figure 1) [1],

$$\begin{array}{c}M\hfill \\ =2\frac{{\mu }_0}{\pi }\sqrt{R_P{R}_S}\underset{0}{\overset{\pi }{\int }}\frac{\left[cos\theta -\frac{d}{R_S}cos\phi \right]\varphi (k)}{k\sqrt{V^3}}d\phi \hfill \end{array}$$

(1)

                          wherein,

$$\begin{array}{cc}\alpha =\frac{R_S}{R_P},\hspace{1em} \hfill & \beta =\frac{c}{R_P}, V\hfill \\ \hfill & =\sqrt{1-{cos}^2\phi {sin}^2\theta -2\frac{d}{R_S}cos\phi cos\theta +\frac{d^2}{R_S^2}}\hfill \end{array}$$

Figure 1. Filamentary circular loops with angular misalignment (axes intersect but not at the center either of one)

Figure 1. Filamentary circular loops with angular misalignment (axes intersect but not at the center either of one)

$${ k}^2=\frac{4\alpha V}{{\left(1+\alpha V\right)}^2+{\zeta }^2} ,\hspace{1em}\hspace{1em}\zeta =\beta -\alpha cos\phi sin\theta$$

$R_P \mathrm{a}\mathrm{n}\mathrm{d} R_S$ - are the radii of the primary and secondary loops in (m), respectively.

$d$ - is the perpendicular distance between the axes of coils in (m).

$c$ - is the axial distance between the centers of coils in (m).

$\theta$ - is angle in radian, between unit vector normal of the inclined loop positioned at its center and the unit vector of the axis z of the primary loop.

${\mu }_0=4\pi ·{10}^{-7}H/m$ is the permeability of vacuum.

In (1) angle $\theta$ takes the value $0\le \theta \le \pi /2$.

In Song at al. gave the modified Grover’s formula (1) (when $\theta =0),$ for calculating the mutual inductance between two non-coaxial loops with the parallel axes

$$\begin{array}{c}\hspace{1em} M=\hfill \\ \frac{{\mu }_0}{\pi }\sqrt{R_P}{\int }_0^{\pi }\frac{r^2+R_s^2-d^2}{k\sqrt{r^3}}\varphi (k)d\phi \hfill \end{array}$$

wherein,

$$r=\sqrt{R_s^2-2R_S{dcos(\phi )+d}^2} , k^2=\frac{4R_{P}r}{{\left(R_P+r\right)}^2+c^2}$$

The restoring (radial) and the repulsive (axial) magnetic force can be obtained by [3],

$$\begin{array}{c}{F}_{restoring}\hfill \\ ={F_r=I}_P{I}_S\frac{\partial M}{\partial d}\hfill \end{array}$$

(3)

$$\begin{array}{c}{F}_{propulsive}=F_a\hfill \\ =I_P{I}_S\frac{\partial M}{\partial c}\hfill \end{array}$$

(4)

According the force between two filaments is one of the attractions if the currents in them are in the same direction around the common axis. If the currents in the two filaments are in opposite directions around the common axis, the force between them is one of repulsion.

Using (2), (3) and (4) we have,

$$\begin{array}{c}{F}_r=\frac{{\mu }_0{I}_P{I}_S{R}_S\sqrt{R_P}}{\pi }\underset{0}{\overset{\pi }{\int }}\left[R_{S}d\right.\left({cos}^2\phi -3\right)+(R_S^2\\ +d^2)\left.cos\phi \right]\frac{\varphi (k)}{k\sqrt{r^7}}d\phi -\end{array}$$

$$\begin{array}{c}\frac{{\mu }_0{I}_P{I}_S{R}_S}{4\pi \sqrt{R_P}}\underset{0}{\overset{\pi }{\int }}(R_{S}cos\phi -d)(R_S-dcos\phi )(R_P^2+c^2\\ -r^2)\frac{k\psi (k)}{\sqrt{r^9}}d\phi \end{array}$$

(5)

$$F_a=-\frac{{\mu }_0{I}_P{I}_{S}c{R}_S}{2\pi \sqrt{R_P}}\underset{0}{\overset{\pi }{\int }}\frac{k\left(R_S-dcos\phi \right)\psi \left(k\right)}{\sqrt{r^5}}d\phi$$

(6)

wherein

$$\psi (k)=\frac{1-\frac{k^2}{2}}{1-k^2}E\left(k\right)-K(k)$$

E(k) and K(k) are the complete elliptic function of the first and the second kind [17,18].

3. Filament Method foe Calculating the Magnetic force F_r and F_a between two non-Coaxial Coils of Rectangular Cross-Section with Parallel Axes

Let us treat two non-coaxial coils of rectangular cross-section with the parallel axes (See Figure 2).

Using the filament method [6], and the formulas for the mutual inductance M as well as for the magnetic forces $F_r$ and $F_a$are as follows,

$$M=\frac{N_1{N}_2}{(2K+1)(2N+1)(2m+1)(2n+1)}\sum _{g=-K}^{g=K}\sum _{h=-N}^{h=N}\sum _{p=-m}^{p=m}\sum _{l=-n}^{l=n}M\left(g,h,p,l\right)$$

(7)

$$F_r=\frac{N_1{N}_2}{(2K+1)(2N+1)(2m+1)(2n+1)}\sum _{g=-K}^{g=K}\sum _{h=-N}^{h=N}\sum _{p=-m}^{p=m}\sum _{l=-n}^{l=n}{F}_r\left(g,h,p,l\right)$$

(8)

$$F_a=\frac{N_1{N}_2}{(2K+1)(2N+1)(2m+1)(2n+1)}\sum _{g=-K}^{g=K}\sum _{h=-N}^{h=N}\sum _{p=-m}^{p=m}\sum _{l=-n}^{l=n}{F}_a\left(g,h,p,l\right)$$

(9)

Figure 2. Configuration of mesh matrix: Two non-coaxial coils of rectangular cross-section with the parallel axes.

Figure 2. Configuration of mesh matrix: Two non-coaxial coils of rectangular cross-section with the parallel axes.

$$\begin{gathered} r=\sqrt{R_s^2-2R_S{dcos(\phi )+d}^2} \\ {R}_I=\frac{R_1+R_2}{2}, R_{II}=\frac{R_3+R_4}{2}, { a=h}_I=R_2-R_1, { b=h}_{II} \\ =R_4-R_3 \\ {R}_P\left(h\right)=R_I+\frac{h_I}{2N+1}h , (h=-N,\dots ,0,\dots ,N) \\ {R}_S\left(l\right)=R_{II}+\frac{h_{II}}{2n+1}l , (l=-n,\dots ,0,\dots ,n) \\ z\left(g,p\right)=c+\frac{a}{2K+1}g-\frac{b}{2m+1}p, (g=-K,\dots ,0,\dots ,K; p \\ =-m,\dots ,0,\dots ,m) \\ { k}^2\left(h,l,p,g\right)=\frac{4R_P\left(h\right)r(l)}{{\left(R_P\left(h\right)+r(l)\right)}^2+z^2\left(g,p\right)} \\ r(l)=\sqrt{R_s^2(l)-2R_S(l){dcos(\phi )+d}^2} \\ M\left(g,h,p,l\right)=\frac{{\mu }_0}{\pi }\sqrt{R_P(h)}\underset{0}{\overset{\pi }{\int }}\frac{r^2(l)+R_s^2(l)-d^2}{k(h,l,p,g)\sqrt{r^3(l)}}\varphi (k)d\phi \\ {F}_r=\frac{{\mu }_0{I}_P{I}_S}{\pi }\underset{0}{\overset{\pi }{\int }}\left[R_S(l)d\right.\left({cos}^2\phi -3\right)+(R_S^2(l) \\ \hspace{1em}\hspace{1em}\hspace{1em}+d^2)\left.cos\phi \right]R_S(l)\sqrt{R_P(h)}\frac{\varphi (k)}{k\sqrt{r^7(l)}}d\phi - \\ \frac{{\mu }_0{I}_P{I}_S}{4\pi }\underset{0}{\overset{\pi }{\int }}(R_S(l)cos\phi -d)(R_S(l)-dcos\phi )(R_P^2(h)+z^2(g,p) \\ \hspace{1em}\hspace{1em}\hspace{1em}-r^2(l))\frac{R_S(l)k\psi (k)}{\sqrt{R_P\left(h\right)}\sqrt{r^9(l)}}d\phi \\ {F}_a=-\frac{{\mu }_0{I}_P{I}_S}{2\pi }\underset{0}{\overset{\pi }{\int }}\frac{{z(g,p)R}_S(l)k\left(R_S-dcos\phi \right)\psi \left(k\right)}{\sqrt{R_P(h)}\sqrt{r^5(l)}}d\varphi \\ \varphi \left(k\right)=\left(1-\frac{k^2}{2}\right)K\left(k\right)-E\left(k\right), \psi \left(k\right)=\frac{1-\frac{k^2}{2}}{\left(1-k^2\right)}E\left(k\right)-K(k) \end{gathered}$$

I₁ – current in the primary coil.

I₂ – current in the secondary coil.

N₁ – number of turns in the primary coil.

N₂ – number of turns in the secondary coil.

R₁ – inner radius of the primary coil of rectangular cross section.

R₂ – outer radius of the primary coil of rectangular cross section.

R₃ – inner radius of the secondary coil of rectangular cross section.

R₄ – outer radius of the secondary coil of rectangular cross section.

a and b – heights of the primary and the secondary coil, respectively.

d – perpendicular distance between axes of coils.

c – distance between the plans of centers of coils.

R_P(h) – average radius of the primary coil positioned in the plane (x, y) whose axis is

‘z’ axis.

R_S(l) – inclined average radius of the secondary coil

K, N, n, and m – the number of the subdivisions of thick coils.

4. Examples

In this example we calculate the mutual inductance, the restoring and the propulsive magnetic force as a function of the displacement of two non-coaxial loops with the parallel axes where we have: R_P = 42.5 mm, R_S = 20 mm. The perpendicular distance between coils axes is d = 3 mm, [10].

Clearly, we obtained identical results because the formulas are the same; however, equation (2) is derived without using the angle of inclination. In both cases the single integration is used. The absolute discrepancy AD is zero in each case. Here, we present another numerical approach to solve equation (2) [5], which is particularly interesting from an engineering perspective. Equation (2) is solved using the summation of small segments of the interval over the range [0, π], thereby avoiding numerical integration. This assumption allows for a considerable reduction in computational time with very high accuracy. In TABLE II, we provide a comparative calculation of equation (2) using both integration and summation methods.

TABLE I. Mutual inductance as a function of the axial displacement c of two non-coaxial loops with the parallel axes, using single integration.

c(m)	M (nH) [1]		M (nH) (2)	A.D. (%)
0.000	$20.48852432867563$		$20.48852432867563$	0.0
0.001	$20.46296134770761$		$20.46296134770761$	0.0
0.002	$20.38668420445834$		$20.38668420445834$	0.0
0.003	$20.26090954763262$		$20.26090954763262$	0.0
0.004	$20.08760437072583$		$20.08760437072583$	0.0
0.005	$19.86940071618871$		$19.86940071618871$	0.0
0.006	$19.60948700160182$		$19.60948700160182$	0.0
0.007	$19.31148475780680$		$19.31148475780680$	0.0
$$\begin{gathered} 0.008 \\ 0.009 \\ 0.010 \\ 0.011 \end{gathered}$$	$$\begin{gathered} 18.97931979155518 \\ \mathrm{18,61709599187806} \\ 18.22897847733389 \\ 17.81909087755921 \end{gathered}$$		$$\begin{gathered} 18.97931979155518 \\ \mathrm{18,61709599187806} \\ 18.22897847733389 \\ 17.81909087755921 \end{gathered}$$	$$\begin{gathered} 0.0 \\ 0.0 \\ 0.0 \\ 0.0 \end{gathered}$$

TABLE I. Mutual inductance as a function of the axial displacement c of two non-coaxial loops with the parallel axes, using single integration.

c(m)	M (nH) [1]		M (nH) (2)	A.D. (%)
0.000	$20.48852432867563$		$20.48852432867563$	0.0
0.001	$20.46296134770761$		$20.46296134770761$	0.0
0.002	$20.38668420445834$		$20.38668420445834$	0.0
0.003	$20.26090954763262$		$20.26090954763262$	0.0
0.004	$20.08760437072583$		$20.08760437072583$	0.0
0.005	$19.86940071618871$		$19.86940071618871$	0.0
0.006	$19.60948700160182$		$19.60948700160182$	0.0
0.007	$19.31148475780680$		$19.31148475780680$	0.0
$$\begin{gathered} 0.008 \\ 0.009 \\ 0.010 \\ 0.011 \end{gathered}$$	$$\begin{gathered} 18.97931979155518 \\ \mathrm{18,61709599187806} \\ 18.22897847733389 \\ 17.81909087755921 \end{gathered}$$		$$\begin{gathered} 18.97931979155518 \\ \mathrm{18,61709599187806} \\ 18.22897847733389 \\ 17.81909087755921 \end{gathered}$$	$$\begin{gathered} 0.0 \\ 0.0 \\ 0.0 \\ 0.0 \end{gathered}$$

From TABLE II we can see a very good agreement between two numerical approaches with the absolute discrepancy around 0.0097 %. For these examples MATLAB is used to calculate numerically the single integral on the interval [0; π] and FORTRAN is used to calculate the summation of the kernel function on the interval [0; π].

In TABLE III and Table IV the restoring and the axial forces calculations are given for (5) and (6) are compared with the results obtained in [10]. The single integration is used in (5) and (6). Clearly, we obtained identical results by two approaches.

Here, we present another numerical approach to solve equations (5) and (6) [5], which is particularly interesting from an engineering perspective. Equations (5) and (6) are solved using the summation of small intervals over the range [0, π], thereby avoiding numerical integration. This assumption allows for a considerable reduction in computational time with very high accuracy. In TABLE V and Table VI, we provide a comparative calculation of equations (5) and (6) using both integration and summation methods.

The results of the restoring and repulsive forces using the numerical integration (MATLAB) and the summation (FORTRAN) are given in TABLE V and TABLE VI. All results are in very good agreement either by the numerical integration or the numerical summation. The average absolute discrepancy is about 0.0097 %.

5. Conclusions

In this paper we give the new formulas for calculating the restoring (radial) and the repulsive (axial) magnetic forces between two non-coaxial coils of rectangular cross-section with the parallel axes. These formulas are derived from modified Grover’s formula for the mutual inductance between two non-coaxial loops with parallel axes. The validity of the presented approach is validated with an already established method. Presented formulas are used for calculating the radial and the axial force between two non-coaxial coils of rectangular cross-section with parallel axes using the filament method. Also, we presented the method to minimize the variables (subdivisions) in the filament method to find the compromise between the satisfactory accuracy and the corresponding small time of the calculation. We mention this method is applicable between non-coaxial conventional coils with parallel axes (massive-loop; massive-disk; massive-solenoid; two disks; disk-loop; disk-solenoid; two solenoids and solenoid-loop). This method can be useful for the engineers which are working in this domain because of its simplicity. The proposed method is comprehensible, fast, and very precise.