Magnetic field and forces in a pair of parallel conductors

Abstract

The article provides a comprehensive idea of the calculating the magnetic field and forces in a pair of parallel solid conductors. The conductors are connected to a sinusoidal voltage source of a frequency that does not exceed 1 MHz. Thus the current density over the conductor cross sections is not constant. Its distribution depends on the distance of the conductors and their resistivity, on the shape of conductor cross sections, and on the voltage source frequency. The distribution of current density over the conductor cross sections affects not only the magnetic field produced by the conductors but also the forces acting on the conductors. For the acting forces, general conclusions are derived that hold for conductors of arbitrary shape of cross section. Specifically, pairs of conductors of rectangular, circular and triangular cross section are examined.

Keywords

Applied classical electromagnetism magnetic induction dynamics of rigid bodies forces and toques numerical simulation

1. Introduction

The calculation of magnetic field and forces in a pair of parallel solid conductors is a classical problem. The magnetic field and forces as well as the distribution of the current density in the conductor cross section, depend on the distance of the conductors and their resistivity, on the shape of conductor cross sections, and on the voltage source frequency. Effect of current density on the magnetic field and acting forces is essential, similar to the effect of current density on the inductance of a pair of parallel conductors [1, 2]. Accurate and simple method of calculating the current density in a pair of conductors was used in calculating the magnetic field and forces, whose cross sections are symmetrical, was recently proposed [1, 3]. For calculation with two non-symmetrical cross section conductors, the integral equation formulated in [4] can be used to calculate the current density, which is quite general, but its numerical solution is not easy. Paper [1] (Subsection 4.7) deals with the methods of calculation of current density via partial differential equations. An original contribution of this work consists in considering the exact current density and in solving the consequent effects in the computation of magnetic field and acting forces in a pair of parallel solid conductors.

Figure 1.

Location of cross sections in the plane of coordinates $x y$ of a pair of parallel conductors of rectangular cross section. The symbol $\odot$ denotes the reference direction of current and current density vector.

Two conductors are considered that are parallel to the axis $z$ of the Cartesian coordinates $x y z$ . The conductor cross sections do not change along the $z$ axis. Specially, pairs of conductors of rectangular (see Fig. 1), circular and triangular cross section are examined. In the plane $z=0$ , the cross section of the right conductor is symmetrical with the cross section of the left conductor with respect to the straight line $x=d/2$ , where $d$ is called the conductor distance.

To enhance the clarity of further relations, the parentheses $($ and $)$ are used when entering functions, points and intervals while the point brackets $\langle$ and $\rangle$ are used when expressing a vector via its components. Underlined symbols denote phasors and complex numbers.

It is assumed in the present paper that voltage, current and current density depend on time $t$ . The potential $V$ across the conductors is constant over the conductor cross section and depends on $z$ , $V=V(z,t)$ . Potential drop along the conductors is characterized by the voltage $U(t)=[V(z_{1},t)-V(z_{2},t)]/(z_{2}-z_{1})$ . The current $I$ in the conductors only depends on $t$ , $I=I(t)$ . The current density $𝑱$ in the conductors does not depend on $z$ , $𝑱$ $=$ $𝑱$ $(x,y,t)$ . In view of these assumptions, the above quantities are slowly varying quantities [5] while related phenomena are called quasi-stationary [6]. It is assumed that the frequency $f$ of the voltage source is in the interval from zero to 1 MHz and the displacement current can therefore be neglected [7]. In the whole of the paper it is assumed that the permeability of both the conductors and the ambient environment is equal to the permeability $\mu_{0}$ of a vacuum. Permeability $\mu$ of conductors or their surroundings can have a different value than vacuum permeability $\mu_{0}$ . The permeability $\mu$ should be taken into consideration during the calculation magnetic field and during the calculation of the magnetic fluxes applied at current density calculation. If $\mu$ is constant, independently of the magnetic field $𝑩$ , complication cannot come. If $\mu$ is function of $𝑩$ , then it is possible to repeat the calculation with a $\mu$ adapted to the particular field $𝑩$ , then one can assume the stabilization of current density and field $𝑩$ .

2. Current density in conductors

The current density $𝑱$ $(x,y,t)=\langle 0,0,J(x,y,t)\rangle$ . It is assumed that the conductor cross sections are symmetrical not only as geometrical figures but both the current density and resistivity of the conductors are symmetrical, i.e. in the plane $z={\rm const}$ it holds

$\displaystyle\varrho_{\mathrm{left}}(x,y)=c\,\varrho_{\mathrm{right}}(d-x,y),$ (1) $\displaystyle J(x,y,t)=-J(d-x,y,t),$ (2)

where $c$ is a non-negative number.

Each of the conductors is formed by infinitely many parallel elementary conductors of infinitesimal cross section. To calculate the current density in the conductors, each conductor is replaced by $N$ parallel partial conductors whose cross sections are of finite magnitude. The cross sections of partial conductors in the plane $x y$ are symmetrical with respect to the straight line $x=d/2$ . Assumed in the $i$ th partial conductor ( $i=1,2,\dots,N$ ) is a constant current density $\underline{J}_{i}=\hat{J}_{i}\exp(\rm{j}\,\varepsilon_{\it i})$ . A current of current density $-\underline{J}_{i}$ is carried by the partial conductor in the right conductor, which is symmetrical with the $i$ th partial conductor in the left conductor. This fact is a consequence of relations Eqs (1) and (2) and the symmetry of cross section with respect to the straight line $x=d/2$ , and is of crucial significance for the calculation of current density. A method for the calculation of current density was described in detail in [3]. The discrete values $\hat{J}_{i}$ and $\varepsilon_{i}$ , $i=1,2,\dots,N$ , are fitted with the continuous functions $\hat{J}(x,y)$ and $\varepsilon(x,y)$ , while it holds

$J(x,y,t)=\hat{J}(x,y)\sin(\omega t+\varepsilon(x,y)),\quad\omega=2\pi f.$ (3)

The method for the computation of current density used in this article has been described in [1, 3], where it is also discussed in more detail the choice of $N$ . Generally speaking the current density converges to a limit with increasing $N$ . The used value $N$ depends on the size of cross section of conductors, on their distance and on the voltage frequency. The results below were obtained with values $N$ in the ranges of thousands to tens of thousands.

3. Magnetic field

The pair of conductors under consideration produce a magnetic field $\mbox{\boldmath$B$}=\langle B_{x},B_{y},0\rangle$ . For the $y$ - (and also $x$ -) component of the field it holds

$\displaystyle B_{y}(x,y,t)=\hat{B}_{y}(x,y)\sin[\omega t+\psi_{y}(x,y)],$ (4) $\displaystyle\underline{B}_{y}(x,y)=\hat{B}_{y}(x,y)\exp[{\rm j}\,\psi_{y}(x,y% )],$ (5) $\displaystyle\underline{B}_{y}(x,y)=\sum_{i=1}^{N}\underline{J}_{i}b_{iy}(x,y),$ (6)

with the quantities $b_{iy}(x,y)$ , and likewise $b_{ix}(x,y)$ in the calculation of $\underline{B}_{x}(x,y)$ , calculated using the the Biot and Savart law

$\displaystyle b_{ix}(x,y)=\frac{\mu_{0}}{4\pi}\int_{A_{i}}\frac{2(\overline{y}% -y)}{(x-\overline{x})^{2}+(y-\overline{y})^{2}}\,{\rm d}\overline{x}\,{\rm d}% \overline{y}$ (7) $\displaystyle b_{iy}(x,y)=\frac{\mu_{0}}{4\pi}\int_{A_{i}}\frac{2(x-\overline{% x})}{(x-\overline{x})^{2}+(y-\overline{y})^{2}}\,{\rm d}\overline{x}\,{\rm d}% \overline{y},$ (8)

where $A_{i}$ is the cross section of the $i$ th partial conductor, $(\overline{x},\overline{y})\in A_{i}$ . The magnitude of the vector $|\mbox{\boldmath$B$}|=B$ is given by the relation

$B(x,y,t)=\sqrt{B_{x}^{2}(x,y,t)+B_{y}^{2}(x,y,t)}.$ (9)

Using Eq. (4) to express $B_{y}$ and similarly also $B_{x}$ , it is possible to derive the Eq. (10), which can be used in the calculation of the dependence of $B(x,y,t)$ on $t$ at the point $(x,y)$ .

$B(x,y,t)=\sqrt{A_{0}(x,y)+|\underline{A}(x,y)|\sin[2\omega t+\arg(\underline{A% }(x,y))]},$ (10)

where

$\displaystyle A_{0}(x,y)=X+Y,$ $\displaystyle\underline{A}(x,y)=A_{\rm real}+{\rm j}\,A_{\rm imag},$ $\displaystyle X=\frac{1}{2}\hat{B}_{x}^{2}(x,y),$ $\displaystyle Y=\frac{1}{2}\hat{B}_{y}^{2}(x,y),$ $\displaystyle A_{\rm real}=X\sin[2\psi_{x}(x,y)]+Y\sin[2\psi_{y}(x,y)],$ $\displaystyle A_{\rm imag}=-X\cos[2\psi_{x}(x,y)]-Y\cos[2\psi_{y}(x,y)].$

In consequence of the validity of relation (2) it holds

$\displaystyle B_{x}(x,y,t)=-B_{x}(d-x,y,t),$ (11) $\displaystyle B_{y}(x,y,t)=B_{y}(d-x,y,t).$ (12)

4. Force acting on conductor

Consider a segment of elementary conductor (parallel to the axis $z$ ) of a cross section d $A$ between the points $z=z_{1}$ and $z=z_{2}$ . In the elementary conductor, a charge d $Q={\rm d}A\,{\rm d}z\,\rho$ is moving at a velocity $𝒗$ , where $\rho$ is the charge density. In the magnetic field $𝑩$ , the charge is being acted on by the force [5]

${\rm d}\mbox{\boldmath$F$}_{Q}={\rm d}Q\,\mbox{\boldmath$v$}\times\mbox{% \boldmath$B$}=\mbox{\boldmath$J$}\times\mbox{\boldmath$B$}\,{\rm d}A\,{\rm d}z,$

while the considered segment of elementary conductor, going through the point $(x,y)$ , that is at time $t$ , is being acted on by the force

${\rm d}\mbox{\boldmath$F$}_{\rm seg}(x,y,t)=\mbox{\boldmath$J$}(x,y,t)\times% \mbox{\boldmath$B$}(x,y,t)\,(z_{2}-z_{1})\,{\rm d}A,$ (13)

since the force d $𝑭$ ${}_{Q}$ does not depend on $z$ . In agreement with the assumptions made, the force acting on 1 m of elementary conductor is

${\rm d}\mbox{\boldmath$F$}(x,y,t)=\langle-J(x,y,t)B_{y}(x,y,t),\,J(x,y,t)B_{x}% (x,y,t),\,0\rangle\,{\rm d}A.$ (14)

Figure 2.

Transfer of the force d $𝑭$ $(x,y,t)$ acting at time $t$ on an elementary conductor going through the point $(x,y)$ in the cross section $A$ of the left cylindrical conductor, to an arbitrary point $(\xi,\eta)$ denoted by $+$ .

The force d $\mbox{\boldmath$F$}(x,y,t)$ acting on the elementary conductor going through the point $(x,y)\in A$ , where $A$ is the cross section of the conductor in the plane $x y$ , is simultaneously a force acting on the whole conductor because the conductor can be considered a rigid body. The force d $𝑭$ is a bound vector: it is bound to the straight line, which it lies on [9]. This straight line is the line of action of d $𝑭$ . The effect of the force d $𝑭$ on the rigid body does not change if the force is shifted arbitrarily along its line of action. The effect of the action of the force d $\mbox{\boldmath$F$}(x,y,t)$ on the whole conductor does not change if at an arbitrary point $(\xi,\eta)\in A$ there are another two acting forces of equal magnitude and opposite orientation d $\mbox{\boldmath$F$}(\xi,\eta,t)$ and $-$ d $\mbox{\boldmath$F$}(\xi,\eta,t)$ , $|{\rm d}\mbox{\boldmath$F$}(\xi,\eta,t)|=|{\rm d}\mbox{\boldmath$F$}(x,y,t)|$ , which are parallel to the force d $𝑭$ $(x,y,t)$ (see Fig. 2). The action of the three forces on the conductor is the same as the action of the force d $𝑭$ $(\xi,\eta,t)$ and the action of the torque d $𝑴$ $(x,y,\xi,\eta,t)$ of the couple d $𝑭$ $(x,y,t)$ and $-$ d $𝑭$ $(\xi,\eta,t)$ [9],

${\rm d}\mbox{\boldmath$M$}(x,y,\xi,\eta,t)=\mbox{\boldmath$r$}\times{\rm d}% \mbox{\boldmath$F$}(x,y,t),$

where $𝒓$ $=\langle x-\xi,y-\eta,0\rangle$ is the position vector of the point $(x,y)$ in the plane $z=0$ with respect to the point $(\xi,\eta)$ . The non-zero component of the vector d $𝑴$ is only d $M_{z}$ ,

${\rm d}M_{z}(x,y,\xi,\eta,t)=(x-\xi)\,{\rm d}F_{y}(x,y,t)-(y-\eta)\,{\rm d}F_{% x}(x,y,t).$ (15)

After a rearrangement of this relation using Eq. (14) it holds

${\rm d}M_{z}(x,y,\xi,\eta,t)=J(x,y,t)\left[(x-\xi)\,B_{x}(x,y,t)+(y-\eta)\,B_{% y}(x,y,t)\right]\,{\rm d}A.$ (16)

Using the above-described procedure, the forces acting on all partial conductors in the given conductor can be transferred to one point $(\xi,\eta)$ and added up. The total force

$\mbox{\boldmath$F$}(t)=\left\langle-\int_{A}J(x,y,t)B_{y}(x,y,t){\rm d}A,\int_% {A}J(x,y,t)B_{x}(x,y,t){\rm d}A,0\right\rangle,$ (17)

does not depend on the position of the point $(\xi,\eta)$ . The torque d $𝑴$ $(x,y,\xi,\eta,t)$ is a free vector that can be transferred to an arbitrary point if its magnitude and orientation in space are preserved [9]. In addition to $𝑭$ $(t)$ there is also acting the torque $𝑴$ $(\xi,\eta,t)$ , whose non-zero component is only

$M_{z}(\xi,\eta,t)=\int_{A}J(x,y,t)\left[(x-\xi)\,B_{x}(x,y,t)+(y-\eta)\,B_{y}(% x,y,t)\right]\,{\rm d}A.$ (18)

In the numerical calculation, the integrals over elementary conductors in relations Eqs (17) and (18) are replaced by the sum for $N$ partial conductors; in the summation the phasors of current density and magnetic field components are used. After rearranging, the resultant relations are

$\displaystyle F_{x}(t)=F_{0x}+\hat{F}_{x}\sin(2\omega t+\varphi_{x}),$ (19) $\displaystyle F_{y}(t)=F_{0y}+\hat{F}_{y}\sin(2\omega t+\varphi_{y}),$ $\displaystyle M_{z}(\xi,\eta,t)=M_{0}+\hat{M}\sin(2\omega t+\mu).$ (20)

Let $F_{x}^{\rm left}(t)$ and $F_{y}^{\rm left}(t)$ denote components of the force acting on the left conductor at the point $(\xi,\eta)\in A$ . Using relations Eqs (2), (11), (12) and the substitutions $x=d-X,X=x$ in integral Eq. (17) it can be shown that there is a force acting on the right conductor at the point $(d-\xi,\eta)$ for the components of which it holds

$\displaystyle F_{x}^{\rm right}(t)=-F_{x}^{\rm left}(t),$ (21) $\displaystyle F_{y}^{\rm right}(t)=F_{y}^{\rm left}(t).$ (22)

The Eqs (21) and (22) express Newton’s law of action and reaction. As has been assumed, each of the conductors is a rigid and provided the conductors do not change their mutual position they can be considered as a rigid body $\mathsf{S}_{2}$ with internal forces. The components $F_{x}^{\rm left}(t)$ and $F_{x}^{\rm right}(t)$ have a common line of action and can therefore be shifted to the point $(d/2,\eta)$ , where their sum is zero. Shifted to the point $(d/2,\eta)$ can also be the components $F_{y}^{\rm left}(t)$ and $F_{y}^{\rm right}(t)$ . Thus the resultant force acting on $\mathsf{S}_{2}$ is

$\mbox{\boldmath$F$}(t)=\left\langle 0,2\,F_{y}^{\rm left}(t),0\right\rangle.$

A rigid body cannot exert force on itself and thus it must hold

$F_{y}^{\rm right}(t)=F_{y}^{\rm left}(t)=0.$ (23)

Using relations Eqs (2), (11), (12) and the substitutions $x=d-X,X=x$ in integral Eq. (18) it can be shown that there is a torque acting on the right conductor whose only non-zero component is

$M_{z}^{\rm right}(\xi,\eta,t)=-M_{z}^{\rm left}(\xi,\eta,t).$ (24)

From Eq. (24) it follows that the total torque acting on $\mathsf{S}_{2}$ is zero.

If the two conductors can move with respect to each other in the plane $y=0$ , then in consequence of the action of internal forces of the system their conductor distance $d$ will increase. Increasing $d$ has two consequences. The first consequence is the decreasing force acting on the conductors due to the increasing mutual distance, as follows from data given below, and this holds for $f\geqslant 0$ . The other consequence is the decreasing amplitude of current density in the conductor cross section (for $f>0$ ) and thus also the decreasing amplitude of current in the conductors, as shown in [3] and as confirmed by data given below. At the end of the action described, the conductor distance $d$ is infinitely large, there is no force acting on the conductors and no current carried by them even if the conductors are connected to a voltage source with $f>0$ .

A magnetic field $𝑩$ produced by a pair of conductors is the sum of fields produced by the left and right conductors

$\mbox{\boldmath$B$}=\mbox{\boldmath$B$}^{\rm left}+\mbox{\boldmath$B$}^{\rm right}.$

From this relation and from Eqs (17) and (18) it follows that the force and torque acting on the left conductor (likewise on the right conductor) is the sum of two forces and two torques

$\displaystyle\mbox{\boldmath$F$}(t)=\mbox{\boldmath$F$}^{\rm own}(t)+\mbox{% \boldmath$F$}^{\rm other}(t),$ (25) $\displaystyle\mbox{\boldmath$M$}(\xi,\eta,t)=\mbox{\boldmath$M$}^{\rm own}(\xi% ,\eta,t)+\mbox{\boldmath$M$}^{\rm other}(\xi,\eta,t).$ (26)

In the following the validity will be proved for the equalities

$\mbox{\boldmath$F$}^{\rm own}(t)=0,\quad M_{z}^{\rm own}(\xi,\eta,t)=0.$ (27)

The validity of these equalities may seem-evident provided the system of elementary or partial conductors replacing only one conductor is considered as a system of rigid bodies with internal forces. The resultant of the internal forces of the system cannot be a non-zero force acting on the system because due to the action of this resultant force the system would accelerate and this is impossible. Elementary or partial conductors replacing one conductor cannot, of course, form a separate system because the current density in the conductor and thus also the magnetic field produced by the conductor and thus also the force acting on each of the partial conductors are affected by the other conductor. A separate system is formed only by partial conductors replacing a solitary solid conductor or a pair of conductors for $f=0$ .

When proving the validity of equalities Eq. (27), it is assumed that the respective conductor is approximated by $N$ partial conductors of square cross section. The sides of the square cross sections lie on the straight lines $x=k\Delta$ and $y=k\Delta$ , where $k=0,\pm 1,\pm 2,\dots$ , and $\Delta$ is the length of the square side. The $i$ th partial conductor is of cross section $A_{i}=[x_{i}-\Delta/2,x_{i}+\Delta/2]\times[y_{i}-\Delta/2,y_{i}+\Delta/2]$ and constant current density $J_{i}(x_{i},y_{i},t)$ . The cross section of each conductor in the pair considered is a Jordan measurable set [10] and can therefore be approximated by squares with arbitrary accuracy. By Eqs (17), (14) and (6) the $i$ th partial conductor is acting on the $j$ th partial conductor with the force

$\displaystyle\mbox{\boldmath$F$}_{ij}(x_{j},y_{j},t)=\left\langle F_{ijx}((x_{% j},y_{j},t),F_{ijy}(x_{j},y_{j},t),0\right\rangle,$ $\displaystyle F_{ijx}(x_{j},y_{j},t)=-J_{j}(x_{j},y_{j},t)\,J_{i}(x_{i},y_{i},% t)\,b_{iy}(x_{j},y_{j})\,A_{i},$ (28) $\displaystyle F_{ijy}(x_{j},y_{j},t)=J_{j}(x_{j},y_{j},t)\,J_{i}(x_{i},y_{i},t% )\,b_{ix}(x_{j},y_{j})\,A_{i}.$ (29)

If $i=j$ , then by Eqs (7) and (8) it holds $b_{ix}(x_{i},y_{i})=b_{iy}(x_{i},y_{i})=0$ and consequently it also holds

$\mbox{\boldmath$F$}_{ii}(x_{i},y_{i},t)=0.$ (30)

if $i\neq j$ , then the $j$ th partial conductor is acting on the $i$ th partial conductor with a force $𝑭$ ${}_{ji}(x_{i},y_{i},t)$ , and it holds

$\mbox{\boldmath$F$}_{ij}(x_{j},y_{j},t)+\mbox{\boldmath$F$}_{ji}(x_{i},y_{i},t% )=0,$ (31)

because by Eqs (7) and (8) it holds

$\displaystyle b_{ix}(x_{j},y_{j})=-b_{jx}(x_{i},y_{i}),$ $\displaystyle b_{iy}(x_{j},y_{j})=-b_{jy}(x_{i},y_{i}).$

Relations Eqs (30) and (31) hold for every $N$ , the total force is

$\displaystyle\mbox{\boldmath$F$}^{\rm own}(t)=\lim_{N\rightarrow\infty}\sum_{i% =1}^{N}\sum_{j=1}^{N}\mbox{\boldmath$F$}_{ij}=\lim_{N\rightarrow\infty}\sum_{i% =1}^{N}\left[\mbox{\boldmath$F$}_{ii}+\sum_{j=i+1}^{N}\left(\mbox{\boldmath$F$% }_{ij}+\mbox{\boldmath$F$}_{ji}\right)\right].$ (32)

The validity of relations Eqs (30) and (31) confirms the validity of the first equality in Eq. (27).

In contrast to the validity of the first equality in Eq. (27), it cannot be proved that the second equality in Eq. (27) holds for every $N$ . The validity of equality Eq. (30) confirms the validity of the equality

$M_{iiz}(x_{i},y_{i},\xi,\eta,t)=0,\quad i=1,2,\dots,N,\ \forall(\xi,\eta)\in A,$

where $M_{iiz}$ is a part of the of torque corresponding to the force $𝑭$ ${}_{ii}$ . Using Eqs (15) and (31) the part $M_{ijz}$ of the torque corresponding to the sum $𝑭$ ${}_{ij}+$ $𝑭$ ${}_{ji}$ in relation Eq. (32) can be calculated, and it holds

$\displaystyle M_{ijz}(x_{i},y_{i},x_{j},y_{j},t)=(x_{j}-x_{i})\,F_{ijy}(x_{j},% y_{j},t)-(y_{j}-y_{i})\,F_{ijx}(x_{j},y_{j},t).$ (33)

It follows from this relation and from Eq. (32) that the component $M_{z}^{\rm own}$ of the total torque with which the conductor is acting on itself does not depend on the choice of the point $(\xi,\eta)$ , and it holds

$\displaystyle M_{z}^{\rm own}(t)=\lim_{N\rightarrow\infty}\sum_{i=1}^{N}\sum_{% j=i+1}^{N}M_{ijz}(x_{i},y_{i},x_{j},y_{j},t).$ (34)

From relations Eqs (33), (28) and (29) it follows

$\displaystyle\lim_{N\rightarrow\infty}M_{ijz}(x_{i},y_{i},x_{j},y_{j},t)=\quad% \lim_{N\rightarrow\infty}J_{j}(x_{j},y_{j},t)\,J_{i}(x_{i},y_{i},t)\,A_{i}% \left[(x_{j}-x_{i})\,b_{ix}(x_{j},y_{j})+(y_{j}-y_{i})\,b_{iy}(x_{j},y_{j})% \right].$

The quantities $b_{ix}(x_{j},y_{j})$ and $b_{iy}(x_{j},y_{j})$ are given by the integrals Eqs (7) and (8) and thus, with $N$ sufficiently large, they can be approximated with sufficient accuracy by the product of $A_{i}$ and the value of the integrand for $x=x_{j}$ and $y=y_{j}$ [11]. It is thus obvious that

$\lim_{N\rightarrow\infty}M_{ijz}(x_{i},y_{i},x_{j},y_{j},t)=0$

with the validity of the second equality in Eq. (34) following from relation Eq. (27).

It follows from the above-proven statements that calculating the forces acting on one of the conductors of the conductor pair under consideration consists in calculating the components $F_{x}^{\rm other}$ and $M_{z}^{\rm other}$ . Numerical computation of the components $F_{y}^{\rm other}$ , $F_{x}^{\rm own}$ and $F_{y}^{\rm own}$ yields a number that can be neglected in the arithmetic used while in double precision computation the number is less than $10^{-12}$ N $\cdot$ m ${}^{-1}$ . Numerical computation of the component $M_{z}^{\rm own}$ gives a low number, which is practically negligible with respect to $M_{z}^{\rm other}$ , but not so in the arithmetic used. With increasing $N$ the magnitude of this number decreases.

The calculation of the force acting on either conductor in a pair of conductors can be made simpler but at the expense of calculation accuracy. The force acting on the right conductor is as large as the force acting on the left conductor but acts in opposite direction. In the simplified calculation of the force acting on the left conductor it is assumed that current in the left conductor, $I(t)=\hat{I}\sin(\omega t+\beta)$ is flowing through a current filament that is identical with the axis $z$ while the component $B_{y}(0,0,t)$ of the magnetic field produced in the axis $z$ by the right conductor is a field produced by a current filament that is parallel to the axis $z$ and goes through the point $(d,0)$ in the plane $z=0$ . By Ampere’s circuital law [5] it holds

$B_{y}(0,0,t)=\frac{\mu_{0}\,I(t)}{2\pi\,d},$

where $I(t)$ is current in the left conductor. By relations Eqs (14) and (17) the right conductor is acting on the left conductor by a force whose only non-zero component is $F_{x}^{\rm sim}(t)=-I(t)B_{y}(0,0,t)$ . After substituting for $I(t)$ and $B_{y}(0,0,t)$ and after rearranging, it holds

$F_{x}^{\rm sim}(t)=\hat{F}^{\rm sim}\left[-1+\sin\left(2\omega t+\varphi^{\rm sim% }\right)\right],$ (35)

where

$\hat{F}^{\rm sim}=10^{-7}\,\hat{I}^{2}/d,\quad\varphi^{\rm sim}=2\beta+\pi/2.$

5. Discussion of computation results

5.1 Conductors of recangular cross section

A pair of conductors are given by the data $w=5$ mm, $h=16$ mm, see Fig. 1. $\varrho_{\rm left}=\varrho_{\rm right}=1.712\times 10^{-8}$ $\Omega\cdot$ m (both conductors are of copper, temperature 25 ${}^{\circ}$ C [12]), $U(t)=\sin\omega t,\ \hat{U}=1$ V $\cdot\mathrm{m}^{-1}$ .

The choice of a specific phasor of the voltage $\underline{U}$ of the source to which the pair of conductors are connected is without loss of generality. If at a voltage $\underline{U}$ the phasor of current density in the left conductor is $\underline{J}(x,y)$ , then at a voltage $\underline{u}\underline{U}$ the phasor of current density in the left conductor is $\underline{u}\underline{J}(x,y)$ , where $\underline{u}$ is an arbitrary complex number [3]. An analogous relation holds for the magnetic field phasors $\underline{B}_{x}(x,y,t)$ and $\underline{B}_{y}(x,y,t)$ .

The cross section of the left (and likewise the right) conductor in the pair under consideration is symmetrical with respect to the axis $x$ . If in addition to relation Eq. (1) it also holds

$\varrho_{\mathrm{left}}(x,y)=\varrho_{\mathrm{left}}(x,-y),$ (36)

then in addition to Eqs (2), (11) and (12) it also holds

$\displaystyle J(x,y,t)=J(x,-y,t),$ (37) $\displaystyle B_{x}(x,y,t)=-B_{x}(x,-y,t),$ (38) $\displaystyle B_{y}(x,y,t)=B_{y}(x,-y,t).$ (39)

Figure 3.

Curves of the zero values of the function $J(x,y,t)$ and regions of identical direction of the current density in the upper half of the left conductor at four instants of the first half of period $T$ of the voltage $U(t)$ ; $d=10$ mm, $f=10$ kHz.

Current density in the conductors depends on the conductor distance and on the frequency of the voltage $U(t)$ . Figure 3 shows the current density distribution over the cross section of the upper half of the left conductor at four instants of the first half of period $T$ of the voltage $U(t)$ , for $d=10$ mm and $f=10$ kHz. At each of the instants considered, the current density is shown via zero value curves of the function $J(x,y,t)$ . These curves, together with the boundaries of the conductor cross section, divide the cross section of the left conductor into several regions. In each region, there is a constant direction of current density, as given in Fig. 1, i.e. the function $J(x,y,t)$ is in each region either positive or negative. The regions are symmetrical with respect to the straight line $y=0$ ; in the right conductor, the regions are symmetrical with regions in the left conductor with respect to the straight line $x=d/2$ . If there is $t\in[T/2,T)$ , then the regions are the same as for $t-T/2$ but the current density in them is of opposite direction to the direction of the current density for $t-T/2$ . The number of regions in the conductor cross section decreases/increases with the frequency $f$ decreasing/increasing.

Figure 4.

Amplitudes $\hat{B}_{y}^{\rm left}$ (solid lines) and $\hat{B}_{y}^{\rm right}$ (dotted lines) in the left conductor of rectangular cross section; $d=$ 10 mm, $f=$ 10 kHz.

The integral on the right side of relation Eq. (18) is the sum of two integrals. The first integral is in the form

$\int_{-w}^{w}(x-\xi)\int_{-h}^{h}J(x,y,t)\,B_{x}(x,y,t)\,{\rm d}y\,{\rm d}x.$ (40)

The integral from $-h$ to $h$ is equal to zero for every value $x\in[-w,w]$ , because Eqs (37) and (38) hold true and thus the expression Eq. (40) is also equal to zero. The other integral on the right side of relation Eq. (18) is for $\eta=0$ in the form

$\int_{-w}^{w}\left[\int_{-h}^{0}y\,J(x,y,t)\,B_{y}(x,y,t)\,{\rm d}y+\int_{0}^{% h}y\,J(x,y,t)\,B_{y}(x,y,t)\,{\rm d}y\right]{\rm d}x.$ (41)

By substituting $y=-Y$ and $Y=y$ , the integral from $-h$ to $0$ is transformed to the integral

$\int_{h}^{0}y\,J(x,-y,t)\,B_{y}(x,-y,t)\,{\rm d}y,$

and because relations Eqs (37) and (39) hold true, the expression Eq. (41) is equal to zero. This proves that each conductor in the considered pair of conductors with rectangular cross section in the plane $y=0$ is acted on by the force $𝑭$ alone because $M_{z}(\xi,0,t)=0$ . The line of action of $𝑭$ is parallel to the axis $x$ and the force $𝑭$ can move arbitrarily along this line of action with its effect remaining unchanged. It follows that the value of $M_{z}(\xi,\mathrm{const},t)$ does not change for $\xi\in[-w,w]$ while $M_{z}(\xi,\mathrm{const},t)\neq 0$ for ${\rm const}\neq 0$ .

Figure 4 gives the dependence of the amplitude of the components $B_{y}^{\rm left}$ and $B_{y}^{\rm right}$ on $x$ for three values of $y$ ; $d=10$ mm, $f=10$ kHz. The dependence of the initial phases $\psi_{y}^{\rm left}$ and $\psi_{y}^{\rm right}$ on $x$ is given in Fig. 5 for three frequencies; $d=10$ mm, $y=8$ mm. If $\zeta$ denotes the point of discontinuity of the function $\psi_{y}^{\rm left}(x,{\rm const})$ , then for $f\rightarrow 0+$ it holds: $\zeta\rightarrow 0$ ; $\psi_{y}^{\rm left}(x,{\rm const})\rightarrow 180^{\circ}-,\ \forall x\in[-w,0]$ and $\psi_{y}^{\rm left}(x,{\rm const})\rightarrow 0-,\ \forall x\in(0,w]$ . The dependence of the initial phase on $x$ for the values $y\in[0,h]$ other than 8 mm scarcely differs from the dependence in Fig. 5. This is due to the large extent of initial phase values in Fig. 5. Figure 6 gives the dependence of the functions $\psi_{y}^{\rm left}(x,y)$ and $\psi_{y}^{\rm right}(x,y)$ on $x$ in the neighbourhood of the point $x=w$ for the values $y\in(0,h)$ in the left conductor. In the neighbourhood of the point $x=-w$ the difference between the maximum and the minimum initial phase values is less than 2.1 ${}^{\circ}$ for $y\in(0,h)$ , $d=10$ mm, $f=1$ kHz.

Table 1

Current in the conductors and parameters determining the force acting on 1 m of the left conductor of rectangular cross section and their dependence on the source frequency, for $d=10$ mm

$f$ (Hz)	$\hat{I}$ (A)	$\beta∼{}(^{\circ})$	$F_{0x}$ (N/m)	$\hat{F}_{x}$ (N/m)	$\varphi_{x}∼{}(^{\circ})$	$\hat{F}^{\rm sim}$ (N/m)
$0$	9345.8	0.0	$-500.0$	500.0	90.0	873.4
$50$	8081.4	$-$ 25.2	$-$ 375.8	375.2	41.6	653.1
$10^{2}$	6204.7	$-$ 39.0	$-$ 224.9	223.7	16.1	385.0
$10^{3}$	1585.8	$-$ 48.9	$-$ 18.76	18.59	0.2	25.15
$10^{4}$	462.80	$-$ 47.4	$-$ 1.867	1.859	$-$ 0.5	2.142
$10^{5}$	140.89	$-$ 47.4	$-$ 0.1854	0.1852	$-$ 5.1	0.1985

Figure 5.

Dependence of the initial phase $\psi_{y}^{\rm left}$ (solid lines) and $\psi_{y}^{\rm right}$ (dotted lines) on $x$ and on frequency $f$ in the left conductor of rectangular cross section; $d=$ 10 mm, $y=$ 8 mm.

For the pair of rectangular cross section conductors under consideration the components $F_{x}(t)$ and $F_{x}^{\rm sim}(t)$ acting on the left conductor were calculated for six frequencies and $d=10$ mm. The calculation results are summarized in Table 1. The mutual effect of the conductors manifests itself the most at the least possible conductor distance and therefore $d=10$ mm was chosen in Table 1. The minimum distance can be interpreted as between conductor is insulation of zero thickness, but electrically conductive connection is not between left and right conductor. Choosing $d=10$ mm causes practically the same results as the choice for example $d=10.0001$ mm with non-zero insulation thickness. The effect of the distance $d$ on the force acting on the left conductor is evident from Table 2.

Table 2

Current in the conductors and the parameters determining the force acting on 1 m of the left conductor of rectangular cross section and their dependence on the conductor distance $d$ , for the source frequency $f=60$ Hz

$d$ (mm)	$\hat{I}$ (A)	$\beta∼{}(^{\circ})$	$F_{0x}$ (N/m)	$\hat{F}_{x}$ (N/m)	$\varphi_{x}∼{}(^{\circ})$	$\hat{F}^{\rm sim}$ (N/m)
10	7679.0	$-$ 28.9	$-$ 340.1	339.3	34.7	589.7
20	5716.2	$-$ 50.0	$-$ 127.6	127.4	$-$ 9.60	163.4
50	3589.3	$-$ 66.2	$-$ 24.33	24.33	$-$ 42.5	25.77
100	2677.4	$-$ 72.5	$-$ 7.058	7.058	$-$ 55.0	7.169
200	2112.5	$-$ 76.2	$-$ 2.222	2.222	$-$ 62.5	2.231
500	1644.1	$-$ 79.3	$-$ 0.5403	0.5403	$-$ 68.7	0.5406
1000	1406.1	$-$ 80.9	$-$ 0.1977	0.1977	$-$ 71.8	0.1977

Figure 6.

Dependence of the initial phase $\psi_{y}^{\rm left}$ (solid lines) and $\psi_{y}^{\rm right}$ (dotted lines) on $x$ in the neighbourhood of the point $x=w$ , for ten values $y\in(0,h)$ , $y=$ 0.125–15.875 mm, in the left conductor of rectangular cross section; $d=$ 10 mm, $f=$ 1 kHz.

The values of the parameter $|F_{0x}|$ differ very little from the values of the parameter $\hat{F}_{x}$ in Tables 1 and 2. Equation (19) can thus be written in the form

$F_{x}(t)\doteq\hat{F}_{x}\left[-1+\sin(2\omega t+\varphi_{x})\right].$

From a comparison of this formula with Eq. (35) it is easy to estimate how accurate the simplified calculation using Eq. (35) is. The accuracy of calculating via Eq. (35) depends not only on the quotient $h/w$ and on $d$ and $f$ , but in particular on the current $I(t)$ . In the calculation of $\hat{F}^{\rm sim}$ in Tables 1 and 2, the respective currents in Tables 1 and 2 were used, which are the integrals of the current density $J(x,y,t)$ . Without an accurate calculation of $J(x,y,t)$ , only the current for $f=0$ can in fact be calculated, which further constrains the utilization of simplified calculation using Eq. (35). It has been verified on a number of specific calculation that for every pair of conductors of rectangular cross section and for a given $f$ it holds

$\lim_{d\rightarrow\infty}\frac{\hat{F}_{x}}{\hat{F}_{x}^{\rm sim}}=1,$ (42)

which is confirmed by the data in Table 2. The speed of the convergence depends on the quotient $h/w$ ; the convergence is the fastest for a pair of conductors of square cross section.

Figure 7.

Location of the cross sections in the plane of the coordinates $x y$ of a pair of parallel cylindrical conductors. The symbol $\odot$ denotes the reference direction of current and currenr density vector.

5.2 Cylindrical conductors

A pair of conductors, see Fig. 7, are given by the data $r=10$ mm, $\varrho_{\rm left}=\varrho_{\rm right}=2.709\times 10^{-8}$ $\Omega\cdot$ m (both conductors are of aluminium, temperature 25 ${}^{\circ}$ C [12]), $U(t)=\sin\omega t,\ \hat{U}=1$ V $\cdot\mathrm{m}^{-1}$ .

Figure 8.

Dependence of the amplitude of current density on $x$ and $y$ in the upper half of the cross section of the left conductor in a pair of cylindrical conductors, for $d=$ 20 mm, $f=$ 50 Hz.

Figure 9.

Dependence of the initial phase of current density on $x$ and $y$ in the upper half of the cross section of the left conductor in a pair of cylindrical conductors, for $d=$ 20 mm, $f=$ 50 Hz.

Figures 8 and 9 illustrate the distribution of current density over the cross section of the upper half of the left conductor, for $d=20$ mm, $f=50$ Hz. The current density is represented using the amplitude $\hat{J}$ (Fig. 8) and the initial phase $\varepsilon$ (Fig. 9). The phasor of the $y$ -component of the magnetic field in the upper half of the left conductor is given in Fig. 10 for the same values of $x$ , $y$ , $d$ and $f$ as in Figs 8 and 9. The phasor is represented by means of the amplitude $\hat{B}_{y}$ and the initial phase $\psi_{y}$ . The dependence of $\psi_{y}$ on $x$ for some $y$ is discontinuous in the neighbourhood of the point $x=-4.3$ mm due to being modified such that its value is in the interval $(-180^{\circ},180^{\circ}]$ .

Figure 10.

Dependence of the amplitude and initial phase of the component $B_{y}$ on $x$ and $y$ in the upper half of the cross section of the left conductor in a pair of cylindrical conductors, for $d=$ 20 mm, $f=$ 50 Hz. The dependence of the initial phase $\psi_{y}$ on $x$ for $y=$ 0.1, 5.1, 7.1 mm in the neighbourhood of the point $x=$ $-$ 4.3 mm is discontinuous.

The pair of cylindrical conductors under consideration meets the conditions Eqs (1) and (36) and thus the only non-zero component of the total force acting on the conductors at the point $(\xi,0)$ , where $\xi\in[-r,r]$ or $\xi\in[d-r,d+r]$ , is $F_{x}(t)$ , and it holds $M_{z}=0$ . This statement is proved in a similar way to the case of conductors of rectangular cross section.

Table 3

Current in the conductors and the parameters determining the force acting on 1 m of the left cylindrical conductor and their dependence on the source frequency, for $d=$ 20 mm

$f$ (Hz)	$\hat{I}$ (A)	$\beta∼{}(^{\circ})$	$F_{0x}$ (N/m)	$\hat{F}_{x}$ (N/m)	$\varphi_{x}∼{}(^{\circ})$	$\hat{F}^{\rm sim}$ (N/m)
$0$	5798.4	$0$	$-$ 168.1	168.1	90.0	168.1
$50$	4652.4	$-$ 33.1	$-$ 110.6	110.0	29.4	108.2
$10^{2}$	3290.5	$-$ 48.7	$-$ 58.40	57.39	2.8	54.14
$10^{3}$	621.12	$-$ 64.1	$-$ 3.500	3.301	$-$ 18.3	1.929
$10^{4}$	114.36	$-$ 66.6	$-$ 2.079e $-1$	1.935e $-1$	$-$ 21.6	6.539e $-2$
$10^{5}$	20.564	$-$ 67.6	$-$ 1.187e $-2$	1.103e $-2$	$-$ 22.9	2.112e $-3$

Table 4

Current in the conductors and the parameters determining the force acting on 1 m of the left cylindrical conductor and their dependence on the conductor distance $d$ , for the source frequencies $f=50$ and $10^{4}$ Hz

$f=50$ Hz
$d$ (mm)	$\hat{I}$ (A)	$\beta∼{}(^{\circ})$	$F_{0x}$ (N/m)	$\hat{F}_{x}$ (N/m)	$\varphi_{x}∼{}(^{\circ})$	$\hat{F}^{\rm sim}$ (N/m)
20	4652.4	$-$ 33.1	$-$ 110.6	110.0	29.4	108.2
40	3702.2	$-$ 49.4	$-$ 34.45	34.44	$-$ 7.6	34.27
100	2739.8	$-$ 61.4	$-$ 7.513	7.512	$-$ 32.7	7.506
200	2255.2	$-$ 66.8	$-$ 2.543	2.543	$-$ 43.6	2.543
$f=10^{4}$ Hz
20	114.36	$-$ 66.6	$-$ 2.079 ${\rm e}-1$	1.935 ${\rm e}-1$	$-$ 21.6	6.539 ${\rm e}-2$
40	29.134	$-$ 87.9	$-$ 2.417 ${\rm e}-3$	2.417 ${\rm e}-3$	$-$ 85.1	2.122 ${\rm e}-3$
100	17.039	$-$ 88.9	$-$ 2.958 ${\rm e}-4$	2.958 ${\rm e}-4$	$-$ 87.8	2.903 ${\rm e}-4$
200	13.109	$-$ 89.2	$-$ 8.632 ${\rm e}-5$	8.632 ${\rm e}-5$	$-$ 88.3	8.593 ${\rm e}-5$
1000	8.5628	$-$ 89.5	$-$ 7.333 ${\rm e}-6$	7.334 ${\rm e}-6$	$-$ 88.9	7.332 ${\rm e}-6$
2000	7.4512	$-$ 89.5	$-$ 2.776 ${\rm e}-6$	2.776 ${\rm e}-6$	$-$ 89.1	2.776 ${\rm e}-6$

For the pair of conductors of circular cross section under consideration the forces $F_{x}$ and $F_{x}^{\rm sim}(t)$ , acting on the left conductor, were calculated for six frequencies. The calculation results are summarized in Table 3. The mutual effect of the conductors menifests itself the most at the least possible conductor distance and therefore $d=20$ mm was chosen in Table 3. The effect of the distance $d$ on the force acting on the left conductor is evident from Table 4. Similar to the conductors of rectangular cross section, relations Eq. (42) and

$\lim_{d\rightarrow\infty}|F_{0x}|=\hat{F}_{x}.$ (43)

hold true here.

Figure 11.

Location of the cross sections in the plane of the coordinates $x y$ of a pair of parallel conductors of triangular cross section. The symbol $\odot$ denotes the reference direction of current and current density vector.

5.3 Conductors of equilateral triangular cross section

A pair of conductors, see Fig. 11, are given by the data $w=10$ mm, $\varrho_{\rm left}=\varrho_{\rm right}=1.712\times 10^{-8}$ $\Omega\cdot$ m (both conductors are of copper, temperature 25 ${}^{\circ}$ C [12]), $U(t)=\sin\omega t,\ \hat{U}=1$ V $\cdot\mathrm{m}^{-1}$ .

Figure 12.

Amplitude $\hat{J}(x,y)$ (solid lines) and initial phase $\varepsilon(x,y)$ (dotted lines) of current density in the triangular cross section of the left conductor, for $d=$ 20 mm, $f=$ 1 kHz; the digits 1 to 5 denote the value of the $y$ -coordinate in mm: 1–0.0866, 2–5.6292, 3–10.1325, 4–15.1554, 5–16.8875.

Figure 13.

Amplitude $\hat{B}_{x}(x,y)$ (solid lines) and the initial phase $\psi_{x}(x,y)$ (dotted lines) of the $x$ -component of magnetic field in the triangular cross section of the left conductor, for $d=$ 20 mm, $f=$ 1 kHz; the digits 1 to 5 denote the value of the $y$ -coordinate in mm: 1–0.0866, 2–5.6292, 3–10.1325, 4–15.1554, 5–16.8875.

Figure 12 gives the dependence of the amplitude $\hat{J}(x,y)$ and initial phase $\varepsilon(x,y)$ of current density on $y$ and $x$ in the triangular cross section of the left conductor, for $d=20$ mm, $f=1$ kHz. Figures 13 and 14 illustrate respectively the dependence of the $x$ - and $y$ -component of magnetic field on $y$ and $x$ in the left conductor, for $d=20$ mm, $f=1$ kHz.

Table 5

Current in the conductors and the parameters determining the component $F_{x}(t)$ (see Eq. (19)) of the force acting on 1 m of the left conductor in a pair of conductors of triangular cross section and their dependence on the source frequency $f$ , for $d=20$ mm

$f$ (Hz)	$\hat{I}$ (A)	$\beta∼{}(^{\circ})$	$F_{0x}$ (N/m)	$\hat{F}_{x}$ (N/m)	$\varphi_{x}∼{}(^{\circ})$	$\hat{F}^{\rm sim}$ (N/m)
0	5058.6	0	$-$ 128.7	128.7	90.0	127.9
50	3225.7	$-$ 46.2	$-$ 53.92	53.52	3.1	52.02
10 ${}^{2}$	2002.4	$-$ 59.4	$-$ 22.19	21.72	$-$ 19.2	20.05
10 ${}^{3}$	320.99	$-$ 72.8	$-$ 8.912e $-1$	8.187e $-1$	$-$ 38.2	5.152e $-1$
10 ${}^{4}$	45.578	$-$ 78.0	$-$ 3.110e $-2$	2.792e $-2$	$-$ 43.1	1.039e $-2$
10 ${}^{5}$	5.9261	$-$ 80.9	$-$ 1.009e $-3$	9.039e $-4$	$-$ 44.9	1.756e $-4$

Table 6

Parameters $M_{0},\hat{M}$ and $\mu$ determining the component $M_{z}(\xi,\eta,t)$ of torque (see Eq. (20)) acting on 1 m of the left conductor in a pair of conductors of triangular cross section and their dependence on the source frequency $f$ , for $d=20$ mm, $\xi=\eta=0$

$f$ (Hz)	$M_{0}$ (N)	$\hat{M}$ (N)	$\mu∼{}(^{\circ})$
0	6.917e $-1$	6.917e $-1$	$-$ 90.0
50	2.902e $-1$	2.908e $-1$	$-$ 177.4
10 ${}^{2}$	1.194e $-1$	1.198e $-1$	158.7
10 ${}^{3}$	3.655e $-3$	3.536e $-3$	123.8
10 ${}^{4}$	6.855e $-5$	6.564e $-5$	110.2
10 ${}^{5}$	1.029e $-6$	9.882e $-7$	103.8

Figure 14.

Amplitude $\hat{B}_{y}(x,y)$ (solid lines) and the initial phase $\psi_{y}(x,y)$ (dotted lines) of the $y$ -component of magnetic field in the triangular cross section of the left conductor, for $d=20$ mm, $f=1$ kHz; the digits 1 to 5 denote the value of the $y$ -coordinate in mm: $1-0.0866$ , 2–5.6292, 3–10.1325, 4–15.1554, 5–16.8875.

For the pair of triangular cross section conductors under consideration the components $F_{x}(t)$ , $F_{x}^{\rm sim}(t)$ and $M_{z}(\xi,\eta,t)$ were calculated for six frequencies. The calculation results are summarized in Tables 5 and 6. The mutual effect of the conductors manifests itself the most at the least possible conductor distance and therefore $d=20$ mm was chosen in Tables 5 and 6. The effect of the distance $d$ of conductor cross sections on the force acting on the left conductor is evident from Tables 7 and 8. The triangular cross section conductors satisfy relations Eqs (42) and (43), in a similar way to the case of rectangular and circular cross section conductors.

Table 7

$d$ (mm)	$\hat{I}$ (A)	$\beta∼{}(^{\circ})$	$F_{0x}$ (N/m)	$\hat{F}_{x}$ (N/m)	$\varphi_{x}∼{}(^{\circ})$	$\hat{F}^{\rm sim}$ (N/m)
20	3225.7	$-$ 46.2	$-$ 53.92	53.52	3.1	52.02
40	2271.6	$-$ 61.7	$-$ 13.00	12.99	$-$ 32.2	12.90
100	1542.2	$-$ 71.4	$-$ 2.381	2.381	$-$ 52.6	2.378
200	1227.3	$-$ 75.3	$-$ 7.533e $-1$	7.533e $-1$	$-$ 60.5	7.531e $-1$
500	962.42	$-$ 78.5	$-$ 1.853e $-1$	1.853e $-1$	$-$ 67.0	1.853e $-1$
1000	826.28	$-$ 80.2	$-$ 6.827e $-2$	6.827e $-2$	$-$ 70.3	6.827e $-2$
2000	723.48	$-$ 81.4	$-$ 2.617e $-2$	2.617e $-2$	$-$ 72.8	2.617e $-2$
5000	621.02	$-$ 82.6	$-$ 7.713e $-3$	7.713e $-3$	$-$ 75.2	7.713e $-3$

Table 8

$d$ (mm)	$M_{0}$ (N)	$\hat{M}$ (N)	$\mu∼{}(^{\circ})$
20	2.902e $-1$	2.908e $-1$	$-$ 177.4
40	7.537e $-2$	7.572e $-2$	152.5
100	1.409e $-2$	1.417e $-2$	133.5
200	4.470e $-3$	4.497e $-3$	125.7
500	1.100e $-3$	1.107e $-3$	119.3
1000	4.055e $-4$	4.080e $-4$	116.0
2000	1.554e $-4$	1.564e $-4$	113.5
5000	4.582e $-5$	4.610e $-5$	111.0

By Eq. (20), the component $M_{z}(\xi,\eta,t)$ has a constant part $M_{0}$ and an alternating part with the amplitude $\hat{M}$ , initial phase $\mu$ and frequency $2f$ . The parameters $M_{0},\hat{M}$ and $\mu$ in Tables 6 and 8 were calculated for $\xi=\eta=0$ . For $\eta\geqslant 0$ the value $M_{0}$ decreases, for $\eta=\eta_{0}$ it is equal to zero and then acquires negative values. The value $\eta_{0}$ depends on $d$ and $f$ , for example $\eta_{0}=$ 5.383 mm for $d=$ 20 mm and $f=$ 50 Hz. The value of the amplitude $\hat{M}$ remains, of course, non-zero also for $\eta=\eta_{0}$ . For a specific lay-out of the pair of conductors and for a specific frequency it is then possible to calculate $\eta_{0}$ and thus eliminate the constant part of torque. In conductors whose cross section and resistivity are symmetrical with respect to the axis $x$ of the coordinates, such as the conductors of rectangular and circular cross section, $\eta_{0}=$ 0 and with the zero value of $M_{0}$ the amplitude $\hat{M}$ of the alternating part of torque is also zero.

6. Conclusion

An analysis was made of the current density, magnetic field and forces in a pair of conductors parallel to the axis $z$ of the Cartesian coordinates. It is assumed that the conductor cross sections and the values of resistivity in the plane $z=0$ are symmetrical with respect to the straight line $x=d/2$ . The conductors are connected to an ideal source of sinusoidal voltage. In the calculation of current density in each conductor the effect of the other conductor is respected [1, 3]. Each conductor is formed by parallel elementary conductors. The force acting on the elementary conductor going through the point $(x,y)$ in the cross section $A$ of the left conductor in the plane $z=0$ is given by the current density at the point $(x,y)$ and by the magnetic field produced by the right conductor at the point $(x,y)$ . The effect of the forces acting on all elementary conductors in the left (likewise the right) conductor can be transferred to an arbitrary point $(\xi,\eta)\in A$ . The result is expressed by the force $𝑭$ $(t)$ and torque $𝑴$ $(\xi,\eta,t)$ .

The force $𝑭$ is a bound vector, whose effect does not change when shifted along the line of action of $𝑭$ , it has a single non-zero component $F_{x}$ , and its magnitude does not depend on the position of the point $(\xi,\eta)$ . The torque $𝑴$ is a free vector, it has a single non-zero component $M_{z}$ , whose magnitude depends on the point $(\xi,\eta)$ . The components $F_{x}(t)$ and $M_{z}(\xi,\eta,t)$ have a constant part and an alternating part with a frequency twice the voltage source frequency. Exact proofs are given for the basic statements about the properties of the vectors $𝑭$ and $𝑴$ . These statements are indubitable and they make easier the calculation of the two vectors. The theoretical part of the paper is complemented with results of particular calculations of current density, magnetic field and forces in conductor pairs of rectangular, circular and triangular cross section.

To prevent any mutual motion of the conductors in consequence of the forces exerted, the conductors need to be fixed in space. The proposed methods for the calculation of vectors $𝑭$ and $𝑴$ make it possible to optimize the position of fixation points and establish the magnitude of the forces that need to be compensated for at the fixation points.

Footnotes

Acknowledgments

This research work has been carried out in the Centre for Research and Utilization of Renewable Energy (CVVOZE). Authors gratefully acknowledge financial support from the Ministry of Education, Youth and Sports of the Czech Republic under NPU I programme (project No. LO1210). Computational resources were provided by the CESNET LM2015042 and the CERIT Scientific Cloud LM2015085, under the programme “Projects of Large Research, Development, and Innovations Infrastructures”.

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