# New Consideration of the Skin Effect in Electrodynamics

Sergey Maslov, Iurii Maslov, Dmitry Maslov

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Sergey Maslov, Iurii Maslov, Dmitry Maslov. New Consideration of the Skin Effect in Electrodynamics. Progress of Theoretical and Experimental Physics [PTEP], In press. hal-04091730

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# New Consideration of the Skin Effect in Electrodynamics

Sergey Maslov, Iurii Maslov and Dmitry Maslov

Freelance researchers-enthusiasts, Chisinau, Moldova; Voronezh, Russia

Corresponding author: Sergey Maslov (sergoloer@gmail.com)

The article proposes an approach to solving the normal skin effect problem based on integral equations derived directly from Ampère's circuital law, Faraday's law of induction and Kirchhoff's circuit laws, in contrast to modern solutions based on Maxwell's equations of electrodynamics in differential form. It is shown that the skin effect is mainly due to the electromotive force of the wire self-induction, which depends on the distance to the surface, and the influence of eddy currents is either very small or absent. New equations, their solutions and formulas for the total current and impedance are proposed for various technically significant cases immediately taking into account the inductance of the external space and in two versions: taking into account only the self-induction of the conductor; taking into account both self-induction and eddy currents. On the example of a solid cylinder, it is shown that, in accordance with the formulas of the existing approach itself, the current density on the surface of the wire does not increase with increasing frequency, as it is stated in words. It is shown that, in fact, the solutions of the existing concept are based not on eddy currents, but on the uneven distribution of elementary self-induction currents over depth and are completely equivalent to the variant of the presented solutions, taking into account only self-induction. Power balances were obtained without the introduction of a frequency-dependent active resistance to alternating current in two versions: based on the total current, voltage and average electromotive force of induction; based on the integration of the balance of specific powers over the entire volume of the wire. The physical meaning of the visual increase of the active resistance with increasing frequency is revealed, but certainly not on the basis of a decrease in the effective cross section of the wire. Based on the symmetry of the distribution of eddy currents, the assumption is made that they are absent in galvanically closed wires, but present in open wires such as antennas. An experiment was carried out with a negative result on the detection of the alleged thermal separation in the wire, an explanation for this was proposed.

## CONTENTS

<table><tr><td>I. Introduction</td><td>2</td></tr><tr><td>II. Physical picture and derivation of equations and solutions</td><td>3</td></tr><tr><td>A. Cylindrical solid wire</td><td>3</td></tr><tr><td>B. Cylindrical tubular wire</td><td>9</td></tr><tr><td>C. Coaxial cable</td><td>11</td></tr><tr><td>D. Flat plate of finite thickness</td><td>17</td></tr><tr><td>E. Infinitely long solenoid</td><td>19</td></tr><tr><td>1. Direct connection to a power supply</td><td>19</td></tr><tr><td>2. Connecting to a power supply in series through a capacitor</td><td>21</td></tr><tr><td>F. Equations and estimation of surface current densities for transformer windings</td><td>23</td></tr><tr><td>III. Analysis of the existing solution for solid cylinder</td><td>27</td></tr><tr><td>IV. Voltage and power balances</td><td>29</td></tr><tr><td>V. Thermal separation effect calculation</td><td>35</td></tr><tr><td>VI. Thermal separation experiment</td><td>37</td></tr><tr><td>VII. Conclusions and outlook</td><td>39</td></tr><tr><td>Acknowledgments</td><td>41</td></tr><tr><td>Explanations for appendices</td><td>41</td></tr><tr><td>Appendix 1: Search and verification of solutions for cylindrical solid wire</td><td>42</td></tr><tr><td>Appendix 2: Graphs of eddy-free and eddy solutions for a solid cylinder without external inductance at large and small skin depths</td><td>44</td></tr><tr><td>Appendix 3: Graphs of eddy-free and purely eddy solutions for a solid cylinder without external inductance at large and small skin depths</td><td>45</td></tr><tr><td>Appendix 4: Graphs of eddy-free and eddy solutions for a solid cylinder in an iron shell and in air</td><td>46</td></tr><tr><td>Appendix 5: Frequency dependences for currents, surface current densities and resistances in the absence of external inductance for eddy-free and eddy solutions for a solid cylinder</td><td>47</td></tr><tr><td>Appendix 6: Frequency dependences for currents, surface current densities and resistances in the presence of external inductance for eddy-free and eddy solutions for a solid cylinder</td><td>48</td></tr><tr><td>Appendix 7: Comparison of resistances as a function of frequency for cylindrical copper wires for eddy-free and eddy solutions</td><td>49</td></tr><tr><td>Appendix 8: Search and verification of solutions for cylindrical tubular wire</td><td>50</td></tr><tr><td>Appendix 9: Search for total currents and impedances for cylindrical tubular wire</td><td>54</td></tr><tr><td>Appendix 10: Comparison of graphs for solid and tubular wire for the eddy case with an optimal ratio of skin depth to wall thickness, providing a lower resistance of tubular wire compared to solid wire</td><td>55</td></tr><tr><td>Appendix 11: Comparison of graphs for solid and tubular wire for the eddy-free case with an optimal ratio of skin</td><td>55</td></tr></table>
<table><tr><td>depth to wall thickness, providing a lower resistance of tubular wire compared to solid wire</td><td>56</td></tr><tr><td>Appendix 12: Comparison of eddy case graphs for the inner wire of a coaxial cable and for a conventional wire, graphs for the outer wire-screen</td><td>57</td></tr><tr><td>Appendix 13: Comparison of eddy-free case graphs for the inner wire of a coaxial cable and for a conventional wire, graphs for the outer wire-screen</td><td>58</td></tr><tr><td>Appendix 14: Search and verification of solutions for a flat plate</td><td>59</td></tr><tr><td>Appendix 15: Comparison of solutions for a flat plate with the existing solution for a semi-infinite space</td><td>60</td></tr><tr><td>Appendix 16: Comparison of solution for a flat plate with solution for a thin-walled tubular cylinder</td><td>61</td></tr><tr><td>Appendix 17: Search and verification of solutions for a long solenoid</td><td>62</td></tr><tr><td>Appendix 18: Search and verification of solutions for a long solenoid with a series-connected capacitor</td><td>63</td></tr><tr><td>Appendix 19: Transition from the balance equation of specific voltages to the equation for general quantities</td><td>65</td></tr><tr><td>Appendix 20: Comparison of vector diagrams of electromotive forces and voltages of the presented and existing solutions for a solid cylinder for the eddy-free variant</td><td>66</td></tr><tr><td>Appendix 21: Vector diagrams of specific electromotive forces and voltages for various distances and comparison of power diagrams of the presented and existing solutions for a solid cylinder for the eddy-free case</td><td>67</td></tr><tr><td>Appendix 22: Power calculation and graphs for a solid cylinder for eddy-free and eddy variants</td><td>68</td></tr><tr><td>Appendix 23: Power calculation and graphs for a solid cylinder with a dedicated part of the eddy specific electromotive force</td><td>69</td></tr><tr><td>Appendix 24: Power calculation and graphs for a long solenoid</td><td>70</td></tr><tr><td>Appendix 25: Power calculation and graphs for a long solenoid for a large frequency</td><td>71</td></tr><tr><td>Appendix 26: Power calculation and graphs for a long solenoid with a capacitor</td><td>72</td></tr><tr><td>References</td><td>73</td></tr></table>

# I. INTRODUCTION

To obtain the existing solutions of the skin effect phenomenon, the method of arbitrary zeroing of the part of the solution that does not fit into the experimentally observed regularity of the current density decrease from the surface to the depth is used. For example, see Ref. [1, p. 392, § 11.02, f. (2), (3)] for a semi-infinite space and Ref. [2, p. 722, f. (A.6), (A.7)] for a cylindrical wire. To justify this action, an argument is usually made about the requirement that the current density be finite in depth for physical reasons. This approach cannot be considered either mathematically or physically rigorous. It seems obvious that the reason for such a forced rejection of the unsuitable part lies in the erroneous physical model of the phenomenon in the existing concept and in the resulting incorrect initial equations. In the presented work, the current density distribution laws are exact solutions of the new initial integral equations without any zeroing of the unsuitable parts. Search and verification of all solutions are implemented in MathCAD and presented in appendices. Designations of phasors and other complex values in the text part of the article are in bold roman type, non-complex variables and constants are in regular italics, standard functions and numbers are in regular roman type. The designations of quantities in the formulas given from various sources, for convenience, are converted into the designations adopted in this article, except for literal quotations.

List of standard functions, mathematical and physical constants and other quantities encountered during the presentation:

$\mu_0 = 4\pi \times 10^{-7}$ H/m is the vacuum permeability (magnetic constant);

$\mu_r$ is the relative magnetic permeability of the wire material;

$\mu = \mu_r \mu_0$ is the magnetic permeability of the wire material;

$\mu_{sr}$ is the relative magnetic permeability of the space around the wire;

$\mu_s = \mu_{sr} \mu_0$ is the magnetic permeability of the space around the wire;

$\rho$ is the resistivity of the wire material;

$f$ is the frequency;

$\omega = 2\pi f$ is the angular frequency;

i is the imaginary unit;

$J_0$ and $J_1$ are Bessel functions of the first kind;

$I_0$ and $I_1$ are modified Bessel functions of the first kind (Infeld functions);

$K_0$ and $K_1$ are modified Bessel functions of the second kind (Macdonald functions). The subscripts

0 and 1 of all the given Bessel functions indicate their order.
## II. PHYSICAL PICTURE AND DERIVATION OF EQUATIONS AND SOLUTIONS

### A. Cylindrical solid wire

Solutions for solid and tubular wires will be presented in different subsections based on universal equations (in contrast to the existing approach), which we will obtain in this subsection. Let there be a cylindrical wire of length $l$ with an outer radius $a$ and an inner radius $b$, as shown in its axial section in Fig. 1. Energy is supplied to the wire by contact from a power source with a voltage phasor $U$ (not shown) with zero internal resistance, outer radius and inner radius, like the wire itself.

<figure><img src="image_2.png" /><figcaption>FIG. 1. Representation of a wire with dimensions $b, r, a, n, l$ as a transformer with magnetic induction $B(r)$, primary $j_1(r)$, secondary $j_{2a}(r)$, $j_{2b}(r)$ and total $j(r)$ current density phasors.</figcaption></figure>

The axial size of the power supply is considered negligible compared to the length of the wire. That is, we will consider the wire closed to the power source integrated into it. We will not take into account the curvature of the wire, assuming its length $l$ is large enough compared to the outer radius $a$. As a consequence of the voltage connected to the wire, we consider the appearance at each of its points of the supply field strength phasor $E = U/l$, parallel to the axis of the wire, with a zero initial phase. Let us imagine a wire in the form of a transformer, consisting of three circuits – primary and two secondary. In turn, it is convenient to represent the primary circuit as consisting of two parts – a constant and a variable. The constant part is a rectangular axial section of the space of length $l$ between the outer radius of the wire $a$ and the boundary radius $n$ of the external magnetic field from our wire. The variable part is a rectangular axial section of a wire with a length $l$ between a variable radial distance $r$ from the axis (hereinafter referred to as distance $r$) and the outer radius $a$. Let the self-induction current density phasor of the primary circuit be $j_1(r)$. The first secondary circuit is the perimeter of the outer rectangular axial section of a cylinder of length $l$, enclosed between the distance $r$ and the outer radius $a$ with the mutual induction current density phasor $j_{2a}(r)$. The second secondary circuit is the perimeter of the inner rectangular axial section of a cylinder of length $l$, enclosed between the inner radius $b$ and the distance $r$ with the mutual induction current density phasor $j_{2b}(r)$. That is, introducing the density of eddy currents in this way, we assumed that the cross sections of elementary eddy currents do not change when the radial position changes. Of course, we do not consider eddy currents as literally flowing along the contours shown in Fig. 1 with blue arrows for $j_{2a}(r)$, and yellow arrows for $j_{2b}(r)$. Rather, these circuits can be considered as the most probable ways of closing the sums of eddy microcurrents, if we consider their resulting action in a certain cylindrical surface at distance $r$ from the axis. As an analogy, we can give an example of currents of free charges in a constant magnetic field, which actually flow along circles of Larmor radius, but which can be represented as a result of the flow of bulk and surface currents of a magnet. The total current density phasor is denoted by $j(r)$. Due to the constant cross-section along the entire wire, including the power source, the current densities $j_1(r)$ and $j(r)$ depend only on the distance $r$. The magnetic field of the flowing current is present both in the wire itself and in the surrounding space coaxial with the wire inside a cylindrical surface with an outer radius $n$. The force lines of the magnetic field induction phasor $B(r)$ are circles centered on the axis. Along the entire length of the wire, including the power supply, the magnetic field does not change at the same distance $r$. To find the original equation, the following are used: Kirchhoff's circuit laws, Faraday's law of induction, Ampère's circuital law. According to the last law, the phasor of magnetic induction at distance $r$ is
$$ \mathbf{B}(r) = \frac{\mu \mathbf{I}(r)}{2\pi r}, $$

where

$$ \mathbf{I}(r) = 2\pi \int_{b}^{r} \mathbf{j}(r) r dr $$

is the current phasor inside the cylinder bounded by the inner radius $b$ and the distance $r$. Substituting $\mathbf{I}(r)$ into $\mathbf{B}(r)$, we have

$$ \mathbf{B}(r) = \frac{\mu}{r} \int_{b}^{r} \mathbf{j}(r) r dr. \qquad (1) $$

The magnetic flux phasor through the axial rectangular section of the cylinder with sides $l$ and $a-r$ is

$$ \Phi_a(r) = l \int_{r}^{a} \mathbf{B}(r) dr . \qquad (2) $$

The total current phasor is

$$ \mathbf{I} = 2\pi \int_{b}^{a} \mathbf{j}(r) r dr . \qquad (3) $$

The magnetic flux phasor through the external space, i.e. through the axial rectangular section of the cylinder with sides $l$ and $n-a$ is

$$ \Phi_s = \mathbf{I} \frac{\mu_s l}{2\pi} \ln\left(\frac{n}{a}\right). \qquad (4) $$

Lines of contours of length $l$, covering magnetic fluxes at different distances $r$, are considered to coincide with lines of current densities at the same distances. According to Faraday's law of induction in complex form, the part of the specific (per unit length of the wire) EMF (electromotive force) phasor of self-induction corresponding to the magnetic flux $\Phi_a(r)$ at distance $r$, is equal to

$$ \mathbf{E}_a(r) = -i\omega \frac{\Phi_a(r)}{l}. \qquad (5) $$

The part of the specific EMF phasor of self-induction corresponding to the magnetic flux $\Phi_s$ is

$$ \mathbf{E}_s = -i\omega \frac{\Phi_s}{l}. \qquad (6) $$

The elementary electric current phasor of the primary circuit at distance $r$ will be determined by Kirchhoff's voltage law as the ratio of the sum of the supply voltage and both found EMF to the elementary DC (direct current) resistance $dR_{DC}(r)$ as follows

$$ d\mathbf{I}_1(r) = \frac{\mathbf{E} + \mathbf{E}_a(r) + \mathbf{E}_s}{dR_{DC}(r)} l . $$

Then, after substituting the obvious relations $dR_{DC}(r) = \rho l/2\pi r dr$ and $d\mathbf{I}_1(r) = \mathbf{j}_1(r)2\pi r dr$ into the last expression, the current density phasor of the primary circuit at distance $r$ is

$$ \mathbf{j}_1(r) = \frac{\mathbf{E} + \mathbf{E}_a(r) + \mathbf{E}_s}{\rho}. \qquad (7) $$
Substituting the values from (5), (6) and their components from (1) – (4) into Eq. (7), we obtain

$$ \mathbf{j}_1(r) = \frac{\mathbf{E}}{\rho} + \frac{-i\omega\mu}{\rho} \int_r^a \frac{1}{r} \int_b^r \mathbf{j}(r) r dr dr + \frac{-i\omega\mu_s}{\rho} \ln\left(\frac{n}{a}\right) \int_b^a \mathbf{j}(r) r dr . \quad (8) $$

Hereinafter the term "phasor" in relation to phasor quantities such as: current, current density, magnetic induction, magnetic flux, EMF, voltage, field strength will not be used for brevity. We also introduce five designations: the skin depth

$$ \delta = \sqrt{\frac{2\rho}{\omega\mu}}; \quad (9) $$

the wave number in the conductor

$$ \mathbf{k} = \sqrt{\frac{-i\omega\mu}{\rho}}; \quad (10) $$

the wave number in the surrounding space

$$ \mathbf{k}_s = \sqrt{\frac{-i\omega\mu_s}{\rho}}; \quad (11) $$

the DC density

$$ \mathbf{j}_{\text{DC}} = \frac{\mathbf{E}}{\rho}; \quad (12) $$

the radius factor of the surrounding space

$$ s = \frac{n}{a}. \quad (13) $$

Here it is necessary to deviate from the process of deriving the equation for the total current density $\mathbf{j}(r)$ in order to analyze the resulting intermediate Eq. (8). Suppose that there are no eddy currents $\mathbf{j}_{2a}(r)$ and $\mathbf{j}_{2b}(r)$ at all, that is, the entire wire current is the self-induction current of the primary circuit. Then Eq. (8), taking into account the designations (10) – (13) and assuming $\mathbf{j}(r)=\mathbf{j}_1(r)$, will be

$$ \mathbf{j}_1(r) = \mathbf{j}_{\text{DC}} + \mathbf{k}^2 \int_r^a \frac{1}{r} \int_b^r \mathbf{j}_1(r) r dr dr + \mathbf{k}_s^2 \ln(s) \int_b^a \mathbf{j}_1(r) r dr . \quad (14) $$

The solution of Eq. (14) in the case of a solid cylinder with $b=0$ (search and verification at the end of Appendix 1) is

$$ \mathbf{j}_1(r) = \frac{\mathbf{j}_{\text{DC}} \mathbf{k} \mathbf{J}_0(r \mathbf{k})}{\mathbf{k} \mathbf{J}_0(a \mathbf{k}) - a \mathbf{k}_s^2 \ln(s) \mathbf{J}_1(a \mathbf{k})}. \quad (15) $$

In the case when there is no magnetic field outside the wire, that is at $n=a$ ($s=1$), the second term in the denominator of solution (15) vanishes, and formula (15) turns into the well-known formula for the current density distribution [1, p. 394, § 11.04, f. (1)]

$$ \mathbf{j}(r) = \mathbf{j}_0 \frac{\mathbf{J}_0(r\mathbf{k})}{\mathbf{J}_0(a\mathbf{k})}, \quad (16) $$

where $\mathbf{j}_0$ is the current density on the surface. It is easy to show that for $n=a$, the current density on the surface $\mathbf{j}_0$ from (16) is in fact exactly equal to the DC density $\mathbf{j}_{\text{DC}}$ (12), which will be done later in Sec. III. Thus, it turns out that even without completing the derivation of the equation for total current density taking into account secondary eddy currents, we have already obtained the solution (15), which absolutely coincides with the existing one. And this means that in fact, nothing remains on the share of the influence of eddy currents, according to the huge array of experimental data on measuring resistance at various frequencies. Or their influence is so small it's almost invisible.
Nevertheless, let us return to the continuation of the process of deriving the equation for the total current density, taking into account the influence of eddy currents. Let's look at the secondary circuits. Obviously, they are short-circuited, since the currents in these circuits are eddy currents and do not create stray magnetic fluxes outside the wire. The magnetic flux penetrating the external secondary closed circuit, limited by the distance *r* and the outer radius of the wire *a*, is determined by the formula (2). Only the specific EMF of such a closed circuit is half the specific EMF determined by formula (5), due to the fact that the perimeter of such a closed circuit is twice the perimeter of the primary circuit for the EMF of self-induction. We neglect the transverse radial dimensions of the perimeters of secondary contours. So, the external eddy specific EMF of mutual induction of the first secondary circuit at distance *r* is

$$ \mathbf{E}_{2a}(r) = -i\omega \frac{\mathbf{\Phi}_a(r)}{2l}. \qquad (17) $$

Then the density of the secondary external eddy current from the specific EMF $\mathbf{E}_{2a}(r)$ at distance *r* is

$$ \mathbf{j}_{2a}(r) = \frac{\mathbf{E}_{2a}(r)}{\rho}. \qquad (18) $$

The magnetic flux through the internal secondary closed circuit (axial rectangular section of the cylinder with sides *l* and *r*−*b*) is

$$ \mathbf{\Phi}_b(r) = l \int_b^r \mathbf{B}(r) dr. \qquad (19) $$

The internal eddy specific EMF of mutual induction from the magnetic flux $\mathbf{\Phi}_b(r)$ at distance *r* is

$$ \mathbf{E}_{2b}(r) = -i\omega \frac{\mathbf{\Phi}_b(r)}{2l}. \qquad (20) $$

The density of the secondary internal eddy current from the specific EMF $\mathbf{E}_{2b}(r)$ at distance *r* is

$$ \mathbf{j}_{2b}(r) = \frac{\mathbf{E}_{2b}(r)}{\rho}. \qquad (21) $$

Due to the fact that at distance *r* the eddy specific EMF found by formulas (17) and (20) create oppositely directed eddy currents (see Fig. 1), we write the Kirchhoff's current law for the total current density at distance *r*, taking into account this fact, i.e.

$$ \mathbf{j}(r) = \mathbf{j}_1(r) + \mathbf{j}_{2a}(r) - \mathbf{j}_{2b}(r). $$

Substituting the current density values from (7), (18), (21) into the last expression, we obtain

$$ \mathbf{j}(r) = \frac{\mathbf{E} + \mathbf{E}_a(r) + \mathbf{E}_s}{\rho} + \frac{\mathbf{E}_{2a}(r)}{\rho} - \frac{\mathbf{E}_{2b}(r)}{\rho}. \qquad (22) $$

Substituting the values from (5), (6), (17), (20), their components from (1) – (4), (19) into Eq. (22), taking into account the designations (10) – (13), we obtain

$$ \mathbf{j}(r) = \mathbf{j}_{DC} + \mathbf{k}^2 \int_{r}^{a} \int_{b}^{r} \mathbf{j}(r) r dr dr + \mathbf{k}_{s}^2 \ln(s) \int_{b}^{a} \mathbf{j}(r) r dr + \frac{\mathbf{k}^2}{2} \int_{r}^{a} \int_{b}^{r} \mathbf{j}(r) r dr dr - \frac{\mathbf{k}^2}{2} \int_{b}^{r} \int_{r}^{a} \mathbf{j}(r) r dr dr. $$
Representing the last term in the last equation as the difference of integrals over other integration limits in the form

$$ \frac{\mathbf{k}^2}{2} \int_{b}^{r} \frac{1}{r} \int_{b}^{r} \mathbf{j}(r) r dr dr = \frac{\mathbf{k}^2}{2} \int_{b}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}(r) r dr dr - \frac{\mathbf{k}^2}{2} \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}(r) r dr dr $$

and immediately grouping the integrals with the same integration limits, we get

$$ \mathbf{j}(r) = \mathbf{j}_{DC} + \frac{4}{2} \mathbf{k}^2 \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}(r) r dr dr + \mathbf{k}_{s}^2 \ln(s) \int_{b}^{a} \mathbf{j}(r) r dr - \frac{1}{2} \mathbf{k}^2 \int_{b}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}(r) r dr dr $$

Instead of a specific number 1/2 in front of the last integral in the last equation, obtained without taking into account the influence of radial dimensions, we introduce the coefficient p, which, in addition to convenience, will also allow us to vary the eddy EMF to approximately account for eddy currents in wires with radial dimensions comparable to axial ones. Then the number 4/2 in front of the first integral can be represented as 1 + 2p and denoted as a square of some factor. Thus, for convenience, we introduce the following coefficients:

$$ p = 0...0.5; \quad q = \sqrt{1+2p} = 1...\sqrt{2}. \qquad (23) $$

Then the final equation taking into account eddy currents is

$$ \mathbf{j}(r) = \mathbf{j}_{DC} + q^2 \mathbf{k}^2 \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}(r) r dr dr + \mathbf{k}_{s}^2 \ln(s) \int_{b}^{a} \mathbf{j}(r) r dr - p \mathbf{k}^2 \int_{b}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}(r) r dr dr, \quad (24) $$

and its solution in the case of a solid cylinder at b=0 (search and verification see Appendix 1) is

$$ \mathbf{j}(r) = \frac{\mathbf{j}_{DC} \mathbf{k} q^2 J_0(r \mathbf{k} q)}{(q^2 - p) \mathbf{k} J_0(a \mathbf{k} q) - a \mathbf{k}_s^2 q \ln(s) J_1(a \mathbf{k} q) + p \mathbf{k}}. \quad (25) $$

The total current for solution (25) is determined by formula (3). The total current for the eddy-free solution (15) is

$$ \mathbf{I}_1 = 2\pi \int_{b}^{a} \mathbf{j}_1(r) r dr \qquad (26) $$

(not to be confused with the I₁ Bessel function). Substituting (15) in (26) and (25) in (3) and assuming that

$$ \int_{0}^{r} J_{0}(r \mathbf{k}) r dr = \frac{r}{\mathbf{k}} J_{1}(r \mathbf{k}) \quad \text{and} \quad \int_{0}^{r} J_{0}(r \mathbf{k} q) r dr = \frac{r}{\mathbf{k} q} J_{1}(r \mathbf{k} q) $$

[3, p. 333, f. (3)], the total current for the eddy-free solution is

$$ \mathbf{I}_1 = \frac{2\pi a \mathbf{j}_{DC} J_1(a\mathbf{k})}{\mathbf{k} J_0(a\mathbf{k}) - a \mathbf{k}_s^2 \ln(s) J_1(a\mathbf{k})}, \quad (27) $$

the total current for the eddy solution is

$$ \mathbf{I} = \frac{2\pi a \mathbf{j}_{DC} q J_1(a\mathbf{k}q)}{(q^2 - p)\mathbf{k} J_0(a\mathbf{k}q) - a\mathbf{k}_s^2 q \ln(s) J_1(a\mathbf{k}q) + p\mathbf{k}}. \quad (28) $$
Then for the eddy-free solution: the impedance, taking into account $I_1$ from (27) and notation (12), is

$$Z_1 = \frac{El}{I_1} = \frac{\rho l[\mathbf{k} \cdot J_0(\mathbf{a}\mathbf{k}) - \mathbf{a}\mathbf{k}_s^2 \ln(s) J_1(\mathbf{a}\mathbf{k})]}{2\pi a J_1(\mathbf{a}\mathbf{k})}; \quad (29)$$

the active AC (alternating current) resistance is

$$R_1 = \mathrm{Re}(Z_1); \quad (30)$$

the inductive resistance is

$$X_1 = \mathrm{Im}(Z_1). \quad (31)$$

Similarly for the eddy solution: the impedance, taking into account $I$ from (28) and notation (12), is

$$Z = \frac{El}{I} = \frac{\rho l[(q^2 - p)\mathbf{k} \cdot J_0(\mathbf{a}\mathbf{k}) - \mathbf{a}\mathbf{k}_s^2 q \ln(s) J_1(\mathbf{a}\mathbf{k}) + p\mathbf{k}]}{2\pi a q J_1(\mathbf{a}\mathbf{k})}; \quad (32)$$

the active AC resistance is

$$R = \mathrm{Re}(Z); \quad (33)$$

the inductive resistance is

$$X = \mathrm{Im}(Z). \quad (34)$$

Formulas (29) – (34) are not direct consequences of the obtained equations and solutions and are presented only for use in further comparison of existing relationships with new ones. Below, in Sec. IV, a new version of the voltage and power balances will be obtained based on solutions (15), (25) and Eqs. (7) and (22).

Calculations according to formulas (15) and (25) for a solid aluminum wire with diameter $2a=6.676$ mm, cross section $S=35$ mm² and a length $l=1000$ m are given in Appendix 2 at frequencies 100 Hz and 100 MHz. The factor taking into account the magnitude of eddy currents was taken as the maximum $p=0.5$. Calculations were performed for a wire without external inductance at $n=a$. All initial data and results are shown in the screenshots (edited to save space) of the calculations. The displayed area of the graphs along the abscissa axis was selected using a scaling factor $k_a$ from $b_1=k_a a$ to $a$. Above the graphs are the values of the amplitudes and arguments of the current densities at the extreme points. The graphs for calculating the current densities show that the eddy-free solution (15) $j_1(r)$, and the eddy solution (25) $j(r)$ differ from each other in the degree of inclination and intersect at some point, the closer to the wire surface, the higher the frequency. The green background on the right shows the values of resistance to direct current $R_{DC}=\rho l/S$ and alternating current $R_1$ according to formula (30) and $R$ according to formula (33). We see from the screenshots that the total currents for the eddy-free solution $I_1$ according to formula (27) and the eddy solution $I$ according to formula (28) are slightly different, as are the resistances calculated by formulas (30) and (33).

It is easy to guess that if instead of the number 1 in the formula for $q$ (23) we substitute the number 0, then in this way we "turn off" the self-induction of the wire itself, leaving only eddy currents. Appendix 3 compares the eddy-free solution (15) $j_1(r)$ with the purely eddy solution (25) $j(r)$ at $q=1$ and $p=0.5$. For the purely eddy solution, the AC resistance $R$ at a small skin depth is half that $R_1$ for the eddy-free one. It can be concluded that the purely eddy solution is not implemented in practice.

The influence of the permeability of the surrounding space is presented in Appendix 4 by comparing graphs (15) and (25) at $n=2a$ ($s=2$) at a frequency of 10 kHz for aluminum wire in a ferromagnetic sheath with $\mu_{sr}=1000$ – at the top and in air (vacuum) with $\mu_{sr}=1$ – at the bottom.

Appendix 5 contains frequency graphs for the case without external inductance at $n=a$ ($s=1$). Above are the modules and arguments of the total currents for the eddy-free (27) and eddy (28) so-
lutions and their differences. It can be seen that the difference between the currents is practically absent, except for a small section in the middle of the range (the scale of the current difference is on the right ordinate). This small section is convenient to use for experiment to answer the question which solution, eddy-free (15) or eddy (25), is realized in practice. Measuring the current at these parameters through a sufficiently long solitary wire in the form of a ring should give the answer to this question. The second row shows graphs of current densities (15) and (25) at $r=a$, i.e., on the surface, and the ratio of their modules to each other and to the DC density (along the right ordinate). The current density on the surface for the eddy-free version (15) did not change with increasing frequency and was equal to the DC density. For the eddy variant (25), the current density on the surface increased with increasing frequency, reaching a limiting value approximately 1.435 times higher than the DC density, then decreased to about 1.333 times and stabilized at this value. The third row shows graphs of impedances (29), (32), active (30), (33) and inductive (31), (34) resistances for eddy-free and eddy solutions. As can be seen from the graphs, there are no visible differences between the eddy-free and eddy variants. For the case without an external inductance at $n=a$ ($s=1$) with an increase in frequency, the modules of active and inductive resistances tended to equalize, and the limiting value of the phase lag of the total current from the voltage tended to $\pi/4$. The third row on the right shows the graphs of inductances for eddy-free $L_1=X_1/\omega$ and eddy $L=X/\omega$ solutions and their ratio. At the bottom of Appendix 5 are graphs of comparison of active (30) and (33) and inductive (31) and (34) resistances for eddy-free and eddy solutions and their difference (along the right ordinate). The tables to the right of the graphs show the numerical differences in resistances, from which it can be seen that at low frequencies (left column) the eddy variant is somewhat smaller than the eddy-free one, and at high frequencies (right column), on the contrary, it is larger, and quite well.

Appendix 6 also contains frequency graphs, as in Appendix 5, only for the case with external inductance at $n=3a$ ($s=3$). In this case, with increasing frequency, the current density on the surface decreases for the eddy-free variant. For the eddy variant, there is a dependence of the magnitude of the maximum surface current density achieved on the magnitude of the external inductance: the greater the inductance, the smaller the maximum achieved. Upon reaching the maximum, the surface current density for the eddy variant also decreases, approaching the eddy-free variant. At $s=3$ for air ($\mu_{sr}=1$), that is, in the case shown in Appendix 6, there is practically no increase in the surface current density compared to the DC density. It can also be seen from the third row of graphs that the inductive reactance increased more than the active resistance, and the limiting value of the total current phase lag from the voltage tended to $\pi/2$.

Comparison of active resistances (30) and (33), inductive resistances (31) and (34) for a copper wire with a diameter of 0.51 mm and 1.3 mm and a length of 0.85 m is given in Appendix 7. These data are qualitatively consistent with the data given in Ref. [4, pp. 61–66] for the same wires, if (30) and (31) are considered as theoretical dependences, and (33) and (34) as experimental ones. These data allow us to conclude that, along with the reasons for the discrepancy between the theoretical and experimental curves presented by the author of Ref. [4], eddy currents can also have some effect on the current density distribution.

## B. Cylindrical tubular wire

For a tubular cylindrical wire, the solution of the eddy-free Eq. (14) for $b>0$ (search and verification see Appendix 8) is

$$ \mathbf{j}_1(r) = \frac{\mathbf{j}_{DC}\mathbf{k}[K_1(b\mathbf{k}\mathbf{i})I_0(r\mathbf{k}\mathbf{i}) + I_1(b\mathbf{k}\mathbf{i})K_0(r\mathbf{k}\mathbf{i})]}{\mathbf{D}_{tA} + \mathbf{D}_{tB}}, \quad (35) $$

where:

$$ \mathbf{D}_{tA} = \mathbf{k} K_1(b\mathbf{k}\mathbf{i}) I_0(\mathbf{a}\mathbf{k}\mathbf{i}) + \mathbf{k} I_1(b\mathbf{k}\mathbf{i}) K_0(\mathbf{a}\mathbf{k}\mathbf{i}); $$

$$ \mathbf{D}_{tB} = i\mathbf{a}\mathbf{k}_s^2 \ln(s)[I_1(\mathbf{a}\mathbf{k}\mathbf{i})K_1(b\mathbf{k}\mathbf{i}) - I_1(b\mathbf{k}\mathbf{i})K_1(\mathbf{a}\mathbf{k}\mathbf{i})]. $$
In the absence of an external magnetic field ($n=a$), solution (35), similar to the situation with the solution for a solid wire (15), completely coincides with the solution from the source "Static and Dynamic Electricity" by W.R. Smythe [1, p. 394, § 11.03, f. (2), (6), (7)]

$$ \mathbf{j}(r) = \mathbf{j}_0 \frac{K_1(b\mathbf{k}\mathbf{i}) I_0(r\mathbf{k}\mathbf{i}) + I_1(b\mathbf{k}\mathbf{i}) K_0(r\mathbf{k}\mathbf{i})}{K_1(b\mathbf{k}\mathbf{i}) I_0(a\mathbf{k}\mathbf{i}) + I_1(b\mathbf{k}\mathbf{i}) K_0(a\mathbf{k}\mathbf{i})}. $$

It should be clarified that the solution from Ref. [1] is not a solution of the Bessel equation [2, p. 722, f. (A.4)]

$$ \frac{d^2 \mathbf{j}(r)}{dr^2} + \frac{1}{r} \frac{d\mathbf{j}(r)}{dr} + \mathbf{k}^2 \mathbf{j}(r) = 0, $$

as it is obtained from Maxwell's equations, but it is a solution of the modified Bessel equation [1, p. 394, § 11.03, f. (3)]

$$ \frac{d^2 \mathbf{j}(r)}{dr^2} + \frac{1}{r} \frac{d\mathbf{j}(r)}{dr} - \mathbf{k}^2 \mathbf{j}(r) = 0, $$

in which the sign before the term $\mathbf{k}^2\mathbf{j}(r)$ has changed as a result of replacing $r$ with $ir$. This method of obtaining the desired solution is no better than the method of discarding the divergent part of the complete solution mentioned in the introduction. Changing the sign of the last term in the Bessel equation, called a modification, is essentially a transformation of the equation for closed eddy currents that do not carry charge through the wire cross section into an equation for a directional current. If we call a spade a spade, then this technique is just a trick, since instead of solving the strictly obtained Bessel equation, a solution of a completely different equation is proposed. Hence, changing the original Bessel equation and obtaining the correct solution is carried out in the existing approach without rigorous mathematical and physical justification. But let's return to our presentation. The formulas for current and impedance for the eddy-free case (35) from Appendix 9 are:

$$ \mathbf{I}_1 = \frac{2i\pi a \mathbf{j}_{DC} [K_1(a\mathbf{k}\mathbf{i}) I_1(b\mathbf{k}\mathbf{i}) - I_1(a\mathbf{k}\mathbf{i}) K_1(b\mathbf{k}\mathbf{i})]}{\mathbf{D}_{tA} + \mathbf{D}_{tB}}; \quad \mathbf{Z}_1 = \frac{\rho l (\mathbf{D}_{tA} + \mathbf{D}_{tB}) \mathbf{i}}{2\pi a [I_1(a\mathbf{k}\mathbf{i}) K_1(b\mathbf{k}\mathbf{i}) - K_1(a\mathbf{k}\mathbf{i}) I_1(b\mathbf{k}\mathbf{i})]} $$

The solution of the eddy Eq. (24) for $b > 0$ (search and verification see Appendix 8) is

$$ \mathbf{j}(r) = \frac{\mathbf{j}_{DC} q^2 \mathbf{k} [K_1(b\mathbf{k}\mathbf{q}\mathbf{i}) I_0(r\mathbf{k}\mathbf{q}\mathbf{i}) + I_1(b\mathbf{k}\mathbf{q}\mathbf{i}) K_0(r\mathbf{k}\mathbf{q}\mathbf{i})]}{\mathbf{D}_{tC} + \mathbf{D}_{tD} + \mathbf{D}_{tE}}, \quad (36) $$

where:

$$ \mathbf{D}_{tC} = (q^2 - p)\mathbf{k} [K_1(b\mathbf{k}\mathbf{q}\mathbf{i}) I_0(a\mathbf{k}\mathbf{q}\mathbf{i}) + I_1(b\mathbf{k}\mathbf{q}\mathbf{i}) K_0(a\mathbf{k}\mathbf{q}\mathbf{i})]; $$

$$ \mathbf{D}_{tD} = i\mathbf{a}\mathbf{k}_s^2 q \ln(s) [I_1(a\mathbf{k}\mathbf{q}\mathbf{i}) K_1(b\mathbf{k}\mathbf{q}\mathbf{i}) - I_1(b\mathbf{k}\mathbf{q}\mathbf{i}) K_1(a\mathbf{k}\mathbf{q}\mathbf{i})]; $$

$$ \mathbf{D}_{tE} = p\mathbf{k} [K_1(b\mathbf{k}\mathbf{q}\mathbf{i}) I_0(b\mathbf{k}\mathbf{q}\mathbf{i}) + I_1(b\mathbf{k}\mathbf{q}\mathbf{i}) K_0(b\mathbf{k}\mathbf{q}\mathbf{i})]. $$

The formulas for current and impedance for the eddy case (36) from Appendix 9 are:

$$ \mathbf{I} = \frac{2i\pi a \mathbf{j}_{DC} q [K_1(a\mathbf{k}\mathbf{q}\mathbf{i}) I_1(b\mathbf{k}\mathbf{q}\mathbf{i}) - I_1(a\mathbf{k}\mathbf{q}\mathbf{i}) K_1(b\mathbf{k}\mathbf{q}\mathbf{i})]}{\mathbf{D}_{tC} + \mathbf{D}_{tD} + \mathbf{D}_{tE}}, \quad \mathbf{Z} = \frac{\rho l (\mathbf{D}_{tC} + \mathbf{D}_{tD} + \mathbf{D}_{tE}) \mathbf{i}}{2\pi a q [I_1(a\mathbf{k}\mathbf{q}\mathbf{i}) K_1(b\mathbf{k}\mathbf{q}\mathbf{i}) - K_1(a\mathbf{k}\mathbf{i}) I_1(b\mathbf{k}\mathbf{q}\mathbf{i})]} $$

As the inner radius tends to zero, both solutions for tubular wire (35) and (36) absolutely coincide with the corresponding solutions for solid wire (15) and (25). The integral transformations given in Appendices 1, 8 and 9 are derived on the basis of formulas from Ref. [3, Ch. XIX].
A comparison of the eddy ($p=0.5$) solution (36) (denoted by $j_{b>0}(r)$) for a tubular cylinder with the eddy solution (25) (denoted by $j_{b=0}(r)$) for a solid wire at 10 MHz is shown in Appendix 10. The outer radius of tubular and solid wire is the same. The graphs for a solid cylinder $j_{b=0}(r)$ are shown only in the range from $b=k_b a$ to $a$, while in reality they naturally occupy the range from $b=0$ to $a$. The value of the inner radius $b$ of the tubular wire is chosen so that its resistance to alternating current $R$ is less than that of a solid wire $R_0$. This means that, under certain parameters, it is more profitable to use tubular wires instead of solid ones, not only in terms of material consumption, but also in terms of throughput. Some theoretical confirmation of this statement, found in the sources, can be considered a quote from Ref. [1, p. 393, § 11.03], which refers to the total current: "This would be increased by removing all conducting matter below a certain depth since, for certain values of $z$, the current is reversed.". The quotation refers to a semi-infinite space with a $z$-coordinate perpendicular to the surface. In our case of a tubular cylinder, this is $r$. The use of tubular wires instead of solid ones in order to save material is a fairly well-known fact. However, no mention has been found (excluding the above assumption from Ref. [1]) that a tubular wire also has less resistance to alternating current at certain parameters than a solid wire with the same outer radius. The minimum value of the resistance $R$ approximately corresponds to the following ratio of the skin depth to the wall thickness $\delta/(a-b) \approx 0.45$, as shown in Appendix 10. For the eddy-free case ($p=0$, and the rest of the initial data, as in Appendix 10), as shown in Appendix 11, the optimal value of the ratio is $\delta/(a-b) \approx 0.637$, and the difference between these values can also serve as a criterion for which of the solutions takes place in practice. The frequency dependences given at the bottom of Appendices 10 and 11 show that for a tubular wire it is always possible to select frequencies at which its active and inductive resistances will be less than those for a solid wire of the same diameter.

### C. Coaxial cable

The analysis revealed the inapplicability of the obtained equations for a cylindrical wire in the presence of a return wire-screen. This is primarily due to the fact that the covered circuit for the specific EMF of self-induction then becomes twice as large (we neglect the radial dimensions), and the increase in magnetic flux due to the appearance of a return wire is not proportional to the increase in the length of the circuit. This, in turn, is due to the fact that the magnetic induction with axial symmetry, even in regions without current, decreases proportionally to $1/r$ with increasing radius, and even faster in the region of the screen. Let's consider the axial cross section of the coaxial cable in Fig. 2. Let the cable length be $l$. All dimensions of the inner wire are marked in the same

<figure><img src="image_3.png" /><figcaption>FIG. 2. Coaxial cable with dimensions $b, r, a, n, u, t, l$ with magnetic induction $B_i(r)$, current densities $j_{1i}(r), j_i(r), j_{ia}(r), j_{ib}(r)$ for inner wire, and magnetic induction $B_o(u)$, current densities $j_{1o}(u), j_o(u), j_{ot}(u), j_{on}(u)$ for outer wire.</figcaption></figure>

way as the conventional wire shown in Fig. 1. Dimensions of the outer wire: inner radius $n$, outer radius $t$, variable distance $u$ from the axis (henceforth as the distance $u$). Let us denote the current densities of the inner wire as $j_{1i}(r)$ for the eddy-free case, $j_i(r)$ for the eddy case, and for the outer wire as $j_{1o}(u)$ for the eddy-free case, $j_o(u)$ for the eddy case. At one end, the inner and outer wires are shorted together. An alternating voltage $U$ is applied to the opposite ends. We will consider the case, when the inner and outer wires have the same magnetic permeability, resistance and resistiv-
ity, and the capacitance between the wires is zero. In this case, the cross sections of the wires S are the same and the DC density for both wires is the same. Then the electric field strength from the power supply is

$$ \mathbf{E} = \frac{\mathbf{U}}{2l} = \frac{\mathbf{j}_{DC} S}{2l} 2 \frac{\rho l}{S} = \mathbf{j}_{DC} \rho . $$

We will assume that the lines of elementary currents at distances $r$ and $u$ close with each other so that $\pi(r^2 - b^2) = \pi(t^2 - u^2)$ and $\pi(a^2 - r^2) = \pi(u^2 - n^2)$, while the elementary cross sections remain equal to each other $2\pi r dr = -2\pi u du$, and the signs of $du$ and $dr$ are opposite. That is, the elementary currents of the wires are closed concentrically so that the inner elementary currents of the inner wire are closed to the outer elementary currents of the outer wire, and the outer currents of the inner wire are closed to the inner currents of the outer wire. That is, in this case, the equalities $\mathbf{j}_i(r) = \mathbf{j}_o(u)$ and $\mathbf{j}_{1i}(r) = \mathbf{j}_{1o}(u)$ will hold. According to Ampère's circuital law, the magnetic induction at distance $r$ in the inner wire is

$$ \mathbf{B}_i(r) = \frac{\mu \mathbf{I}_i(r)}{2\pi r}, $$

where

$$ \mathbf{I}_i(r) = 2\pi \int_b^r \mathbf{j}_i(r) r dr $$

is the current of the inner wire inside the cylinder bounded by the inner radius $b$ and the distance $r$. Substituting $\mathbf{I}_i(r)$ into $\mathbf{B}_i(r)$, we get

$$ \mathbf{B}_i(r) = \frac{\mu}{r} \int_b^r \mathbf{j}_i(r) r dr . \qquad (37) $$

The magnetic induction at distance $u$ in the outer wire is

$$ \mathbf{B}_o(u) = \frac{\mu[\mathbf{I}_i - \mathbf{I}_o(u)]}{2\pi u}, \qquad (38) $$

where

$$ \mathbf{I}_i = 2\pi \int_b^a \mathbf{j}_i(r) r dr \qquad (39) $$

is the total current of the inner wire,

$$ \mathbf{I}_o(u) = 2\pi \int_n^u \mathbf{j}_o(u) u du \qquad (40) $$

is the current inside the cylinder of the outer wire, bounded by the inner radius $n$ and the distance $u$. When writing this integral in the range from $n$ to $u$, the increment $du > 0$ and the "-" sign before it is not needed to ensure that the signs of the currents of both wires are equal. The total current of the outer wire is

$$ \mathbf{I}_o = 2\pi \int_n^t \mathbf{j}_o(u) u du . \qquad (41) $$

Obviously, $\mathbf{I}_i = \mathbf{I}_o$. Substituting $\mathbf{I}_i$ from (39) and $\mathbf{I}_o(u)$ from (40) into (38), replacing the interpretation of $\mathbf{I}_i$ with the interpretation of $\mathbf{I}_o$ from (41), and taking the resulting difference, we obtain

$$ \mathbf{B}_o(u) = \frac{2\pi\mu}{2\pi u} \int_b^a \mathbf{j}_i(r) r dr - \frac{2\pi\mu}{2\pi u} \int_n^u \mathbf{j}_o(u) u du = \frac{\mu}{u} \int_n^t \mathbf{j}_o(u) u du - \frac{\mu}{u} \int_n^u \mathbf{j}_o(u) u du = \frac{\mu}{u} \int_u^t \mathbf{j}_o(u) u du . \qquad (42) $$
The magnetic flux through the axial rectangular sections of the inner wire cylinders: with sides $l$ and $a-r$ is equal to

$$ \mathbf{\Phi}_a(r) = l \int_r^a \mathbf{B}_i(r) dr; \qquad (43) $$

with sides $l$ and $r-b$ is equal to

$$ \mathbf{\Phi}_b(r) = l \int_b^r \mathbf{B}_i(r) dr. \qquad (44) $$

The magnetic flux through the axial rectangular sections of the outer wire cylinders: with sides $l$ and $u-n$ is equal to

$$ \mathbf{\Phi}_n(u) = l \int_n^u \mathbf{B}_o(u) du; \qquad (45) $$

with sides $l$ and $t-u$ is equal to

$$ \mathbf{\Phi}_t(u) = l \int_u^t \mathbf{B}_o(u) du. \qquad (46) $$

Let's assume that the line enclosing the magnetic flux through an axial rectangular section with dimensions $l$ and $u-r$ coincides with the elementary current line and passes at distances $r$ and $u$ so that $\mathbf{I}_i - \mathbf{I}_i(r) = \mathbf{I}_o(u)$. The parts of the specific EMFs (electromotive forces) of self-induction corresponding to the magnetic fluxes $\mathbf{\Phi}_a(r)$ and $\mathbf{\Phi}_n(u)$ at distances $r$ and $u$ are

$$ \mathbf{E}_i(r) = \mathbf{E}_o(u) = -i\omega \frac{\mathbf{\Phi}_a(r) + \mathbf{\Phi}_n(u)}{2l}. \qquad (47) $$

The eddy specific EMFs of mutual induction from magnetic fluxes $\mathbf{\Phi}_a(r)$ and $\mathbf{\Phi}_n(u)$ at distances $r$ and $u$ are

$$ \mathbf{E}_{ia}(r) = \mathbf{E}_{on}(u) = -i\omega \frac{\mathbf{\Phi}_a(r) + \mathbf{\Phi}_n(u)}{4l}. \qquad (48) $$

The densities of secondary eddy currents from eddy specific EMFs $\mathbf{E}_{ia}(r)$ and $\mathbf{E}_{on}(u)$ at distances $r$ and $u$ (see Fig. 2) are

$$ \mathbf{j}_{ia}(r) = \frac{\mathbf{E}_{ia}(r)}{\rho} = \mathbf{j}_{on}(u) = \frac{\mathbf{E}_{on}(u)}{\rho}. \qquad (49) $$

The eddy specific EMFs of mutual induction from magnetic fluxes $\mathbf{\Phi}_b(r)$ and $\mathbf{\Phi}_t(u)$ at distances $r$ and $u$ are

$$ \mathbf{E}_{ib}(r) = \mathbf{E}_{ot}(u) = -i\omega \frac{\mathbf{\Phi}_b(r) + \mathbf{\Phi}_t(u)}{4l}. \qquad (50) $$

The densities of secondary eddy currents from eddy specific EMFs $\mathbf{E}_{ib}(r)$ and $\mathbf{E}_{ot}(u)$ at distances $r$ and $u$ (see Fig. 2) are

$$ \mathbf{j}_{ib}(r) = \frac{\mathbf{E}_{ib}(r)}{\rho} = \mathbf{j}_{ot}(u) = \frac{\mathbf{E}_{ot}(u)}{\rho}. \qquad (51) $$

The magnetic flux through the axial rectangular section of the space between the wires with sides $l$ and $n-a$ is

$$ \mathbf{\Phi}_s = \mathbf{I}_i \frac{\mu_s l}{2\pi} \ln\left(\frac{n}{a}\right). \qquad (52) $$
The part of the specific EMF of self-induction corresponding to the magnetic flux $\Phi_s$ is

$$ \mathbf{E}_s = -i\omega\frac{\mathbf{\Phi}_s}{2l}. \qquad (53) $$

The primary elementary currents at distances $r$ and $u$ according to Kirchhoff's voltage law are

$$ d\mathbf{I}_{1i}(r) = \frac{\mathbf{E} + \mathbf{E}_i(r) + \mathbf{E}_s}{dR_{DCi}(r) + dR_{DCo}(u)} 2l = d\mathbf{I}_{1o}(u) = \frac{\mathbf{E} + \mathbf{E}_o(u) + \mathbf{E}_s}{dR_{DCi}(r) + dR_{DCo}(u)} 2l, \qquad (54) $$

where:

$$ dR_{DCi}(r) = \frac{\rho l}{2\pi r dr} = dR_{DCo}(u) = \frac{\rho l}{-2\pi u du}; \quad d\mathbf{I}_{1i}(r) = \mathbf{j}_{1i}(r)2\pi rdr = d\mathbf{I}_{1o}(u) = -\mathbf{j}_{1o}(u)2\pi u du. \qquad (55) $$

After substituting the values from (55) into Eq. (54), the primary current densities at distances $r$ and $u$ are

$$ \mathbf{j}_{1i}(r) = \frac{\mathbf{E} + \mathbf{E}_i(r) + \mathbf{E}_s}{\rho} = \mathbf{j}_{1o}(u) = \frac{\mathbf{E} + \mathbf{E}_o(u) + \mathbf{E}_s}{\rho}. $$

Substituting the values from (47), (53), their components from (37), (39), (42), (43), (45), (52) into the last equation, taking into account the designations (10) – (13), we obtain for the primary current densities

$$ \mathbf{j}_{1i}(r) = \mathbf{j}_{1o}(u) = \mathbf{j}_{DC} + \frac{\mathbf{k}^2}{2} \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}_{i}(r) r dr dr + \frac{\mathbf{k}^2}{2} \int_{n}^{u} \frac{1}{u} \int_{u}^{t} \mathbf{j}_{o}(u) u du du + \frac{\mathbf{k}^2}{2} \ln(s) \int_{b}^{a} \mathbf{j}_{i}(r) r dr. \qquad (56) $$

Let's write the Kirchhoff's current law for the total current densities at distances $r$ and $u$

$$ \mathbf{j}_i(r) = \mathbf{j}_{1i}(r) + \mathbf{j}_{1a}(r) - \mathbf{j}_{ib}(r) = \mathbf{j}_o(u) = \mathbf{j}_{1o}(u) + \mathbf{j}_{on}(u) - \mathbf{j}_{ot}(u). $$

Substituting the values from (49), (51), (56), their components from (37), (42) – (46), (48), (50) into the last expression, taking into account the designations (10) – (13), (23) and immediately grouping the integrals, we obtain for the eddy case

$$ \mathbf{j}_i(r) = \mathbf{j}_o(u) = \mathbf{j}_{DC} + q^2 \frac{\mathbf{k}^2}{2} \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}_i(r) r dr dr + q^2 \frac{\mathbf{k}^2}{2} \int_{n}^{u} \frac{1}{u} \int_{u}^{t} \mathbf{j}_o(u) u du du + \\ + \frac{\mathbf{k}_s^2}{2} \ln(s) \int_{b}^{a} \mathbf{j}_i(r) r dr - p \frac{\mathbf{k}_s^2}{2} \int_{b}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}_i(r) r dr dr - p \frac{\mathbf{k}_s^2}{2} \int_{n}^{t} \frac{1}{u} \int_{u}^{t} \mathbf{j}_o(u) u du du. \qquad (57) $$

The search for exact solutions of Eqs. (56) and (57) is apparently not an easy task. To find an approximate equation for the inner wire, instead of Eq. (56), we replace the third term on its right side so that in the integrands it is identical to the second term, but with some coefficient. Since both terms are proportional to the magnetic flux, which depends on induction and area, the magnetic flux through the outer wire can be represented as the magnetic flux through the inner wire multiplied by some factor depending on both the ratio of the average radii and the ratio of the areas of both fluxes. We will assume that the magnetic flux of the inner wire $\Phi_a(r)$ is concentrated in a layer with a skin depth $\delta$ by (9) from the surface with radius $a$, and the magnetic flux of the outer wire $\Phi_n(u)$ is concentrated in a layer $\Delta = \sqrt{2a \min(\delta, a-b) - \min(\delta, a-b)^2 + n^2} - n$ from the surface with radius $n$. The formula for $\Delta$ is the solution of the quadratic equation $\pi[(n+\Delta)^2 - n^2] = \pi[a^2 - (a-\delta)^2]$, which describes the equality of the cross sections of the currents of the outer and inner wires in the zones adjacent to the space between them. The maximum limiting depths of layers with magnetic flux are
determined by the differences $t-n$ for the outer and $a-b$ for the inner wires, that is, by the wall thicknesses. Based on the axial symmetry, we assume that the average inductions in the outer wire $\mathbf{B}_{om} = \mathbf{\Phi}_n(u)/\min(\Delta, t-n)l$ and the inner wire $\mathbf{B}_{im} = \mathbf{\Phi}_a(r)/\min(\delta, a-b)l$ obey the law $1/r$, that is

$$ \frac{\mathbf{B}_{om}}{\mathbf{B}_{im}} = \frac{2a - \min(\delta, a-b)}{2n + \min(\Delta, t-n)}. $$

Then the ratio between the magnetic fluxes of the outer and inner wires, determined through their average inductions, is equal to

$$ k_{\phi} = \frac{\mathbf{\Phi}_n(u)}{\mathbf{\Phi}_a(r)} = \frac{\mathbf{B}_{om}}{\mathbf{B}_{im}} \frac{\min(\Delta, t-n)l}{\min(\delta, a-b)l} = \frac{2a - \min(\delta, a-b)}{2n + \min(\Delta, t-n)} \frac{\min(\Delta, t-n)}{\min(\delta, a-b)}. $$

Similar to the above relation between magnetic fluxes, the same relation, equal to $k_\phi$, will be between the second and third terms on the right side of Eq. (56), which are proportional to the magnetic fluxes $\mathbf{\Phi}_a(r)$ and $\mathbf{\Phi}_n(u)$, respectively. Then, expressing the third term in terms of the second, we immediately write Eq. (56) for the inner wire for the eddy-free case, setting also $\mathbf{j}_i(r) = \mathbf{j}_{1i}(r)$

$$ \mathbf{j}_{1i}(r) = \mathbf{j}_{DC} + \frac{\mathbf{k}^2}{2} \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}_{1i}(r) r dr dr + k_{\phi} \frac{\mathbf{k}^2}{2} \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}_{1i}(r) r dr dr + \frac{\mathbf{k}_s^2}{2} \ln(s) \int_{b}^{a} \mathbf{j}_{1i}(r) r dr . $$

Grouping the integrals in the last equation and introducing the notation

$$ \mathbf{k}_i = \sqrt{\frac{\mathbf{k}^2}{2} + k_{\phi} \frac{\mathbf{k}^2}{2}} \quad \text{and} \quad \mathbf{k}_{sc} = \sqrt{\frac{\mathbf{k}_s^2}{2}}, $$

we finally get the eddy-free equation for the inner wire

$$ \mathbf{j}_{1i}(r) = \mathbf{j}_{DC} + \mathbf{k}_i^2 \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}_{1i}(r) r dr dr + \mathbf{k}_{sc}^2 \ln(s) \int_{b}^{a} \mathbf{j}_{1i}(r) r dr . \quad (58) $$

Then the solution of Eq. (58), similarly to solution (35), is

$$ \mathbf{j}_{1i}(r) = \frac{\mathbf{j}_{DC} \mathbf{k}_i [K_1(b \mathbf{k}_i \mathbf{i}) I_0(r \mathbf{k}_i \mathbf{i}) + I_1(b \mathbf{k}_i \mathbf{i}) K_0(r \mathbf{k}_i \mathbf{i})]}{\mathbf{D}_{iA} + \mathbf{D}_{iB}}, \quad (59) $$

where:

$$ \mathbf{D}_{iA} = \mathbf{k}_i K_1(b \mathbf{k}_i \mathbf{i}) I_0(a \mathbf{k}_i \mathbf{i}) + \mathbf{k}_i I_1(b \mathbf{k}_i \mathbf{i}) K_0(a \mathbf{k}_i \mathbf{i}); \\ \mathbf{D}_{iB} = i a \mathbf{k}_{sc}^2 \ln(s) [I_1(a \mathbf{k}_i \mathbf{i}) K_1(b \mathbf{k}_i \mathbf{i}) - I_1(b \mathbf{k}_i \mathbf{i}) K_1(a \mathbf{k}_i \mathbf{i})]. $$

Let's immediately give an equation, taking into account eddy currents for the inner wire, similar to (24)

$$ \mathbf{j}_i(r) = \mathbf{j}_{DC} + q^2 \mathbf{k}_i^2 \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}_i(r) r dr dr + \mathbf{k}_{sc}^2 \ln(s) \int_{b}^{a} \mathbf{j}_i(r) r dr - p \mathbf{k}_i^2 \int_{b}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}_i(r) r dr dr, \quad (60) $$

and its solution is similar to solution (36)

$$ \mathbf{j}_i(r) = \frac{\mathbf{j}_{DC} \mathbf{k}_i q^2 [K_1(b \mathbf{k}_i q \mathbf{i}) I_0(r \mathbf{k}_i q \mathbf{i}) + I_1(b \mathbf{k}_i q \mathbf{i}) K_0(r \mathbf{k}_i q \mathbf{i})]}{\mathbf{D}_{iC} + \mathbf{D}_{iD} + \mathbf{D}_{iE}}, \quad (61) $$
where:

$$ \mathbf{D}_{iC} = (q^2 - p)\mathbf{k}_i [K_1(b\mathbf{k}_i q\mathbf{i}) I_0(a\mathbf{k}_i q\mathbf{i}) + I_1(b\mathbf{k}_i q\mathbf{i}) K_0(a\mathbf{k}_i q\mathbf{i})]; $$

$$ \mathbf{D}_{iD} = i\mathbf{a}\mathbf{k}_{sc}^2 q \ln(s) [I_1(a\mathbf{k}_i q\mathbf{i}) K_1(b\mathbf{k}_i q\mathbf{i}) - I_1(b\mathbf{k}_i q\mathbf{i}) K_0(a\mathbf{k}_i q\mathbf{i})]; $$

$$ \mathbf{D}_{iE} = p\mathbf{k}_i [I_0(b\mathbf{k}_i q\mathbf{i}) K_1(b\mathbf{k}_i q\mathbf{i}) + I_1(b\mathbf{k}_i q\mathbf{i}) K_0(b\mathbf{k}_i q\mathbf{i})]. $$

Since in the derivation we took $\mathbf{j}_{1i}(r) = \mathbf{j}_{1o}(u)$ and $\mathbf{j}_i(r) = \mathbf{j}_o(u)$, the solutions for the wire-screen can be obtained by mirroring solutions (59) and (61) for the inner wire. That is, we just need to replace $\mathbf{j}_{1i}(r)$ with $\mathbf{j}_{1o}(u)$, $\mathbf{j}_i(r)$ with $\mathbf{j}_o(u)$ and $r$ with $\sqrt{t^2 - u^2 + b^2}$. Here are the solutions: eddy-free

$$ \mathbf{j}_{1o}(u) = \frac{\mathbf{j}_{DC}\mathbf{k}_i [K_1(b\mathbf{k}_i\mathbf{i}) I_0(\sqrt{t^2 - u^2 + b^2}\mathbf{k}_i\mathbf{i}) + I_1(b\mathbf{k}_i\mathbf{i}) K_0(\sqrt{t^2 - u^2 + b^2}\mathbf{k}_i\mathbf{i})]}{\mathbf{D}_{iA} + \mathbf{D}_{iB}}, \quad (62) $$

and eddy

$$ \mathbf{j}_o(u) = \frac{\mathbf{j}_{DC}\mathbf{k}_i q^2 [K_1(b\mathbf{k}_i q\mathbf{i}) I_0(\sqrt{t^2 - u^2 + b^2}\mathbf{k}_i q\mathbf{i}) + I_1(b\mathbf{k}_i q\mathbf{i}) K_0(\sqrt{t^2 - u^2 + b^2}\mathbf{k}_i q\mathbf{i})]}{\mathbf{D}_{iC} + \mathbf{D}_{iD} + \mathbf{D}_{iE}}, \quad (63) $$

The currents and, under the assumptions made, the impedances for the solutions for the inner (59) and outer (62) wires are equal and similar to those for solution (35). Then for the eddy-free case:

$$ \mathbf{I}_1 = \frac{2i\pi a \mathbf{j}_{DC} [K_1(a\mathbf{k}_i\mathbf{i}) I_1(b\mathbf{k}_i\mathbf{i}) - I_1(a\mathbf{k}_i\mathbf{i}) K_1(b\mathbf{k}_i\mathbf{i})]}{\mathbf{D}_{iA} + \mathbf{D}_{iB}}, \quad \mathbf{Z}_1 = \frac{\rho l (\mathbf{D}_{iA} + \mathbf{D}_{iB})\mathbf{i}}{2\pi a [I_1(a\mathbf{k}_i\mathbf{i}) K_1(b\mathbf{k}_i\mathbf{i}) - K_1(a\mathbf{k}_i\mathbf{i}) I_1(b\mathbf{k}_i\mathbf{i})]} $$

The currents and, under the assumptions made, the impedances for the solutions for the inner (61) and outer (63) wires are equal and similar to those for solution (36). Then for the eddy case:

$$ \mathbf{I} = \frac{2i\pi a \mathbf{j}_{DC} q [K_1(a\mathbf{k}_i q\mathbf{i}) I_1(b\mathbf{k}_i q\mathbf{i}) - I_1(a\mathbf{k}_i q\mathbf{i}) K_1(b\mathbf{k}_i q\mathbf{i})]}{\mathbf{D}_{iC} + \mathbf{D}_{iD} + \mathbf{D}_{iE}}, $$

$$ \mathbf{Z} = \frac{\rho l (\mathbf{D}_{iC} + \mathbf{D}_{iD} + \mathbf{D}_{iE})\mathbf{i}}{2\pi a q [I_1(a\mathbf{k}_i q\mathbf{i}) K_1(b\mathbf{k}_i q\mathbf{i}) - K_1(a\mathbf{k}_i q\mathbf{i}) I_1(b\mathbf{k}_i q\mathbf{i})]}, $$

It is obvious that the eddy-free Eq. (58) and its solution (59) for the inner wire of the coaxial cable do not coincide with Eq. (14) and the solution (35) for conventional wire, which coincides with the solution of the existing approach. For the existing concept of the skin effect and the proximity effect, the presence of a return wire in the form of a screen is absolutely indifferent. In the inner wire of a coaxial cable, the existing concept will show the same distribution of current density as in the same conventional wire, if you do not resort to any artificial tricks to take into account the effect of the screen on the inner wire.

The graph of the current density distribution in the inner wire of the coaxial cable $\mathbf{j}_i(r)$ in accordance with the formula (61) is shown in Appendix 12. For comparison, a graph of $\mathbf{j}(r)$ is shown for a conventional wire with a length of $l$ according to formula (36). The voltage applied to the conventional wire is half that applied to the coaxial cable to ensure the same field strength from the power supply $E$ in both cases. The resistances are given in Appendix 12: for a conventional wire $R$, for an inner wire $R_i$, for an outer screen $R_o$, as well as the DC resistance $R_{DC}$ for each of the coaxial cable cores and conventional wire. It can be seen that the presence of the screen changes the distribution of the current density, so that the total current becomes greater and the resistance is less. For the outer wire, the solution graphs according to formula (63) $\mathbf{j}_o(u)$ are also given in Appendix 12 in the middle. When $s=1$, i.e. when the shield is close to the inner wire, the graphs $\mathbf{j}_i(r)$ for the inner wire
and $\mathbf{j}(r)$ for the conventional wire are the closer to each other the smaller the thickness of the skin depth $\delta$, and practically coincide (see Appendix 12 below). Appendix 13 shows graphs similar to Appendix 12, but for the eddy-free version ($p=0$).

### D. Flat plate of finite thickness

In the existing solution for a semi-infinite space [1, p. 392, § 11.02, f. (2)], the current density distribution does not depend at all on the real thickness of the plate, which makes this solution hardly suitable for practice. There is also an existing solution for a flat bar, given in Ref. [2, p. 678], which takes into account the real thickness. We will consider the simplest case that can be verified experimentally. To bring the boundary conditions as close as possible to physically realizable ones, imagine that our plate, shown in Fig. 3 is part of the wall of a thin-walled cylinder with a very large radius, so that it can be considered almost flat. Let us introduce the following designations: $t$ – thickness of the plate, $l$ – vertical dimension, $z$ – dimension in the direction perpendicular to the plane of Fig. 3, $t_s$ – depth of the space surrounding the plate. Place the origin 0 on the inner surface of the cylinder. Let the distance from the origin 0 be $x$, the self-induction current density $\mathbf{j}_1(x)$, the current density including eddy currents $\mathbf{j}(x)$. By virtue of Ampère's circuital law in the case of axial

<figure><img src="image_4.png" /><figcaption>FIG. 3. Representation of a flat plate as a part of the wall of a thin-walled cylinder of very large radius with dimensions $t, x, l, z, t_s$ with magnetic induction $\mathbf{B}(x)$ and current densities $\mathbf{j}_1(x), \mathbf{j}(x)$.</figcaption></figure>

symmetry, it is assumed that the inner cavity of the cylinder is free from the magnetic field. That is, we have reduced the plane problem to a cylindrical one with all the advantages due to axial symmetry, but at the same time we got rid of the dependence on the radius. The acceptability of these assumptions can be assessed by comparing the calculation by the expected formula with the calculation by the exact formula for a tubular cylinder (35) with a very large radius and a thin wall. It seems that such a model is still more favorable for experimental studies compared to the mentioned existing solutions. Let us perform all actions similar to those in formulas (1) – (7). The magnetic induction (directed from us perpendicular to the plane of Fig. 3) at distance $x$ is

$$ \mathbf{B}(x) = \frac{\mu \mathbf{I}(x)}{z}, $$

where

$$ \mathbf{I}(x) = z \int_{0}^{x} \mathbf{j}(x) dx $$

is the current of a parallelepiped bounded by surface 0 and a surface with distance $x$. Substituting $\mathbf{I}(x)$ into $\mathbf{B}(x)$, we get

$$ \mathbf{B}(x) = \mu \int_{0}^{x} \mathbf{j}(x) dx. \qquad (64) $$

The total current is

$$ \mathbf{I} = z \int_{0}^{t} \mathbf{j}(x) dx. \qquad (65) $$

The magnetic flux through the section with sides $l$ and $t-x$ is
$$ \Phi_t(x) = l \int_x^t \mathbf{B}(x) dx. \qquad (66) $$

The magnetic flux through the cross section of the external space with sides $l$ and $t_s$ is

$$ \Phi_s = \mathbf{I} \frac{\mu_s t_s l}{z}. \qquad (67) $$

The part of the specific EMF of self-induction corresponding to the magnetic flux $\Phi_t(x)$ at distance $x$ is

$$ \mathbf{E}_t(x) = -i\omega \frac{\Phi_t(x)}{l}. \qquad (68) $$

The part of the specific EMF of self-induction corresponding to the magnetic flux $\Phi_s$ is

$$ \mathbf{E}_s = -i\omega \frac{\Phi_s}{l}. \qquad (69) $$

The primary current density is

$$ \mathbf{j}_1(x) = \frac{\mathbf{E} + \mathbf{E}_t(x) + \mathbf{E}_s}{\rho}. $$

Substituting the values from (68), (69), their components from (64) – (67) into the last formula, taking into account the designations (10) – (12) and assuming $\mathbf{j}(x)=\mathbf{j}_1(x)$, we obtain the equation for the eddy-free case

$$ \mathbf{j}_1(x) = \mathbf{j}_{DC} + \mathbf{k}^2 \int_0^t \int_x^t \mathbf{j}_1(x) dx dx + t_s \mathbf{k}_s^2 \int_0^t \mathbf{j}_1(x) dx. \qquad (70) $$

Then the solution of Eq. (70) (search and verification at the end of Appendix 14) is

$$ \mathbf{j}_1(x) = \frac{\mathbf{j}_{DC} \mathbf{k} \cos(x\mathbf{k})}{\mathbf{k} \cos(t\mathbf{k}) - t_s \mathbf{k}_s^2 \sin(t\mathbf{k})}. \qquad (71) $$

The formulas for current and impedance for the eddy-free case (71) are:

$$ \mathbf{I}_1 = z \int_0^t \mathbf{j}_1(x) dx = \frac{\mathbf{j}_{DC} z \sin(t\mathbf{k})}{\mathbf{k} \cos(t\mathbf{k}) - t_s \mathbf{k}_s^2 \sin(t\mathbf{k})}; \quad \mathbf{Z}_1 = \frac{\mathbf{E}/l}{\mathbf{I}_1} = \frac{\rho l [\mathbf{k} \cos(t\mathbf{k}) - t_s \mathbf{k}_s^2 \sin(t\mathbf{k})]}{z \sin(t\mathbf{k})}. $$

If we take into account eddy currents (we omit the derivation, since it is similar to the search for Eq. (24)), then the equation will be

$$ \mathbf{j}(x) = \mathbf{j}_{DC} + q^2 \mathbf{k}^2 \int_0^t \int_x^t \mathbf{j}(x) dx dx + t_s \mathbf{k}_s^2 \int_0^t \mathbf{j}(x) dx - p \mathbf{k}^2 \int_0^t \int_0^t \mathbf{j}(x) dx dx, \qquad (72) $$

and its solution (search and verification see Appendix 14) will be

$$ \mathbf{j}(x) = \frac{\mathbf{j}_{DC} \mathbf{k} q^2 \cos(x \mathbf{k} q)}{(q^2 - p) \mathbf{k} \cos(t \mathbf{k} q) - t_s \mathbf{k}_s^2 q \sin(t \mathbf{k} q) + p \mathbf{k}}. \qquad (73) $$

The formulas for current and impedance for the eddy case (73) are:
$$ \mathbf{I} = z \int_{0}^{t} \mathbf{j}(x) dx = \frac{\mathbf{j}_{DC} qz \sin(t\mathbf{k}q)}{(q^2 - p)\mathbf{k} \cos(t\mathbf{k}q) - t_s \mathbf{k}_s^2 q \sin(t\mathbf{k}q) + p\mathbf{k}}; $$

$$ \mathbf{Z} = \frac{\mathbf{E}l}{\mathbf{I}} = \frac{\rho l [(q^2 - p)\mathbf{k} \cos(t\mathbf{k}q) - t_s \mathbf{k}_s^2 q \sin(t\mathbf{k}q) + p\mathbf{k}]}{qz \sin(t\mathbf{k}q)}. $$

Eqs. (70), (72) and their solutions (71), (73) are also valid in the range from 0 to t for the idealized flat bar described in Ref. [2, pp. 677–679], with its thickness 2t and mirror-like symmetric distribution of current density relative to the central plane 0. The eddy-free solution (71), as in the previous situations with solutions (15) and (35) for a cylindrical wire, completely coincides with the solution given by L.A. Bessonov in Ref. [2, p. 678]

$$ \mathbf{j}(x) = \frac{i\mathbf{I}_1 \mathbf{k} \cosh(x\mathbf{k}i)}{z \sinh(t\mathbf{k}i)}. $$

Appendix 15 compares the obtained solutions (71) $j_1(x)$ and (73) $j(x)$ at $t_s=0$ with the existing solution for the semi-infinite space $j_{hs}(x)=j_0 e^{-x\mathbf{k}i}$ from Ref. [1, p. 392, § 11.02, f. (2)], where we have taken on the right side $j_0=j_{DC}$ and $x=t-x$ (to match the origins), i.e. $j_{hs}(x)=j_{DC} e^{(x-t)\mathbf{k}i}$. The top row shows graphs at 1 kHz, the bottom row shows graphs at 10 MHz. The difference between the found (71) and the existing solution $j_{hs}(x)$ is significant approximately in the region $\delta/t>0.3$, i.e., at a low frequency. For high frequency, more precisely for skin depth $\delta$ small compared to thickness $t$, the curves $j_1(x)$ and $j_{hs}(x)$ almost coincide, as seen at the bottom of Appendix 15.

Appendix 16 compares the solution for a flat plate (71) $j_1(x)$ (shown in red in the bottom row of graphs) with solution (35) (indicated by $j_{WS}(r)$ and shown in red in the top row of graphs) for a thin-walled tubular cylinder at $(a-b)/a=0.01$, as well as graphs for $j_{hs}(x)$. As expected, when making the assumptions for deriving Eq. (70), the solutions $j_1(x)$ and $j_{WS}(r)$ practically coincide, i.e., for a thin-walled cylinder, instead of the exact formula (35), formula (71) can be used for any frequencies. The existing $j_{hs}(x)$ solution for low frequencies runs below the $j_1(x)$ curve, resulting in its total current being less.

## E. Infinitely long solenoid

### 1. Direct connection to a power supply

Let's consider an infinitely long solenoid (without a magnetic field outside) with a dense single-layer multi-turn winding of a rectangular wire. We neglect the interturn capacitance. In Fig. 4 shows

<figure><img src="image_5.png" /><figcaption>FIG. 4. Coil turn of solenoid with dimensions $t, x, z/N, H_s, W_s$ with magnetic induction $\mathbf{B}(x)$, current densities $j_1(x), j(x)$.</figcaption></figure>

a coil turn of the solenoid. Let us assume that the thickness of the wire $t$ is negligible compared to the dimensions $W_s$ and $H_s$ of the magnetic core. The cross-sectional area of the magnetic circuit is $S_s=W_sH_s$, and its perimeter is $P_s=2(W_s+H_s)$. Let's $N$ is the number of winding turns, $z$ is the length of the magnetic core, $l=NP_s$ is the length of the wire, $x$ is the distance from the origin 0, $j_1(x)$ is the self-induction current density, $j(x)$ is the total current density. The magnetic induction (directed to us perpendicular to the plane of Fig. 4) at distance $x$ is
$$ \mathbf{B}(x) = \frac{\mu N \mathbf{I}(x)}{z}, \qquad (74) $$

where

$$ \mathbf{I}(x) = \frac{z}{N} \int_{0}^{x} \mathbf{j}(x) dx \qquad (75) $$

is the current inside the ring of the width $z/N$ bounded by the outer surface 0 and the surface at distance $x$. Substituting (75) into (74), we get

$$ \mathbf{B}(x) = \mu \int_{0}^{x} \mathbf{j}(x) dx. \qquad (76) $$

The total current is

$$ \mathbf{I} = \frac{z}{N} \int_{0}^{t} \mathbf{j}(x) dx. \qquad (77) $$

The magnetic flux through the annular section with dimensions $P_s$ and $t-x$ is

$$ \mathbf{\Phi}_t(x) = P_s \int_x^t \mathbf{B}(x) dx. \qquad (78) $$

The magnetic flux through the cross section of the magnetic circuit of the solenoid $S_s$ is

$$ \mathbf{\Phi}_s = \mathbf{I} \frac{\mu_s N S_s}{z}. \qquad (79) $$

The part of the specific EMF of self-induction corresponding to the magnetic flux $\mathbf{\Phi}_t(x)$ at distance $x$ is

$$ \mathbf{E}_t(x) = -i\omega \frac{\mathbf{\Phi}_t(x)N}{P_s N}. \qquad (80) $$

The part of the specific EMF of self-induction corresponding to the magnetic flux $\mathbf{\Phi}_s$ is

$$ \mathbf{E}_s = -i\omega \frac{\mathbf{\Phi}_s N}{P_s N}. \qquad (81) $$

The primary current density at distance $x$ is

$$ \mathbf{j}_1(x) = \frac{\mathbf{E} + \mathbf{E}_t(x) + \mathbf{E}_s}{\rho}. $$

Substituting the values from (80), (81), their components from (76) – (79) into the last formula, taking into account the designations (10) – (12) and assuming $\mathbf{j}(x)=\mathbf{j}_1(x)$, we obtain the equation for the eddy-free case

$$ \mathbf{j}_1(x) = \mathbf{j}_{DC} + \mathbf{k}^2 \int_{x_0}^{t} \int_{0}^{x} \mathbf{j}_1(x) dx dx + \frac{S_s}{P_s} \mathbf{k}_s^2 \int_{0}^{t} \mathbf{j}_1(x) dx, \qquad (82) $$

and its solution (see at the end of Appendix 17) is
$$ \mathbf{j}_1(x) = \frac{\mathbf{j}_{DC} \mathbf{k} \cos(x\mathbf{k})}{\mathbf{k} \cos(t\mathbf{k}) - \frac{S_s}{P_s} \mathbf{k}_s^2 \sin(t\mathbf{k})} . \qquad (83) $$

The formulas for current and impedance for the eddy-free case (83) are:

$$ \mathbf{I}_1 = \frac{z}{N} \int_0^t \mathbf{j}_1(x) dx = \frac{z}{N} \frac{\mathbf{j}_{DC} \sin(t\mathbf{k})}{\mathbf{k} \cos(t\mathbf{k}) - \frac{S_s}{P_s} \mathbf{k}_s^2 \sin(t\mathbf{k})} ; \quad \mathbf{Z}_1 = \frac{\mathbf{E}l}{\mathbf{I}_1} = \frac{\rho l N \left[ \mathbf{k} \cos(t\mathbf{k}) - \frac{S_s}{P_s} \mathbf{k}_s^2 \sin(t\mathbf{k}) \right]}{z \sin(t\mathbf{k})} . $$

If eddy currents are taken into account, then the equation is

$$ \mathbf{j}(x) = \mathbf{j}_{DC} + q^2 \mathbf{k}^2 \int_x^t \int_0^x \mathbf{j}(x) dx dx + \frac{S_s}{P_s} \mathbf{k}_s^2 \int_0^t \mathbf{j}(x) dx - p \mathbf{k}^2 \int_0^t \int_0^x \mathbf{j}(x) dx dx, \qquad (84) $$

and its solution (see Appendix 17) is

$$ \mathbf{j}(x) = \frac{\mathbf{j}_{DC} \mathbf{k} q^2 \cos(x \mathbf{k} q)}{(q^2 - p) \mathbf{k} \cos(t \mathbf{k} q) - \frac{S_s}{P_s} \mathbf{k}_s^2 q \sin(t \mathbf{k} q) + p \mathbf{k}}. \qquad (85) $$

The formulas for current and impedance for the eddy case (85) are:

$$ \mathbf{I} = \frac{z}{N} \int_0^t \mathbf{j}(x) dx = \frac{z}{N} \frac{\mathbf{j}_{DC} q \sin(t \mathbf{k} q)}{(q^2 - p) \mathbf{k} \cos(t \mathbf{k} q) - \frac{S_s}{P_s} \mathbf{k}_s^2 q \sin(t \mathbf{k} q) + p \mathbf{k}} ; $$

$$ \mathbf{Z} = \frac{\mathbf{E}l}{\mathbf{I}} = \frac{\rho l N \left[ (q^2 - p) \mathbf{k} \cos(t \mathbf{k} q) - \frac{S_s}{P_s} \mathbf{k}_s^2 q \sin(t \mathbf{k} q) + p \mathbf{k} \right]}{qz \sin(t \mathbf{k} q)} . $$

As can be seen, the solutions and equations for an infinitely long solenoid completely coincide with the solutions and equations for a flat plate from the previous paragraph, if we take $S_s/P_s=t_s$. Solution (83), as in the previous cases for eddy-free variants, coincides with the existing solution for an infinite solenoid presented in Ref. [5, p. 172, § 21, f. (6)]. But a detailed analysis of this solution was not carried out.

## 2. Connecting to a power supply in series through a capacitor

Let us consider also the case of a series connection of an infinitely long solenoid (L and $R_{DC}$) with a capacitor C and a power source U, as shown in Fig. 5. We will assume that both the capacitor and the power source are built into the wire so that their dimensions along the length of the wire l

<figure><img src="image_6.png" /><figcaption>FIG. 5. Series connection of the wire (L and $R_{DC}$) with capacitor C and power supply U.</figcaption></figure>
are negligibly small. In addition, their cross section is the same as that of the wire. These assumptions allow us to consider this case as similar to the case considered earlier and shown in Fig.4. The equations and solutions are derived in Appendix 18. Let's immediately rewrite the final results, the equation itself for the eddy-free case (see the end of Appendix 18)

$$ \mathbf{j}_1(x) = \mathbf{j}_{DC} + \mathbf{k}^2 \int\limits_{x=0}^{t} \int\limits_{0}^{x} \mathbf{j}_1(x) dx dx + \frac{S_s}{P_s} \mathbf{k}_s^2 \int\limits_{0}^{t} \int\limits_{0}^{x} \mathbf{j}_1(x) dx - z \mathbf{k}_c^2 \int\limits_{0}^{t} \int\limits_{0}^{x} \mathbf{j}_1(x) dx, \quad (86) $$

where

$$ \mathbf{k}_c = \sqrt{\frac{-\mathbf{i}}{\rho P_s N^2 \omega C}}, $$

and solution for Eq. (86)

$$ \mathbf{j}_1(x) = \frac{\mathbf{j}_{DC} \mathbf{k} \cos(x \mathbf{k})}{\mathbf{k} \cos(t \mathbf{k}) - \frac{S_s}{P_s} \mathbf{k}_s^2 \sin(t \mathbf{k}) + z \mathbf{k}_c^2 \sin(t \mathbf{k})}. \quad (87) $$

The formulas for current and impedance for the eddy-free case (87) are:

$$ \mathbf{I}_1 = \frac{z}{N} \int_{0}^{t} \int_{0}^{x} \mathbf{j}_1(x) dx = \frac{z}{N} \frac{\mathbf{j}_{DC} \sin(t \mathbf{k})}{\mathbf{k} \cos(t \mathbf{k}) - \frac{S_s}{P_s} \mathbf{k}_s^2 \sin(t \mathbf{k}) + z \mathbf{k}_c^2 \sin(t \mathbf{k})}; $$

$$ \mathbf{Z}_1 = \frac{\mathbf{E}l}{\mathbf{I}_1} = \frac{\rho l N \left[ \mathbf{k} \cos(t \mathbf{k}) - \frac{S_s}{P_s} \mathbf{k}_s^2 \sin(t \mathbf{k}) + z \mathbf{k}_c^2 \sin(t \mathbf{k}) \right]}{z \sin(t \mathbf{k})}. $$

If eddy currents are taken into account, then the equation is

$$ \mathbf{j}(x) = \mathbf{j}_{DC} + q^2 \mathbf{k}^2 \int\limits_{x=0}^{t} \int\limits_{0}^{x} \mathbf{j}(x) dx dx + \frac{S_s}{P_s} \mathbf{k}_s^2 \int\limits_{0}^{t} \int\limits_{0}^{x} \mathbf{j}(x) dx - p \mathbf{k}^2 \int\limits_{0}^{t} \int\limits_{0}^{x} \mathbf{j}(x) dx dx - z \mathbf{k}_c^2 \int\limits_{0}^{t} \int\limits_{0}^{x} \mathbf{j}(x) dx, \quad (88) $$

and its solution is

$$ \mathbf{j}(x) = \frac{\mathbf{j}_{DC} \mathbf{k} q^2 \cos(x \mathbf{k} q)}{(q^2 - p) \mathbf{k} \cos(t \mathbf{k} q) - \frac{S_s}{P_s} \mathbf{k}_s^2 q \sin(t \mathbf{k} q) + p \mathbf{k} + z \mathbf{k}_c^2 q \sin(t \mathbf{k} q)}. \quad (89) $$

The formulas for current and impedance for the eddy case (89) are:

$$ \mathbf{I} = \frac{z}{N} \int_{0}^{t} \int \mathbf{j}(x) dx = \frac{z}{N} \frac{\mathbf{j}_{DC} q \sin(t \mathbf{k} q)}{(q^2 - p) \mathbf{k} \cos(t \mathbf{k} q) - \frac{S_s}{P_s} \mathbf{k}_s^2 q \sin(t \mathbf{k} q) + p \mathbf{k} + z \mathbf{k}_c^2 q \sin(t \mathbf{k} q)}; $$

$$ \mathbf{Z} = \frac{\mathbf{E}l}{\mathbf{I}} = \frac{\rho l N \left[ (q^2 - p) \mathbf{k} \cos(t \mathbf{k} q) - \frac{S_s}{P_s} \mathbf{k}_s^2 q \sin(t \mathbf{k} q) + p \mathbf{k} + z \mathbf{k}_c^2 q \sin(t \mathbf{k} q) \right]}{qz \sin(t \mathbf{k} q)}. $$

Obviously, in the absence of a capacitor, i.e. if $C=\infty$, then $\mathbf{k}_c=0$ and all the formulas of this subsection are converted to the formulas of the previous subsection.
## F. Equations and estimation of surface current densities for transformer windings

Let's find the equations for the current densities for the primary and secondary windings of the transformer in Fig. 6 (only a fragment is shown), wound tightly on top of each other in one layer with a rectangular wire on a magnetic circuit, as in Fig. 4, i.e. with the same perimeter $P_s$, cross section $S_s$ and length $z$. Inter-turn and inter-winding capacitances are not taken into account. The thickness of the inner winding is $n-t$, the outer one is $t$. The current density of the inner winding is $\mathbf{j}_i(u)$

<figure><img src="image_7.png" /><figcaption>FIG. 6. A fragment of a transformer with a core (denoted $\mu_s$) with dimensions $x, t, u, n, z/N_i, z/N_o$ with induction $\mathbf{B}_i(u)$, current density $\mathbf{j}_i(u)$ for the inner winding and induction $\mathbf{B}_o(x)$, current density $\mathbf{j}_o(x)$ for the outer winding.</figcaption></figure>

at distance $u$, the outer one is $\mathbf{j}_o(x)$ at distance $x$. The number of turns of the inner winding is $N_i$, the outer one is $N_o$. The magnetic flux through the cross section of the magnetic circuit $S_s$ is

$$ \Phi_s = (\mathbf{I}_i N_i + \mathbf{I}_o N_o) \frac{\mu_s S_s}{z}, \qquad (90) $$

where

$$ \mathbf{I}_i = \frac{z}{N_i} \int_{t}^{n} \mathbf{j}_i(u) du \qquad (91) $$

is the current of the inner coil,

$$ \mathbf{I}_o = \frac{z}{N_o} \int_{0}^{t} \mathbf{j}_o(x) dx \qquad (92) $$

is the current of the outer coil. The magnetic induction in the inner winding (directed from us perpendicular to the plane of Fig. 6) at distance $u$ is

$$ \mathbf{B}_i(u) = \frac{\mu N_i \mathbf{I}_i(u) + \mu N_o \mathbf{I}_o}{z}, \qquad (93) $$

where

$$ \mathbf{I}_i(u) = \frac{z}{N_i} \int_{t}^{u} \mathbf{j}_i(u) du \qquad (94) $$

is the current of a turn of the inner winding of width $z/N_i$ inside the ring bounded by its outer surface $t$ and a surface with distance $u$. Substituting (94) and (92) into (93), we have

$$ \mathbf{B}_i(u) = \mu \int_{t}^{u} \mathbf{j}_i(u) du + \mu \int_{0}^{t} \mathbf{j}_o(x) dx. \qquad (95) $$

The magnetic induction in the outer winding (directed to us perpendicular to the plane of Fig. 6) at distance $x$ is

$$ \mathbf{B}_o(x) = \frac{\mu N_o \mathbf{I}_o(x)}{z}, \qquad (96) $$

where

$$ \mathbf{I}_o(x) = \frac{z}{N_o} \int_{0}^{x} \mathbf{j}_o(x) dx \qquad (97) $$

is the current of the outer winding turn with the width $z/N_o$ inside the ring, bounded by its outer surface 0 and the surface with the distance $x$. Substituting (97) into (96), we have
$$ \mathbf{B}_o(x) = \mu \int_0^x \mathbf{j}_o(x) dx. \qquad (98) $$

The magnetic flux through the entire annular section of the inner winding with dimensions $P_s$ and $n-t$ is

$$ \mathbf{\Phi}_i = P_s \int_t^n \mathbf{B}_i(u) du. \qquad (99) $$

The magnetic flux through the annular section of the inner winding with dimensions $P_s$ and $n-u$ is

$$ \mathbf{\Phi}_{in}(u) = P_s \int_u^n \mathbf{B}_i(u) du. \qquad (100) $$

The magnetic flux through the annular section of the inner winding with dimensions $P_s$ and $u-t$ is

$$ \mathbf{\Phi}_{it}(u) = P_s \int_t^u \mathbf{B}_i(u) du. \qquad (101) $$

The magnetic flux through the annular section of the outer winding with dimensions $P_s$ and $t-x$ is

$$ \mathbf{\Phi}_{ot}(x) = P_s \int_x^t \mathbf{B}_o(x) dx. \qquad (102) $$

The magnetic flux through the annular section of the outer winding with dimensions $P_s$ and $x$ is

$$ \mathbf{\Phi}_{o0}(x) = P_s \int_0^x \mathbf{B}_o(x) dx. \qquad (103) $$

The specific EMF of induction of the inner and outer windings from the magnetic flux $\mathbf{\Phi}_s$ is

$$ \mathbf{E}_s = -i\omega \frac{\mathbf{\Phi}_s N_i}{P_s N_i} = -i\omega \frac{\mathbf{\Phi}_s N_o}{P_s N_o}. \qquad (104) $$

The specific EMF of induction of the inner winding from the magnetic flux $\mathbf{\Phi}_{in}(u)$ at distance $u$ is

$$ \mathbf{E}_{in}(u) = -i\omega \frac{\mathbf{\Phi}_{in}(u) N_i}{P_s N_i}. \qquad (105) $$

The specific eddy EMF of induction of the inner winding from the magnetic flux $\mathbf{\Phi}_{in}(u)$ in a short-circuited circuit between the distance $u$ and the surface $n$ is

$$ \mathbf{E}_{in2}(u) = -i\omega \frac{\mathbf{\Phi}_{in}(u) N_i}{2P_s N_i} = \frac{\mathbf{E}_{in}(u)}{2}. \qquad (106) $$

The specific eddy EMF of induction of the inner winding from the magnetic flux $\mathbf{\Phi}_{it}(u)$ in a short-circuited circuit between the distance $u$ and the surface $t$ is
$$ \mathbf{E}_{it2}(u) = -i\omega \frac{\mathbf{\Phi}_{it}(u)N_i}{2P_s N_i}. \qquad (107) $$

The specific EMF of induction of the outer winding from magnetic fluxes $\mathbf{\Phi}_{ot}(x)$ and $\mathbf{\Phi}_i$ at distance $x$ is

$$ \mathbf{E}_{ot}(x) = -i\omega \frac{\mathbf{\Phi}_{ot}(x)N_o}{P_s N_o} - i\omega \frac{\mathbf{\Phi}_i N_o}{P_s N_o}. \qquad (108) $$

The specific eddy EMF of induction of the outer winding from the magnetic flux $\mathbf{\Phi}_{ot}(x)$ in a short-circuited circuit between the distance $x$ and the surface $t$ is

$$ \mathbf{E}_{ot2}(x) = -i\omega \frac{\mathbf{\Phi}_{ot}(x)N_o}{2P_s N_o}. \qquad (109) $$

The specific eddy EMF of induction of the outer winding from the magnetic flux $\mathbf{\Phi}_{o0}(x)$ in a short-circuited circuit between the distance $x$ and the surface $0$ is

$$ \mathbf{E}_{o02}(x) = -i\omega \frac{\mathbf{\Phi}_{o0}(x)N_o}{2P_s N_o}. \qquad (110) $$

The current density of inner winding if acting as primary is

$$ \mathbf{j}_i(u) = \frac{\mathbf{E} + \mathbf{E}_{in}(u) + \mathbf{E}_s}{\rho} + \frac{\mathbf{E}_{in2}(u)}{\rho} - \frac{\mathbf{E}_{it2}(u)}{\rho}. $$

The current density of outer winding if acting as secondary is

$$ \mathbf{j}_o(x) = \frac{\mathbf{E}_{ot}(x) + \mathbf{E}_s - \mathbf{E}_L}{\rho} + \frac{\mathbf{E}_{ot2}(x)}{\rho} - \frac{\mathbf{E}_{o02}(x)}{\rho}, $$

where $\mathbf{E}_L$ is the field strength in the secondary winding from the voltage drop across the external load. Substituting the values from (104) – (110), their components from (90) – (92), (95), (98) – (103) into the last two equations, taking into account the designations (10) – (12), (23), introducing the notation $\mathbf{j}_L = \mathbf{E}_L/\rho$ and immediately grouping the integrals, we obtain the system of equations

$$ \left\{ \begin{aligned} \mathbf{j}_i(u) &= \mathbf{j}_{DC} + q^2 \mathbf{k}^2 \iint_{u,t}^{n} \mathbf{j}_i(u) du du + q^2 \mathbf{k}^2 \iint_{u,0}^{n} \mathbf{j}_o(x) dx du - p \mathbf{k}^2 \iint_{t,t}^{n} \mathbf{j}_i(u) du du - \\ & \quad - p \mathbf{k}^2 \iint_{t,0}^{n} \mathbf{j}_o(x) dx du + \frac{S_s}{P_s} \mathbf{k}_s^2 \iint_{t}^{n} \mathbf{j}_i(u) du + \frac{S_s}{P_s} \mathbf{k}_s^2 \iint_{0,t}^{n} \mathbf{j}_o(x) dx \\ \mathbf{j}_o(x) &= -\mathbf{j}_L + q^2 \mathbf{k}^2 \iint_{x,0}^{t} \mathbf{j}_o(x) dx dx + \mathbf{k}^2 \iint_{t,t}^{n} \mathbf{j}_i(u) du du + \mathbf{k}^2 \iint_{t,0}^{n} \mathbf{j}_o(x) dx du - \\ & \quad - p \mathbf{k}^2 \iint_{0,0}^{t} \mathbf{j}_o(x) dx dx + \frac{S_s}{P_s} \mathbf{k}_s^2 \iint_{t}^{n} \mathbf{j}_i(u) du + \frac{S_s}{P_s} \mathbf{k}_s^2 \iint_{0,t}^{n} \mathbf{j}_o(x) dx \end{aligned} \right. . \qquad (111) $$

Let us simplify the distance-dependent second integral (third term) in the first equation in (111)

$$ q^2 \mathbf{k}^2 \iint_{u,0}^{n} \mathbf{j}_o(x) dx du = q^2 \mathbf{k}^2 \iint_{u}^{n} du \iint_{0}^{t} \mathbf{j}_o(x) dx = q^2 \mathbf{k}^2 (n-u) \mathbf{H}_o, $$
where we denote $\mathbf{H}_o = \int_0^t \mathbf{j}_o(x)dx = \frac{\mathbf{I}_o N_o}{z}$, and also introduce the notation for distance-independent integrals:

$$ \mathbf{j}_{Ci} = -p\mathbf{k}^2 \iint_{t}^{n} \int_{t}^{u} \mathbf{j}_{i}(u)dudu - p\mathbf{k}^2 \iint_{t}^{n} \int_{0}^{t} \mathbf{j}_{o}(x)dxdu; \quad \mathbf{j}_{S} = \frac{S_s}{P_s} \mathbf{k}_s^2 \iint_{t}^{n} \mathbf{j}_{i}(u)du + \frac{S_s}{P_s} \mathbf{k}_s^2 \iint_{0}^{t} \mathbf{j}_{o}(x)dx; $$

$$ \mathbf{j}_{Co1} = \mathbf{k}^2 \iint_{t}^{n} \int_{t}^{u} \mathbf{j}_{i}(u)dudu + \mathbf{k}^2 \iint_{t}^{n} \int_{0}^{t} \mathbf{j}_{o}(x)dxdu; \quad \mathbf{j}_{Co2} = -p\mathbf{k}^2 \iint_{0}^{t} \int_{0}^{x} \mathbf{j}_{o}(x)dxdx. $$

Let us rewrite system (111) by substituting the above notation

$$ \left\{ \begin{aligned} \mathbf{j}_i(u) &= \mathbf{j}_{DC} + q^2 \mathbf{k}^2 \iint_{\substack{u \\ t}}^{n} \int \mathbf{j}_i(u) dudu + q^2 \mathbf{k}^2 (n-u) \mathbf{H}_o + \mathbf{j}_{Ci} + \mathbf{j}_S \\ \mathbf{j}_o(x) &= -\mathbf{j}_L + q^2 \mathbf{k}^2 \iint_{\substack{x \\ 0}}^{t} \int \mathbf{j}_o(x) dx dx + \mathbf{j}_{Co1} + \mathbf{j}_{Co2} + \mathbf{j}_S \end{aligned} \right. . \quad (112) $$

It is obvious that the form of solution of the second equation in (112) will be as follows

$$ \mathbf{j}_o(x) = (-\mathbf{j}_L + \mathbf{j}_{Co1} + \mathbf{j}_{Co2} + \mathbf{j}_S) \frac{\cos(x\mathbf{k}q)}{\cos(t\mathbf{k}q)}. $$

Let us estimate the surface current densities for both equations from (112). Substituting into the limits of integration of external integrals first $u=n$ and $x=t$, then $u=t$ and $x=0$, we get:

$$ \left\{ \begin{aligned} \mathbf{j}_i(n) &= \mathbf{j}_{DC} + \mathbf{j}_{Ci} + \mathbf{j}_S \\ \mathbf{j}_o(t) &= -\mathbf{j}_L + \mathbf{j}_{Co1} + \mathbf{j}_{Co2} + \mathbf{j}_S \end{aligned} \right. ; \quad \left\{ \begin{aligned} \mathbf{j}_i(t) &= \mathbf{j}_{DC} + q^2 \mathbf{k}^2 \iint_{\substack{t \\ t}}^{n} \int \mathbf{j}_i(u) dudu + q^2 \mathbf{k}^2 (n-t) \mathbf{H}_o + \mathbf{j}_{Ci} + \mathbf{j}_S \\ \mathbf{j}_o(0) &= (-\mathbf{j}_L + \mathbf{j}_{Co1} + \mathbf{j}_{Co2} + \mathbf{j}_S) \frac{1}{\cos(t\mathbf{k}q)} \end{aligned} \right. . $$

The main power source of the secondary winding is the specific EMF of induction $E_s$, and the role of the DC density from the power source is played by the current density $\mathbf{j}_S$, which is counteracted by the load current density $\mathbf{j}_L$. Since the directions of magnetic fluxes in the core and in the secondary winding are opposite, then, unlike a conventional choke, its inductance during a short circuit ($\mathbf{j}_L=0$) will be determined only by the inductance of the wire itself. It follows that with increasing frequency in the eddy case, the current density of the outer secondary winding on the surface $t$ will behave similarly to the previously considered case for a cylindrical wire, described in Appendix 5 in the absence of external inductance. Namely, it will grow, reaching a certain peak, then fall and stabilize at a certain value, which is greater in relation to the DC density of the power supply, the role of which is now played by $\mathbf{j}_S$. In the presence of external inductance, as shown in Appendix 6, the surface current density decreases very rapidly with increasing frequency. It can be said that the outer secondary winding of the transformer is the only option for studying the skin effect without external inductance, since with a normal power connection, the space around the wire cannot be eliminated (can be compensated at a certain frequency by connecting in series a capacitive reactance equal to the inductive reactance of the external space). The success of high-frequency hardening is precisely determined by the fact that the hardened part is always a short-circuited secondary winding. It can also be assumed that during hardening of products there is a limit to the increase in
frequency due to the presence of a peak, after which the surface current density slightly decreases and stabilizes (talking about the eddy case). It would also be very interesting to test this experimentally. It is possible to measure the secondary winding temperature at surface t at a fixed voltage and increasing frequency in the primary winding. The presence of a temperature peak at a certain frequency will confirm the presence of eddy currents, and the temperature constancy, on the contrary, will disprove their presence. Due to the practically unchanged magnetic flux of the transformer core and the preservation of the difference in ampere-turns of the windings when the load changes, it can be concluded that the current density in the primary winding will be similar to the secondary.

If the power source is connected to an outer winding, then $\mathbf{j}_L$ and $\mathbf{j}_{DC}$ in the equations from (111) are interchanged and then the surface current densities will take the form:

$$ \left\{ \begin{array}{l} \mathbf{j}_i(n) = -\mathbf{j}_L + \mathbf{j}_{Ci} + \mathbf{j}_S \\ \mathbf{j}_o(t) = \mathbf{j}_{DC} + \mathbf{j}_{Co1} + \mathbf{j}_{Co2} + \mathbf{j}_S \end{array} \right. ; \quad \left\{ \begin{array}{l} \mathbf{j}_i(t) = -\mathbf{j}_L + q^2 \mathbf{k}^2 \iint_{t}^{n u} \mathbf{j}_i(u) du du + q^2 \mathbf{k}^2 (n-t) \mathbf{H}_o + \mathbf{j}_{Ci} + \mathbf{j}_S \\ \mathbf{j}_o(0) = (\mathbf{j}_{DC} + \mathbf{j}_{Co1} + \mathbf{j}_{Co2} + \mathbf{j}_S) \frac{1}{\cos(t \mathbf{k} q)} \end{array} \right. $$

With a short circuit of the secondary (now the inner) winding ($\mathbf{j}_L=0$) and the absence of a magnetic circuit ($\mathbf{j}_S=0$), the current density of the secondary winding on the surface $n$ is less than on the surface $t$. If we also neglect eddy currents ($p=0, q=1, \mathbf{j}_{Ci}=0, \mathbf{j}_{Co2}=0$), then $\mathbf{j}_i(n)=0$. But there is nothing paradoxical in this, since now the maximum current densities in both windings will be observed at distance $t$, that is, at the place where the windings adjoin each other. This is the same as in the coaxial cable discussed earlier. It can be concluded that no matter which of the windings is primary and which is secondary, in the outer winding the maximum current density is always at the inner surface $t$. In the inner winding, the maximum current density is determined by the magnitude of the magnetic induction on the surface and is maximum on the surface with maximum induction. If the magnetic induction in the area between the surfaces $t$ and $n$ is less than its values on these surfaces, then the current density amplitude curve will have a U-shaped form.

In addition to the transformer, the resulting system (111) can also be applied to a winding with parallel-connected layers with an equal number of turns, if $\mathbf{j}_L = -\mathbf{j}_{DC}$ is taken in its second equation.

### III. ANALYSIS OF THE EXISTING SOLUTION FOR SOLID CYLINDER

Let's analyze the solution of the existing concept (16) for a solid cylindrical wire for the case where there is no external inductance, i.e., for $n=a$. We will act on the basis of the relations of the existing approach. We rewrite from Ref. [2, p. 723, f. (A.10)] the existing solution in the form

$$ \mathbf{j}(r) = \frac{\mathbf{I} \mathbf{k} \mathbf{J}_0(r \mathbf{k})}{2 \pi a \mathbf{J}_1(\mathbf{k})} $$

We transform it into a solution in the form (16), for which we find the current $\mathbf{I}$ from Ohm's law in complex form $\mathbf{I} = \mathbf{E} // \mathbf{Z}$, where the impedance $\mathbf{Z}$ for a wire of length $l$ from Ref. [5, p. 169, § 20, f. (23)] is

$$ \mathbf{Z} = \frac{\rho l \mathbf{k} \mathbf{J}_0(\mathbf{k})}{2 \pi a \mathbf{J}_1(\mathbf{k})} $$

Then we substitute the resulting $\mathbf{Z}$ into the formula for the current

$$ \mathbf{I} = \frac{\mathbf{E} l}{\mathbf{Z}} = \mathbf{E} l \frac{2 \pi a \mathbf{J}_1(\mathbf{k})}{\rho l \mathbf{k} \mathbf{J}_0(\mathbf{k})} = \frac{\mathbf{E} 2 \pi a \mathbf{J}_1(\mathbf{k})}{\rho \mathbf{k} \mathbf{J}_0(\mathbf{k})} $$
The last expression completely coincides with the previously obtained formula (27) if $n=a$ and the notation from (12) is taken into account. Substituting the resulting expression for $\mathbf{I}$ into the formula for the current density from Ref. [2, p. 723, f. (A.10)] we obtain

$$ \mathbf{j}(r) = \frac{\mathbf{I}\mathbf{k} \, J_0(r\mathbf{k})}{2\pi a \, J_1(a\mathbf{k})} = \frac{\mathbf{E} \, 2\pi a \, J_1(a\mathbf{k})}{\rho \, \mathbf{k} \, J_0(a\mathbf{k})} \frac{\mathbf{k} \, J_0(r\mathbf{k})}{2\pi a \, J_1(a\mathbf{k})} = \frac{\mathbf{E} \, J_0(r\mathbf{k})}{\rho \, J_0(a\mathbf{k})}. $$

It is quite obvious that the last expression absolutely coincides with the formula (15) obtained by us, i.e., $\mathbf{E}/\rho = \mathbf{j}_{DC} = \mathbf{j}_0$ if $n=a$. There is no direct confirmation in the existing concept of the fact that the current density on the surface is the DC density in the absence of external inductance. Probably from the considerations that the current density on the surface should be a function of frequency, but in fact it is a constant. And this fact cannot be logically explained in terms of the physical essence of current displacement to the surface due to its decrease in the inner layers and increase in the outer ones as a result of the action of eddy currents. It is quite obvious that the existing concept, which explains the physical essence of the skin effect through eddy currents, requires the dependence of the current density on the surface on the frequency. The higher the frequency, the more current must be displaced to the surface from the depth. Hence, with increasing frequency, the current density on the surface should increase, and at a low frequency it should be slightly higher than the DC density. But the formal mathematical results of the existing approach of this growth do not reflect at all. With increasing frequency, the current density decreases in depth, while on the surface it remains unchanged and equal to the DC density (see Appendix 5).

Let's see how the existing concept of the skin effect circumvents or masks this contradiction. The analysis of the behavior of the current density on the surface was found in the source "Electrodynamics: Lectures on Theoretical Physics, Vol. 3" by A. Sommerfeld. This analysis is based on the formula from Ref. [5, p. 168, § 20, f. (21)]

$$ \frac{\mathbf{j}(r)}{\mathbf{j}_0} = \frac{ka \, J_0(r\mathbf{k})}{2 \, J_1(a\mathbf{k})}. $$

But there is an error (rather a trick) in the definition of $\mathbf{j}_0$ in this formula. Namely, the average value of the current density, obtained by dividing the total current by the cross-sectional area of the wire $\mathbf{j}_0 = \mathbf{I}/\pi a^2$, decreasing with increasing frequency, was simply called the DC density, literally written as $J_0 = I/(\pi a^2)$ is the direct-current value of $J$" [5, p. 168, § 20, below f. (21)]. Or, in other words, the total current $\mathbf{I}$ decreasing with increasing frequency was assigned a fixed value. That is, the formula from Ref. [5, p. 168, § 20, f. (21)] with this definition of $\mathbf{j}_0$ cannot be considered as some variant of formula (16), as, for example, the above formula from Ref. [2, p. 723, f. (A.10)]. This is a completely different formula, which has nothing to do with the usual distribution of current density along the radius at a fixed frequency-independent voltage. Then for the surface of the wire the above formula from Ref. [5, p. 168, § 20, f. (21)] at a constant value of $\mathbf{j}_0$, naturally, will lead to an increase in the surface current density $\mathbf{j}(a)$ in accordance with the expression from Ref. [5, p. 169, § 20, f. (22a)] by a factor of about $ka/2$ for significant frequencies, when $J_0(a\mathbf{k})/J_1(a\mathbf{k}) \approx \mathbf{i}$, that is, in accordance with

$$ \frac{\mathbf{j}(a)}{\mathbf{j}_0} = \frac{ka \, J_0(a\mathbf{k})}{2 \, J_1(a\mathbf{k})} \approx \frac{ka}{2} \mathbf{i}. $$

(remember that $k \propto \sqrt{f}$). Decoding $\mathbf{j}_0$ by analogy with the above chain for the interpretation of the formula from Ref. [2, p. 723, f. (A.10)] immediately puts everything in its place, as it should be, i.e.

$$ \mathbf{j}(r) = \mathbf{j}_0 \frac{ka \, J_0(r\mathbf{k})}{2 \, J_1(a\mathbf{k})} = \frac{\mathbf{I} \, ka \, J_0(r\mathbf{k})}{\pi a^2 \, 2 \, J_1(a\mathbf{k})} = \frac{\mathbf{E} \, 2\pi a \, J_1(a\mathbf{k})}{\rho \, \mathbf{k} \, J_0(a\mathbf{k})} \frac{1}{\pi a^2 \, 2 \, J_1(a\mathbf{k})} = \mathbf{j}_{DC} \frac{J_0(r\mathbf{k})}{J_0(a\mathbf{k})}. $$
It is possible that the analysis of the behavior of the current density on the surface in Ref. [5] was specially made for the case of the invariance of the total current with increasing frequency, and then it is not erroneous or a trick. That is, the frequency increase in AC resistance was compensated by a proportional increase in the supply voltage. But then, firstly, it had to be specifically stipulated, and secondly, it was impossible to use the name "DC density" for the value of $j_0$ in Ref. [5, p. 168, § 20, f. (21)], and, thirdly, to explain in general the meaning of constructing such an unusual dependence of the voltage increase with increasing frequency, apparently, on the basis of which sources claim that the surface current density increases with increasing frequency. For example, in Ref. [2, p. 679] when calculating the current density on the surface of a flat bar, the constancy of the total current is used. The relations for a cylindrical wire in Ref. [5, p. 167] are given even without taking into account the external inductance, that is, in the simplest case, when the current density on the surface simply has nowhere to go, except to be equal to the DC density. Taking into account the external inductance at $n>a$, of course, will lead to a dependence of the surface current density on the frequency, but only in the direction of its decrease with increasing frequency in accordance with (15) at $r=a$ (see Appendix 6), i.e.

$$j_0 = j_1(a) = \frac{j_{DC} k J_0(a k)}{k J_0(a k) - a k_s^2 \ln(s) J_1(a k)}.$$

Hence, the formal mathematical apparatus of the existing concept does not leave the slightest possibility of increasing the current density on the surface with increasing frequency with or without taking into account the external inductance. Thus, we have proved that the existing solution (16) and the eddy-free solution (15) we obtained are the same solution. It is also obvious that the existing solution (16) is based not on eddy currents, but on the uneven distribution of elementary self-induction currents over depth, as well as the solution (15) we obtained. It is clear that existing solutions for other types of conductors are also based on self-inductive current and not on eddy currents.

## IV. VOLTAGE AND POWER BALANCES

Let's consider the voltage balance for a cylindrical solid wire shown in Fig. 1. Let's rewrite Eq. (22) as

$$\mathbf{j}(r)\rho = \mathbf{E} + \mathbf{E}_a(r) + \mathbf{E}_s + \mathbf{E}_{2a}(r) - \mathbf{E}_{2b}(r). \quad (113)$$

Let's reshape all specific EMFs in this equation according to Eq. (24)

$$\mathbf{j}(r)\rho = \mathbf{E} + \mathbf{E}_{ab}(r) + \mathbf{E}_s + \mathbf{E}_w, \quad (114)$$

where:

$$\mathbf{E}_{ab}(r) = \rho q^2 \mathbf{k}^2 \int_{r}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}(r) r dr dr; \quad \mathbf{E}_s = \rho \mathbf{k}_s^2 \ln(s) \int_{b}^{a} \mathbf{j}(r) r dr; \quad \mathbf{E}_w = -\rho p \mathbf{k}^2 \int_{b}^{a} \frac{1}{r} \int_{b}^{r} \mathbf{j}(r) r dr dr.$$

All analytical formulas for specific EMFs at $b=0$ are found in Appendix 1. Let's rewrite them from there:

$$\mathbf{E}_{ab}(r) = - \frac{\rho \mathbf{j}_{DC} \mathbf{k} q^2 [\mathbf{J}_0(\mathbf{a} \mathbf{k} q) - \mathbf{J}_0(\mathbf{r} \mathbf{k} q)]}{(q^2 - p) \mathbf{k} \mathbf{J}_0(\mathbf{a} \mathbf{k} q) - \mathbf{a} \mathbf{k}_s^2 q \ln(s) \mathbf{J}_1(\mathbf{a} \mathbf{k} q) + p \mathbf{k}};$$

$$\mathbf{E}_s = \frac{\rho \mathbf{j}_{DC} \mathbf{a} \mathbf{k}_s^2 q \ln(s) \mathbf{J}_1(\mathbf{a} \mathbf{k} q)}{(q^2 - p) \mathbf{k} \mathbf{J}_0(\mathbf{a} \mathbf{k} q) - \mathbf{a} \mathbf{k}_s^2 q \ln(s) \mathbf{J}_1(\mathbf{a} \mathbf{k} q) + p \mathbf{k}}; \quad \mathbf{E}_w = \frac{\rho \mathbf{j}_{DC} \mathbf{k} p [\mathbf{J}_0(\mathbf{a} \mathbf{k} q) - 1]}{(q^2 - p) \mathbf{k} \mathbf{J}_0(\mathbf{a} \mathbf{k} q) - \mathbf{a} \mathbf{k}_s^2 q \ln(s) \mathbf{J}_1(\mathbf{a} \mathbf{k} q) + p \mathbf{k}}.$$

Passing from specific values to general ones, we represent the voltage balance (114) in the form (transition see Appendix 19)

$$\mathbf{I}_{DC} = \mathbf{E}_l + \mathbf{E}_{am} l + \mathbf{E}_s l + \mathbf{E}_w l, \quad (115)$$
where

$$ \mathbf{E}_{am} = \frac{\rho \mathbf{j}_{DC} [2q \mathbf{J}_1(\mathbf{a}\mathbf{k}\mathbf{q}) - \mathbf{a}\mathbf{k}\mathbf{q}^2 \mathbf{J}_0(\mathbf{a}\mathbf{k}\mathbf{q})]}{(q^2 - p)\mathbf{a}\mathbf{k}\mathbf{J}_0(\mathbf{a}\mathbf{k}\mathbf{q}) - a^2 \mathbf{k}_s^2 q \ln(s) \mathbf{J}_1(\mathbf{a}\mathbf{k}\mathbf{q}) + a \mathbf{p}\mathbf{k}} $$

is the average value of the specific EMF of the wire over the distance r (from Appendix 19). For the eddy-free case at p=0: q=1, see (23) and $\mathbf{E}_w=0$, so the Eq. (115) will be (see Appendix 19)

$$ \mathbf{I}_1 R_{DC} = \mathbf{E} l + \mathbf{E}_{am} l + \mathbf{E}_s l. \qquad (116) $$

Let's compare Eq. (116) with Ohm's law in a complex form used in the existing concept of the skin effect for a circuit containing a voltage source $\mathbf{U}=\mathbf{E}l$, active (30) and inductive (31) resistances in the form

$$ \mathbf{E} l = \mathbf{I}_1 \mathbf{R}_1 + i \mathbf{I}_1 X_1. \qquad (117) $$

The vector diagrams in Appendix 20 show that the terms $\mathbf{I}_1 \mathbf{R}_1$ and $i \mathbf{I}_1 X_1$ from Eq. (117), as they should be, are perpendicular to each other, but the terms $\mathbf{I}_1 \mathbf{R}_{DC}$ and $\mathbf{E}_{am} l$ from Eq. (116) are not. This means that there is a non-zero projection of the EMF vector of the self-induction of the wire $\mathbf{E}_{am} l$ on the direction of the voltage drop vector $\mathbf{I}_1 \mathbf{R}_{DC}$. That is, along the voltage drop vector $\mathbf{I}_1 \mathbf{R}_{DC}$ in the opposite direction, the projection of the average self-induction EMF of the wire $\mathbf{E}_{am} l$ acts, which, together with the real voltage drop $\mathbf{I}_1 \mathbf{R}_{DC}$, makes the value $\mathbf{I}_1 \mathbf{R}_1$ artificially assigned in the existing concept. When determining the resistances according to formulas (30) and (31), the angle $\pi/2$ between the components of the voltage drops across the active and inductive resistances is essentially assigned on the assumption that everything is the same in the wire as in a conventional active-inductive circuit with lumped parameters. It may seem that the lack of perpendicularity between the EMF $\mathbf{E}_{am} l$ and the total current $\mathbf{I}_1$ contradicts Faraday's law of induction. But there is no contradiction. Perpendicularity, in accordance with this law in our case, exists between the magnetic flux and the EMF that occurs in the circuit covering this particular flux, and not between the total current $\mathbf{I}_1$ and the average EMF $\mathbf{E}_{am} l$. The current density argument diagrams in Appendix 2 show that the phase changes from surface to depth so that the current density vector rotates clockwise. Moreover, the higher the frequency, the larger the rotation angle. For high frequencies, the current density vector makes even several turns. The vector diagrams in Appendix 21 at the top show the balance of specific EMFs and voltage drop according to Eq. (7) for different radii. It is seen that the vector $\mathbf{j}_1(r)\rho$, as the radius decreases, rotates and decreases in magnitude. The projection of the vector of the specific EMF of self-induction of the wire $\mathbf{E}_a(r)$ on the direction of the vector $\mathbf{j}_1(r)\rho$ changes its sign to the opposite when the radius decreases. It can be seen that from the surface to a radius of about $r_4=0.965a$ its projection is directed against the vector $\mathbf{j}_1(r)\rho$, and then in depth it coincides with it. In the region of radius $r_4=0.965a$, the angle between vectors $\mathbf{j}_1(r)\rho$ and $\mathbf{E}_a(r)$ is $\pi/2$.

To compare the power balances, we multiply the left and right sides of Eqs. (116) and (117) by the conjugate current complex $\overline{\mathbf{I}_1}$. Then the power balance for Eq. (116) is

$$ \mathbf{I}_1 R_{DC} \overline{\mathbf{I}_1} = \mathbf{E} l \overline{\mathbf{I}_1} + \mathbf{E}_{am} l \overline{\mathbf{I}_1} + \mathbf{E}_s l \overline{\mathbf{I}_1}, \qquad (118) $$

the power balance for Eq. (117) is

$$ \mathbf{E} l \overline{\mathbf{I}_1} = \mathbf{I}_1 \mathbf{R}_1 \overline{\mathbf{I}_1} + i \mathbf{I}_1 X_1 \overline{\mathbf{I}_1}. \qquad (119) $$

Comparison diagrams of Eqs. (118) and (119) are shown at the bottom of Appendix 21. At first glance, it may seem that since the Joule heating power $\mathbf{I}_1 \mathbf{R}_{DC} \overline{\mathbf{I}_1}$ from Eq. (118) is less than $\mathbf{I}_1 \mathbf{R}_1 \overline{\mathbf{I}_1}$ from Eq. (119), less heat will be released in the first case. Naturally this will not happen. It's just that now all the heat evolved in Eq. (118) consists of two parts: the actual Joule heating (Ohmic losses) $\mathbf{I}_1 \mathbf{R}_{DC} \overline{\mathbf{I}_1}$ and the power losses to overcome the back-EMF by the power supply, equal to the
projection $E_{am}l$ on the direction of the current $I_1$. The physical essence of these losses can be remotely considered analogous to the physical essence of the heat evolved in the hot junction of dissimilar metals when current flows through the junction (Peltier effect). As is well known, if the direction of the contact EMF is opposite to the direction of the current through the contact, then heat is evolved. Conversely, when the directions of the contact EMF and current coincide, heat is absorbed, as it happens in a cold junction. The power balance for the eddy variant (115) is

$$IR_{DC} \bar{\mathbf{I}} = \mathbf{E}/\bar{\mathbf{I}} + \mathbf{E}_{am}/\bar{\mathbf{I}} + \mathbf{E}_s/\bar{\mathbf{I}} + \mathbf{E}_w/\bar{\mathbf{I}}. \quad (120)$$

There is another way to approach the consideration of the energy balance with the skin effect. Let us multiply Eq. (7) by the conjugate complex of the current density $\overline{\mathbf{j}}_1(r)$, that is

$$\mathbf{j}_1(r) \rho \overline{\mathbf{j}}_1(r) = \mathbf{E} \overline{\mathbf{j}}_1(r) + \mathbf{E}_a(r) \overline{\mathbf{j}}_1(r) + \mathbf{E}_s \overline{\mathbf{j}}_1(r), \quad (121)$$

and Eq. (114) by the conjugate complex of the current density $\overline{\mathbf{j}}(r)$, that is

$$\mathbf{j}(r) \rho \overline{\mathbf{j}}(r) = \mathbf{E} \overline{\mathbf{j}}(r) + \mathbf{E}_{ab}(r) \overline{\mathbf{j}}(r) + \mathbf{E}_s \overline{\mathbf{j}}(r) + \mathbf{E}_w \overline{\mathbf{j}}(r). \quad (122)$$

Appendix 22 shows graphs of the real and imaginary parts of all terms of Eq. (121) (on the upper graphs) and Eq. (122) (on the lower graphs) at $f=1$ MHz, $n=1.01a$ ($s=1.01$) and other initial data from Appendix 2. Brown color in Appendix 22 shows the values of all terms according to Eqs. (118) and (120). To go to the balances of total powers, it is necessary to integrate Eqs. (121) and (122) over the entire volume of the wire. The green lines show the terms obtained by integrating Eq. (121) over the entire volume of the wire according to

$$2\pi d \int_{b}^{a} \mathbf{j}_{1}(r) \rho \overline{\mathbf{j}}_{1}(r) r dr = 2\pi d \int_{b}^{a} \mathbf{E} \overline{\mathbf{j}}_{1}(r) r dr + 2\pi d \int_{b}^{a} \mathbf{E}_{a}(r) \overline{\mathbf{j}}_{1}(r) r dr + 2\pi d \int_{b}^{a} \mathbf{E}_{s} \overline{\mathbf{j}}_{1}(r) r dr, \quad (123)$$

and integrating Eq. (122) over the entire volume of the wire according to

$$2\pi d \int_{b}^{a} \mathbf{j}(r) \rho \overline{\mathbf{j}}(r) r dr = 2\pi d \int_{b}^{a} \mathbf{E} \overline{\mathbf{j}}(r) r dr + 2\pi d \int_{b}^{a} \mathbf{E}_{ab}(r) \overline{\mathbf{j}}(r) r dr + 2\pi d \int_{b}^{a} \mathbf{E}_{s} \overline{\mathbf{j}}(r) r dr + 2\pi d \int_{b}^{a} \mathbf{E}_{w} \overline{\mathbf{j}}(r) r dr. \quad (124)$$

Let's look at a comparison of balances according to Eqs. (118) and (123) for the eddy-free case. As can be seen from Appendix 22, in both cases (118) and (123) there is a balance, but the terms to the left of the equal sign are different – Joule heating. Also, the second terms to the right of the equal sign are not equal – the power from the EMF of the self-induction of the wire, and more specifically – the real parts of these terms are not equal. A seemingly unpleasant situation when two strict formulas (118) and (123) give a different distribution in terms of the energy balance components. Maybe one of the formulas is correct, and the other is not? In fact, both formulas are correct. The current from Eq. (116) is a consequence of the vector sum of multidirectional elementary currents (see Appendix 21 where vector $\mathbf{j}_1(r)\rho$ rotates clockwise along skin depth), which, if aligned in one direction, will result in a much higher current. In the presence of the dependence of the phase of the current density on the coordinate, the formula for the Joule heating cannot be expressed in terms of the product of the square of the current measured by the ammeter and the resistance. It is necessary to use a more general expression in the form of the left side of formula (123). Eq. (123) acts, figuratively speaking, from the inside, immediately summing up the Joule losses from differently directed elementary currents. In formula (118), there is a total current, which is the geometric sum of elementary currents, which (total current) can be measured with an ammeter. In the existing concept, the Joule heating are calculated in terms of the measurable current and the fitted resistance to
correspond to the experiment. For clarity, we will give the following example. There are two wires side by side of the same resistance $2R_{DC}$ each with direct currents $I_1$ and $I_2$ in the opposite directions. Let only the difference of currents from the sum of their magnetic fields, as well as the total resistance of parallel wires, be available for measurement. This situation is almost the same as for the surface and deep parts of the real component of alternating current with the skin effect. If the Joule heating of parallel wires is calculated using the formula $(I_1 - I_2)^2 R_{DC}$, then the difference compared to the real Joule losses equal to $I_1^2 2R_{DC} + I_2^2 2R_{DC}$ will be $(I_1 + I_2)^2 R_{DC}$. To avoid this difference, in the existing concept, $(I_1 - I_2)^2$ is left, and $R_{DC}$ is turned into $R(f) > R_{DC}$, so that the equality $I_1^2 2R_{DC} + I_2^2 2R_{DC} = (I_1 - I_2)^2 R(f)$ is observed. This is convenient for calculations, but does not reflect the essence of the physical picture. The real heating current in this case is not equal to $I_1 - I_2$ available for measurement, but is equal to $\sqrt{2(I_1^2 + I_2^2)}$. We can say that physically there is no increase in active resistance with the skin effect. This is clearly confirmed by Eqs. (116), (118) and (123). There are no artificially adjusted resistances in them. If in the frame of reference of the measured current in Eq. (118) there appears a seemingly not quite usual term in the form of the real part of $E_{am}l\overline{I_1}$, which together with $I_1 R_{DC} \overline{I_1}$ makes up the total heat losses, then in Eq. (123) everything is where it should be. That is, Eq. (123) is correct from a physical point of view, and Eq. (118) is a variant of considering the problem in the frame of reference of the current measured by an ammeter, as is done in the existing concept. It can also be concluded that the explanation of the existing concept of increasing resistance by reducing the effective cross-section of the wire is incorrect.

In Appendix 22 for Eq. (121) (on the top graphs), the value of the radius $r=3.227$ mm (let's call the point $c$) is the point of intersection of the abscissa axis with the graph of the specific active power supplied from the power source $\text{Re}[E_j(r)]$. With values of the radius from $c$ to $a$ in the upper layers of the wire, this power is positive, and with a radius from 0 to $c$ in the lower layers it is negative. The components of the power balance according to Eq. (123) with a radius from 0 to $c$ are highlighted in yellow in Appendix 22, with a radius from $c$ to $a$ in blue. A wire in the range from 0 to $c$ is a generator of active electric power, one part of which, minus 76 W, is given to the power source, and the other part, 15 W, is spent on heating in the form of Joule losses. But where does it need to come from to spend it? What is the generator of electricity in this area? The active components of the self-induction EMF of both the wire itself and the surrounding space in this area coincide with the direction of the active component of the current density. Therefore, as in the Peltier effect, this active power $51+40=91$ (W) will be taken from the internal energy of the wire material, that is, the wire in the radius from 0 to $c$ will be cooled. Then the balance of active powers in this section is $15 = -76+51+40$ (W). The reverse picture will be observed in the range from $c$ to $a$. In this section, the source delivers active power of 275 W – more than the entire output power of 199 W. This power is consumed for Joule heating 184 W and for additional heating in the same way as for cooling in the lower layers $-51-40=-91$ (W). Then the balance of active powers in the range from $c$ to $a$ will be $184=275-51-40$ (W). The total balance of active powers in the range from 0 to $a$ will be $15+184=-76+275+51-51+40-40$ (W), i.e. $199=199$ (W).

It should be noted that the probable thermal separation will take place only if it is possible to transfer electricity to the power source in the cold layer and receive it from the source in the hot one. That is, this very source must be able to receive and give additional energy. To do this, the field strength inside the source with linear size $l_p$ must satisfy the relation $E_p = -E/l_p$. At the very beginning, we assumed that the linear size of the source is negligible compared to the length of the wire. Then, neglecting the active and inductive resistances of the source, the source power balance for the eddy-free variant is

$$ \mathbf{P}_{ba} = -2\pi d_p \int_b^a \mathbf{E}_p \overline{\mathbf{j}_1(r)} r dr = \mathbf{P}_{bc} + \mathbf{P}_{ca} = -2\pi d_p \int_b^c \mathbf{E}_p \overline{\mathbf{j}_1(r)} r dr - 2\pi d_p \int_c^a \mathbf{E}_p \overline{\mathbf{j}_1(r)} r dr, $$
where $\mathbf{P}_{ba}$ is the power of external forces of the source (chemical, electromagnetic, etc.), $\mathbf{P}_{bc}$ and $\mathbf{P}_{ca}$ are its parts in the cold and hot zones, respectively. Also, of course, there is a balance between the power of the source and the load wire both as a whole, i.e.

$$-2\pi d_p \int_b^a \mathbf{E}_p \overline{\mathbf{j}_1(r)} r dr = 2\pi d \int_b^a \mathbf{E} \overline{\mathbf{j}_1(r)} r dr,$$

and in the cold and hot zones separately, i.e.

$$-2\pi d_p \int_b^c \mathbf{E}_p \overline{\mathbf{j}_1(r)} r dr = 2\pi d \int_b^c \mathbf{E} \overline{\mathbf{j}_1(r)} r dr \quad \text{and} \quad -2\pi d_p \int_c^a \mathbf{E}_p \overline{\mathbf{j}_1(r)} r dr = 2\pi d \int_c^a \mathbf{E} \overline{\mathbf{j}_1(r)} r dr.$$

But to find a real source with such properties is hardly possible. EMFs, created by all known sources, works in such a way that in real closed circuits there are no zones with opposite field strengths $\mathbf{E}_p$ and $\mathbf{E}$. That is, in the entire closed circuit, when a voltage $\mathbf{U}$ is applied to it, the power field strength $\mathbf{E} = \mathbf{U}/(l+l_p)$ appears, which is essentially the field strength of external forces. In other words, the voltage of the source, as its power characteristic, is distributed throughout the circuit, and then the power balance equation of the source and receiver will have the form

$$\mathbf{P}_{ba} = 2\pi d \int_b^a \mathbf{E} \overline{\mathbf{j}_1(r)} r dr = \mathbf{P}_{bc} + \mathbf{P}_{ca} = 2\pi d \int_b^c \mathbf{E} \overline{\mathbf{j}_1(r)} r dr + 2\pi d \int_c^a \mathbf{E} \overline{\mathbf{j}_1(r)} r dr.$$

It can be said that in a closed circuit, the external force of the source does not work to separate charges against the strength of the electric field inside the source, but to overcome all possible obstacles in the entire circuit, including active resistance and induction EMF. That is, even if the power source is a separate object from the load wire, it still acts like an EMF of mutual induction in the secondary winding of the transformer, when it is impossible to localize the area that is the source of energy. To the question, what, for example, in this case, we then measure with a voltmeter at the terminals of the secondary winding of the transformer connected to the load? We always measure, as is known, only the current that branches off into the measuring device. And already by the measured current, together with the known internal resistance of the device, we calibrate it in units of voltage. Thus, we can formulate the following conditions for the existence of the thermal separation. Firstly, it is the very presence of a source in the form of an object separate from the load wire, the field strength inside of which is opposite to the field strength in the load. Secondly, a conduction current must flow through the source, which can change its direction depending on the behavior of the external load, and either do work on the source or receive additional work from it. If such a source is nevertheless possible, then thermal separation is easily detected. In the meantime, we can conclude that the electricity produced by reducing the internal energy in the cold zone in the same zone will turn into thermal energy, and will not be transferred to the source. The same goes for the hot zone. Electricity, which is converted into heat in this zone, is taken not from the source, but from the wire itself in the same zone by reducing its internal energy. As a result, naturally, there will be neither cooling in the cold zone, nor heating in the hot one. It is possible to proceed in another way to obtain a balance of specific powers, namely, to immediately add all strengths in Eqs. (7) and (114), and then multiply the resulting equations by the conjugate complex of current density. In this case, there is no need to talk about any additional release or absorption of heat at all.

In Appendix 22 for the eddy Eqs. (122) and (124), the value of the radius $c=3.259$ mm (under the lower graphs) is larger than for the eddy-free case, which is a consequence of the additional displacement of the current to the surface due to eddy currents. But what is surprising here is that at a lower power consumption from a source of 195 W, the Joule losses of 260 W exceed the actual power consumption and exceed the Joule losses in the eddy-free case of 199 W. This is very similar to a heat pump, or a Peltier effect refrigerator, in which the thermal power released exceeds the en-
ergy consumed due to the removal of heat from the environment. Let's consider the energy balances by isolating the eddy specific EMF. We return to Eq. (113) as follows

$$ \mathbf{j}(r)\rho = \mathbf{E} + \mathbf{E}_a(r) + \mathbf{E}_s + \mathbf{E}_e(r), \qquad (125) $$

where:

$$ \mathbf{E}_a(r) = \frac{\mathbf{E}_{ab}(r)}{q^2} = - \frac{\rho \mathbf{j}_{DC} \mathbf{k} [J_0(\mathbf{a} \mathbf{k} \mathbf{q}) - J_0(\mathbf{r} \mathbf{k} \mathbf{q})]}{(q^2 - p) \mathbf{k} J_0(\mathbf{a} \mathbf{k} \mathbf{q}) - a \mathbf{k}_s^2 q \ln(s) J_1(\mathbf{a} \mathbf{k} \mathbf{q}) + p \mathbf{k}}; $$

$$ \mathbf{E}_e(r) = \mathbf{E}_{2a}(r) - \mathbf{E}_{2b}(r) = \mathbf{E}_{ab}(r) + \mathbf{E}_w - \frac{\mathbf{E}_{ab}(r)}{q^2} = \frac{\rho \mathbf{j}_{DC} \mathbf{k} [2p J_0(\mathbf{r} \mathbf{k} \mathbf{q}) - p - p J_0(\mathbf{a} \mathbf{k} \mathbf{q})]}{(q^2 - p) \mathbf{k} J_0(\mathbf{a} \mathbf{k} \mathbf{q}) - a \mathbf{k}_s^2 q \ln(s) J_1(\mathbf{a} \mathbf{k} \mathbf{q}) + p \mathbf{k}} $$

(from Appendix 1). We will rewrite all the equations of the balance of specific and total capacities with this distribution of specific EMFs. The balance of total capacities in the reference frame of the measured current is

$$ \mathbf{I}_{DC} \bar{\mathbf{I}} = \mathbf{E} / \bar{\mathbf{I}} + \mathbf{E}_m / \bar{\mathbf{I}} + \mathbf{E}_s / \bar{\mathbf{I}} + \mathbf{E}_{em} / \bar{\mathbf{I}}, \qquad (126) $$

where (see Appendix 19):

$$ \mathbf{E}_m = \frac{\mathbf{E}_{am}}{q^2} = \frac{\rho \mathbf{j}_{DC} [2 J_1(\mathbf{a} \mathbf{k} \mathbf{q}) - \mathbf{a} \mathbf{k} \mathbf{q} J_0(\mathbf{a} \mathbf{k} \mathbf{q})]}{a q [(q^2 - p) \mathbf{k} J_0(\mathbf{a} \mathbf{k} \mathbf{q}) - a \mathbf{k}_s^2 q \ln(s) J_1(\mathbf{a} \mathbf{k} \mathbf{q}) + p \mathbf{k}]}; $$

$$ \mathbf{E}_{em} = \mathbf{E}_{am} + \mathbf{E}_w - \frac{\mathbf{E}_{am}}{q^2} = \frac{\rho \mathbf{j}_{DC} [(2 - 2q^2) J_1(\mathbf{a} \mathbf{k} \mathbf{q}) + \mathbf{a} \mathbf{k} \mathbf{p} \mathbf{q} + \mathbf{a} \mathbf{k} \mathbf{q} J_0(\mathbf{a} \mathbf{k} \mathbf{q})(q^2 - p - 1)]}{a q [(p - q^2) \mathbf{k} J_0(\mathbf{a} \mathbf{k} \mathbf{q}) + a \mathbf{k}_s^2 q \ln(s) J_1(\mathbf{a} \mathbf{k} \mathbf{q}) - p \mathbf{k}]}. $$

The balance of specific capacities is

$$ \mathbf{j}(r) \rho \overline{\mathbf{j}(r)} = \mathbf{E} \overline{\mathbf{j}(r)} + \mathbf{E}_a(r) \overline{\mathbf{j}(r)} + \mathbf{E}_s \overline{\mathbf{j}(r)} + \mathbf{E}_e \overline{\mathbf{j}(r)}. \qquad (127) $$

The balance of total capacities by integrating Eq. (127) over the entire volume of the wire is

$$ 2\pi d \int_{b}^{a} \mathbf{j}(r) \rho \overline{\mathbf{j}(r)} r dr = 2\pi d \int_{b}^{a} \mathbf{E} \overline{\mathbf{j}(r)} r dr + 2\pi d \int_{b}^{a} \mathbf{E}_{a}(r) \overline{\mathbf{j}(r)} r dr + 2\pi d \int_{b}^{a} \mathbf{E}_{s} \overline{\mathbf{j}(r)} r dr + 2\pi d \int_{b}^{a} \mathbf{E}_{e}(r) \overline{\mathbf{j}(r)} r dr. \qquad (128) $$

Appendix 23 shows the calculated values of the total powers according to the Eqs. (126), (128) and $\mathbf{I} \mathbf{Z} \bar{\mathbf{I}} = \mathbf{E} / \bar{\mathbf{I}}$ and graphs of specific powers (127) for initial data similar to Appendix 22. As can be seen from the calculation, the eddy specific EMF $\mathbf{E}_e(r)$ is a generator of electricity in both the hot and cold zones $22 + 43 = 65$ (W). To generate this power, most likely, the internal energy of the wire itself is used. All this energy in the wire itself is released in the form of heat.

Let us consider in more detail what we encountered with the appearance of eddy currents in our power balance equations. At the very beginning, we compared eddy currents with Larmor currents of free charges in a constant magnetic field, essentially following the hypothesis accepted in modern theory about the occurrence of microcurrents in an alternating magnetic field, leading to a greater consumption of active power from the source and its release in the form of heat. Comparing the results of calculations according to Eqs. (122) and (124) in Appendix 22, we see that taking into account eddy currents does not lead to an increase in the power consumed from the source, but to its decrease. That is, for the eddy-free version, it amounted to 199 W, and for the eddy one it became 195 W. In the well-known manifestations of eddy currents, such as the braking of a metal disk in a constant magnetic field, the occurrence of short-circuited currents in the magnetic circuit of a transformer, energy is dissipated. At the same time, a feedback is clearly traced with the energy source, due to which this dissipation is paid. For a disk, this is the cost of mechanical work for its movement against the resultant Ampère forces acting on eddy currents. For a transformer, this is an addi-
tional increase in the primary winding current, reflecting the additional consumption of electricity from the source. And what happens with the eddy current with the skin effect, if the energy consumed by the source does not increase, but decreases? From Appendix 23 we saw that the isolated eddy specific EMF $E_e(r)$ is an active power generator, equal to 65 W. But then it is very similar to the violation of the second law of thermodynamics. Probably, eddy currents in the form in which we formalized them at the beginning do not arise in conductors. Such a conclusion suggests itself on the basis of a contradiction with the second law of thermodynamics, since their energy in the form of the last terms on the right in (127) and (128) does not lead to additional dissipation, but, on the contrary, reduces the energy consumption from the source. In addition to the calculated reduction in power consumption in the presence of eddy currents, the following argument can be made to justify their absence. At any distance, the directions of the supposed microcurrents are always opposite from the adjacent sections located along the length of the wire, and the magnitudes are equal. Therefore, in fact, the currents are not closed across the length of the wire. In the cylindrical wire in Fig. 1, the upper and lower parts of the considered eddy current circuits overlap and cancel each other out. And since there is no short circuit between the inner and outer layers, then part of the eddy current in any outer layer no longer covers the magnetic flux inside the layer, but covers the magnetic flux outside the layer. Therefore, the current outside will be determined by the supply voltage, internal and external self-induction and resistance in accordance with the equation without taking into account eddy currents. And this is true for any distance, since any cylindrical layer, arbitrarily close to the wire axis, will always be external with respect to layers at smaller radii. The situation is somewhat similar to the situation with unipolar electric machines, in which, due to symmetry, eddy currents are also absent. For galvanically open circuits, such as dipole antennas, these considerations do not apply, since they have loops for eddy currents to pass through the ends of the antenna. It is possible that there are no eddy currents in galvanically closed circuits, but they are in open circuits. That is, the question of the existence of eddy currents in the skin effect requires experimental study. If the effect of eddy currents, if any, is only a slight decrease in power consumption (195 W instead of 199 W from Appendix 22), then there is nothing to worry about the second law of thermodynamics, because all the 65 W generated by the eddy currents from the internal energy of the wire is dissipated as heat in the wire itself. That is, the total Joule losses of 260 W during the calorific measurement of heat will decrease by 65 W due to the mutual compensation of the input and output of heat from eddy currents, that is, the heat losses available for measurement will be equal to the active power supplied by the source, i.e., equal to 195 W. To test the second law of thermodynamics, you can take an axisymmetric U-shaped resonator and place a winding around the eddy current circuit on it, similar to the winding on a torus, only with a core with a U-shaped cross section. Next, we connect power to the ends of the resonator with a resonance frequency, and connect an adjustable load to the secondary winding. With an increase in the load connected to the toroidal winding, a decrease in the temperature of the resonator material will indicate a violation of second law. But this is unlikely, because it would have been discovered long ago if it existed.

## V. THERMAL SEPARATION EFFECT CALCULATION

We cannot leave without experimental verification the conclusions of the previous section about the absence of thermal separation due to the impossibility of the existence of the required power source. To calculate the experiment to establish the fact of the presence or absence of thermal separation, we will assume that the power source satisfies the previously stated criteria for the existence of thermal separation. Let's look at the energy relationships for the infinitely long solenoid shown in Fig. 4. The balance of specific powers for the eddy case (84) is

$$ \mathbf{j}(x)\rho\overline{\mathbf{j}(x)} = \mathbf{E}\overline{\mathbf{j}(x)} + \mathbf{E}_t(x)\overline{\mathbf{j}(x)} + \mathbf{E}_s\overline{\mathbf{j}(x)} + \mathbf{E}_w\overline{\mathbf{j}(x)}, \quad (129) $$

where the formulas for $E_t(x)$, $E_s$, $E_w$ are given in Appendix 17. The balance of total powers, obtained by integrating Eq. (129) over the entire volume of the wire is
$$P_s z \int_0^t \overline{\mathbf{j}(x)} \rho \overline{\mathbf{j}(x)} dx = P_s z \int_0^t \overline{\mathbf{E} \cdot \mathbf{j}(x)} dx + P_s z \int_0^t \overline{\mathbf{E}_t(x)} \overline{\mathbf{j}(x)} dx + P_s z \int_0^t \overline{\mathbf{E}_s \cdot \mathbf{j}(x)} dx + P_s z \int_0^t \overline{\mathbf{E}_w \cdot \mathbf{j}(x)} dx. \quad (130)$$

Solution plots (85) for a single-turn toroidal coil with a rectangular cross-section magnetic circuit with all the initial data are given in Appendix 24 above. In the calculation, an assumption was made that the infinitely long and toroidal solenoids are identical for the accepted sizes. Graphs according to the formula for specific powers (129) and calculated values of all quantities according to the formula for total powers (130) are given in Appendix 24 in the middle, where, to save space, we introduced the notation for the terms of Eq. (130) $Q_{0t} = P_{0t} + T_{0t} + S_{0t} + W_{0t}$, as well as its parts in the cold zone $Q_{0c} = P_{0c} + T_{0c} + S_{0c} + W_{0c}$ and the hot zone $Q_{ct} = P_{ct} + T_{ct} + S_{ct} + W_{ct}$. At the bottom of Appendix 24 are graphs of magnetic induction. The power balance clearly indicates that with a high permeability of the core $\mu_s$, the amount of heat absorbed in the range from 0 to c (the balance line is colored yellow) and evolved in the range from c to t (the balance line is colored blue) significantly exceeds the Joule heating. This means that under such conditions it is possible to expect noticeable cooling of the wire in the range from 0 to c, and heating in the range from c to t. In this design, the maximum current density and heat release, as can be seen from Appendix 24, will be observed in the wire layer adjacent to the magnetic core. Heat absorption, on the contrary, will be observed at the outer surface. Therefore, to establish the fact of cooling, we can simply measure the temperature of the outer surface of the wire. It is possible that it will be necessary to select some optimal turn-on time in order to prevent the establishment of thermal equilibrium between the layers. The main factor equalizing the temperature of the cold and hot zones along the x coordinate is still thermal conductivity. It may seem that in order to prevent heat transfer between the hot and cold zones due to thermal conductivity, it is possible to make a winding in the form of a two-layer winding with parallel connection of the layers to the power source, and place thin thermal insulation on the c coordinate between the layers. However, solution (85) for Eq. (84) was obtained from the condition of invariance of resistivity $\rho$. With a jump in resistivity at the location of the thermal insulation, the solution will be completely different. Evaluating the picture qualitatively, we can say that in each of the layers there will be approximately the same current density distribution. Also, each layer will have its own hot and cold zones, and the average temperatures of each layer will be approximately the same. Therefore, division into layers probably will not give anything. Without giving calculations and graphs, because everything is similar to what was previously given in Appendices 10 and 11, an optimal ratio of the skin depth to the wire thickness is also acceptable for the solenoid, providing minimal resistance to alternating current. For the eddy-free case it is $\delta/t \approx 0.637$, for the eddy case it is $\delta/t \approx 0.45$, i.e. the same as for the round wire.

Appendix 25 shows the same graphs as Appendix 24, but at an order of magnitude higher frequency. Since the two main zones of heat release and absorption are located near the surface with a higher current density, following each other, it is much more difficult to detect a temperature drop under such conditions. In this case, there is no heat absorption near the surface with a lower current density, and the decrease in temperature can only be a consequence of heat conduction from the cold layer to the outer surface. But the same thermal conductivity equalizes the temperature of the hot and cold layers. It can be concluded that if you do not specifically look for an option with the best possibility of detecting heat absorption, then you may not notice this absorption in a design that is not particularly suitable or with the wrong parameters, unless you accidentally stumble upon it.

The balance of specific powers for the eddy case of a series connection of a wire with a capacitor by Eq. (88) is

$$\mathbf{j}(x) \rho \overline{\mathbf{j}(x)} = \mathbf{E} \cdot \overline{\mathbf{j}(x)} + \mathbf{E}_t(x) \cdot \overline{\mathbf{j}(x)} + \mathbf{E}_s \cdot \overline{\mathbf{j}(x)} + \mathbf{E}_w \cdot \overline{\mathbf{j}(x)} + \mathbf{E}_c \cdot \overline{\mathbf{j}(x)}, \quad (131)$$

where the formulas for $E_t(x)$, $E_s$, $E_w$, $E_c$ are given in Appendix 18. The balance of total powers, obtained by integrating Eq. (131) over the entire volume of the wire is
$$P_s z \int_0^t \overline{\mathbf{j}(x)} \rho \overline{\mathbf{j}(x)} dx = P_s z \int_0^t \overline{\mathbf{E} \cdot \mathbf{j}(x)} dx + P_s z \int_0^t \overline{\mathbf{E}_t(x)} \overline{\mathbf{j}(x)} dx + P_s z \int_0^t \overline{\mathbf{E}_s \cdot \mathbf{j}(x)} dx + P_s z \int_0^t \overline{\mathbf{E}_w \cdot \mathbf{j}(x)} dx + P_s z \int_0^t \overline{\mathbf{E}_c \cdot \mathbf{j}(x)} dx. \quad (132)$$

The last term on the right side of Eqs. (131) and (132) is responsible for the action of the capacitor, which creates a field strength $E_c$ in the wire in the opposite direction to the direction of its field strength in the dielectric between the capacitor plates. Solution plots (89) with all the initial data are given in Appendix 26 above. All calculations and graphs based on formulas (131), (132) are given in Appendix 26 in the middle, where we used the notation for the terms of Eq. (132) $Q_{0t} = P_{0t} + T_{0t} + S_{0t} + W_{0t} + C_{0t}$, as well as its parts in the cold zone $Q_{0c} = P_{0c} + T_{0c} + S_{0c} + W_{0c} + C_{0c}$ and the hot zone $Q_{ct} = P_{ct} + T_{ct} + S_{ct} + W_{ct} + C_{ct}$. At the bottom of Appendix 26 are graphs of magnetic induction. The initial data are similar to those given in Appendix 24. The capacitance is taken from the voltage resonance condition. The power supply voltage is chosen so that the reactive voltage drops on the inductance and capacitance are equal to the supply voltage from Appendix 24. As we remember, for the case of a direct connection of the power supply, most of the active power generated by the inductances of the wire and the surrounding space in the cold zone was transferred to the source (see Appendix 24). In the hot zone, on the contrary, active power was transferred from the source to the inductances. In the case just discussed with the capacitor, the power supply operates in the usual way, namely, it generates active power in both zones. Due to the action of inductive specific EMF, as well as without a capacitor, heat is released in the hot zone and consumed in the cold zone. In Appendix 26, from the graphs of Eq. (131), it can be seen that the capacitor has now actually taken over the role of the power source. More precisely, not the capacitor itself, but the circuit external to it, i.e. inductor, since $E_c$, as previously stated, is the field strength of the capacitor in the external circuit. In the hot zone, the active energy $Re[E_c \cdot \overline{\mathbf{j}(x)}]$ coming from the capacitor is positive, and in the cold zone, it is negative. We write out from Appendix 26 the total current $I = I_{0c} + I_{ct} = 65.061$ A and its parts in the cold $I_{0c} = 21.648 - 21.696i$ A and hot zones $I_{ct} = 43.413 + 21.696i$ A. It is obvious that the imaginary parts of the currents of the hot and cold zones should close with each other within the wire, since the capacitor will let through only real components that lead the voltage on the capacitor by an angle $\pi/2$. It is possible that the closure of the imaginary parts of the currents will be carried out along the surface of the capacitor plates by moving the accumulated charges, if this path has less resistance. At least, the coincidence of the phases of the imaginary components of the currents and the charge on the capacitor allow us to assume this. With regard to the supposed thermal separation, although there is a field inside the capacitor with a strength opposite to the wire, there is no conduction current capable of doing or receiving additional work. Therefore, most likely, there will be no thermal separation in a circuit with a capacitor. Whether this is actually the case, the experiment will show.

## VI. THERMAL SEPARATION EXPERIMENT

According to the calculation presented in Appendix 24, the test sample was made in the form of a toroidal inductor with one turn, as shown in the axial section in Fig. 7. A hollow ferrite cylinder (pos. 1) with an outer diameter $D_e = 20$ mm, an inner diameter $D_i = 12$ mm and an axial size $H_s = 12$ mm (two rings M100NM 20×12×6) is once wrapped with copper foil (pos. 2) with a thickness

<figure><img src="image_8.png" /><figcaption>FIG. 7. Design of the test sample with dimensions t, $H_s$, $D_i$, $D_e$: pos. 1 – hollow ferrite cylinder, pos. 2 – copper foil.</figcaption></figure>
$t=0.2$ mm. Relative permeability of ferrite $\mu_{sr}=100$. Calculated values of inductance $L=1.2 \times 10^{-7}$ H, DC resistance $R_{DC}=5.47 \times 10^{-5}$ $\Omega$, AC resistance $R=9.56 \times 10^{-5}$ $\Omega$ (see Appendix 24). The generator was a simple push-pull Colpitts oscillator based on two KT940A type VT transistors with the test sample ($L$ and two $R_{DC}/2$) as an inductance and 48 parallel-connected ceramic capacitors of the K10-62 type with a nominal value of 47 nF each as a capacitance $C=48 \times 47 \times 10^{-9}=2.256 \times 10^{-6}$ F as shown in Fig. 8. All 48 capacitors with the minimum lead lengths were evenly distributed in a circle

<figure><img src="image_9.png" /><figcaption>FIG. 8. Schematic diagram of the experiment.</figcaption></figure>

along the ends of the inductor coil under study and soldered to them, as shown in Fig. 7 (one capacitor out of 48 is shown). The total cross section of the capacitor terminals was approximately 94 % of the average cross section of the inductance wire $S=10.05 \text{ mm}^2$, which, with their small length of about 3 mm, is considered quite acceptable. Since the tested inductance consists of one turn, the inner cylindrical surface with an axial size of about 1 mm was chosen as the midpoint for connecting the positive power wire $+U$, diagonally opposite to the place where the capacitors are connected (see Fig. 7). Variable resistors $R_{op}=0...100 \text{ k}\Omega$ are used to select the optimal position of the operating point of transistors and are coaxial parts of a dual variable resistor. The feedback capacitors $C_{fb}=47 \text{ nF}$ are selected empirically based on the fact of the stable appearance of an alternating voltage at the generator output. As a constant voltage source, a JN-84W-420200 type charger was used, connected in parallel through a 33 $\text{ k}\Omega$ variable resistor as a voltage divider-regulator (not shown in the diagram) with a voltage range of $U=0...42 \text{ V}$. The calculated value of the generator frequency was determined from the condition of equality of the reactances of the inductance and capacitance according to the formula $f=1/2\pi\sqrt{LC}$ and was expected to be about 305.826 kHz (see Appendix 24). The impedance of the resonant circuit when connected in parallel is purely active and is defined as $Z_r=L/CR \approx 557 \ \Omega$. The calculated load current through the transistors will be taken equal to the resonant current at an applied voltage $U=15 \text{ V}$, that is, $I_r=U/Z_r=0.027 \text{ A}$. As we can see, the value of this current exactly coincides with the real part of the total current $I=0.027 - 65.06i$ A (see Appendix 24) when the test sample is directly connected to an alternating voltage source of 15 V. As shown at the end of the previous section, for an inductor, the actual sink-source reacting to its processes in the cold and hot zones is the capacitor. Therefore, it was concluded that there is no difference whether the resonant circuit is made in series or in parallel in terms of finding the fact of thermal separation. Measurements of the voltage on the oscillatory circuit and temperature were carried out using an M838 multimeter. The fact of the presence of fluctuations, in addition to measuring the alternating voltage, was also recorded by the appearance of pickups in the form of a characteristic squeak in the speakers connected to the computer. When the supply voltage of the oscillator changed, the loudness of the squeak in the speakers changed. At a measured voltage of 18...40 V, the temperature of both the outer and inner surfaces of the copper winding of the inductor remained unchanged and equal to the room temperature of 25 °C. When the calculated value of the voltage amplitude on the oscillatory circuit is 15 V, the calculated thermal power released in the hot zone and absorbed in the cold zone is about 325 W (see Appendix 24). If such power were indeed released and absorbed, this, of course, would immediately become noticeable by changing the temperature of the outer surfaces of the cold and hot zones. It can be concluded that thermal separation is not fixed. If a temperature separation existed, it would have been detected long ago because of the very large excess of calculated values of absorbed and released heat over Joule losses. Nevertheless, even such an experiment, one might say, "on the knee", is still better than mere assumptions.
# VII. CONCLUSIONS AND OUTLOOK

Eddy currents are not the main cause of the skin effect. Its main reason is the distance-dependent EMF of the self-induction of the wire, which increases from zero at the surface to a maximum in depth. Eddy currents only redistribute the current density to some extent, increasing it on the surface and decreasing it in depth, if any. In the existing concept, the EMF of self-induction actually appears when the divergent part of the solution of the differential equation is discarded or when the equation itself is modified. But it seems that modern theory is not aware of this and continues to insist on eddy currents as the cause of the skin effect. This incorrect statement of the existing concept inevitably led to the need either to discard the divergent part of the complete solution of the differential equation, or to modify the equation itself to bring it into line with the experimental data.

Double differentiation of the integral equations obtained here for the eddy-free case will, of course, lead to differential equations strictly following from Maxwell's equations. That is, in the case of a tubular cylinder, for example, not to the modified Bessel equation, but to the original Bessel equation. It would seem that since the correct solution gives a modified equation of the existing concept, then the differentiation of the integral equation obtained here, which also gives the correct solution, should lead precisely to the modified one. We believe that the explanation of this contradiction lies in the fact that the process of differentiation of integral equations degenerates the real circuits of elementary self-induction currents into infinitely small circuits of currents, which, in fact, become local short-circuited eddy currents. Since eddy currents as a whole do not transfer charge through the cross section of the conductor, the solution of such differential equations will always contain two oppositely directed components, one of which is simply discarded in the existing concept. A modification of the original Bessel equation is, as indicated above in subsection II.B, its artificial transformation into an equation for directional current by changing the sign of one of the terms. The directional current, of course, cannot spontaneously arise as a result of solving the original Bessel equation, which was derived for closed eddy currents. Apparently, on the basis of the foregoing, we can conclude that differential equations are fundamentally inapplicable for an accurate description of the skin effect.

The current density on the surface in existing solutions for the case of the absence of external inductance is actually the DC density, independent of frequency. Hence, the physical picture of the current displacement to the surface due to its decrease in the inner layers was not reflected in the existing solution. That is, the existing physical explanation of the skin effect contradicts its own mathematical description. To mask this in the existing concept, the idea arose to show the dependence of surface current density versus frequency at a fixed total current, i.e., with a simultaneous increase in voltage and frequency, which ensures that the current remains unchanged and the surface current density increases. Of course, such methods cannot cause anything but regret.

The division in the existing concept of the phenomenon under consideration into the actual skin effect and the proximity effect looks contradictory and artificial. That is, it is proposed to choose only a "foreign" field for calculating the proximity effect and one's "own" field for calculating the skin effect from the formed general picture of the magnetic field, which is the sum of the magnetic fields of all wires. This completely contradicts the concept accepted in modern electrodynamics of taking into account the common field from different sources, which we were guided by when deriving our equations. In addition to taking into account the superposition of all magnetic fields, it is necessary to pay attention to the fact that the specific EMFs of both self-induction and mutual induction, acting at any point of the conductor, depends on the length of the lines covering the magnetic fluxes. Based on the principle of current flow along the path of least resistance, it can be assumed that the maximum specific EMF, which determines the value of the current corresponding to it, will be determined by the maximum possible value of the loop length due to the decrease in the influence of the dimensions of the transverse parts of the wires with an increase in the loop length. This means that when deriving equations for a specific section of a wire that is inhomogeneous along the length, all parts of the circuit with their actual dimensions must be taken into account, as we did when deriving equations for a coaxial cable. When deriving the equations of a circuit with a
series-connected capacitor, we simplified the task by taking the transverse dimensions of the capacitor equal to the dimensions of the conductor, although, of course, this cannot be done for a real capacitor. To summarize: to obtain accurate equations for a part of a conductor, it is necessary to consider not only all the magnetic fields in this part, both own and neighboring wires, but also all the series-connected circuit elements with their magnetic fields, real sizes and other parameters. The existing concept of the skin effect as a division into the skin effect proper and the proximity effect does not take this into account, in contrast to the formulas obtained by us, from which the dependence of the active AC resistance of a coaxial cable on the distance between its wires follows.

The representation of the skin effect as a consequence of the attenuation of the amplitude of the incident wave when it penetrates the conductor is also not correct, since it is based on the same incorrect statement about the dissipation of energy due to the action of eddy currents. The very involvement of the wave theory when considering the skin effect is completely contrived. This is especially true for the case when the current is supplied by contact to the conductor. To consider a conductor as if receiving its own radiation, which also hardly penetrates into it, is generally a permutation of cause and effect. It can be said that the external radiation most likely freely penetrates the conductor, causing the appearance of a field strength over the entire depth, identical to the field strength from a contact power source. Figuratively speaking, all the rough work on the current density distribution, as shown in this article, is done by the distance-dependent EMF of self-induction of the wire. This statement fundamentally contradicts the currently accepted (actually since 1883) concept of the skin effect in the form of a consequence of difficulties in the penetration of waves into a conducting substance. As far as we know, the skin effect has never been considered analytically, i.e. in mathematical form, on the basis of the basic laws of Faraday, Ampère and Kirchhoff. The first consideration of the skin effect was apparently made in an article by H. Lamb [6], which theoretically investigated currents in a sphere located in an electromagnetic field. This study was largely motivated by a desire to prove the correctness of Maxwell's theory, as is evident from the quotation: "The investigation was undertaken some time ago in illustration of Maxwell's theory of Electricity. This theory is so remarkable..." [6, p. 519]. One of the confirmations of this then new theory was considered to be the skin effect, which at that time had not yet received its modern name. In the work of H. Lamb, the skin effect was immediately interpreted as a decrease in the amplitude of the incident wave as it penetrates the conductor, as can be seen from the quotation: "It appears from this that the disturbance inside the sphere consists of a series of waves propagated inwards from the surface with rapidly decreasing amplitude." [6, p. 536]. Since then, this has not been questioned, although it has been quite clear that all the methods that legitimize this statement in the form of discarding the divergent part of the solution or changing (modifying) the Bessel equation itself are not strict, but are justified by some "physical considerations". A physically correct explanation of the skin effect at a qualitative level was found in Ref. [7, p. 30]: "This action, termed *skin effect*, is a result of magnetic flux lines that circle part but not all of the conductor. Those parts of the cross section which are circled by the largest number of flux lines have higher inductance than other parts of the conductor, and hence a greater reactance.". That is, in official electrodynamics, not everything is so simple, if those who deal with applied issues do not even mention eddy currents when explaining the skin effect. It is true that eddy currents are mentioned in the paragraph above in Ref. [7, p. 30], but as one of the causes of the increase in AC versus DC losses, along with the skin effect, not as its cause. Recall that our proposed model for accounting eddy currents leads not to an increase but to a decrease in the consumption of active power by the power supply.

It is also possible to substantiate the inconsistency of the concept of attenuation when fields penetrate the conductor from the outside with the help of mathematics. To do this, it is enough to integrate the original (not modified) differential equations of the existing concept twice. As a result, a term will appear in the equation for the desired current density, equal to the ratio of the field strength from the applied (no matter how) supply voltage to the resistivity, namely the DC density. That is, we come to the eddy-free integral equations obtained in this article. The DC density is a constant (distance-independent) part of the integral equation, present at any depth, and not just on the surface. It turns out that the field strength from the power source is present at every point of the
conductor. Together with the terms responsible for the EMF of the self-induction of the wire and the surrounding space, it fully participates in the formation of the final picture of the current density distribution over depth. And, finally, if it is experimentally confirmed that for a tubular wire the AC resistance can be less than for a solid wire of the same diameter (for a solenoid, the resistance of thin wires compared to thick ones), then this will be an experimental refutation of the concept of the skin effect as attenuation when waves pass through a conductor. According to the logic of the penetration concept, the thicker the sample, the more it should absorb the energies of the radiation penetrating into it, and the greater should be the current through it. But if the current can be greater for a thinner sample, then the whole concept of penetration fails. This conclusion has nothing to do with our results, but is in fact the result of yet another contradiction in the existing approach itself. The total current of the solution for a semi-infinite space is not a strictly increasing function of the "thickness" *t* of the semi-infinite space, but has a pronounced maximum at $\delta/t \approx 0.438$, upon reaching which the total current, oscillating, drops to a certain value, which is about 0.935 of the previously reached maximum, and stabilizes. Thus, the concept of the skin effect as wave attenuation is refuted not even by the presence of minimum resistance at a certain thickness in the solution for a finite-thickness plate, but by the presence of minimum resistance in the most popular solution for a semi-infinite space, this is what is used in the sources to illustrate the concept of wave attenuation.

In the model of the phenomenon proposed in this article, we believe that everything falls into place. We also believe that the patterns obtained should be tested experimentally to determine, first of all, the resistance of the coaxial cable and the presence or absence of eddy currents. All criteria for this are given in the course of the presentation. We hope that our work will arouse interest in the scientific community and experiments will not be long in coming. We wish success to all experimenters who will respond to our call. Well, let's wish theorists to painlessly accept the new, and in fact the old, pre-Maxwellian reality in relation to solving the problem, so far only of the skin effect.

## ACKNOWLEDGEMENTS

The authors express their deep gratitude to their mother and grandmother Nadezhda Ivanovna Maslova for financial and moral support, to Tatiana Baramzina (spouse of author I. Maslov) and Evgeniya Semynina (spouse of author D. Maslov) for help in translating the article into English.

## EXPLANATIONS FOR APPENDICES

All 26 appendices are formed by taking screenshots from symbolic and digital calculations, the files of which are written in the MathCAD 15 environment, with subsequent graphic processing. Careful layout work was done to squeeze the bulky results into the A4 book format, while maintaining readability and logical integrity. This presentation of appendices allows you to check them in other programs, adapting them. In symbolic appendices 1, 8, 9, 14, 17, 18 and 19, functions, variables, and program commands are in italics, and explanatory text is in roman type. In order to save space, voluminous repeating fragments associated with the subsequent assignment of a variable to its previously obtained symbolic value have been removed. Vertical blocks of commands were edited by transferring their parts to the next line, also parts of long expressions were transferred to the next line, which, as a rule, were numerators and denominators of fractions. In such places we would put one or two red asterisks and give comments in red font at the end of the appendix as a footnote. Also, parts of the expressions after the operational arrow were transferred to the next line with the repetition of the arrow written on it. Such places were not commented on, because, without this, everything, we hope, is clear. We hope that these edits did not greatly contradict the rules of writing mathematical expressions, but their readability and fit into the A4 book format were worth it. The remaining appendices containing numerical calculations, initial data, graphs, vector and power diagrams are made in the usual plain font and without textual explanations, since everything, we hope, is quite understandable. The cumbersome formulas by which the calculation was made are not given in the text of the appendices to save space, since they are in the text of the article.
APPENDIX 1: Search and verification of solutions for cylindrical solid wire

Finding a solution to an equation :

$$
j(r) = j_{DC} + q^2 \cdot k^2 \cdot \int_{r}^{a} \frac{1}{r} \cdot \int_{b}^{r} j(r) \cdot r \, dr \, dr - p \cdot k^2 \cdot \int_{b}^{a} \frac{1}{r} \cdot \int_{b}^{r} j(r) \cdot r \, dr \, dr + k_{s}^{2} \cdot \ln(s) \cdot \int_{b}^{a} j(r) \cdot r \, dr
$$

$$
\text{where } q^2 = 1 + 2p
$$

$$
\text{where } q^2 = 1 + 2p \\
\text{we denote the constant terms: } -p \cdot k^2 \cdot \int_{b}^{a} \frac{1}{r} \cdot \int_{b}^{r} j(r) \cdot r \, dr \, dr = j_{W} \cdot k_{s}^{2} \cdot \ln(s) \cdot \int_{b}^{a} j(r) \cdot r \, dr = j_{S}
$$

$$
\text{When } b = 0 \text{ we represent the solution in the form: } j(r) = \frac{j_{DC} + j_{W} + j_{S}}{J_{0}(a \cdot q \cdot k)} \cdot J_{0}(r \cdot q \cdot k)
$$

$$
\text{Table integrals: } \int_{0}^{r} J0(r \cdot q \cdot k) \cdot r \, dr = \frac{r}{q \cdot k} \cdot J1(r \cdot q \cdot k) \qquad \int_{0}^{a} J0(r \cdot q \cdot k) \cdot r \, dr = \frac{a}{q \cdot k} \cdot J1(a \cdot q \cdot k)
$$

Finding the values of constants jW and jS:

$$
\begin{align*}
& -p \cdot k^2 \cdot \int_0^a \frac{1}{r} \cdot \int_0^r j(r) \cdot r \, dr \, dr = j_W \\
& \rightarrow \frac{p \cdot (J0(a \cdot k \cdot q) - 1) \cdot (j_S + j_{DC})}{p - p \cdot J0(a \cdot k \cdot q) + q^2 \cdot J0(a \cdot k \cdot q)}
\end{align*}
\[
\text{Then let: } j_{WI} := \frac{p \cdot (J0(a \cdot k \cdot q) - 1) \cdot (j_S + j_{DC})}{p - p \cdot J0(a \cdot k \cdot q) + q^2 \cdot J0(a \cdot k \cdot q)}
\]
$$

$$
\begin{align*}
& k_s^2 \cdot \ln(s) \cdot \int_0^a j(r) \cdot r \, dr = j_S \\
& \rightarrow \frac{a \cdot k_s^2 \cdot \ln(s) \cdot J1(a \cdot k \cdot q) \cdot (j_W + j_{DC})}{k \cdot q \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot \ln(s) \cdot J1(a \cdot k \cdot q)} \quad \text{Then let: } j_{S1} := \frac{a \cdot k_s^2 \cdot \ln(s) \cdot J1(a \cdot k \cdot q) \cdot (j_W + j_{DC})}{k \cdot q \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot \ln(s) \cdot J1(a \cdot k \cdot q)} \\
& \quad \left( \begin{array}{@{}l@{}}
    \left. \begin{array}{l} \text{simplify, } j_S \cdot j_W \\ \text{simplify} \end{array} \right> \\
    \left. \begin{array}{l} a \cdot j_{DC} \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q) \\ k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q) \end{array} \right\rangle \\
    \left. \begin{array}{l} j_{DC} \cdot k \cdot p \cdot (J0(a \cdot k \cdot q) - 1) \\ k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q) \end{array} \right]
\end{array} \right)
\end{align*}
\[
j_S := \frac{a \cdot j_{DC} \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q)}{k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q)}
\]
\[
j_W := \frac{j_{DC} \cdot k \cdot p \cdot (J0(a \cdot k \cdot q) - 1)}{k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q)}
\]
$$

$$
\text{Substituting } j_W \text{ and } j_S \text{ into: } j(r) = (j_{DC} + j_W + j_S) \cdot \frac{J0(r \cdot q \cdot k)}{J0(a \cdot q \cdot k)} \quad \text{simplify} \rightarrow
$$

$$
\rightarrow j(r) = \frac{j_{DC} \cdot k \cdot q^2 \cdot J0(k \cdot q \cdot r)}{k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q)!}
$$
APPENDIX 1 (continued)

Testing a solution:

$$j(r) = A \cdot J0(r \cdot k \cdot q) \text{ where: } A := \frac{j_{DC} + jW + jS}{J0(a \cdot k \cdot q)} \quad \text{Substituting the solution into the expression :}$$

$$-j(r) + j_{DC} + q^2 \cdot k^2 \cdot \int_{r}^{a} \frac{1}{r} \cdot \int_{0}^{r} j(r) \cdot r \, dr \, dr - p \cdot k^2 \cdot \int_{0}^{a} \frac{1}{r} \cdot \int_{0}^{r} j(r) \cdot r \, dr \, dr + k_s^2 \cdot \ln(s) \cdot \int_{0}^{a} j(r) \cdot r \, dr$$

we get:

$$-A \cdot J0(r \cdot k \cdot q) + j_{DC} + q^2 \cdot k^2 \cdot \int_{r}^{a} \frac{1}{r} \cdot \int_{0}^{r} A \cdot J0(r \cdot k \cdot q) \cdot r \, dr \, dr - p \cdot k^2 \cdot \int_{0}^{a} \frac{1}{r} \cdot \int_{0}^{r} A \cdot J0(r \cdot k \cdot q) \cdot r \, dr \, dr +$$

$$+ k_s^2 \cdot \ln(s) \cdot \int_{0}^{a} A \cdot J0(r \cdot k \cdot q) \cdot r \, dr$$

$$\text{Substituting } \int_{0}^{r} A \cdot J0(r \cdot k \cdot q) \cdot r \, dr = A \cdot \frac{r}{q \cdot k} \cdot J1(r \cdot q \cdot k) \int_{0}^{a} A \cdot J0(r \cdot k \cdot q) \cdot r \, dr = A \cdot \frac{a}{q \cdot k} \cdot J1(a \cdot q \cdot k) \\ \text{we get:}$$

$$-A \cdot J0(r \cdot k \cdot q) + j_{DC} + q^2 \cdot k^2 \cdot \int_{r}^{a} \frac{1}{r} \cdot A \cdot \frac{r}{q \cdot k} \cdot J1(r \cdot q \cdot k) \, dr - p \cdot k^2 \cdot \int_{0}^{a} \frac{1}{r} \cdot A \cdot \frac{r}{q \cdot k} \cdot J1(r \cdot q \cdot k) \, dr +$$

$$+ k_s^2 \cdot \ln(s) \cdot A \cdot \frac{a}{q \cdot k} \cdot J1(a \cdot q \cdot k) \xrightarrow{\text{simplify} \to 0!}$$

$$E := j_{DC} \cdot \rho \qquad j(r) := \frac{j_{DC} \cdot k \cdot q^2 \cdot J0(k \cdot q \cdot r)}{k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q)}$$

Then the balance of voltages and EMF is $j(r) \cdot \rho = E + E_{ab}(r) + E_s + E_w$

or $j(r) \cdot \rho = E + E_a(r) + E_s + E_e(r)$ where

$$E_s = \rho \cdot k_s^2 \cdot \ln(s) \cdot A \cdot \frac{a}{q \cdot k} \cdot J1(a \cdot q \cdot k) \xrightarrow{\text{simplify} \to} E_s := \frac{a \cdot \rho \cdot j_{DC} \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot q \cdot k)}{\left[ k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q) \right]^*}$$

$$E_{ab}(r) = \rho \cdot q^2 \cdot k^2 \cdot \int_{r}^{a} \frac{1}{r} \cdot A \cdot \frac{r}{q \cdot k} \cdot J1(r \cdot q \cdot k) \, dr \xrightarrow{\text{simplify} \to} E_{ab}(r) := \frac{\rho \cdot j_{DC} \cdot k \cdot q^2 \cdot (J0(a \cdot k \cdot q) - J0(k \cdot q \cdot r))}{\left[ k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q) \right]^*}$$

$$E_w = -\rho \cdot p \cdot k^2 \cdot \int_{0}^{a} \frac{1}{r} \cdot A \cdot \frac{r}{q \cdot k} \cdot J1(r \cdot q \cdot k) \, dr \xrightarrow{\text{simplify} \to} E_w := \frac{\rho \cdot j_{DC} \cdot k \cdot p \cdot (J0(a \cdot k \cdot q) - 1)}{k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q)}$$

$$E_e(r) := E_{ab}(r) + E_w - \frac{E_{ab}(r)}{q^2} \qquad E_a(r) := \frac{E_{ab}(r)}{q^2}$$

$$E_a(r) \xrightarrow{\text{simplify} \to} - \frac{\rho \cdot j_{DC} \cdot k \cdot (J0(a \cdot k \cdot q) - J0(k \cdot q \cdot r))}{k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q)}$$

$$E_e(r) \Big|_{\text{factor}} \xrightarrow{\text{simplify}} \frac{\rho \cdot j_{DC} \cdot k \cdot (q^2 \cdot J0(k \cdot q \cdot r) - p + p \cdot J0(a \cdot k \cdot q) + J0(a \cdot k \cdot q) - J0(k \cdot q \cdot r) - q^2 \cdot J0(a \cdot k \cdot q))}{k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q)}$$

$$-j(r) \cdot \rho + E + E_{ab}(r) + E_s + E_w \xrightarrow{\text{simplify} \to 0}$$

$$-j(r) \cdot \rho + E + E_a(r) + E_s + E_e(r) \xrightarrow{\text{simplify} \to 0}$$

Finding a solution for the eddy-free equation at: $p := 0$

$$q := \sqrt{2 \cdot p + 1} \qquad q = 1$$

$$j_1(r) = j_{DC} + k^2 \cdot \int_{r}^{a} \frac{1}{r} \cdot \int_{b}^{r} j_1(r) \cdot r \, dr \, dr + k_s^2 \cdot \ln(s) \cdot \int_{b}^{a} j_1(r) \cdot r \, dr$$

$$j_1(r) = \frac{j_{DC} \cdot k \cdot q^2 \cdot J0(k \cdot q \cdot r)}{k \cdot p + k \cdot q^2 \cdot J0(a \cdot k \cdot q) - k \cdot p \cdot J0(a \cdot k \cdot q) - a \cdot k_s^2 \cdot q \cdot \ln(s) \cdot J1(a \cdot k \cdot q)} \xrightarrow{\text{simplify} \to} j_1(r) = \frac{j_{DC} \cdot k \cdot J0(k \cdot r)}{k \cdot J0(a \cdot k) - a \cdot k_s^2 \cdot \ln(s) \cdot J1(a \cdot k)}!$$

<footnote>*Long expressions have been edited by moving the terms to the next line and placing them in square brackets.</footnote>
APPENDIX 2: Graphs of eddy-free and eddy solutions for a solid cylinder without external induc-
tance at large and small skin depths

$$
\begin{align*}
& \mathbf{f} \equiv 1 \cdot 10^2 \qquad \mathbf{\omega} \equiv 2\pi\mathbf{f} \qquad \mathbf{E} \equiv 0.11508 \, \mathbf{s} \equiv 1.0 \qquad \mathbf{k}_a \equiv 0.1 \quad 1 \equiv 1000 \rho \equiv 2.877 \times 10^{-8} \, \mathbf{a} \equiv 3.338 \times 10^{-3} \, \mathbf{S} \equiv \pi \cdot \mathbf{a}^2 \\
& \mu_T \equiv 1 \qquad \mu \equiv \mu_0 \cdot \mu_T \qquad \mu_{sr} \equiv 10^0 \qquad \mu_s \equiv \mu_0 \cdot \mu_s \qquad j_{DC} \equiv \frac{E}{\rho} R_{DC} \equiv \rho \cdot \frac{1}{S} \qquad \delta \equiv \sqrt{\frac{2\rho}{\mu \cdot \omega}} \qquad k \equiv \sqrt{\frac{-i \cdot \mu \cdot \omega}{\rho}} \qquad k_s \equiv \sqrt{\frac{-i \cdot \mu_s \cdot \omega}{\rho}} \\
& \mathbf{p} \equiv 0.5 \qquad \mathbf{q} \equiv \sqrt{1 + 2 \cdot \mathbf{p}} \qquad \mathbf{n} \equiv \mathbf{s} \cdot \mathbf{a} \qquad \mathbf{b} \equiv k_b \cdot \mathbf{a} \qquad \mathbf{j}_{DC} \equiv \frac{E}{\rho} R_{DC} \equiv \rho \cdot \frac{1}{S} \qquad \delta \equiv \sqrt{\frac{2\rho}{\mu \cdot \omega}} \qquad k \equiv \sqrt{\frac{-i \cdot \mu \cdot \omega}{\rho}} \qquad k_s \equiv \sqrt{\frac{-i \cdot \mu_s \cdot \omega}{\rho}} \\
& b1 \equiv k_a \cdot \mathbf{a} \quad \mathbf{S} \equiv 3.5 \times 10^{-5} \qquad n \equiv 3.338 \times 10^{-3} \qquad \frac{\delta}{a} \equiv 2.557 \qquad \delta \equiv 8.537 \times 10^{-3} \qquad r \equiv b1, b1 + \frac{a - b1}{100} \quad \mathbf{a}
\end{align*}
$$

$$
\mu_T = 1 \quad \mu = \mu_0 \cdot \mu_T \quad \mu_{sr} = 10^0 \quad \mu_s = \mu_0 \cdot \mu_s \quad j_{\text{DC}} = \frac{E}{\rho} R_{\text{DC}} = \rho \cdot \frac{1}{S} \quad \delta = \sqrt{\frac{2\rho}{\mu \cdot \omega}} \quad k = \sqrt{\frac{-i \cdot \mu \cdot \omega}{\rho}} \quad k_s = \sqrt{\frac{-i \cdot \mu_s \cdot \omega}{\rho}}
$$

$$
b1 = k_a \cdot a \quad S = 3.5 \times 10^{-5} \qquad n = 3.338 \times 10^{-3} \qquad \frac{\delta}{a} = 2.557 \qquad \delta = 8.537 \times 10^{-3} \qquad r = b1, b1 + \frac{a - b1}{100} \quad \mathbf{a}
$$

$$
|j_1(\mathbf{b})| = 3.99 \times 10^6 \quad |j_1(\mathbf{a})| = 4 \times 10^6 \quad |I_1| = 139.85 \quad \arg(I_1) = -0.04 \quad \arg(j_1(\mathbf{b})) = -0.08 \quad \arg(j_1(\mathbf{a})) = 0
$$

$$
|j(\mathbf{b})| = 3.99 \times 10^6 |j(\mathbf{a})| = 4.01 \times 10^6 |I| = 139.85 \arg(I) = -0.04 \arg(j(\mathbf{b})) = -0.11 \arg(j(\mathbf{a})) = 0.04
$$

$$
b1 = 0 \qquad I_{\text{DC}} = 140.02 \qquad j_{\text{DC}} = 4 \times 10^6 \qquad R_{\text{DC}} = 0.8219 \quad R_1 = 0.8223 \qquad R = 0.8223
$$

<figure><img src="image_10.png" /></figure>

$$
\begin{align*}
& \mathbf{f} \equiv 1 \cdot 10^8 \qquad \mathbf{\omega} \equiv 2\pi\mathbf{f} \qquad \mathbf{E} \equiv 0.11508 \, \mathbf{s} \equiv 1.0 \qquad \mathbf{k}_a \equiv 0.991 \quad 1 \equiv 1000 \rho \equiv 2.877 \times 10^{-8} \, \mathbf{a} \equiv 3.338 \times 10^{-3} \, \mathbf{S} \equiv \pi \cdot \mathbf{a}^2 \\
& \mu_T \equiv 1 \qquad \mu \equiv \mu_0 \cdot \mu_T \qquad \mu_{sr} \equiv 10^0 \qquad \mu_s \equiv \mu_0 \cdot \mu_s \qquad j_{DC} \equiv \frac{E}{\rho} R_{DC} \equiv \rho \cdot \frac{1}{S} \qquad \delta \equiv \sqrt{\frac{2\rho}{\mu \cdot \omega}} \qquad k \equiv \sqrt{\frac{-i \cdot \mu \cdot \omega}{\rho}} \qquad k_s \equiv \sqrt{\frac{-i \cdot \mu_s \cdot \omega}{\rho}}
\end{align*}
$$

$$
\mu_T = 1 \quad \mu = \mu_0 \cdot \mu_T \quad \mu_{sr} = 10^0 \quad \mu_s = \mu_0 \cdot \mu_s \quad j_{\text{DC}} = \frac{E}{\rho} R_{\text{DC}} = \rho \cdot \frac{1}{S} \quad \delta = \sqrt{\frac{2\rho}{\mu \cdot \omega}} \quad k = \sqrt{\frac{-i \cdot \mu \cdot \omega}{\rho}} \quad k_s = \sqrt{\frac{-i \cdot \mu_s \cdot \omega}{\rho}}
$$

$$
b1 = k_a \cdot a \quad S = 3.5 \times 10^{-5} \qquad n = 3.338 \times 10^{-3} \qquad \frac{\delta}{a} = 2.557 \times 10^{-3} \qquad \delta = 8.537 \times 10^{-6} \qquad r = b1, b1 + \frac{a - b1}{100} \quad \mathbf{a}
$$

$$
|j_1(\mathbf{b})| = 0 \quad |j_1(\mathbf{a})| = 4 \times 10^6 \quad |I_1| = 0.51 \quad \arg(I_1) = -0.78 \quad \arg(j_1(\mathbf{b})) = -1.07 \quad \arg(j_1(\mathbf{a})) = 0
$$

$$
|j(\mathbf{b})| = 0 |j(\mathbf{a})| = 5.33 \times 10^6 |I| = 0.48 \arg(I) = -0.78 \arg(j(\mathbf{b})) = 0.33 \arg(j(\mathbf{a})) = 0
$$

$$
b1 = 3.305 \times 10^{-3} \qquad I_{\text{DC}} = 140.02 \qquad j_{\text{DC}} = 4 \times 10^6 \qquad R_{\text{DC}} = 0.8219 \quad R_1 = 160.89373 \qquad R = 170.58963
$$

<figure><img src="image_11.png" /></figure>
APPENDIX 3: Graphs of eddy-free and purely eddy solutions for a solid cylinder without external inductance at large and small skin depths

$$
\begin{align*}
& f \equiv 1 \cdot 10^2 \qquad \omega \equiv 2\pi f \qquad E \equiv 0.11508 \qquad s \equiv 1 \qquad k_a \equiv 0, \quad 1 \equiv 1000 \qquad \rho \equiv 2.877 \times 10^{-8} \qquad a \equiv 3.338 \times 10^{-3} \qquad S \equiv \pi \cdot a^2 \\
& \mu_r \equiv 1 \qquad \mu \equiv \mu_0 \cdot \mu_r \qquad \mu_{sr} \equiv 10^0 \qquad \mu_s \equiv \mu_0 \cdot \mu_s \qquad j_{DC} \equiv \frac{E}{\rho} R_{DC} \equiv \rho \cdot \frac{1}{S} \qquad \delta \equiv \sqrt{\frac{2\rho}{\mu \cdot \omega}} \qquad k \equiv \sqrt{\frac{-i \cdot \mu \cdot \omega}{\rho}} \qquad k_s \equiv \sqrt{\frac{-i \cdot \mu_s \cdot \omega}{\rho}} \\
& p \equiv 0.5 \qquad q \equiv \sqrt{0 + 2 \cdot p} \qquad n \equiv s \cdot a \qquad b \equiv k_b \cdot a \\
& b1 \equiv k_a \cdot a \qquad S \equiv 3.5 \times 10^{-5} \qquad n \equiv 3.338 \times 10^{-3} \qquad \frac{\delta}{a} \equiv 2.557 \qquad \delta \equiv 8.537 \times 10^{-3} \qquad r \equiv b1, b1 + \frac{a - b1}{100} \quad \dots \text{a} \\
& |j_1(b)| \equiv 3.99 \times 10^6 \qquad |j_1(a)| \equiv 4 \times 10^6 \qquad |I_1| \equiv 139.85 \qquad \arg(j_1(b)) \equiv -0.04 \qquad \arg(j_1(a)) \equiv 0 \\
& |j_1(b)| \equiv 4 \times 10^6 \qquad |j_1(a)| \equiv 4.01 \times 10^6 \qquad |I_1| \equiv 140.05 \qquad \arg(I_1) \equiv -6.2 \times 10^{-6} \qquad \arg(j_1(b)) \equiv -0.04 \arg(j_1(a)) \equiv 0.04 \\
& b1 \equiv 0 \qquad I_{DC} \equiv 140.02 \qquad j_{DC} \equiv 4 \times 10^6 \qquad R_{DC} \equiv 0.8218969 \qquad R_1 \equiv 0.8223 \qquad R \equiv 0.8217
\end{align*}
$$

<figure><img src="image_12.png" /></figure>

$$
\begin{align*}
& f \equiv 1 \cdot 10^8 \qquad \omega \equiv 2\pi f \qquad E \equiv 0.11508 \qquad s \equiv 1 \qquad k_a \equiv 0.991 \equiv 1000 \qquad \rho \equiv 2.877 \times 10^{-8} \qquad a \equiv 3.338 \times 10^{-3} \qquad S \equiv \pi \cdot a^2 \\
& \mu_r \equiv 1 \qquad \mu \equiv \mu_0 \cdot \mu_r \qquad \mu_{sr} \equiv 10^0 \qquad \mu_s \equiv \mu_0 \cdot \mu_s \qquad j_{DC} \equiv \frac{E}{\rho} R_{DC} \equiv \rho \cdot \frac{1}{S} \qquad \delta \equiv \sqrt{\frac{2\rho}{\mu \cdot \omega}} \qquad k \equiv \sqrt{\frac{-i \cdot \mu \cdot \omega}{\rho}} \qquad k_s \equiv \sqrt{\frac{-i \cdot \mu_s \cdot \omega}{\rho}} \\
& p \equiv 0.5 \qquad q \equiv \sqrt{0 + 2 \cdot p} \qquad n \equiv s \cdot a \qquad b \equiv k_b \cdot a \\
& b1 \equiv k_a \cdot a \qquad S \equiv 3.5 \times 10^{-5} \qquad n \equiv 3.338 \times 10^{-3} \qquad \frac{\delta}{a} \equiv 2.557 \times 10^{-3} \qquad \delta \equiv 8.537 \times 10^{-6} \qquad r \equiv b1, b1 + \frac{a - b1}{100} \quad \dots \text{a} \\
& |j_1(b)| \equiv 0 \qquad |j_1(a)| \equiv 4 \times 10^6 \qquad |I_1| \equiv 0.51 \qquad \arg(j_1(b)) \equiv -0.78 \qquad \arg(j_1(a)) \equiv 0 \\
& |j_1(b)| \equiv 0 \qquad |j_1(a)| \equiv 8 \times 10^6 \qquad |I_1| \equiv 1.01 \qquad \arg(I_1) \equiv -0.78 \qquad \arg(j_1(b)) \equiv -1.07 \arg(j_1(a)) \equiv 0 \\
& b1 \equiv 3.305 \times 10^{-3} \qquad I_{DC} \equiv 140.02 \qquad j_{DC} \equiv 4 \times 10^6 \qquad R_{DC} \equiv 0.8218969 \qquad R_1 \equiv 160.89373 \qquad R \equiv 80.44686
\end{align*}
$$

<figure><img src="image_13.png" /></figure>
APPENDIX 4: Graphs of eddy-free and eddy solutions for a solid cylinder in an iron shell and in air

$$
\begin{align*}
& f \equiv 1 \cdot 10^{-4} \qquad \omega \equiv 2\pi f \qquad E \equiv 0.11508 \qquad s \equiv 2.0 \qquad k_a \equiv 0.1 \qquad 1 \equiv 1000 \rho \equiv 2.877 \times 10^{-8} a \equiv 3.338 \times 10^{-3} S \equiv \pi \cdot a^2 \\
& \mu_T \equiv 1 \qquad \mu \equiv \mu_0 \cdot \mu_T \qquad \mu_{sr} \equiv 10^3 \qquad \mu_s \equiv \mu_0 \cdot \mu_s \\
& p \equiv 0.5 \qquad q \equiv \sqrt{1 + 2 \cdot p} \qquad n \equiv s \cdot a \qquad b \equiv k_b \cdot a \\
& j_{DC} \equiv \frac{E}{\rho} R_{DC} \equiv \rho \cdot \frac{1}{S} \qquad \delta \equiv \sqrt{\frac{2\rho}{\mu \cdot \omega}} \qquad k \equiv \sqrt{\frac{-i \cdot \mu \cdot \omega}{\rho}} \qquad k_s \equiv \sqrt{\frac{-i \cdot \mu_s \cdot \omega}{\rho}} \\
& b1 \equiv k_a \cdot a \qquad S = 3.5 \times 10^{-5} \qquad n = 6.676 \times 10^{-3} \qquad \frac{\delta}{a} = 0.256 \qquad \delta = 8.537 \times 10^{-4} \qquad r \equiv b1, b1 + \frac{a - b1}{100} \quad \therefore a
\end{align*}
$$

$$
\begin{align*}
|j_1(\text{b})| &= 129.22 \quad |j_1(\text{a})| = 1.11 \times 10^3 \quad |I_1| = 0.01 \quad \arg(j_1(\text{b})) = -1.57 \quad \arg(j_1(\text{a})) = -0.86 \\
|j_1(\text{b})| &= 42.43 \quad |j_1(\text{a})| = 1.54 \times 10^3 \quad |I_1| = 0.01 \quad \arg(j_1(\text{b})) = 0.32 \quad \arg(j_1(\text{a})) = -0.83 \\
b1 = 0 \qquad I_{\text{DC}} = 140.02 \qquad j_{\text{DC}} = 4 \times 10^6 \qquad R_{\text{DC}} = 0.8219 \quad R_1 = 1.832 \qquad R = 1.861
\end{align*}
$$

<figure><img src="image_14.png" /></figure>

$$
\begin{align*}
& f \equiv 1 \cdot 10^{-4} \quad \omega \equiv 2\pi f \quad E \equiv 0.11508 \quad s \equiv 2.0 \quad k_a \equiv 0.1 \quad 1 \equiv 1000 \rho \equiv 2.877 \times 10^{-8} a \equiv 3.338 \times 10^{-3} S \equiv \pi \cdot a^2 \\
& \mu_T \equiv 1 \quad \mu \equiv \mu_0 \cdot \mu_T \quad \mu_{sr} \equiv 10^0 \quad \mu_s \equiv \mu_0 \cdot \mu_s \\
& p \equiv 0.5 \quad q \equiv \sqrt{1 + 2 \cdot p} \quad n \equiv s \cdot a \quad b \equiv k_b \cdot a \\
& j_{DC} \equiv \frac{E}{\rho} R_{DC} \equiv \rho \cdot \frac{1}{S} \quad \delta \equiv \sqrt{\frac{2\rho}{\mu \cdot \omega}} \quad k \equiv \sqrt{\frac{-i \cdot \mu \cdot \omega}{\rho}} \quad k_s \equiv \sqrt{\frac{-i \cdot \mu_s \cdot \omega}{\rho}} \\
& b1 \equiv k_a \cdot a \quad S = 3.5 \times 10^{-5} \quad n = 6.676 \times 10^{-3} \quad \frac{\delta}{a} = 0.256 \quad \delta = 8.537 \times 10^{-4} \quad r \equiv b1, b1 + \frac{a - b1}{100} \quad \therefore a
\end{align*}
$$

$$
\begin{align*}
|j_1(\text{b})| &= 1.08 \times 10^5 |j_1(\text{a})| = 9.26 \times 10^5 |I_1| = 11.01 \arg(I_1) = -1.39 \arg(j_1(\text{b})) = 2.1 \arg(j_1(\text{a})) = -0.68 \\
|j_1(\text{b})| &= 3.49 \times 10^4 |j_1(\text{a})| = 1.27 \times 10^6 |I_1| = 10.87 \arg(I_1) = -1.39 \arg(j_1(\text{b})) = 0.5 \arg(j_1(\text{a})) = -0.66 \\
b1 = 0 \qquad I_{\text{DC}} = 4 \times 10^6 \qquad I_{\text{DC}} = 140.02 \qquad R_{\text{DC}} = 0.8219 \quad R_1 = 1.832 \qquad R = 1.861
\end{align*}
$$

<figure><img src="image_15.png" /></figure>
APPENDIX 5: Frequency dependences for currents, surface current densities and resistances in the absence of external inductance for eddy-free and eddy solutions for a solid cylinder

$$s = 1 \qquad \mu = 1.257 \times 10^{-6} \qquad l = 1 \times 10^3 R_{DC} = 0.8219 \qquad \mu_s = 1.257 \times 10^{-6} \qquad \rho = 2.877 \times 10^{-8} \qquad a = 3.338 \times 10^{-3} \\ \mathrm{d}m\mathrm{I}(\mathrm{f}) := |\mathrm{I}(\mathrm{f})| - |\mathrm{I}_1(\mathrm{f})| \qquad \mathrm{d}a\mathrm{I}(\mathrm{f}) := \arg(\mathrm{I}(\mathrm{f})) - \arg(\mathrm{I}_1(\mathrm{f}))$$

<figure><img src="image_16.png" /></figure>

<figure><img src="image_17.png" /></figure>

$$dmj(f) := \frac{|j_a(f)|}{|j_{1a}(f)|} \qquad dmj_{DC}(f) := \frac{|j_a(f)|}{j_{DC}}$$

$$dmj_{1DC}(f) := \frac{|j_{1a}(f)|}{j_{DC}} \qquad daj(f) := \arg(j_a(f)) - \arg(j_{1a}(f))$$

<figure><img src="image_18.png" /></figure>

<figure><img src="image_19.png" /></figure>

<figure><img src="image_20.png" /></figure>

<figure><img src="image_21.png" /></figure>

<figure><img src="image_22.png" /></figure>

<figure><img src="image_23.png" /></figure>

<figure><img src="image_24.png" /></figure>

$$dR(f) := R(f) - R_1(f)$$

$$dR(f) = \begin{bmatrix} -1.945 \cdot 10^{-7} \\ -4.883 \cdot 10^{-4} \\ -4.394 \cdot 10^{-3} \end{bmatrix}$$

<figure><img src="image_25.png" /></figure>

<figure><img src="image_26.png" /></figure>

$$dX(f) := X(f) - X_1(f)$$

<figure><img src="image_27.png" /></figure>
APPENDIX 6: Frequency dependences for currents, surface current densities and resistances in the presence of external inductance for eddy-free and eddy solutions for a solid cylinder

$$
\begin{array}{l}
s = 3 \qquad \mu = 1.257 \times 10^{-6} \quad 1 = 1 \times 10^3 R_{\text{DC}} = 0.8219 \quad \mu_s = 1.257 \times 10^{-6} \quad \rho = 2.877 \times 10^{-8} \quad a = 3.338 \times 10^{-3} \\
\text{dmI}(f) := |I(f)| - |I_1(f)| \qquad \text{d}\text{aI}(f) := \arg(I(f)) - \arg(I_1(f))
\end{array}
$$

<figure><img src="image_28.png" /></figure>

<figure><img src="image_29.png" /></figure>

$$
\mathrm{dmj}(f) := \frac{|j_a(f)|}{|j_{1a}(f)|} \qquad \mathrm{dmj}_{\mathrm{DC}}(f) := \frac{|j_a(f)|}{j_{\mathrm{DC}}} \qquad \mathrm{dmj}_{1\mathrm{DC}}(f) := \frac{|j_{1a}(f)|}{j_{\mathrm{DC}}} \qquad \mathrm{daj}(f) := \arg(j_a(f)) - \arg(j_{1a}(f))
$$

<figure><img src="image_30.png" /></figure>

<figure><img src="image_31.png" /></figure>

<figure><img src="image_32.png" /></figure>

<figure><img src="image_33.png" /></figure>

<figure><img src="image_34.png" /></figure>

<figure><img src="image_35.png" /></figure>

$$
\begin{array}{c}
\mathrm{dR}(f) = \\
\begin{array}{@{}c@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r}
-1.945 \cdot 10^{-7} & 9.351 \\
-4.883 \cdot 10^{-4} & 9.351 \\
-4.394 \cdot 10^{-3} & 9.351 \\
-0.013 & 9.351 \\
-0.025 & 9.352 \\
-0.036 & 9.352 \\
-0.043 & 9.352 \\
-0.045 & 9.352 \\
-0.042 & 9.352
\end{array}
\\
\text{dR}(f) =
\begin{array}{@{}c@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r}
-1.945 \cdot 10^{-7} & 9.351 \\
-4.883 \cdot 10^{-4} & 9.351 \\
-4.394 \cdot 10^{-3} & 9.351 \\
-0.013 & 9.351 \\
-0.025 & 9.352 \\
-0.036 & 9.352 \\
-0.043 & 9.352 \\
-0.045 & 9.352 \\
-0.042 & 9.352
\end{array}
\end{array}
$$

<figure><img src="image_36.png" /></figure>

$$
\begin{array}{c}
\mathrm{dX}(f) = \\
\begin{array}{@{}c@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r}
-2.54 \cdot 10^{-6} & 9.447 \\
-7.993 \cdot 10^{-4} & 9.447 \\
-2.943 \cdot 10^{-3} & 9.447 \\
-2.906 \cdot 10^{-3} & 9.447 \\
2.834 \cdot 10^{-3} & 9.447 \\
0.015 & 9.447 \\
0.031 & 9.447 \\
0.049 & 9.447 \\
0.067 & 9.447
\end{array}
\\
\text{dX}(f) =
\begin{array}{@{}c@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r@{\,}r}
-2.54 \cdot 10^{-6} & 9.447 \\
-7.993 \cdot 10^{-4} & 9.447 \\
-2.943 \cdot 10^{-3} & 9.447 \\
-2.906 \cdot 10^{-3} & 9.447 \\
2.834 \cdot 10^{-3} & 9.447 \\
0.015 & 9.447 \\
0.031 & 9.447 \\
0.049 & 9.447 \\
0.067 & 9.447
\end{array}
\end{array}
$$
APPENDIX 7: Comparison of resistances as a function of frequency for cylindrical copper wires for eddy-free and eddy solutions

<figure><img src="image_37.png" /></figure>

<figure><img src="image_38.png" /></figure>
APPENDIX 8 (continued)

$$
j_W=j_{W1} \quad \left|
\begin{array}{l}
\textbf{**} \\
\text{substitute, } \int_b^a \mathcal{I} \ell(r \cdot k \cdot i \cdot q) dr = \frac{1}{k \cdot i \cdot q} \cdot \mathcal{I} 0(a \cdot k \cdot i \cdot q) - \frac{1}{k \cdot i \cdot q} \cdot \mathcal{I} 0(b \cdot k \cdot i \cdot q) \\
\text{substitute, } \int_b^a K \ell(r \cdot k \cdot i \cdot q) dr = \frac{-1}{k \cdot i \cdot q} \cdot K 0(a \cdot k \cdot i \cdot q) + \frac{1}{k \cdot i \cdot q} \cdot K 0(b \cdot k \cdot i \cdot q) \\
\text{substitute, } A_1 = \frac{K \ell(b \cdot k \cdot i \cdot q)}{\mathcal{I} 0(a \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(a \cdot k \cdot i \cdot q)} \\
\text{substitute, } A_2 = \frac{\mathcal{I} 1(b \cdot k \cdot i \cdot q)}{\mathcal{I} 0(a \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(a \cdot k \cdot i \cdot q)} \\
\text{p} \cdot (j_S + j_{DC}) \cdot \left[ \begin{array}{l} \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ - \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot K 0(b \cdot k \cdot q \cdot i) \end{array} \right]^*
\end{array}
\right| \quad
\begin{array}{l}
\text{substitute } j_{DCWS} = j_{DC} + j_W + j_S \\
\text{solve } j_W \\
\text{simplify} \\
\rightarrow \\
\text{p} \cdot (j_S + j_{DC}) \cdot \left[ \begin{array}{l} \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ - \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + \mathcal{I} 1(b \cdot k \cdot q \cdot i) + \\ + \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot K 0(b \cdot k \cdot q \cdot i) \end{array} \right]^*
\end{array}
$$

$$
\rightarrow j_{W2} := \frac{\left[ \begin{array}{l} q^2 \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + q^2 \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ -p \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - p \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) + \\ +p \cdot \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + p \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot K 0(b \cdot k \cdot q \cdot i) \end{array} \right]^*}{\left[ \begin{array}{l} q^2 \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + q^2 \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ -p \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - p \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) + \\ +p \cdot \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + p \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot K 0(b \cdot k \cdot q \cdot i) \end{array} \right]^*}
$$

$$
j_S = j_{S1} \quad \left|
\begin{array}{l}
\text{substitute, } j_W = j_{W2} \rightarrow j_{S2} := - \frac{i \cdot a \cdot k_5^2 \cdot \ln(s) \cdot q \cdot j_{DC} \cdot \left[ \begin{array}{l} (\mathcal{I} 1(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - \\ -K \ell(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i)) \end{array} \right]^*}{-k \cdot p \cdot \mathcal{I} 0(a \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) - k \cdot p \cdot K 0(a \cdot k \cdot i \cdot q) \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) + \\
\text{solve, } j_S \\
\text{simplify} \\
\text{factor} \qquad \left[ \begin{array}{l} +k \cdot p \cdot \mathcal{I} 0(b \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + k \cdot p \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(b \cdot k \cdot i \cdot q) + \\ +k \cdot q^2 \cdot \mathcal{I} 0(a \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + k \cdot q^2 \cdot K 0(a \cdot k \cdot i \cdot q) \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) - \\ -a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot K \ell(a \cdot k \cdot i \cdot q) \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot i + \\ +a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot \mathcal{I} 1(a \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) \cdot i \end{array} \right]^* \\
\text{factor} \qquad \left[ \begin{array}{l} -k \cdot p \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ -\mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot K 0(b \cdot k \cdot q \cdot i) \end{array} \right]^* \\
\text{substitute, } j_S = j_{S2} \quad j_{DC} \cdot p \cdot \left[ \begin{array}{l} \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ -\mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot K 0(b \cdot k \cdot q \cdot i) \end{array} \right]^* \\
\text{solve, } j_W \rightarrow j_W := \left[ \begin{array}{l} -k \cdot p \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - k \cdot p \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) + \\ +k \cdot p \cdot \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + k \cdot p \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(b \cdot k \cdot i \cdot q) + \\ +k \cdot q^2 \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot i \cdot q) + k \cdot q^2 \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) - \\ -a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot K \ell(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot i + \\ +a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot \mathcal{I} 1(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) \cdot i \end{array} \right]^* \\
\text{simplify} \qquad \left[ \begin{array}{l} -k \cdot p \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - k \cdot p \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) + \\ +k \cdot p \cdot \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + k \cdot p \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(b \cdot k \cdot i \cdot q) + \\ +k \cdot q^2 \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + k \cdot q^2 \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ -a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot K \ell(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot i + \\ +a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot \mathcal{I} 1(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) \cdot i \end{array} \right]^* \\
\text{j_S := j_{S2}} \quad j_0(r) := (j_{DC} + j_W + j_S) \cdot \frac{\mathcal{I} 0(r \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + K 0(r \cdot k \cdot i \cdot q) \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q)}{\mathcal{I} 0(a \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(a \cdot k \cdot i \cdot q)}
\end{array}
\right| \quad
$$

$$
j_W = j_{W2} \quad \left|
\begin{array}{l}
\text{substitute, } j_S = j_{S2} \quad j_{DC} \cdot k \cdot p \cdot \left[ \begin{array}{l} \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ -\mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot K 0(b \cdot k \cdot q \cdot i) \end{array} \right]^* \\
\text{solve, } j_W \rightarrow j_W := \left[ \begin{array}{l} -k \cdot p \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - k \cdot p \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) + \\ +k \cdot p \cdot \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + k \cdot p \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(b \cdot k \cdot i \cdot q) + \\ +k \cdot q^2 \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + k \cdot q^2 \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ -a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot K \ell(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot i + \\ +a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot \mathcal{I} 1(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) \cdot i \end{array} \right]^* \\
\text{simplify} \quad \left[ \begin{array}{l} -k \cdot p \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - k \cdot p \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) + \\ +k \cdot p \cdot \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + k \cdot p \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(b \cdot k \cdot i \cdot q) + \\ +k \cdot q^2 \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + k \cdot q^2 \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ -a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot K \ell(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot i + \\ +a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot \mathcal{I} 1(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) \cdot i \end{array} \right]^* \\
\text{j_S := j_{S2}} \quad j_0(r) := (j_{DC} + j_W + j_S) \cdot \frac{\mathcal{I} 0(r \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + K 0(r \cdot k \cdot i \cdot q) \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q)}{\mathcal{I} 0(a \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(a \cdot k \cdot i \cdot q)}
\end{array}
\right| \quad
$$

$$
j(r) = j_0(r) \text{simplify} \rightarrow j(r) = \left[ \begin{array}{l} j_{DC} \cdot k \cdot q^2 \cdot (\mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot K 0(a \cdot k \cdot q \cdot i) + K \ell(b \cdot k \cdot q \cdot i) \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i)) \\ -k \cdot p \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) - k \cdot p \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) + \\ +k \cdot p \cdot \mathcal{I} 0(b \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + k \cdot p \cdot \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(b \cdot k \cdot i \cdot q) + \\ +k \cdot q^2 \cdot \mathcal{I} 0(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) + k \cdot q^2 \cdot K 0(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) - \\ -a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot K \ell(a \cdot k \cdot q \cdot i) \cdot \mathcal{I} 1(b \cdot k \cdot q \cdot i) \cdot i + \\ +a \cdot k_5^2 \cdot q \cdot \ln(s) \cdot \mathcal{I} 1(a \cdot k \cdot q \cdot i) \cdot K \ell(b \cdot k \cdot q \cdot i) \cdot i \end{array} \right]!
$$

Testing a solution: $j(r) = j_{DCWS} \cdot A_1 \cdot \mathcal{I} 0(r \cdot k \cdot i \cdot q) + j_{DCWS} \cdot A_2 \cdot K 0(r \cdot k \cdot i \cdot q)$

where: $j_{DCWS} = j_{DC} + j_W + j_S$

$$
A_1 = \frac{K \ell(b \cdot k \cdot i \cdot q)}{\mathcal{I} 0(a \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(a \cdot k \cdot i \cdot q)}
$$

$$
A_2 = \frac{\mathcal{I} 1(b \cdot k \cdot i \cdot q)}{\mathcal{I} 0(a \cdot k \cdot i \cdot q) \cdot K \ell(b \cdot k \cdot i \cdot q) + \mathcal{I} 1(b \cdot k \cdot i \cdot q) \cdot K 0(a \cdot k \cdot i \cdot q)}
$$

Substituting the solution into the expression :

$$
zero1 = -j(r) + j_{DC} + q^2 \cdot k^2 \cdot \int_{r}^{a} \frac{1}{r} \cdot \int_{b}^{r} j(r) \cdot r dr dr - p \cdot k^2 \cdot \int_{b}^{a} \frac{1}{r} \cdot \int_{b}^{r} j(r) \cdot r dr dr + k_5^2 \cdot \ln(s) \cdot \int_{b}^{a} j(r) \cdot r dr
$$
APPENDIX 8 (continued)

$$
\text{we get: } \text{zero1} := -(j_{\text{DCWS}} \cdot A_1 \cdot I_0(r \cdot k \cdot i \cdot q) + j_{\text{DCWS}} \cdot A_2 \cdot K_0(r \cdot k \cdot i \cdot q)) + j_{\text{DC}} + \\
+ q^2 \cdot k^2 \cdot \int_r^a \frac{1}{r} \cdot \int_b^r (j_{\text{DCWS}} \cdot A_1 \cdot I_0(r \cdot k \cdot i \cdot q) + j_{\text{DCWS}} \cdot A_2 \cdot K_0(r \cdot k \cdot i \cdot q)) \cdot r \, dr \, dr - \\
- p \cdot k^2 \cdot \int_b^a \frac{1}{r} \cdot \int_b^r (j_{\text{DCWS}} \cdot A_1 \cdot I_0(r \cdot k \cdot i \cdot q) + j_{\text{DCWS}} \cdot A_2 \cdot K_0(r \cdot k \cdot i \cdot q)) \cdot r \, dr \, dr + \\
+ k_s^2 \cdot \ln(s) \cdot \int_b^a (j_{\text{DCWS}} \cdot A_1 \cdot I_0(r \cdot k \cdot i \cdot q) + j_{\text{DCWS}} \cdot A_2 \cdot K_0(r \cdot k \cdot i \cdot q)) \cdot r \, dr
$$

$$
\text{zero} = \text{zero1 expand} \rightarrow \text{zero2} := j_{\text{DC}} - A_1 \cdot j_{\text{DCWS}} \cdot I_0(k \cdot q \cdot r \cdot i) - A_2 \cdot j_{\text{DCWS}} \cdot K_0(k \cdot q \cdot r \cdot i) + \\
+ A_1 \cdot j_{\text{DCWS}} \cdot k_s^2 \cdot \ln(s) \cdot \int_b^a r \cdot I_0(k \cdot q \cdot r \cdot i) \, dr + A_2 \cdot j_{\text{DCWS}} \cdot k_s^2 \cdot \ln(s) \cdot \int_b^a r \cdot K_0(k \cdot q \cdot r \cdot i) \, dr + \\
+ A_1 \cdot j_{\text{DCWS}} \cdot k^2 \cdot q^2 \cdot \int_r^a \frac{\int_b^r r \cdot I_0(k \cdot q \cdot r \cdot i) \, dr}{r} + A_2 \cdot j_{\text{DCWS}} \cdot k^2 \cdot q^2 \cdot \int_r^a \frac{\int_b^r r \cdot K_0(k \cdot q \cdot r \cdot i) \, dr}{r} - \\
- A_1 \cdot j_{\text{DCWS}} \cdot k^2 \cdot p \cdot \int_b^a \frac{\int_r^a r \cdot I_0(k \cdot q \cdot r \cdot i) \, dr}{r} - A_2 \cdot j_{\text{DCWS}} \cdot k^2 \cdot p \cdot \int_b^a \frac{\int_r^a r \cdot K_0(k \cdot q \cdot r \cdot i) \, dr}{r} dr
$$

$$
\text{zero = zero2} \quad \left|
\begin{array}{l}
\text{substitute, } \int_b^a r \cdot I_0(r \cdot k \cdot i \cdot q) \, dr = \frac{a}{k \cdot i \cdot q} \cdot \mathrm{II}(a \cdot k \cdot i \cdot q) - \frac{b}{k \cdot i \cdot q} \cdot \mathrm{II}(b \cdot k \cdot i \cdot q) \\
\text{substitute, } \int_b^a r \cdot K_0(r \cdot k \cdot i \cdot q) \, dr = \frac{-a}{k \cdot i \cdot q} \cdot \mathrm{Kl}(a \cdot k \cdot i \cdot q) + \frac{b}{k \cdot i \cdot q} \cdot \mathrm{Kl}(b \cdot k \cdot i \cdot q) \\
\text{substitute, } \int_b^r r \cdot I_0(r \cdot k \cdot i \cdot q) \, dr = \frac{r}{k \cdot i \cdot q} \cdot \mathrm{II}(r \cdot k \cdot i \cdot q) - \frac{b}{k \cdot i \cdot q} \cdot \mathrm{II}(b \cdot k \cdot i \cdot q) \\
\text{substitute, } \int_b^r r \cdot K_0(r \cdot k \cdot i \cdot q) \, dr = \frac{-r}{k \cdot i \cdot q} \cdot \mathrm{Kl}(r \cdot k \cdot i \cdot q) + \frac{b}{k \cdot i \cdot q} \cdot \mathrm{Kl}(b \cdot k \cdot i \cdot q) \\
\text{simplify}
\end{array}
\right.
\rightarrow
$$

$$
\begin{align*}
& \rightarrow \text{zero3} := j_{\text{DC}} - A_1 \cdot j_{\text{DCWS}} \cdot I_0(k \cdot q \cdot r \cdot i) - A_2 \cdot j_{\text{DCWS}} \cdot K_0(k \cdot q \cdot r \cdot i) - \frac{A_2 \cdot j_{\text{DCWS}} \cdot k \cdot p \cdot \int_b^a K_1(k \cdot q \cdot r \cdot i) \, dr \cdot i}{q} \\
& \qquad - A_2 \cdot b \cdot j_{\text{DCWS}} \cdot k \cdot q \cdot \ln(a) \cdot K_1(b \cdot k \cdot q \cdot i) \cdot i + \frac{A_2 \cdot a \cdot j_{\text{DCWS}} \cdot k_s^2 \cdot \ln(s) \cdot K_1(a \cdot k \cdot q \cdot i) \cdot i}{k \cdot q} \\
& \qquad - \frac{A_2 \cdot b \cdot j_{\text{DCWS}} \cdot k_s^2 \cdot \ln(s) \cdot K_1(b \cdot k \cdot q \cdot i) \cdot i}{k \cdot q} + \frac{A_2 \cdot b \cdot j_{\text{DCWS}} \cdot k \cdot p \cdot \ln(a) \cdot K_1(b \cdot k \cdot q \cdot i) \cdot i}{q} \\
& \qquad - \frac{A_2 \cdot b \cdot j_{\text{DCWS}} \cdot k \cdot p \cdot \ln(b) \cdot K_1(b \cdot k \cdot q \cdot i) \cdot i}{q} + \frac{A_1 \cdot j_{\text{DCWS}} \cdot k \cdot p \cdot \int_b^a II(k \cdot q \cdot r \cdot i) \, dr \cdot i}{q} \\
& \qquad - A_1 \cdot b \cdot j_{\text{DCWS}} \cdot k \cdot q \cdot \ln(r) \cdot II(b \cdot k \cdot q \cdot i) \cdot i + \frac{A_1 \cdot b \cdot j_{\text{DCWS}} \cdot k_s^2 \cdot \ln(s) \cdot II(b \cdot k \cdot q \cdot i) \cdot i}{k \cdot q} \\
& \qquad + \frac{A_1 \cdot b \cdot j_{\text{DCWS}} \cdot k \cdot p \cdot \ln(b) \cdot II(b \cdot k \cdot q \cdot i) \cdot i}{q} - \frac{A_1 \cdot b \cdot j_{\text{DCWS}} \cdot k \cdot p \cdot \ln(a) \cdot II(b \cdot k \cdot q \cdot i) \cdot i}{q} \\
& \qquad - \frac{A_1 \cdot a \cdot j_{\text{DCWS}} \cdot k_s^2 \cdot \ln(s) \cdot II(a \cdot k \cdot q \cdot i) \cdot i}{k \cdot q} - A_1 \cdot j_{\text{DCWS}} \cdot k \cdot q \cdot \int_r^a II(k \cdot q \cdot r \cdot i) \, dr \cdot i + \\
& \qquad + A_2 \cdot b \cdot j_{\text{DCWS}} \cdot k \cdot q \cdot \ln(r) \cdot K_1(b \cdot k \cdot q \cdot i) \cdot i + A_2 \cdot j_{\text{DCWS}} \cdot k \cdot q \cdot \int_r^a K_1(k \cdot q \cdot r \cdot i) \, dr \cdot i + \\
& \qquad + A_1 \cdot b \cdot j_{\text{DCWS}} \cdot k \cdot q \cdot \ln(a) \cdot II(b \cdot k \cdot q \cdot i) \cdot i
\end{align*}
$$