Simple calculation of eddy currents in a long passive conductor

Abstract

A method is proposed for the calculation of current density in a long solid metallic conductor placed in an external variable magnetic field. The external magnetic field is perpendicular to the conductor and constant on straight lines parallel to the conductor. The conductor is passive, connected to neither a voltage source nor a current source. It is assumed that the external magnetic field is quasi-stationary and is not affected by the magnetic field of the currents induced in the conductor under examination. The frequency of external field does not exceed 1 MHz and the displacement current is neglected. The proposed method enables the calculation of current density not only in the steady state but also in the case that the external field has an arbitrary time-dependent course. Results are given of the calculation of current density in conductors with symmetrical and unsymmetrical cross section, with constant and non-constant resistivity in the conductor cross section, for both homogeneous and non-homogeneous external steady-state sinusoidal field, and for switching the external sinusoidal field on and off.

Keywords

Applied classical electromagnetism magnetic induction eddy currents numerical simulation

1. Introduction

In the conductor placed in a time-variable magnetic field, an electric current is induced the current density of which is not constant and depends on time [1–3]. The induced current is generally referred to as eddy current and its existence is the consequence of Faraday’s law of electromagnetic induction, and the quantitative description of eddy current, too, is given by this law [4]. Instead of the term eddy current the term induced current will be used in the following.

The theory of induced currents and their application in practice represents an extensive set of problems. The literature quoted is thus just a random selection from a wealth of published papers. The field B ^ex can be produced by a macroscopic electric current [5,6] or by the motion of conductor [7] or permanent magnets [8]. The conductor examined can vary in length, shape [9], cross section [10], resistivity, and permeability. The conductor can be connected to a source, in which case the field B ^co is the magnetic field of the induced current and of the current produced by the source. In addition to the generation of electric power in generators, the existence of induced currents can be made practical use of heating [11], brakes [12,13], dampers [10,14,15], non-destructive testing [5,14,16], etc. but in some devices their action is undesirable due to Joule’s heating.

The subject of the paper is the calculation of current density in a long solid conductor. The conductor under examination is a passive conductor (only conductor in the following), i.e. it is not connected to a source and is placed in an external magnetic field B ^ex. In the conductor examined, the induced current produces a magnetic field B ^co. The field B ^ex is assumed not to be affected by the field B ^co.

The Cartesian coordinate system xyz is chosen such that the axis z is parallel to the conductor under examination. The conductor cross section does not change along the conductor. The permeability of the conductor material is constant and is equal to the vacuum permeability 𝜇₀, its resistivity does not depend on z, it is a function of x and y, 𝜚 = 𝜚(x, y). Figure 1 illustrates the cross section A of a conductor in the plane xy. The cross section is a rectangle with the sides w and h. Generally, A can be an arbitrary closed continuous set [17,18]. The same as its components, the external magnetic field $\boldsymbol{B}^{\text{ex}}=\langle B_{x}^{\text{ex}},B_{y}^{\text{ex}},0\rangle$ , does not depend on z and is a function of x, y and time t. The field B ^ex is assumed to be quasi-stationary (slowly varying) [4] and at any instant t it is continuous on A.

Fig. 1.

Cross section A of a conductor with marked cross sections A _i and A _j of two partial conductors.

By Faraday’s law of electromagnetic induction, the voltage U _ind is induced along an arbitrary closed continuous curve C, which lies in a magnetic field B $\begin{eqnarray}\displaystyle U_{\text{ind}}=\displaystyle \frac{\text{d}{\Phi}}{\text{d}t}, & & \displaystyle\end{eqnarray}$ (1) where $\Phi$ is linked flux to the curve C. It is the flux of the vector B through a continuous surface S _C bounded by the curve C. $\Phi$ does not depend on the shape of the surface S _C. If the whole curve C lies in an electrically conductive medium (1∕𝜚 > 0), then U _ind will excite a conductive electric current, i.e. induced current. The non-zero component of the density of this current is only the z-component, $\begin{eqnarray}\displaystyle \boldsymbol{J}=\langle 0,0,J(x,y,t)\rangle , & & \displaystyle \nonumber\end{eqnarray}$ because the conductor is parallel to the axis z and the field B ^ex is perpendicular to the conductor, as assumed.

2. Calculation of current density in conductor

The (not only rectangular) cross section A of a real conductor is a Jordan measurable set [17,18] and therefore any solid conductor can with arbitrary precision be replaced by partial conductors of rectangular cross section A _i, ∀i (notation ∀i will be used instead of writing i = 1,2…, N). Let there be $\begin{eqnarray}\displaystyle A^{N}=\mathop{\bigcup }_{i=1}^{N}A_{i},\quad A_{i}\cap A_{j}=∅∼\text{for}∼i\neq j. & & \displaystyle \nonumber\end{eqnarray}$ The rectangles A _i, ∀i, form the decomposition D _N of the cross section A ^N. The decomposition D _N depends on N and a sequence of decompositions $\{D_{N}\}$ can be chosen such that it holds $\begin{eqnarray}\displaystyle A^{N}\rightarrow A\text{ and }\Vert D_{N}\Vert \rightarrow 0\quad \text{for }N\rightarrow \infty , & & \displaystyle \nonumber\end{eqnarray}$ where $\begin{eqnarray}\displaystyle \Vert D_{N}\Vert =\max _{\forall i}∼\text{dia}A_{i}, & & \displaystyle \nonumber\end{eqnarray}$ where diaA _i is the length of the diagonal of the rectangle A _i. It can be assumed that the conductor under consideration is formed by N parallel partial conductors, with the ith partial conductor having the cross section A _i. Inside A _i, ∀i, a point (X _i, Y _i) is chosen. In each partial conductor, the constant resistivity 𝜚_i = 𝜚(X _i, Y _i) and current density J _i(t) independent of x and y are assumed.

In the paper the sets A _i are assumed to be rectangles but they can also have a different shape; in [19], for example, they are sectors of circular ring.

Consider a segment of the conductor examined between the planes z = z ₁ and z = z ₂, z ₂ > z ₁. Together with z ₁ and z ₂, the points (X _i, Y _i) and (X _j, Y _j) (see Fig. 1) determine the loop C _ij, which is formed by four line segments formed by the points (x, y, z): $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle x=X_{i},∼y=Y_{i},∼z_{1}\leq z\leq z_{2}.\end{eqnarray}$ (2) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle z=z_{2},∼X_{i}\leq x\leq X_{j},∼y=Y_{j}+(x-X_{j})(Y_{j}-Y_{i})/(X_{j}-X_{i});\end{eqnarray}$ (3) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle x=X_{j},∼y=Y_{j},∼z_{1}\leq z\leq z_{2};\end{eqnarray}$ (4) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle z=z_{1},∼X_{i}\leq x\leq X_{j},∼y=Y_{j}+(x-X_{j})(Y_{j}-Y_{i})/(X_{j}-X_{i}).\end{eqnarray}$ (5) The loop C _ij, ∀i, j, forms the boundary of a continuous surface S _ij.

Voltage is induced in the loop C _ij according to formula (1), in which $\Phi$ is the flux of the vector $\begin{eqnarray}\displaystyle \boldsymbol{B}=\boldsymbol{B}^{\text{ex}}+\boldsymbol{B}^{\text{co}} & & \displaystyle \nonumber\end{eqnarray}$ through the surface S _ij. The flux $\Phi$ does not depend on the shape of the surface S _ij because the magnetic field is solenoidal [20] and thus the surface S _ij can be considered a rectangle the sides of which form the loop C _ij. B ^co is a magnetic field produced by the currents in the partial conductors. If segment (2) of the loop C _ij is replaced by a segment of the ith partial conductor and segment (4) by the jth partial conductor, the loop C _ij can then be replaced by a lumped-elements circuit.

The points (X _i, Y _i), ∀i, can be regarded as the vertices of the graph G. The edge in G between the vertices (X _i, Y _i) and (X _j, Y _j), i ≠ j, edge ij for short, is an intersection of the rectangle S _ij with the plane z = 0 and in G it represents the loop C _ij. Between the vertices (X _i, Y _i) and (X _j, Y _j) in G there is thus just one edge, the line segment joining the two vertices. The graph G is undirected because the loop C _ij is the same as the loop C _ji. No current is induced in the loops C _ii, ∀i, and therefore it is no use considering the edges ii; G is thus a simple graph. Any two different partial conductors can form a loop and G can thus be a complete graph with K _N = N (N −1)∕2 edges ij, ∀i, j; i ≠ j, [21]. For N > 3 it holds K _N > N and some of loops are therefore dependent on the other loops. A dependent loop can be formed by the composition of independent loops. A dependent loop together with the loops it depends on will in G show as a closed path, which is a consequence of the magnetic field being solenoidal. In the graph G with the above properties there is a subgraph G _tree, which has N vertices, N −1 edges, and does not contain any closed path of the graph G [21]. The edges of the subgraph G _tree have their correspondent independent loops in the conductor examined.

Figure 2 gives four examples of choosing a subgraph G _tree ⊂ G in the cross section of a conductor subdivided into partial conductors of equally large rectangular cross sections. It follows from Fig. 2 that there can exist several subgraphs G _tree that have the same vertices and differ in the position of the edges. However, the existence of at least one subgraph G _tree ensures that there exist N −1 independent loops for N partial conductors. The formally simplest is the selection on the assumption that G _tree is a path graph with the end-vertices (X ₁, Y ₁) and (X _N, Y _N). The graph in Fig. 2a is not a path graph, in contrast to the remaining graphs in Fig. 2.

Fig. 2.

Conductor cross section A = [0, $w$ ] × [0, h], with partial conductor cross sections marked out (dotted lines). The points (X _i, Y _i) marked with $\circ$ are the vertices of the graph G _tree, the line segments connecting the graph vertices are its edges.

Further in the exposition it is assumed that the graph G _tree is a path graph, i.e. the index j is equal to i +1. The index j, if it is together with the index i, can thus be omitted. The loop currents method can be applied to N −1 independent loops in the path graph. In Fig. 3 is a lumped-elements circuit which replaces the ith loop ∀i, i < N. In Fig. 3 there is also loop current in the (i −1)th and (i +1)th loops and current in the ith and (i +1)th partial conductors. Applying Kirchhoff’s voltage law to the circuit in Fig. 3 will yield the equation $\begin{eqnarray}\displaystyle R_{i}∼(I_{i}-I_{i-1})+R_{i+1}∼(I_{i}-I_{i+1})+U_{i,\text{ind}}=0, & & \displaystyle\end{eqnarray}$ (6) where the voltage U _{i, ind} is given by relation (1). The differences in the loop currents in (6) can be expressed in terms of current densities in the partial conductors, as follows from Fig. 3, with $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \text{m}(A_{1})∼J_{1}=I_{1},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \text{m}(A_{k})∼J_{k}=I_{k}-I_{k-1},∼k=2,3,\ldots ,N-1,\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \text{m}(A_{N})∼J_{N}=-I_{N-1},\nonumber\end{eqnarray}$ where the symbol m(A _i) denotes the measure of the cross section A _i. Equation (6) written for N −1 independent loops forms, after rewriting, a system of N −1 equations $\begin{eqnarray}\displaystyle (z_{2}-z_{1})[{\varrho}_{i}J_{i}(t)-{\varrho}_{i+1}J_{i+1}(t)]+\frac{\text{d}}{\text{d}t}[{\Phi}_{i}^{\text{co}}(t)+{\Phi}_{i}^{\text{ex}}(t)]=0,\quad \forall i,i<N, & & \displaystyle\end{eqnarray}$ (7) where ${\Phi}_{i}^{\text{co}}$ and ${\Phi}_{i}^{\text{ex}}$ are linked fluxes to C _i of the vectors B ^co and B ^ex, respectively.

Fig. 3.

Replacement of the ith loop by a lumped-elements circuit, and loop currents in the (i −1)th and (i +1)th loops.

The flux ${\Phi}_{i}^{\text{co}}$ is the sum of the fluxes produced by the partial conductors $\begin{eqnarray}\displaystyle {\Phi}_{i}^{\text{co}}(t)=\mathop{\sum }_{k=1}^{N}J_{k}(t)∼(z_{2}-z_{1})∼\frac{{\mu}_{0}}{4{\pi}}∼{\phi}_{ik}, & & \displaystyle\end{eqnarray}$ (8) where J _k(t) (z ₂ − z ₁) 𝜇₀ 𝜙_ik∕(4π) is the contribution of the kth partial conductor to the flux through the surface S _i. The coefficients 𝜙_ik, ∀i, k, do not depend on t and their calculation is given in the Appendix. For the flux of the vector B ^ex through the surface S _i it holds $\begin{eqnarray}\displaystyle {\Phi}_{i}^{\text{ex}}(t)=(z_{2}-z_{1})∼{\phi}_{i}^{\text{ex}}\mathscr{T}(t). & & \displaystyle\end{eqnarray}$ (9) The quantity ${\phi}_{i}^{\text{ex}}$ does not depend on t and its calculation is given in the Appendix. $\mathscr{T}(t)$ is the given function of t. After specifying the fluxes ${\Phi}_{i}^{\text{co}}$ and ${\Phi}_{i}^{\text{ex}}$ and after rearranging, equation (7) has the form $\begin{eqnarray}\displaystyle {\varrho}_{i}∼J_{i}(t)-{\varrho}_{i+1}∼J_{i+1}(t)+\frac{{\mu}_{0}}{4{\pi}}\mathop{\sum }_{k=1}^{N}{\phi}_{ik}∼\frac{\text{d}}{\text{d}t}J_{k}(t)=-{\phi}_{i}^{\text{ex}}∼\frac{\text{d}}{\text{d}t}\mathscr{T}(t). & & \displaystyle\end{eqnarray}$ (10) (10) is a system of N −1 ordinary differential first-order equation with unknown N current densities J ₁(t), J ₂(t), …, J _N(t). To determine all the current densities, it is necessary to add one more equation. This equation is the condition that the total current through the conductor examined is zero, i.e. $\begin{eqnarray}\displaystyle \mathop{\sum }_{k=1}^{N}\text{m}(A_{k})∼J_{k}(t)=0. & & \displaystyle\end{eqnarray}$ (11)

3. Calculation results and discussion

The proposed method for calculating the current density in a passive conductor is described by a system of Eqs (10) and (11). The application of the method will be demonstrated via solving several problems in both the steady state and the transient state.

Without loss of generality, it will be assumed that all the partial conductors are of the same rectangular cross section, i.e. m(A _i) = m (A _j), ∀i, j. The cross sections of the partial conductors form in the rectangular cross section of the conductor n layers in the direction of the axis y, with each layer containing m cross sections of the partial conductors. Let there be Δx = $w/m$ and Δy = h∕n. The cross section of the ith partial conductor is the rectangle [x _i1, x _i2] × [y _i1, y _i2], where $\begin{eqnarray}\displaystyle x_{i1}=i_{x}{\Delta}x,\quad x_{i2}=x_{i1}+{\Delta}x,\quad y_{i1}=i_{y}{\Delta}y,\quad y_{i2}=y_{i1}+{\Delta}y, & & \displaystyle \nonumber\end{eqnarray}$ where $\begin{eqnarray}\displaystyle i_{x}=(i-1)\text{ mod }m,\quad i_{y}=(i-1)\div m. & & \displaystyle \nonumber\end{eqnarray}$ The centre filament of the ith partial conductor passes through the point (X _i, Y _i) in the plane xy, $\begin{eqnarray}\displaystyle X_{i}=(x_{i1}+x_{i2})/2,\quad Y_{i}=(y_{i1}+y_{i2})/2. & & \displaystyle \nonumber\end{eqnarray}$

The calculated current density J _i, ∀i, is constant in the cross section of the ith partial conductor and is thus piecewise constant, which is an approximation of the real state. The calculation result is the more precise, the larger the value of N. In the Figures given below, the discontinuous values of the amplitude ${\hat{J}}_{i}$ and the initial phase ϵ_i of the current density are approximated by the continuous curves ${\hat{J}}(x,y)$ and ϵ(x, y), respectively. For a constant y the curves begin at the point x = X ₁ and terminate at the point x = X _m. With increasing N these points get closer to the conductor surface.

In addition to the current density, the Joule dissipation power is also determined. The total current passing through the passive conductor is zero. In one part of the cross section the current is positive while in the rest of the cross section the current is negative. The two currents are of the same magnitude but opposite sign.

3.1. Sinusoidal field B ^ex

Of considerable significance among the possible dependence relations between the field B ^ex and t is the sinusoidal dependence, i.e. $\begin{eqnarray}\displaystyle B_{x}^{\text{ex}}(x,y,t)=\hat{B}_{x}^{\text{ex}}(x,y)\sin ({\omega}t+{\varphi}), & & \displaystyle \nonumber\\ \displaystyle B_{y}^{\text{ex}}(x,y,t)=\hat{B}_{y}^{\text{ex}}(x,y)\sin ({\omega}t+{\varphi}), & & \displaystyle \nonumber\end{eqnarray}$ where ω = 2πf is the angular frequency. A certain time after the field B ^ex starts acting, the current induced in the conductor gets steady and it can be assumed that the current density in the conductor is also sinusoidal, with the angular frequency ω, $\begin{eqnarray}\displaystyle J_{i}(t)={\hat{J}}_{i}\sin ({\omega}t+{\varepsilon}_{i}),\quad \forall i. & & \displaystyle \nonumber\end{eqnarray}$ The flux of the vector B ^ex through the rectangle S _i is also sinusoidal $\begin{eqnarray}\displaystyle {\Phi}_{i}^{\text{ex}}(t)=\hat{{\Phi}}_{i}^{\text{ex}}\sin ({\omega}t+{\varphi}),\quad \forall i. & & \displaystyle \nonumber\end{eqnarray}$

It follows from the above that the system of Eqs (10) and (11) can be solved in the complex domain if the complex external field and current density $\begin{eqnarray}\displaystyle \text{}\underline{B}_{x}^{\text{ex}}(x,y)\exp (\text{j}{\omega}t),\quad \text{}\underline{B}_{y}^{\text{ex}}(x,y)\exp (\text{j}{\omega}t),\quad \text{}\underline{J}_{i}\exp (\text{j}{\omega}t),∼\forall i,j, & & \displaystyle \nonumber\end{eqnarray}$ are considered, where the underlined symbols denote the phasors $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \text{}\underline{B}_{x}^{\text{ex}}(x,y)=\hat{B}_{x}^{\text{ex}}(x,y)\exp (\text{j}{\varphi}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \text{}\underline{B}_{y}^{\text{ex}}(x,y)=\hat{B}_{y}^{\text{ex}}(x,y)\exp (\text{j}{\varphi}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \text{}\underline{J}_{i}={\hat{J}}_{i}\exp (\text{j}{\varepsilon}_{i}),\quad \forall i.\nonumber\end{eqnarray}$ Substituting the complex quantities into equation (10) and (11) and rewriting will yield a system of N linear algebraic equations in the complex domain $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\varrho}_{i}\text{}\underline{J}_{i}-{\varrho}_{i+1}\text{}\underline{J}_{i+1}+\frac{\text{j}∼{\mu}_{0}∼{\omega}}{4{\pi}}\mathop{\sum }_{k=1}^{N}{\phi}_{ik}∼\text{}\underline{J}_{k}=-\text{j}∼{\omega}∼\text{}\underline{{\Phi}}_{i}^{\text{ex}},\quad \forall i,i<N,\end{eqnarray}$ (12) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle \mathop{\sum }_{k=1}^{N}\text{}\underline{J}_{k}=0.\end{eqnarray}$ (13) Solving the system of Eqs (12) and (13) gives the current density phasors J _i, ∀i.

If J _i, ∀i, is the solution of the system of equations for the external field B ^ex, then u J _i is the solution of the system of Eqs (12) and (13) for the field u B ^ex, where u is an arbitrary complex number. The choice of the phasor B ^ex is thus arbitrary to a certain extent.

The left side of equation (13) can, after being multiplied by the term ΔxΔy, be replaced by the sum of two complex numbers, I ⁺ + I ⁻, which in a steady state only depend on N, where $\begin{eqnarray}\displaystyle \text{}\underline{I}^{+}(N)={\Delta}x∼{\Delta}y\mathop{\sum }_{\stackrel{{\varepsilon}_{i}\geq 0}{i=1}}^{N}\text{}\underline{J}_{i},\quad \text{}\underline{I}^{-}(N)=\text{}\underline{I}^{+}(N)\exp (\text{j}∼{\pi}). & & \displaystyle \nonumber\end{eqnarray}$ In a steady state, the mean Joule power $\bar{P}$ in one metre of conductor only depends on N $\begin{eqnarray}\displaystyle \bar{P}(N)=\displaystyle \frac{1}{2}{\Delta}x∼{\Delta}y\mathop{\sum }_{i=1}^{N}{\varrho}_{i}{\hat{J}}_{i}^{2}. & & \displaystyle \nonumber\end{eqnarray}$ With increasing N the values | I ⁺| and $\bar{P}$ get steady. The convergence is getting worse with increasing frequency f of the field B ^ex and with increasing dimensions w and h in the conductor cross section. The degree of stability of these quantities can serve as a criterion in the choice of the magnitude of N.

The time of current density calculation depends on the magnitude of N, where N = m × n for a conductor of rectangular cross section. The criterion for the choice of an appropriate magnitude N is generally the stabilization of the current density norm as a piecewise constant function on the conductor cross section or the norm stabilization of the vector 〈| J ₁|, | J ₂|, …, | J _N|〉. In the present paper, N is considered to be appropriately chosen if for N ₁ ≥ 2N it holds $\begin{eqnarray}\displaystyle \displaystyle \frac{|\text{}\underline{I}^{+}(N)-\displaystyle \text{}\underline{I}^{+}(N_{1})|}{|\text{}\underline{I}^{+}(N)|}<0.01,\quad \displaystyle \frac{|P(N)-P(N_{1})|}{P(N)}<0.01. & & \displaystyle \nonumber\end{eqnarray}$ N depends in the first place on f, with increasing f N increases too. N also depends on the shape of the conductor cross section, on its resistivity 𝜚(x, y) and on the choice of the cross section of partial conductors. The most convenient is to choose Δx and Δy such that their values differ from each other as little as possible.

3.1.1. Example 1

The dimensions of the cross section of the Cu conductor are $w$ = 10 mm, h = 16 mm, (see Fig. 1). The conductor resistivity at a temperature of 25 °C 𝜚 = 1.712 ×10⁻⁸ Ω ⋅ m [22] does not depend on x and y. The field B ^ex, has just one non-zero component $B_{y}^{\text{ex}}=\hat{B}_{y}^{\text{ex}}\sin {\omega}t$ , where $\hat{B}_{y}^{\text{ex}}=∼10^{-6}$ T, ω = 2πf, f = 50 Hz; m = 50, n = 80.

Figure 4 illustrates the current density phasor from Example 1. In Fig. 4 the curves for y = 0.1 mm and y = 7.9 mm are identical to the curves for y = 15.9 mm and y = 8.1 mm, respectively. The cause of the symmetry of the current density amplitude is the symmetry of the external field and of the conductor cross section with respect to the straight line y = 8 mm. The external field and the conductor cross section are also symmetrical with respect to the straight line x = 5 mm. Symmetrical with respect to this straight line is also the amplitude ${\hat{J}}$ , in contrast to the initial phase, for which it holds $\begin{eqnarray}\displaystyle -{\varepsilon}(x,y)+{\varepsilon}(w-x,y)=180^{\circ },\quad 0\leq x<w/2. & & \displaystyle\end{eqnarray}$ (14) It follows from equality (14) that the current in the left half of the conductor cross section is opposite to the current in the right half of the conductor cross section, and equation (13) holds. The simple dependence of the current density on x and y in Fig. 4 is conditional on small cross section dimensions of the conductor and on low external field frequency.

Fig. 4.

Dependence of the current density phasor on x for a constant y in Example 1. The thick lines give the amplitude ${\hat{J}}$ , the thin lines give the initial phase ϵ; the numbers at the curves are the values of y expressed in mm.

3.1.2. Example 2

The conductor cross section dimensions are ten times the dimensions in Example 1, $w$ = 100 mm, h = 160 mm; the other data are the same as in Example 1. In view of the symmetry of the external field and the conductor cross section the data in Fig. 5 on the current density phasor in Example 2 are only given in the left bottom quarter of the conductor cross section. The curve ϵ (x, 79 mm) in Fig. 5 is discontinuous because the initial phase has been chose such that its value is in the interval (−180°,180°].

Fig. 5.

Dependence of the current density phasor on x for a constant y in Example 2. The thick lines give the amplitude ${\hat{J}}$ , the thin lines give the initial phase ϵ; the numbers at the curves are the values of y expressed in mm.

3.1.3. Example 3

The conductor resistivity is not constant. The conductor cross section dimensions are the same as in Example 1. The conductor is composed of m layers that are parallel to the plane yz. The cross section of the ℓth layer is the rectangle $\begin{eqnarray}\displaystyle [(\ell -1){\Delta}x,\ell ∼{\Delta}x)\times [0,h],\quad \forall \ell ,∼{\Delta}x=w/m. & & \displaystyle \nonumber\end{eqnarray}$ The layers are of Al+Cu alloy at a temperature of 300 K and they differ in the Al content in the alloy. The Al content in the alloy changes linearly in dependence on x (mm), mass% Al = 9.091 x + 9.545. The Al content in the first layer is 10 mass% Al, 𝜚 = 1.11 ×10⁻⁷ Ω ⋅ m, then with increasing x the value 𝜚 increases to a maximum value of 1.76 ×10⁻⁷ Ω ⋅ m, which corresponds to 25 mass% Al in the alloy. With increasing x, 𝜚 decreases from the maximum value to the value 𝜚 = 2.733 ×10⁻⁸ Ω ⋅ m for x = $w$ and 100 mass% Al. The data on resistivity were published in the form of Tables in [22], p. 12-39 and p. 12-41. The field B ^ex has just one non-zero component $B_{y}^{\text{ex}}=\hat{B}_{y}^{\text{ex}}\sin {\omega}t$ , where $\hat{B}_{y}^{\text{ex}}=∼10^{-6}$ T, f = 10⁴ Hz; m = 100, n = 160.

The resistivity is not symmetrical with respect to the straight line x = $w/2$ but, the same as the conductor cross section and the field B ^ex, it is symmetrical with respect to the straight line y = h∕2. Figures 6 and 7 give the dependence of the current density phasor on x on the straight lines y = const in the lower half of the conductor cross section. In Fig. 6 it can be seen how the dependence of $\lg {\hat{J}}$ on x “copies” the linear segments of the dependence of 𝜚 on x, which illustrate the linear interpolation in the Table of 𝜚 values.

Fig. 6.

Dependence of the current density amplitude ${\hat{J}}$ on x for a constant y and dependence of the resistivity 𝜚 on x in Example 3. The numbers at the curves are the values of y expressed in mm.

Fig. 7.

Dependence of the initial phase ϵ of the current density on x for a constant y in Example 3. The numbers at the curves are the values of y expressed in mm.

3.1.4. Example 4

The dimensions of the Al conductor cross section are $w$ = 10 mm, h = 16 mm (see Fig. 1). Its resistivity at a temperature of 25 °C 𝜚 = 2.709 ×10⁻⁸ Ω ⋅ m [22] does not depend on x and y. The components of the field B ^ex are $B_{x}^{\text{ex}}=\hat{B}_{x}^{\text{ex}}\sin {\omega}t,B_{y}^{\text{ex}}=\hat{B}_{y}^{\text{ex}}\sin {\omega}t$ , kde $\hat{B}_{x}^{\text{ex}}=\hat{B}_{y}^{\text{ex}}=∼10^{-6}$ T, f = 10⁶ Hz; m = 128, n = 200.

The external field, conductor cross section and resistivity are symmetrical with respect to the symmetry centre ( $w/2$ , $h/2$ ), which is the centre of the conductor cross section. If the point (x ^′, y ^′) ∈ A is symmetrical to the point (x, y) with respect to the symmetry centre, then for the current density it holds $\begin{eqnarray}\displaystyle {\hat{J}}(x^{\prime },y^{\prime })={\hat{J}}(x,y),\quad |{\varepsilon}(x^{\prime },y^{\prime })-{\varepsilon}(x,y)|=180^{\circ }. & & \displaystyle \nonumber\end{eqnarray}$ With increasing frequency the extent of the current density amplitude increases, as can be seen in Figs 4–6. Figure 8 gives the dependence of ${\hat{J}}$ on x for several values of y in Example 4. It can be seen that low amplitude values are loaded with a considerable error and thus cannot be considered correct. The extent of the values of the amplitude ${\hat{J}}$ in which the calculated values can be considered correct depends on the number of the bits representing a rational number in the computer. In the present work, the calculation results are given in double precision and the values ${\hat{J}}$ are considered precise in the range of eight orders. This means that ${\hat{J}}$ values lower than ${\hat{J}}_{\text{min}}$ are considered to be zero values. Chosen in Example 4 was ${\hat{J}}_{\text{min}}=∼10^{-4}$ A ⋅ m⁻². In Fig. 9 a region of the conductor cross section is shown marked with the numeral 1 in which the current density amplitude is considered zero in Example 4. For comparison, a zero amplitude region is shown in Fig. 9 marked with the numeral 2 for a Cu conductor, f = 10⁵ Hz and ${\hat{J}}_{\text{min}}=∼10^{-5}$ A ⋅ m⁻². The region takes up 53.09% of the conductor cross section. If it held in the whole of region 1 that ${\hat{J}}={\hat{J}}_{\text{min}}$ , then there would be a current 8.5 ×10⁻⁹ A flowing through region 1 and there would be a Joule power of 1.2 ×10⁻²⁰ W ⋅ m⁻¹ in this region but in the rest of the cross section the amplitude of the phasor I ⁺ is equal to 3.892 ×10⁻² A and $\bar{P}=∼2.044\times 10^{-5}$ W ⋅ m⁻¹.

Fig. 8.

Dependence of the current density amplitude on x for a constant y in Example 4. The numbers at the curves are the values of y expressed in mm.

Fig. 9.

Regions of the zero current density amplitude in the conductor cross section in Example 4. Region 1: Al conductor, f = 10⁶ Hz, ${\hat{J}}_{\text{min}}=∼10^{-4}$ A ⋅ m⁻²; region 2: Cu conductor, f = 10⁵ Hz, ${\hat{J}}_{\text{min}}=∼10^{-5}$ A ⋅ m⁻².

3.1.5. Example 5

The dimensions of the Cu conductor cross section are $w$ = h = 100 mm (see Fig. 1). Its resistivity at a temperature of 25 °C 𝜚 = 1.712 ×10⁻⁸ Ω ⋅ m [22] does not depend on x and y. The field B ^ex has just one non-zero component $B_{x}^{\text{ex}}=\hat{B}_{x}^{\text{ex}}\sin {\omega}t$ , where $\hat{B}_{x}^{\text{ex}}=∼10^{-6}$ T, f = 10³ Hz; m = n = 200.

The external field, conductor cross section and resistivity are symmetrical with respect to the straight lines y = $h/2$ , x = $w/2$ , y = x and the centre ( $w/2$ , $h/2$ ). The sign of the current density J (x, y) and the directions of the currents in the conductor cross section are given by the phasors J _i, ∀i. Regions in the left bottom quarter of the conductor cross section in Example 5, in which the current density J (x, y) has the same sign, are illustrated in Fig. 10 for four instants in a period T = 1∕f. If there is t ∈ [T∕2, T), then the regions are the same as for t − T∕2 but the current density in them is of opposite direction to the direction of the current density for t − T∕2. For the given parameters of the conductor the number of regions in the conductor cross section decreases/increases with the frequency f decreasing/increasing.

Fig. 10.

Regions in the left bottom quarter of the conductor cross section in Example 5, in which the current density J (x, y) has the same sign. In the region marked with the numeral 0 the current density is of negligible magnitude ${\hat{J}}<10^{-5}$ A ⋅ m⁻².

3.1.6. Example 6

The conductor cross section A is non-symmetrical, its shape and dimensions are given in Fig. 11. At a temperature of 25 °C the Al conductor resistivity is 𝜚 = 2.709 ×10⁻⁸ Ω ⋅ m [22], which does not depend on x and y. The field B ^ex has just one non-zero component $B_{x}^{\text{ex}}=\hat{B}_{x}^{\text{ex}}\sin {\omega}t$ , where $\hat{B}_{x}^{\text{ex}}=∼10^{-6}$ T, f = 10³ Hz.

Fig. 11.

Conductor cross section A in Example 6.

Figures 12 and 13 give the dependence of the current density phasor on x on the straight lines y = const. The partial conductors are of square cross section with the side Δx = 0.1 mm.

Fig. 12.

Dependence of the current density amplitude on x for a constant y in Example 6. The numbers at the curves are the values of y expressed in mm. The curves for y = 4.05 mm and y = 6.05 mm are dotted.

Fig. 13.

Dependence of the initial current density phase on x for a constant y in Example 6. The numbers at the curves are the values of y expressed in mm. The curve for y = 4.05 mm almost merges with the curve for y = 3.95 mm.

3.1.7. Example 7

The dimensions of the Al conductor cross section are $w$ = h = 10 mm, its resistivity at a temperature of 25 °C 𝜚 = 2.709 ×10⁻⁸ Ω ⋅ m [22] does not depend on x and y. The field B ^ex is produced by a current filament, which is parallel to the axis z and passes through the point (12,0) mm. The current passing through the filament is $I_{\text{f}}=I_{\text{f}}(t)=\hat{I} _{\text{f}}\sin (2{\pi}ft)$ , where $\hat{I} _{\text{f}}=∼1$ A, f = 10³ Hz.

An equation for the calculation of the current density in a conductor which is in the external field produced by a current filament that is parallel to the conductor and passes through the point (x _f, y _f) is obtained by substituting for ${\Phi}_{i}^{\text{ex}}(t)$ in equation (7). It holds $\begin{eqnarray}\displaystyle {\Phi}_{i}^{\text{ex}}(t)=\displaystyle \frac{{\mu}_{0}∼I_{\text{f}}(t)∼(z_{2}-z_{1})}{4{\pi}}∼{\phi}_{i}^{\text{ex}}, & & \displaystyle\end{eqnarray}$ (15) where the value ${\phi}_{i}^{\text{ex}}$ is obtained by a procedure described in the Appendix. After substituting in (7) N −1 equations are obtained, which together with equation (11) form a system of N equations for determining the current density $\begin{eqnarray}\displaystyle {\varrho}_{i}∼J_{i}(t)-{\varrho}_{i+1}∼J_{i+1}(t)+\displaystyle \frac{{\mu}_{0}}{4{\pi}}\mathop{\sum }_{k=1}^{N}{\phi}_{ik}∼\displaystyle \frac{\text{d}}{\text{d}t}J_{k}(t)=-\frac{{\mu}_{0}}{4{\pi}}{\phi}_{i}^{\text{ex}}\frac{\text{d}}{\text{d}t}I_{\text{f}}(t),\quad \forall i,i\lt N. & & \displaystyle\end{eqnarray}$ (16) Equation (16) for the phasors is after rewriting in the form $\begin{eqnarray}\displaystyle \displaystyle \frac{4{\pi}∼\text{j}}{{\mu}_{0}∼{\omega}}[{\varrho}_{i+1}∼\text{}\underline{J}_{i+1}(t)-{\varrho}_{i}∼\text{}\underline{J}_{i}(t)]+\mathop{\sum }_{k=1}^{N}{\phi}_{ik}∼\text{}\underline{J}_{k}(t)=-{\phi}_{i}^{\text{ex}}∼\text{}\underline{I}_{\text{f}}(t),\quad \forall i,i\lt N, & & \displaystyle\end{eqnarray}$ (17) and together with (13) it forms a system of N equations, the solution of which gives the current density phasors.

The current density phasor in Example 7 is illustrated in Figs 14 and 15.

Fig. 14.

Dependence of the current density amplitude on x for a constant y in Example 7. The numbers at the curves are the values of y expressed in mm.

Fig. 15.

Dependence of the initial current density phase on x for a constant y in Example 7. The numbers at the curves are the values of y expressed in mm.

3.2. Transient current density in a conductor

In the transition of the external field B ^ex from one steady state to another steady state the current densities in the partial conductors of the conductor under examination are transient. The transient current densities are the solution of the system of N −1 ordinary differential equations of the first order (10) and one algebraic equation (11). In this subsection, results will be given of calculating the transient current density when switching the external sinusoidal field B ^ex on and off. Prior to the numerical solution, the system of Eqs (10) and (11) needs to be rearranged. The differentiation of a function with respect to time t will be denoted by a dot over the symbol of the function.

As has been assumed, the partial conductor cross sections are identical and equation (11) can thus be replaced by the equation $\begin{eqnarray}\displaystyle \mathop{\sum }_{k=1}^{N}\dot{J}_{k}(t)=0. & & \displaystyle\end{eqnarray}$ (18) The system of Eqs (10) and (18) can after rewriting be replaced by a single equation $\begin{eqnarray}\displaystyle \mathsf{F}\dot{\mathsf{J}}(t)=\mathsf{U}(t,\mathsf{J}), & & \displaystyle\end{eqnarray}$ (19) where F = (F _ik) is a matrix of the type of (N, N), for the elements of which it holds $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle F_{ik}={\phi}_{ik},\quad \forall i,k;∼i\lt N,\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle F_{Nk}=1,\quad \forall k,\nonumber\end{eqnarray}$ the matrices J = J (t) = (J _i(t)), ${\dot{\mathsf{J}}}$ (t) = ( $\dot{J}_i$ (t)) and U = (U _i(t)) are of the type of (N,1), with J (t) being the matrix of the current densities in the partial conductors at time t, $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle U_{i}(t)=\frac{4{\pi}}{{\mu}_{0}}∼\left[{\varrho}_{i+1}∼J_{i+1}(t)-{\varrho}_{i}∼J_{i}(t)-{\phi}_{i}^{\text{ex}}∼\dot{\mathscr{T}}(t)\right],\quad \forall i,∼i\lt N,\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle U_{N}(t)=0.\nonumber\end{eqnarray}$ It is still necessary to complete equation (19) with the initial condition $\begin{eqnarray}\displaystyle \mathsf{J}(0)=\mathsf{J}_{0},\quad \mathsf{J}_{0}=(J_{i}(0)). & & \displaystyle\end{eqnarray}$ (20) The matrix J (t) is the solution of equation (19) at time t, which satisfies initial condition (20). Similar to the steady state, the current density is directly proportional to the magnitude of the external magnetic field, which follows from equation (10).

Multiplying equation (19) by the inverse matrix F ⁻¹ from the left yields the equation $\begin{eqnarray}\displaystyle \dot{\mathsf{J}}=\mathsf{F}^{-1}\mathsf{U}, & & \displaystyle\end{eqnarray}$ (21) which is in a form suitable for applying the numerical method of solution [18,23].

The positive current in the transient state depends on N and t $\begin{eqnarray}\displaystyle I^{+}(N,t)={\Delta}x∼{\Delta}y\mathop{\sum }_{\stackrel{J_{i}(t)\geq 0}{i=1}}^{N}J_{i}(t). & & \displaystyle \nonumber\end{eqnarray}$ In the transient state, the instantaneous Joule power P in one metre of conductor depends on N and t $\begin{eqnarray}\displaystyle P(N,t)={\Delta}x∼{\Delta}y\mathop{\sum }_{i=1}^{N}{\varrho}_{i}∼J_{i}(t)^{2}. & & \displaystyle \nonumber\end{eqnarray}$

3.2.1. Example 8

This example concerns switching on and off the external field with the components $\hat{B}_{x}^{\text{ex}}\sin {\omega}t$ and $\hat{B}_{y}^{\text{ex}}\sin {\omega}t$ , where $\hat{B}_{x}^{\text{ex}}=\hat{B}_{y}^{\text{ex}}=∼10^{-6}$ T; f = 10³ Hz. Data about the conductor are the same as in Example 1, n = 100, m = 160.

The period of the field B ^ex is T = 10⁻³ s. Figure 16 gives the transient current densities, for t ∈ [0, 4T], in four partial conductors, whose cross sections are cut by the diagonal of the conductor cross section. If (x, y) and (x ^′, y ^′) are points symmetrical with respect to the cross section centre, then it holds J (x, y) = −J (x ^′, y ^′). The transient current density attains the highest values in the neighbourhood of the right bottom and left upper corner of the conductor cross section. It can be seen in Fig. 16 that from a certain instant t _ste the transient current density J (x, y, t) can be considered steady. The value t _ste depends on the conductor parameters, on the external field B ^ex and its frequency, and on the choice of the criterion under whose fulfilment the current density can be considered steady. In Fig. 16 the values ${\hat{J}}$ and $-{\hat{J}}$ in the steady state that correspond to curve 1 in Fig. 16 are shown as a dotted line, ${\hat{J}}=∼2335.19$ A ⋅ m⁻². The extreme values of the transient current density on curve 1 in Fig. 16 for t > 3T are 2334 A ⋅ m⁻² and −2336 A ⋅ m⁻².

Fig. 16.

Dependence of the transient current density on time in four partial conductors, whose cross sections are cut by the diagonal of the conductor cross section in Example 8; the cross section centres of partial conductors are for curve 1: (9.95, 0.05) mm, 2: (9.35, 1.05) mm, 3: (8.35, 2.65) mm, 4: (5.05, 7.95) mm. The external magnetic field has a period T and its time dependence is illustrated by the thin line. The steady-state amplitude ${\hat{J}}$ pertaining to curve 1 is shown as a thin line segment.

Figure 17 gives the dependence of the global quantities I ⁺ and P on t in Example 8. The dotted line segment represents the value | I ⁺| = 30.344 mA, the thin line segment is the mean Joule power $\bar{P}=∼4.569\times 10^{-7}$ W ⋅ m⁻¹.

Fig. 17.

Dependence of I ⁺ (thick line) and P (thin line) on t in Example 8. The dotted line gives the value | I ⁺|, the line segment shown as a thin line is the mean Joule power $\bar{P}$ .

In Fig. 16 the dependence is given of the transient current density on t from the instant t = 0 (switching the external field on), when the external field passes through zero because its initial phase is zero. Figure 18 gives the dependence of the transient current density in partial conductors 1, 2, and 3 (see the caption to Fig. 16) when switching on at the instant t = 0 and when switching off at times t = 2.25 T and t = 2.4 T. The external magnetic field differs from the field specified in Example 8 in that its non-zero initial phase is 45°.

Fig. 18.

Dependence of the transient current density in partial conductors 1, 2, and 3 (see the caption to Fig. 16) when switching on at time t = 0 and off at times t = 2.25 T and t = 2.4 T. The dependence B ^ex(t) is shown as a thin line.

3.3. Comments on the calculation of current density

When solving some of the examples, the symmetry of the current density was mentioned. This symmetry is a consequence of the symmetry of the conductor cross section, external field and resistivity. Symmetry can be made use of when constructing the system of equations (10), (11) and to reduce their number to a half or quarter since due to the symmetry the number of unknown current densities gets lower. The time necessary to solve the system of equations (10), (11) is directly proportional to the third power of the number of equations; reducing their number can significantly reduce the calculation time, in particular when calculating the steady current density and when the values of m and n are high.

Fig. 19.

Various possible mutual positions of the points (X _i, Y _i) − $\bullet$ and (X _j, Y _j) −× and the choice of the rectangles O ^x and O ^y, their orientation, and the orientation of the rectangle S .

All the areas through which the relevant fluxes of the magnetic field vector flow lie inside the conductor examined and are thus affected by the magnetic properties of the conductor material. In the paper it is assumed that the conductor permeability is equal to the vacuum permeability 𝜇₀. The vacuum permeability is sufficiently exact for diamagnetic and paramagnetic materials as well as for ferromagnetic materials at temperatures higher than their Curie temperature. In a linear material it holds B = 𝜇 H , where 𝜇 is a constant and H is the vector of the magnetic intensity. In such a case it is sufficient in equation (10) to replace 𝜇₀ by 𝜇. Ferromagnetic materials are usually non-linear and, in principle, equations can be formulated for them that are similar to equations (10) and (11). It is, however, hard to guess a priori whether such equations could be solved numerically. Another problem consists in the knowledge of full and precise magnetization characteristics of ferromagnetic materials.

In equations (10) and (11) and in all the Examples the flux ${\Phi}_{ij}^{\text{ex}}$ of the external field was assumed to be determined by relation (9), in which the dependence on t is given by the function $\mathscr{T}(t)$ . In more general terms, there can be $\begin{eqnarray}\displaystyle {\Phi}_{ij}^{\text{ex}}=(z_{2}-z_{1})\mathop{\sum }_{\ell =1}^{s}{\phi}_{ij}^{\text{ex},\ell }∼\mathscr{T}_{\ell }(t). & & \displaystyle \nonumber\end{eqnarray}$ In the steady state, the functions $\mathscr{T}_ℓ(t)$ , ∀ℓ, must be sinusoidal and of the same frequency while in the transient state they may be sinusoidal functions with different frequencies, or some or all of them need not be sinusoidal.

In Example 7, the external field is produced by a single current filament. The external field flux is determined by relation (15), in which the dependence on t is given by the function I _f(t). Similar to the preceding paragraph, the external field can be formed by a finite number of current filaments with different dependence relations I _f(t).

4. Comments on calculation method

On the assumptions as given in the Introduction section, the method for calculating current density in a passive conductor placed in time-variable magnetic field was exactly derived in a physically clear way. The derivation of the method is based on Faraday’s law of electromagnetic induction and is an application of the Biot and Savart law, the loop current method, Kirchhoff’s voltage law, Ohm’s law, the Jordan measure theory, and graph theory. These starting points cannot be called in question, the same as the result that follows from them. Current density is generally the solution of a system of equations formed by one algebraic equation (11) and N −1 ordinary differential equations of the first order (10). These equations allow calculating the current density in both the transient and the steady state. The numerical calculation is performed in two parts. The first part consists of calculating the equation coefficients, using the formulae given in the paper. The second part depends on whether the steady or the transient state is being solved. In the steady state the external field is sinusoidal and what is solved is a system of linear algebraic equations (12) and (13) in the complex domain. In the transient state the external field is of arbitrary given course and what is solved is a system of ordinary linear differential equations of the first order with initial conditions (20), using one of the known methods, e.g. the Runge–Kutta method [18,23].

The source of electromagnetic field is the electric charge. In the paper this field is a quasi-stationary magnetic field produced by a moving electric charge. The result of calculating by the proposed method is the current density. The current density directly determines the magnetic field [24] and Joule’s dissipated power.

The calculation of induced currents has been the subject of many publications. The methods proposed in these papers are based on Maxwell’s equations in differential form. By means of these equations, a partial differential equation of the second order is derived whose solution is the vector potential A for the steady state. Paper [19] (Subsection 4.7) deals with methods for the calculation of current density via partial differential equations. Partial differential equations cannot be solved without the knowledge of boundary conditions, which constitutes quite a problem, and for the numerical solution the respective method (the finite elements method in most cases but not always [25]) must be available. The current density is the result of numerical differentiation of the vector potential and thus it is loaded with a non-negligible error [23]. “The avoid leaving a false impression, namely, that the vector potential is as useful as the electrostatic potential in computing simple fields, it must be noted that there are essentially no cases where A can be computed in simple closed form (although it could always be done numerically for bounded current distributions) [4], p. 208.”

In comparison with the published methods, the method for solving the problem as formulated in the Introduction section is general (it includes both the steady and the transient state). It does not require the knowledge of boundary conditions while the numerical calculation is very simple.

5. Conclusion

A simple method has been proposed for calculating the current density in a long solid conductor of arbitrary cross section which is not connected to a source and occurs in an external magnetic field. It is assumed that the external magnetic field is quasi-stationary and is unaffected by the magnetic fields of the currents induced in the conductor under examination. The frequency of the external field does not exceed 1 MHz and the displacement current is neglected. The permeability of the conductor material is constant and is equal to the vacuum permeability 𝜇₀.

The essence of the proposed method consists in the conductor being replaced by N partial conductors of rectangular cross section and constant current density. The derivation of the method is based, in the first place, on Faraday’s law of electromagnetic induction and is an application of the Biot and Savart law, loop current method, Kirchhoff’s voltage law, Ohm’s law, the Jordan measure theory, and graph theory. The current density is generally the solution of a system of equations formed by one algebraic equation (11) and N −1 ordinary differential equations of the first order (10).

The application of the proposed method is demonstrated via solving eight examples. Results are given of the calculation of the current density in a conductor with symmetrical and non-symmetrical cross section, with constant and non-constant resistivity in the conductor cross section, for homogeneous and non-homogeneous external steady sinusoidal field, and for the case of switching the external sinusoidal field on and off.

Footnotes

Acknowledgements

This research work has been carried out in the Centre for Research and Utilization of Renewable Energy (CVVOZE). Authors gratefully acknowledge financial support from the Ministry of Education, Youth and Sports of the Czech Republic under OP VVV Programme (project No. CZ.02.1.01/0.0/0.0/16_013/0001638 CVVOZE Power Laboratories-Modernization of Research Infrastructure).

Appendix

According to formulae (8) and (9) the fluxes $\Phi^{\rm co}_{i}(t)$ (t) and $\Phi_{i}^{\rm ex}(t)$ through the rectangle S _i depend on the quantities 𝜙_ik and 𝜙_i ^ex(t). In these quantities and in S _i the index i is for the sake of simplicity left out in the following. The boundary of the rectangle S is formed by the points (x, y, z) which satisfy relations (2)–(5). The rectangle is determined by the coordinates z ₁, z ₂ and the points (X _i, Y _i) and (X _j, Y _j). When calculating the quantities 𝜙_k and 𝜙^ex(t) it holds j = i +1, but the undermentioned formulae hold for any arbitrary point (X _j, Y _j) in the plane xy.

Magnetic field of a conductor of rectangular cross section

The magnetic field B ^co of the conductor examined is the sum of the fields of the partial conductors. The cross section of the kth partial conductor is the rectangle [x _k1, x _k2] × [y _k1, y _k2] in the plane z = 0, the conductor is parallel with the axis z. The current density J _k = 〈0,0, J _k〉 is constant in the partial conductor cross section. The vector of the magnetic field B _k produced by the kth partial conductor has the zero component B _kz, B _k = 〈B _kx, B _ky〉, $\begin{eqnarray}\displaystyle \boldsymbol{B}_{k}(x,y)=\frac{{\mu}_{0}}{4{\pi}}J_{k}∼\boldsymbol{b}_{k}(x,y), & & \displaystyle \nonumber\end{eqnarray}$ where (x, y) is the point in the plane z = 0 in which the calculation of the field is carried out, b _k = 〈b _kx, b _ky〉, (A.1) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle b_{kx}(x,y)=\int _{y_{k1}}^{y_{k2}}\int _{x_{k1}}^{x_{k2}}\frac{2({\eta}-y)}{(x-{\xi})^{2}+(y-{\eta})^{2}}∼\text{d}{\eta}∼\text{d}{\xi},\end{eqnarray}$ (A.2) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle b_{ky}(x,y)=\int _{y_{k1}}^{y_{k2}}\int _{x_{k1}}^{x_{k2}}\displaystyle \frac{2(x-{\xi})}{(x-{\xi})^{2}+(y-{\eta})^{2}}∼\text{d}{\xi}∼\text{d}{\eta}.\end{eqnarray}$ The value of integral (22) can be expressed using the function (A.3) $\begin{eqnarray}\displaystyle f_{x}(x,y,{\xi},{\eta})=(x-{\xi})\ln [(x-{\xi})^{2}+(y-{\eta})^{2}]+2(y-{\eta})\arctan \displaystyle \frac{x-{\xi}}{y-{\eta}}. & & \displaystyle\end{eqnarray}$ On the right side of relation (24) the first term is equal to zero for x = 𝜉 and the second term is equal to zero for y = 𝜂. It holds (A.4) $\begin{eqnarray}\displaystyle b_{x}(x,y)=-f_{x}(x,y,x_{k2},y_{k2})+f_{x}(x,y,x_{k2},y_{k1})+f_{x}(x,y,x_{k1},y_{k2})-f_{x}(x,y,x_{k1},y_{k1}). & & \displaystyle\end{eqnarray}$ The value of integral (23) can be expressed using the function (A.5) $\begin{eqnarray}\displaystyle f_{y}(x,y,{\xi},{\eta})=(y-{\eta})\ln [(x-{\xi})^{2}+(y-{\eta})^{2}]+2(x-{\xi})\arctan \displaystyle \frac{y-{\eta}}{x-{\xi}}. & & \displaystyle\end{eqnarray}$ On the right side of relation (26) the first term is equal to zero for y = 𝜂 and the second term is equal to zero for x = 𝜉. It holds (A.6) $\begin{eqnarray}\displaystyle b_{y}(x,y)=f_{y}(x,y,x_{k2},y_{k2})-f_{y}(x,y,x_{k2},y_{k1})-f_{y}(x,y,x_{k1},y_{k2})+f_{y}(x,y,x_{k1},y_{k1}). & & \displaystyle\end{eqnarray}$ The value of the component b _y(x, y) can be expressed not only by means of f _y(x, y, 𝜉, 𝜂) but also by means of f _x(x, y, 𝜉, 𝜂) because $\begin{eqnarray}\displaystyle f_{y}(x,y,{\xi},{\eta})=-f_{x}(y,x,{\eta},{\xi}). & & \displaystyle \nonumber\end{eqnarray}$ Similarly, the value b _x(x, y) can be expressed by means of f _y(x, y, 𝜉, 𝜂).

The flux of the vector of the partial conductor magnetic field

The flux through the rectangle S is determined by means of the fluxes of the vector b _k through the rectangles O ^x and O ^y formed by the points (x, y, z), $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle O^{x}:∼x=x_{\text{c}},\quad Y_{\text{beg}}\leq y\leq Y_{\text{end}},\quad z_{1}\leq z\leq z_{2},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle O^{y}:∼X_{\text{beg}}\leq x\leq X_{\text{end}},\quad y=y_{\text{c}},\quad z_{1}\leq z\leq z_{2},\nonumber\end{eqnarray}$ where $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle X_{\text{beg}}=\min (X_{i},X_{j}),\quad X_{\text{end}}=\max (X_{i},X_{j}),\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle Y_{\text{beg}}=\min (Y_{i},Y_{j}),\quad Y_{\text{end}}=\max (Y_{i},Y_{j}).\nonumber\end{eqnarray}$ The rectangle O ^x is parallel to the plane yz, the rectangle O ^y is parallel to the plane xz. The cross sections of the rectangles O ^x and O ^y and their orientation are illustrated in Fig. 19.

For the fluxes 𝜙(O ^x) and 𝜙(O ^y) of the vector b _k through the rectangles O ^x and O ^y, respectively, it holds (A.7) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\phi}(O^{x})=\int _{Y_{\text{beg}}}^{Y_{\text{end}}}b_{x}(x_{\text{c}},y)∼\text{d}y,\end{eqnarray}$ (A.8) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\phi}(O^{y})=\int _{X_{\text{beg}}}^{X_{\text{end}}}b_{y}(x,y_{\text{c}})∼\text{d}x.\end{eqnarray}$ Using relations (25) and (27), the value of the integrals on the right side of relations (28) and (29) can be expressed by means of the function $\begin{eqnarray}\displaystyle F(X,Y,{\xi},{\eta}) & = & \displaystyle (X-{\xi})(Y-{\eta})\ln [(X-{\xi})^{2}+(Y-{\eta})^{2}]\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +∼(X-{\xi})^{2}\arctan \displaystyle \frac{Y-{\eta}}{X-{\xi}}+(Y-{\eta})^{2}\arctan \displaystyle \frac{X-{\xi}}{Y-{\eta}}.\nonumber\end{eqnarray}$ It holds $\begin{eqnarray}\displaystyle {\phi}(O^{x}) & = & \displaystyle +F(x_{\text{c}},Y_{\text{beg}},x_{k2},y_{k2})-F(x_{\text{c}},Y_{\text{end}},x_{k2},y_{k2})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle -F(x_{\text{c}},Y_{\text{beg}},x_{k2},y_{k1})+F(x_{\text{c}},Y_{\text{end}},x_{k2},y_{k1})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle -F(x_{\text{c}},Y_{\text{beg}},x_{k1},y_{k2})+F(x_{\text{c}},Y_{\text{end}},x_{k1},y_{k2})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +F(x_{\text{c}},Y_{\text{beg}},x_{k1},y_{k1})-F(x_{\text{c}},Y_{\text{end}},x_{k1},y_{k1}),\nonumber\\ \displaystyle {\phi}(O^{y}) & = & \displaystyle +F(X_{\text{end}},y_{\text{c}},x_{k2},y_{k2})-F(X_{\text{beg}},y_{\text{c}},x_{k2},y_{k2})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle -F(X_{\text{end}},y_{\text{c}},x_{k2},y_{k1})+F(X_{\text{beg}},y_{\text{c}},x_{k2},y_{k1})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle -F(X_{\text{end}},y_{\text{c}},x_{k1},y_{k2})+F(X_{\text{beg}},y_{\text{c}},x_{k1},y_{k2})\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle +F(X_{\text{end}},y_{\text{c}},x_{k1},y_{k1})-F(X_{\text{beg}},y_{\text{c}},x_{k1},y_{k1}).\nonumber\end{eqnarray}$

The flux 𝜙(S) of the vector b _k through the rectangle S depends on the mutual position of the points (X _i, Y _i) and (X _j, Y _j), see Fig. 19, (A.9) $\begin{eqnarray}\displaystyle {\phi}(S)=\left\{\begin{array}{@{}ll@{}}-{\phi}(O^{x})+{\phi}(O^{y}) & \text{for Fig. 19a},∼x_{\text{c}}=X_{j},y_{\text{c}}=Y_{i},\\ +{\phi}(O^{x})+{\phi}(O^{y}) & \text{for Fig. 19b},∼x_{\text{c}}=X_{i},y_{\text{c}}=Y_{j},\\ +{\phi}(O^{x})-{\phi}(O^{y}) & \text{for Fig. 19c},∼x_{\text{c}}=X_{j},y_{\text{c}}=Y_{i},\\ -{\phi}(O^{x})-{\phi}(O^{y}) & \text{for Fig. 19d},∼x_{\text{c}}=X_{i},y_{\text{c}}=Y_{j},\end{array}\right. & & \displaystyle\end{eqnarray}$

The flux of the vector of the external homogeneous magnetic field

It follows from relation (9) that for a given function $\mathscr{T}(t)$ the flux $\Phi^{\rm ex}(t)$ of the vector $\boldsymbol{B}^{\text{ex}}=\langle B_{x}^{\text{ex}},B_{y}^{\text{ex}}\rangle$ through the rectangle S is determined by the quantity 𝜙^ex. The same as in the case of the partial conductor, the flux through the rectangle S is determined by the fluxes through the rectangles O ^x and O ^y. If the external field is homogeneous, i.e. $B_{x}^{\text{ex}}$ and $B_{y}^{\text{ex}}$ do not depend on x and y, then $\begin{eqnarray}\displaystyle {\phi}^{\text{ex}}=B_{x}^{\text{ex}}\int _{Y_{\text{beg}}}^{Y_{\text{end}}}\text{d}y+B_{y}^{\text{ex}}\int _{X_{\text{beg}}}^{X_{\text{end}}}\text{d}x=B_{x}(Y_{i}-Y_{j})+B_{y}(X_{j}-X_{i}). & & \displaystyle \nonumber\end{eqnarray}$

The field and flux of an infinitely long filament

The current filament which produces the external field is parallel to the axis z and passes through the point (x _f, y _f) in the plane xy. The current filament must lie outside the conductor examined and thus it cannot pass through the point (0,0). The magnetic field of the current filament at the point (x, y) can be calculated using Ampere’s circuital law; it holds $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle B_{\text{f}}^{\text{ex}}(x,y)={\mu}_{0}∼I_{\text{f}}/(2{\pi}∼p),\nonumber\end{eqnarray}$ (A.10) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle B_{\text{f}x}^{\text{ex}}(x,y)=-B_{\text{f}}^{\text{ex}}(x,y)∼(y-y_{\text{f}})/p,\end{eqnarray}$ $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle B_{\text{f}y}^{\text{ex}}(x,y)=B_{\text{f}}^{\text{ex}}(x,y)∼(x-x_{\text{f}})/p,\nonumber\end{eqnarray}$ where I _f is the current through the filament and $\begin{eqnarray}\displaystyle p=\sqrt{(x-x_{\text{f}})^{2}+(y-y_{\text{f}})^{2}}. & & \displaystyle \nonumber\end{eqnarray}$

The flux $\Phi^{\rm ex}(t)$ of the current filament field is determined by relation (15) and for a given current I _f(t) the flux is determined by the quantity 𝜙^ex. The same as in the preceding cases the flux through the rectangle S is determined by the fluxes through the rectangles O ^x and O ^y. This means that 𝜙^ex(S) is determined by 𝜙^ex(O ^x) and 𝜙^ex(O ^y) analogously to 𝜙(S) being determined by means of 𝜙(O ^x) and 𝜙(O ^y), the same as in formula (30). It holds $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\phi}^{\text{ex}}(O^{x})=\int _{Y_{\text{beg}}-y_{\text{f}}}^{Y_{\text{end}}-y_{\text{f}}}\frac{-2∼{\eta}}{(x_{\text{c}}-x_{\text{f}})^{2}+{\eta}^{2}}∼\text{d}{\eta},\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle {\phi}^{\text{ex}}(O^{y})=\int _{X_{\text{beg}}-x_{\text{f}}}^{X_{\text{end}}-x_{\text{f}}}\frac{2∼{\xi}}{{\xi}^{2}+(y_{\text{c}}-y_{\text{f}})^{2}}∼\text{d}{\xi},\nonumber\end{eqnarray}$ where 𝜉 = x − x _f, 𝜂 = y − y _f. After integration (A.11) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\phi}^{\text{ex}}(O^{x})=\ln \displaystyle \frac{(x_{\text{c}}-x_{\text{f}})^{2}+(Y_{\text{beg}}-y_{\text{f}})^{2}}{(x_{\text{c}}-x_{\text{f}})^{2}+(Y_{\text{end}}-y_{\text{f}})^{2}},\end{eqnarray}$ (A.12) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\phi}^{\text{ex}}(O^{y})=\ln \displaystyle \frac{(X_{\text{end}}-x_{\text{f}})^{2}+(y_{\text{c}}-y_{\text{f}})^{2}}{(X_{\text{beg}}-x_{\text{f}})^{2}+(y_{\text{c}}-y_{\text{f}})^{2}}.\end{eqnarray}$

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