## 4 Analysis of Magnetic Field from 400 kV AC Power Transmission Line  
In this challenging task regarding magnetic fields, the vectors of magnetic flux density \(B\) or magnetic field strength \(\mathbf{H}\) are found to be functions of coordinates when the dependency of the current density vector \(\mathbf{J}\) on coordinates is known. As the magnetic field strength or magnetic flux density is to be found in areas in which there are no currents, LaPlace's equation is solved for the scalar magnetic potential. Such a magnetic field will also be of a potential nature [7; 8; 9; 10; 11; 12; 13; 14; 15; 23]. For the purpose of solving this task, the method of reflections is used. The magnetic field of the conductor system in the double-circuit 400 kV overhead transmission line is examined when the conductor system is comprised of long, thin, round cylindrical conductors running in parallel and with flat conductive surfaces with the symmetrical current systems \(i_{11}\) (\(i_{1}\) = A\({}_{1}\), B\({}_{1}\), C\({}_{1}\)) and \(i_{2}\) (\(i_{2}\) = A\({}_{2}\), B\({}_{2}\), C\({}_{2}\)) (see Figure 2). The radii of current leads \(r_{i}\ll h_{i}\) are significantly lower than the distance from the ground surface to the current leads. The sag of the double-circuit overhead power line is not taken into consideration. Following the method of reflections, the magnetic field in which the distribution of currents over the flat surface of the conductor is known is replaced with the magnetic field of currents \(i_{in}\) and that of their reflections-\(i_{in}\). It is assumed that the distances from the conductors \(i_{in}\) to their respective reflections-\(i_{in}\) are equal; i.e., \(y_{in}\) (\(i\) = A, B, C) (\(n\) = 1, 2). With the values of the currents \(i_{A1}\), \(i_{B1}\), \(i_{C1}\), \(i_{A2}\), \(i_{B2}\), \(i_{C2}\) being known, the values of reflections of these loads-\(i_{A1}^{*}\), \(i_{B1}^{*}\), -\(i_{C1}^{*}\), -\(i_{A2}^{*}\), -\(i_{B2}^{*}\), -\(i_{C2}^{*}\) are assumed to have the opposite signs. In order to determine how the values of currents change in a single period of sinusoidal quantity, calculations are made by varying the angle of the phase current at steps of \(10^{0}\); i.e., \(\bigtriangleup wt=10^{0}\). Instantaneous phase current values of a symmetrical current system are interconnected, as follows:  
\[ \begin{cases}i_{\text{A}}{=}I_{\text{m}}\sin(wt),\\ i_{\text{B}}{=}I_{\text{m}}\sin(wt{\text{-}}120^{0}),\\ i_{\text{C}}{=}I_{\text{m}}\sin(wt{\text{+}}120^{0});\end{cases}  \tag{3}\]  
where \(I_{\text{m}}\) is the maximum amplitude value of the phase current, in A. The 400 kV double-circuit power transmission line has a maximum effective linear current \(I_{\text{l}}=I_{\text{f}}=2500\) A; thus, the value of the amplitude phase current \(I_{\text{mf}}\) is as follows: \(I_{\text{m}}=I_{\text{l}}\cdot\sqrt{2}=3540\) A. Using Equation (3), instantaneous current values are calculated at steps of \(\triangle wt\) = 10\({}^{0}\). \[H_{in}=\frac{\pm i_{in}}{2\pi r_{in}}; \tag{4}\]  
where \(i_{in}\) (\(i\) = A, B, C) (\(n\) = 1, 2) shows the phase current flowing through the conductor, which is calculated with Equation (3), A, \(r_{in}\) (\(i\) = A, B, C) (\(n\) = 1, 2) are the lengths of distances from phases and their reflections to the reference point M, in m., which are calculated from Figure 2, and \(H_{in}\) is the magnetic field strength, in A/m. The magnetic field strength generated by the double-circuit three-phase 400 kV overhead transmission line in the plane in Figure 2 is calculated as follows:  
\[H_{\text{M}}=\sqrt{H_{\text{M}inx}^{2}+H_{\text{M}iny}^{2}}; \tag{5}\]  
where \(H_{\text{M}inx}\) and \(H_{\text{M}iny}\) are the constituents of the magnetic field strength vectors with respect to \(x\) and \(y\) axes. Next, we must examine how the values of \(H\) vary with the measurement point M moving between the phase conductors of the power line.