## 2 Theoretical Background and Mathematical Models
### Rusck Closed Formulation and Rusck Coupling Model
#### 2.5.2 Rusck Formulation  
Rusck illustrates the LEMP phenomenon with a vertical and straight channel, as shown in Figure 7. To explain the LEMP effect, a negative charge dissipates along the LEMP path before the return stroke begins. The return stroke may be a current surge within the framework of a step function that moves upward throughout the LEMP channel with a constant velocity, neutralizing the charge. The Rusck return impact model is valid for a vertical and straight LEMP channel initially loaded with an evenly distributed load, as shown in Figure 7[28]. The geometry of the problem for the evaluation of the area is depicted in Figure 8. The return stroke current is a step current that is spread along the channel in an undistorted and undiminished state. The expression portraying the current distribution for the model is shown in Equation (8). \[\mathrm{i}(\mathrm{z}^{\prime},\mathrm{t})=\mathrm{I}_{0}\cdot\mathrm{u} \left(\mathrm{t}-\frac{\mathrm{z}^{\prime}}{v}\right)\,\mathrm{(A)} \tag{8}\]  
As it travels through the channel, the return current is initially charged with a favorable charge distribution, (\(\mathrm{q}_{0}\)), which neutralizes the negative charge distribution. The length of the initially loaded channel is limited and is expected to be (\(\mathrm{h}_{c}\)), the distance at any point in  
Figure 8: The geometry of the problem according to the Rusck formulation. Figure 7: The mathematical model of the LEMP according to the Rusck model. (\(v\): the velocity of the pulse along the LEMP channel; the length of the initially loaded channel is limited and is expected to be (\(\mathrm{h}_{c}\)). \(\mathrm{h}\): the height of the line from the ground). three-dimensional space (R) (Figure 8). The relationship between the current and charge distributions \(q_{0}\) is represented in Equation (9):  
\[I_{0}=q_{0}\cdot v \tag{9}\]  
\(\varphi\) is the scalar potential. \(\overset{\rightarrow}{A}\) defines the vector potential. \(\beta\) is the value related to the relative velocity. \(\Gamma\) is the Lorentz factor. \[\varphi(r,z,t)=\frac{1}{4\pi\varepsilon_{0}}\int_{s_{1}}^{h_{c}}\frac{q\left( z^{\prime},t-\frac{R}{c}\right)}{R}dz^{\prime} \tag{10}\]  
\[\overset{\rightarrow}{A}(r,z,t)=\frac{\mu_{0}}{4\pi\int_{0}^{s_{1}}\frac{i \left(z^{\prime},t-\frac{R}{c}\right)}{R}dz^{\prime}} \tag{11}\]  
\[\varphi(r,z,t)=\frac{-q_{0}}{4\pi\varepsilon_{0}}\int_{s_{1}}^{h_{c}}\frac{dz ^{\prime}}{\sqrt{(z_{s}-z)^{2}+r^{2}}} \tag{12}\]  
The potential created by the integral at the value of \(z=s_{1}\) for the lower limit value \(s_{1}\) and the ground parameter \(z\) value is shown in Equation (13), and the solution for \(s_{1}\) is depicted in Equation (14):  
\[t=\frac{s_{1}}{v}+\frac{\sqrt{(s_{1}-z)^{2}+r^{2}}}{c} \tag{13}\]  
Figure 8.0  
If the scalar potential formulation is rewritten after the solution, we obtain the solution presented in Equation (15):  
Figure 8.1  
The vector potential can be expressed as follows, given by Equations (13) and (16):  
Figure 8.2  
By substituting the general expression \(s_{1}\) into the vector potential in Equation (12), it is also possible to acquire the vector potential's vertical component, as denoted in Equation (17). Equation (18) expresses the magnetic field \"\(\overset{\rightarrow}{B}(\overset{\rightarrow}{b})\)\" and the electric field \"\(\overset{\rightarrow}{E}(\overset{\rightarrow}{e})\)\". Figure 8.3  
According to the Rusck formulation, the electric field is divided into two parts. The first part is expressed as the scalar potential (\(e_{\varphi}\)) and the vector potential (\(\overset{\rightarrow}{e_{\Delta}}\)). \[e_{\varphi}(r,z,t)=\frac{Z_{0}I_{0}}{4\pi\beta}\left( \begin{array}{c}1\\ \sqrt{\left(vt-z\right)^{2}+\left(\frac{r}{\Gamma}\right)^{2}}\\ -\frac{1}{\sqrt{(h_{c}-z)^{2}+r^{2}}}\end{array}