Since we consider the transmission line in the form of a coaxial (cylindrical) capacitor, here we can use the well-known classical formulas for its calculation [1].

The electric field strength in a coaxial capacitor is found by the following formula: \[E = {q \over 2 \pi \varepsilon \varepsilon_0 r l} \qquad (1.8) \] where \(r \) is the radius, which varies from the value of the radius of the conductor \(r_1 \), to the value of the radius of the outer plate of the capacitor \(r_2 \), and \(l \) - capacitor length.

Let us dwell on the electric charge in a little more detail. It is found like this: \(q = C U \), where \(C \) is the capacitance of the capacitor, and \(U \) is the voltage between its plates, which, in general complex form, looks like this: \(U = U_0 \mathrm {e} ^ {i \omega t} \), where \(\omega = 2 \pi f \), and \(U_0 \) is the amplitude value of the voltage. In this paper, we consider the propagation of exclusively sinusoidal signals in power lines.

Recall the formula for the capacity of a coaxial capacitor: \[C = {2 \pi \varepsilon \varepsilon_0 l \over \ln (r_2 / r_1)} \qquad (1.9) \] Now we find the time derivative of (1.8), substituting this capacity there: \[{\partial E \over \partial t} = {1 \over r \ln (r_2 / r_1)} {\partial U \over \partial t} = {U_0 i \omega \mathrm {e} ^ {i \omega t} \over r \ln (r_2 / r_1)} \qquad (1.10) \] From here we can find the electrical component of the energy flow \[S_E = \varepsilon \varepsilon_0 E {\partial E \over \partial t} l = C U_0 ^ 2 {i \omega \mathrm {e} ^ {i 2 \omega t } \over 2 \pi r ^ 2 \ln (r_2 / r_1)} \qquad (1.11) \] and the electric power component according to (1.7): \[P_E = \int \limits_ {s} S_E \, ds = C U_0 ^ 2 {i \omega \mathrm {e} ^ {i 2 \omega t} \over 2 \pi \ln (r_2 / r_1)} \int \limits_ {r_1} ^ {r_2} {d (\pi r ^ 2) \over r ^ 2} = C U_0 ^ 2 \, i \omega \mathrm {e} ^ {i 2 \omega t} \qquad (1.12) \] Let's select only the real part of this power: \[P_E = - \omega C U_0 ^ 2 \sin (2 \omega t) \qquad (1.13) \] This part of the power is created by the bias current and must be carried by the E-wave, which has a longitudinal component [2].

To search for the magnetic component of the power, we will go the same way and first find the magnetic field strength inside the coaxial cable (Fig. 5) according to the well-known formula [1]: \[H = {I \over {2 \pi r}} \qquad (1.14) \] where \(I \) is the conduction current through the central core, which in general complex form looks like this: \(I = I_0 \mathrm {e} ^ {i \omega t} \), in this case \(I_0 \) is the amplitude value of this current.

Recall how the inductance of a coaxial capacitor is found: \[L = {\mu \mu_0 l \over 2 \pi} \ln (r_2 / r_1) \qquad (1.15) \] Now we find the time derivative of (1.14), substituting this inductance there: \[{\partial H \over \partial t} = {1 \over \mu \mu_0} {L I_0 i \omega \mathrm {e} ^ {i \omega t} \over rl \ln (r_2 / r_1)} \qquad (1.16) \] From here we can find the electrical component of the energy flow \[S_H = \mu \mu_0 H {\partial H \over \partial t} l = \omega L I_0 ^ 2 {i \mathrm {e} ^ {i 2 \omega t } \over 2 \pi r ^ 2 \ln (r_2 / r_1)} \qquad (1.17) \] and the magnetic power component according to (1.7): \[P_H = \int \limits_ {s} S_H \, ds = \omega L I_0 ^ 2 {i \mathrm {e} ^ {i 2 \omega t} \over 2 \pi \ln (r_2 / r_1)} \int \limits_ {r_1} ^ {r_2} {d (\pi r ^ 2) \over r ^ 2} \qquad (1.18) \] Finally: \[P_H = \omega L I_0 ^ 2 \, i \mathrm {e} ^ {i 2 \omega t} \qquad (1.19) \] Here we also need to select only the real part of this power: \[P_H = - \omega L I_0 ^ 2 \sin (2 \omega t) \qquad (1.20) \] Obviously, unlike the previous electrical power component, the magnetic one is created due to the electromagnetic wave with the H-component. In real ROES, this power component is relatively small and in many cases may not be taken into account.