A METHOD OF SOLVING CLOSED FORM EXACT SOLUTION FOR THE FIELD GENERATED BY A FINITE-LENGTH HARMONIC LINEAR CURRENT SOURCE IN A WHOLE SPACE

Abstract

A method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in a whole space. The vector magnetic potential formula of the finite-length harmonic linear current source containing a source point position vector is listed, the uniform current is subjected to cosine processing, and the current in the vector magnetic potential formula of the harmonic linear current source is expressed by a cosine function. The vector magnetic potential formula can be subjected to quadrature by an elementary function to obtain the closed form exact solution for the field generated by the finite-length harmonic linear current source in whole space. The cosine expression of the linear current source can better reflect the fundamental attributes of the electric dipole and harmonic current of linear current source in the conductive whole space. The obtained closed form exact solution is applicable in the all zone.

Claims

1-6. (canceled)

7. A method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in whole space, the method comprising: step 1, listing integral formula for vector magnetic potential containing a source point position vector of a finite-length harmonic linear current source; step 2, performing a cosine processing for the uniform current; step 3, expressing current in the integral formula for vector magnetic potential containing the source point position vector by a cosine function; and step 4, solving the integral containing the source point position vector to obtain the closed form exact solution for the field generated by the finite-length harmonic linear current source in the whole space.

8. The method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in whole space according to claim 7, wherein, taking a cylindrical-coordinate system, a middle point of the linear current source coincides with coordinates origin O and is placed along the z axis; in a uniform, linear, isotropic and time-invariant unbounded medium, a vector magnetic potential formula A(r) of the harmonic linear current source containing the source point position vector is listed: $\begin{array}{cc}A\left(r\right)=\hat{z}{A}_z\left(\rho ,z\right)=\hat{z}\frac{\mu }{4\pi }{\int }_{-l\text{/}2}^{l\text{/}2}\frac{I\left(z^{\prime }\right)e^{-\mathrm{jk}.Math.r-r^{\prime }.Math.}}{.Math.r-r^{\prime }.Math.}{\mathrm{dz}}^{\prime }=\hat{z}\frac{\mu }{4\pi }\stackrel{l\text{/}2}{\underset{-l\text{/}2}{\int }}\frac{I\left(z^{\prime }\right)e^{-\mathrm{jk}\sqrt{{\rho }^2+{\left(z-z^{\prime }\right)}^2}}}{\sqrt{{\rho }^2+{\left(z-z^{\prime }\right)}^2}}{\mathrm{dz}}^{\prime }=\hat{z}\frac{\mu }{4\pi }\stackrel{l\text{/}2}{\underset{-l\text{/}2}{\int }}\frac{I\left(z^{\prime }\right)e^{-\mathrm{jkR}}}{R}{\mathrm{dz}}^{\prime }& \end{array}$ wherein A.sub.z(ρ,z) is the z component of the vector magnetic potential A(r); due to the symmetry, A.sub.z is only the function of (ρ,z) among three coordinate variables (ρ,φ,z) of the cylindrical-coordinate system, {circumflex over (z)} is the unit vector along the z-axis direction, r is the field point position vector, r′ is the source point position vector, z′ is the source point position coordinate, μ is the permeability, l is the length of the linear current source, I(z) is the current distribution function, j is the imaginary unit, R is the distance from the source point to the field point, k is the wave number, and p is the radial distance in the cylindrical-coordinate system.

9. The method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in whole space according to claim 8, wherein the wave number is:
k=√{square root over (ω.sup.2με−jωμσ)}=√{square root over (2πfμ(2πfε−jσ))} wherein ω is the angular frequency of the source current, f is the frequency of the source current, ε is the permittivity, and σ is the conductivity.

10. The method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in a whole space according to claim 7, wherein, the current at the middle point of the linear current source is set to I.sub.0, and the uniform current is subjected to a cosine processing to obtain: $I\left(z^{\prime }\right)=I_0\cos \left({\mathrm{kz}}^{\prime }\right),-\frac{l}{2}\le z^{\prime }\le \frac{l}{2}\text{;}$ wherein I(z) is a current distribution function, I.sub.0 is a peak value of the harmonic current, k is a wave number, and z′ is a source point position coordinate.

11. The method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in whole space according to claim 7, wherein the current in integral for the vector magnetic potential of the harmonic linear current source containing the source point position vector is expressed by cosine function as following: $A\left(r\right)=\hat{z}\frac{\mu I_0}{4\pi }\stackrel{l\text{/}2}{\underset{-l\text{/}2}{\int }}\cos \left({\mathrm{kz}}^{\prime }\right)\frac{e^{-\mathrm{jk}.Math.r-r^{\prime }.Math.}}{.Math.r-r^{\prime }.Math.}{\mathrm{dz}}^{\prime }=\hat{z}\frac{\mu I_0}{4\pi }\stackrel{l\text{/}2}{\underset{-l\text{/}2}{\int }}\cos \left({\mathrm{kz}}^{\prime }\right)\frac{e^{-\mathrm{jkR}}}{R}{\mathrm{dz}}^{\prime }.$

12. The method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in whole space according to claim 8, wherein the current in integral for the vector magnetic potential of the harmonic linear current source containing the source point position vector is expressed by cosine function as following: $A\left(r\right)=\hat{z}\frac{\mu I_0}{4\pi }\stackrel{l\text{/}2}{\underset{-l\text{/}2}{\int }}\cos \left({\mathrm{kz}}^{\prime }\right)\frac{e^{-\mathrm{jk}.Math.r-r^{\prime }.Math.}}{.Math.r-r^{\prime }.Math.}{\mathrm{dz}}^{\prime }=\hat{z}\frac{\mu I_0}{4\pi }\stackrel{l\text{/}2}{\underset{-l\text{/}2}{\int }}\cos \left({\mathrm{kz}}^{\prime }\right)\frac{e^{-\mathrm{jkR}}}{R}{\mathrm{dz}}^{\prime }.$

13. The method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in whole space according to claim 7, wherein the step of solving the exact solution of the integral containing the source point position vector comprises the following steps: step 4.1, the relation formula between the vector magnetic potential A(r) containing the source point position vector and the magnetic field intensity H(r) is expanded in the cylindrical-coordinate system to obtain the following magnetic field intensity H(r): $H\left(r\right)=\frac{1}{\mu }\nabla \times A\left(r\right)=-\hat{\phi }\frac{1}{\mu }\frac{\partial A_z\left(\rho ,z\right)}{\partial \rho }=\hat{\phi }{H}_{\phi }\text{;}$ in the formula, {circumflex over (φ)} is the unit vector along the go direction; step 4.2, the vector magnetic potential A(r) containing the source point position vector is put into the magnetic field intensity H(r) to obtain: $H_{\phi }=-\frac{I_0}{4\pi }\frac{\partial }{\partial \rho }\underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}\cos \left({\mathrm{kz}}^{\prime }\right)\frac{e^{-\mathrm{jkR}}}{R}{\mathrm{dz}}^{\prime }\text{;}$ step 4.3, the exact solution is obtained by using Euler formula and homogeneous Maxwell equation: $\begin{array}{c}\begin{array}{c}{H}_{\phi }=\frac{I_0}{4\pi \rho }[\left(\frac{l\text{/}2-z}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{l\text{/}2+z}{R_2}{e}^{-{\mathrm{jkR}}_2}\right)\cos \left(\frac{\mathrm{kl}}{2}\right)+\\ j\left(e^{-{\mathrm{jkR}}_1}+e^{-{\mathrm{jkR}}_2}\right)\sin \left(\frac{\mathrm{kl}}{2}\right)]\text{;}\\ E_{\rho }=\frac{{\mathrm{jI}}_0}{4\pi \rho }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[\left(\frac{1}{{\mathrm{kR}}_1^3}+\frac{j}{R_1^2}\right){\left(\frac{l}{2}-z\right)}^2{e}^{-{\mathrm{jkR}}_1}-\frac{1}{{\mathrm{kR}}_1}{e}^{-{\mathrm{jkR}}_1}-\\ \left(\frac{1}{{\mathrm{kR}}_2^3}+\frac{j}{R_2^2}\right){\left(\frac{l}{2}+z\right)}^2{e}^{-{\mathrm{jkR}}_2}+\frac{1}{{\mathrm{kR}}_2}{e}^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-\\ \left[\frac{1}{R_1}\left(\frac{l}{2}-z\right)e^{-{\mathrm{jkR}}_1}-\frac{1}{R_2}\left(\frac{l}{2}+z\right)e^{-{\mathrm{jkR}}_2}\right]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\text{;}\end{array}\\ \begin{array}{c}{E}_z=\frac{{\mathrm{jI}}_0}{4\pi }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[\left(\frac{j}{R_1^2}+\frac{1}{{\mathrm{kR}}_1^3}\right)\left(\frac{l}{2}-z\right)e^{-{\mathrm{jkR}}_1}+\\ \left(\frac{j}{R_2^2}+\frac{1}{{\mathrm{kR}}_2^3}\right)\left(\frac{l}{2}+z\right)e^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-\\ \left[\frac{1}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{1}{R_2}{e}^{-{\mathrm{jkR}}_2}\right]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\text{;}\end{array}\end{array}$ wherein R.sub.1=√{square root over (ρ.sup.2+(z−l/2).sup.2)} and R.sub.2=√{square root over (ρ.sup.2+(z+l/2).sup.2)}.

14. The method of solving closed form exact solution for the field generated by a finite-Docket No. 451-001 US length harmonic linear current source in whole space according to claim 8, wherein the step of solving the exact solution of the integral containing the source point position vector comprises the following steps: step 4.1, the relation formula between the vector magnetic potential A(r) containing the source point position vector and the magnetic field intensity H(r) is expanded in the cylindrical-coordinate system to obtain the following magnetic field intensity H(r): $H\left(r\right)=\frac{1}{\mu }\nabla \times A\left(r\right)=-\hat{\phi }\frac{1}{\mu }\frac{\partial A_z\left(\rho ,z\right)}{\partial \rho }=\hat{\phi }{H}_{\phi }\text{;}$ in the formula, {circumflex over (φ)} is the unit vector along the go direction; step 4.2, the vector magnetic potential A(r) containing the source point position vector is put into the magnetic field intensity H(r) to obtain: $H_{\phi }=-\frac{I_0}{4\pi }\frac{\partial }{\partial \rho }\underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}\cos \left({\mathrm{kz}}^{\prime }\right)\frac{e^{-\mathrm{jkR}}}{R}{\mathrm{dz}}^{\prime }\text{;}$ step 4.3, the exact solution is obtained by using Euler formula and homogeneous Maxwell equation: $\begin{array}{c}\begin{array}{c}{H}_{\phi }=\frac{I_0}{4\pi \rho }[\left(\frac{l\text{/}2-z}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{l\text{/}2+z}{R_2}{e}^{-{\mathrm{jkR}}_2}\right)\cos \left(\frac{\mathrm{kl}}{2}\right)+\\ j\left(e^{-{\mathrm{jkR}}_1}+e^{-{\mathrm{jkR}}_2}\right)\sin \left(\frac{\mathrm{kl}}{2}\right)]\text{;}\\ E_{\rho }=\frac{{\mathrm{jI}}_0}{4\pi \rho }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[\left(\frac{1}{{\mathrm{kR}}_1^3}+\frac{j}{R_1^2}\right){\left(\frac{l}{2}-z\right)}^2{e}^{-{\mathrm{jkR}}_1}-\frac{1}{{\mathrm{kR}}_1}{e}^{-{\mathrm{jkR}}_1}-\\ \left(\frac{1}{{\mathrm{kR}}_2^3}+\frac{j}{R_2^2}\right){\left(\frac{l}{2}+z\right)}^2{e}^{-{\mathrm{jkR}}_2}+\frac{1}{{\mathrm{kR}}_2}{e}^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-\\ \left[\frac{1}{R_1}\left(\frac{l}{2}-z\right)e^{-{\mathrm{jkR}}_1}-\frac{1}{R_2}\left(\frac{l}{2}+z\right)e^{-{\mathrm{jkR}}_2}\right]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\text{;}\end{array}\\ \begin{array}{c}{E}_z=\frac{{\mathrm{jI}}_0}{4\pi }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[\left(\frac{j}{R_1^2}+\frac{1}{{\mathrm{kR}}_1^3}\right)\left(\frac{l}{2}-z\right)e^{-{\mathrm{jkR}}_1}+\\ \left(\frac{j}{R_2^2}+\frac{1}{{\mathrm{kR}}_2^3}\right)\left(\frac{l}{2}+z\right)e^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-\\ \left[\frac{1}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{1}{R_2}{e}^{-{\mathrm{jkR}}_2}\right]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\text{;}\end{array}\end{array}$ wherein R.sub.1=√{square root over (ρ.sup.2+(z−l/2).sup.2)} and R.sub.2=√{square root over (ρ.sup.2+(z+l/2).sup.2)}.

Description

IV. BRIEF DESCRIPTION OF DRAWINGS

[0019] FIG. 1 shows a linear current source and a coordinate system;

[0020] FIG. 2 shows the current distribution of a finite-length harmonic linear current source in a free space;

[0021] FIG. 3 shows the current distribution of a finite-length harmonic linear current source in the wet soil environment;

[0022] FIG. 4 shows the current distribution of a finite-length harmonic linear current source in the seawater environment; and

[0023] FIG. 5 shows a flow diagram of a method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in a whole space.

V. DETAILED DESCRIPTION OF THE INVENTION

[0024] To clarify the purpose, technical solutions and advantages of the present invention, the present invention is further described below in conjunction with the drawings and examples. It should be understood that the examples described herein are only used to explain the present invention, not to limit the present invention.

[0025] As shown in FIG. 5, a method of solving closed form exact solution for the field generated by a finite-length harmonic linear current source in a whole space includes:

[0026] step 1, listing integral formula for vector magnetic potential containing a source point position vector of a finite-length harmonic linear current source;

[0027] step 2, performing a cosine processing for the uniform current;

[0028] step 3, expressing current in the integral formula for vector magnetic potential containing the source point position vector by a cosine function; and

[0029] step 4, solving the integral containing the source point position vector to obtain the closed form exact solution for the field generated by the finite-length harmonic linear current source in the whole space.

[0030] Further, taking a cylindrical-coordinate system, a middle point of the linear current source coincides with coordinates origin O and is placed along the z axis; as shown in FIG. 1, in a uniform, linear, isotropic and time-invariant unbounded medium, a vector magnetic potential formula A(r) of the harmonic linear current source containing the source point position vector is listed:

[00010]$\begin{array}{cc}A\left(r\right)=\hat{z}{A}_z\left(\rho ,z\right)=\hat{z}\frac{\mu }{4\pi }{\int }_{-l\text{/}2}^{l\text{/}2}\frac{I\left(z^{\prime }\right)e^{-\mathrm{jk}.Math.r-r^{\prime }.Math.}}{.Math.r-r^{\prime }.Math.}{\mathrm{dz}}^{\prime }=\hat{z}\frac{\mu }{4\pi }\stackrel{l\text{/}2}{\underset{-l\text{/}2}{\int }}\frac{I\left(z^{\prime }\right)e^{-\mathrm{jk}\sqrt{{\rho }^2+{\left(z-z^{\prime }\right)}^2}}}{\sqrt{{\rho }^2+{\left(z-z^{\prime }\right)}^2}}{\mathrm{dz}}^{\prime }=\hat{z}\frac{\mu }{4\pi }\underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}\frac{I\left(z^{\prime }\right)e^{-\mathrm{jkR}}}{R}{\mathrm{dz}}^{\prime }& \left(1a\right)\end{array}$

where A.sub.z(ρ,z) is the z component of the vector magnetic potential A(r); due to the symmetry, A.sub.z is only the function of (ρ,z) among three coordinate variables (ρ,φ,z) of the cylindrical-coordinate system, {circumflex over (z)} is the unit vector along the z-axis direction, r is the field point position vector, r′ is the source point position vector, z′ is the source point position coordinate, μ is the permeability, l is the length of the linear current source, I(z) is the current distribution function, j is the imaginary unit, R is the distance from the source point to the field point, k is the wave number, and co is the radial distance in the cylindrical-coordinate system.

k=√{square root over (ω2με−jωμσ=)}=√{square root over (2πfμ(2πfε−jσ))}  (1b)

where ω is the angular frequency (in the unit of rad/s) of the source current, f is the frequency (in the unit of Hz) of the source current, ε is the permittivity (in the unit of F/m), and σ is the conductivity (in the unit of S/m).

[0031] Further, the uniform current is subjected to cosine processing as following: setting the current at the middle point of the linear current source to I.sub.0;

[00011]$\begin{array}{cc}I\left(z^{\prime }\right)=I_0\cos \left({\mathrm{kz}}^{\prime }\right),-\frac{l}{2}\le z^{\prime }\le \frac{l}{2}\text{;}& \left(2\right)\end{array}$

where I.sub.0 is the peak value of harmonic current.

[0032] Further, the formula (2) is put into the formula (1a) to obtain the integral expression containing the source point position vector:

[00012]$\begin{array}{cc}A\left(r\right)=\hat{z}\frac{\mu I_0}{4\pi }\underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}\cos \left({\mathrm{kz}}^{\prime }\right)\frac{e^{-\mathrm{jk}.Math.r-r^{\prime }.Math.}}{.Math.r-r^{\prime }.Math.}{\mathrm{dz}}^{\prime }=\hat{z}\frac{\mu I_0}{4\pi }\underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}\cos \left({\mathrm{kz}}^{\prime }\right)\frac{e^{-\mathrm{jkR}}}{R}{\mathrm{dz}}^{\prime }\text{;}& \left(3\right)\end{array}$

[0033] Further, a process for obtaining the exact solution includes:

step 4.1, the relation formula between the vector magnetic potential A(r) containing the source point position vector and the magnetic field intensity H(r) is expanded in the cylindrical-coordinate system to obtain the following magnetic field intensity H(r):

[00013]$\begin{array}{cc}H\left(r\right)=\frac{1}{\mu }\nabla \times A\left(r\right)=-\hat{\phi }\frac{1}{\mu }\frac{\partial A_z\left(\rho ,z\right)}{\partial \rho }=\hat{\phi }{H}_{\phi }\text{;}& \left(4\right)\end{array}$

in the formula, {circumflex over (φ)} is the unit vector along the co direction.
step 4.2, the integral formula (3) of the vector magnetic potential A(r) containing the source point position vector is put into the formula (4) of the magnetic field intensity H(r), to obtain:

[00014]$\begin{array}{cc}{H}_{\phi }=-\frac{I_0}{4\pi }\frac{\partial }{\partial \rho }\underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}\cos \left({\mathrm{kz}}^{\prime }\right)\frac{e^{-\mathrm{jkR}}}{R}{\mathrm{dz}}^{\prime }\text{;}& \left(5a\right)\end{array}$

step 4.3, the exact solution is obtained by using Euler formula and homogeneous Maxwell equation: according to the Euler formula:

[00015]$\begin{array}{cc}\cos \left({\mathrm{kz}}^{\prime }\right)=\frac{e^{j\left({\mathrm{kz}}^{\prime }\right)}+e^{-j\left({\mathrm{kz}}^{\prime }\right)}}{2}& \left(5b\right)\end{array}$

formula (5b) is put into the formula (5a) for expansion to obtain

[00016]$\begin{array}{cc}\begin{array}{c}{H}_{\phi }=-\frac{I_0}{8\pi }\stackrel{l\text{/}2}{\underset{-l\text{/}2}{\int }}\left[\frac{\partial }{\partial \rho }\frac{e^{-\mathrm{jk}\left(R-z^{\prime }\right)}}{R}+\frac{\partial }{\partial \rho }\frac{e^{-\mathrm{jk}\left(R+z^{\prime }\right)}}{R}\right]{\mathrm{dz}}^{\prime }\\ =-\frac{I_0}{8\pi }\{\underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}[\frac{1}{R}\frac{\partial }{\partial \rho }{e}^{-\mathrm{jk}\left(R-z^{\prime }\right)}+e^{-\mathrm{jk}\left(R-z^{\prime }\right)}\frac{\partial }{\partial \rho }\frac{1}{R}{\mathrm{dz}}^{\prime }+\\ \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}\left[\frac{1}{R}\frac{\partial }{\partial \rho }{e}^{-\mathrm{jk}\left(R+z^{\prime }\right)}+e^{-\mathrm{jk}\left(R+z^{\prime }\right)}\frac{\partial }{\partial \rho }\frac{1}{R}\right]{\mathrm{dz}}^{\prime }\}\\ =-\frac{I_0}{8\pi }\{\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}\left[-\mathrm{jk}\frac{e^{-\mathrm{jk}\left(R-z^{\prime }\right)}}{R^2}-\frac{e^{-\mathrm{jk}\left(R-z^{\prime }\right)}}{R^3}\right]{\mathrm{dz}}^{\prime }+\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}[-\mathrm{jk}\frac{e^{-\mathrm{jk}(R+z^{\prime }}}{R^2}-\\ \frac{e^{-\mathrm{jk}\left(R+z^{\prime }\right)}}{R^3}]{\mathrm{dz}}^{\prime }\}\\ =-\frac{I_0}{8\pi }\{\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}{e}^{-\mathrm{jk}\left(R-z^{\prime }\right)}\left(-\mathrm{jk}\frac{1}{R^2}-\frac{1}{R^3}\right){\mathrm{dz}}^{\prime }+\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}{e}^{-\mathrm{jk}\left(R+z^{\prime }\right)}(-\mathrm{jk}\frac{1}{R^2}-\\ \frac{1}{R^3}){\mathrm{dz}}^{\prime }\}\\ =-\frac{I_0}{8\pi }\{\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}{e}^{-\mathrm{jk}\left(R-z^{\prime }\right)}[-\mathrm{jk}\frac{1}{R^2}{\mathrm{dz}}^{\prime }+\frac{z^{\prime }-z}{R^3\left(R-z^{\prime }+z\right)}{\mathrm{dz}}^{\prime }-\\ \frac{1}{R^2\left(R-z^{\prime }+z\right)}{\mathrm{dz}}^{\prime }]+\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}{e}^{-\mathrm{jk}\left(R+z^{\prime }\right)}[-\mathrm{jk}\frac{1}{R^2}{\mathrm{dz}}^{\prime }-\\ \frac{z^{\prime }-z}{R^3\left(R+z^{\prime }-z\right)}{\mathrm{dz}}^{\prime }-\frac{1}{R^2\left(R+z^{\prime }-z\right)}{\mathrm{dz}}^{\prime }]\}\\ =-\frac{I_0}{8\pi }\{\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}{e}^{-\mathrm{jk}\left(R-z^{\prime }\right)}[\mathrm{jk}\frac{d\left(R-z^{\prime }\right)}{R\left(R-z^{\prime }+z\right)}-\frac{{\mathrm{dR}}^{-1}}{\left(R-z^{\prime }+z\right)}{\mathrm{dz}}^{\prime }-\\ \frac{{d\left(R-z^{\prime }+z\right)}^{-1}}{R}]+\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}{e}^{-\mathrm{jk}\left(R+z^{\prime }\right)}[-\mathrm{jk}\frac{d\left(R+z^{\prime }\right)}{R\left(R+z^{\prime }-z\right)}+\\ \frac{{\mathrm{dR}}^{-1}}{\left(R+z^{\prime }-z\right)}+\frac{{d\left(R+z^{\prime }-z\right)}^{-1}}{R}]\}\\ =\frac{I_0}{8\pi }\left\{\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}d\left[\frac{e^{-\mathrm{jk}\left(R-z^{\prime }\right)}}{R\left(R-z^{\prime }+z\right)}\right]-\rho \underset{-l\text{/}2}{\stackrel{l\text{/}2}{\int }}d\left[\frac{e^{-\mathrm{jk}\left(R+z^{\prime }\right)}}{R\left(R+z^{\prime }-z\right)}\right]\right\}\\ =\frac{I_0\rho }{8\pi }\left[\frac{e^{-\mathrm{jk}\left(R-z^{\prime }\right)}}{R\left(R-z^{\prime }+z\right)}+\frac{e^{-\mathrm{jk}\left(R+z^{\prime }\right)}}{R\left(R+z^{\prime }-z\right)}\right]{|}_{-l\text{/}2}^{l\text{/}2}\\ =\frac{I_0\rho }{8\pi }[\frac{e^{-\mathrm{jk}\left(R_1-l\text{/}2\right)}}{R_1\left(R_1-l\text{/}2+z\right)}-\frac{e^{-\mathrm{jk}\left(R_2+l\text{/}2\right)}}{R_2\left(R_2+l\text{/}2-z\right)}-\frac{e^{-\mathrm{jk}\left(R_1+l\text{/}2\right)}}{R_1\left(R_1+l\text{/}2-z\right)}+\\ \frac{e^{-\mathrm{jk}\left(R_2-l\text{/}2\right)}}{R_2\left(R_2-l\text{/}2-z\right)}]\\ =\frac{I_0\rho }{8\pi }\{\frac{e^{-{\mathrm{jkR}}_1}}{R_1}\left[\frac{e^{\mathrm{jk}l\text{/}2}}{\left(R_1-l\text{/}2+z\right)}-\frac{e^{-\mathrm{jk}l\text{/}2}}{\left(R_1+l\text{/}2-z\right)}\right]+\\ \frac{e^{-{\mathrm{jkR}}_2}}{R_2}\left[\frac{e^{\mathrm{jk}l\text{/}2}}{\left(R_2-l\text{/}2-z\right)}-\frac{e^{-\mathrm{jk}l\text{/}2}}{\left(R_2+l\text{/}2-z\right)}\right]\}\\ =\frac{I_0\rho }{8\pi }\{\frac{e^{-{\mathrm{jkR}}_1}}{R_1}\left[\frac{\cos \left(\mathrm{kl}\text{/}2\right)+j\sin \left(\mathrm{kl}\text{/}2\right)}{\left(R_1+l\text{/}2+z\right)}-\frac{\cos \left(\mathrm{kl}\text{/}2\right)-j\sin \left(\mathrm{kl}\text{/}2\right)}{\left(R_1+l\text{/}2-z\right)}\right]-\\ \frac{e^{-{\mathrm{jkR}}_2}}{R_2}\left[\frac{\cos \left(\mathrm{kl}\text{/}2\right)+j\sin \left(\mathrm{kl}\text{/}2\right)}{\left(R_2-l\text{/}2-z\right)}-\frac{\cos \left(\mathrm{kl}\text{/}2\right)-j\sin \left(\mathrm{kl}\text{/}2\right)}{\left(R_2+l\text{/}2-z\right)}\right]\}\\ =\frac{I_0}{4\pi \rho }[\left(\frac{l\text{/}2-z}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{l\text{/}2+z}{R_2}{e}^{-{\mathrm{jkR}}_2}\right)\cos \left(\frac{\mathrm{kl}}{2}\right)+\\ j\left(e^{-{\mathrm{jkR}}_1}+e^{-{\mathrm{jkR}}_2}\right)\sin \left(\frac{\mathrm{kl}}{2}\right)]\text{;}\end{array}& \left(6\right)\end{array}$

according to the homogeneous Maxwell equation:

[00017]$\begin{array}{cc}E\left(r\right)=\frac{1}{j\omega \mathrm{.Math.}+\sigma }\nabla \times H\left(r\right)& \left(7\right)\end{array}$

an electric field expression is obtained:

[00018]$\begin{array}{cc}\begin{array}{c}{E}_{\rho }=-\frac{1}{j\omega \mathrm{.Math.}+\sigma }\frac{\partial H_{\phi }}{\partial z}\\ =\frac{j{I}_0}{4\pi \rho }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[(\frac{1}{{\mathrm{kR}}_1^3}+\frac{j}{R_1^2}{\left(\frac{l}{2}-z\right)}^2{e}^{-{\mathrm{jkR}}_1}-\frac{1}{{\mathrm{kR}}_1}{e}^{-{\mathrm{jkR}}_1}-\\ \left(\frac{1}{{\mathrm{kR}}_2^3}+\frac{j}{R_2^2}\right){\left(\frac{l}{2}+z\right)}^2{e}^{-{\mathrm{jkR}}_2}+\frac{1}{{\mathrm{kR}}_2}{e}^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-\\ \left[\frac{1}{R_1}\left(\frac{l}{2}-z\right)e^{-{\mathrm{jkR}}_1}-\frac{1}{R_2}\left(\frac{l}{2}+z\right)e^{-{\mathrm{jkR}}_2}\right]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\text{;}\end{array}& \left(8\right)\\ \begin{array}{c}{E}_z=\frac{1}{j\mathrm{\omega .Math.}+\sigma }\frac{1}{\rho }\frac{\partial \left(\rho H_{\phi }\right)}{\partial \rho }=\frac{1}{j\omega \mathrm{.Math.}+\sigma }\left(\frac{\partial H_{\phi }}{\partial \rho }+\frac{1}{\rho }{H}_{\phi }\right)\\ =\frac{j{I}_0}{4\pi }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[\left(\frac{j}{R_1^2}+\frac{1}{{\mathrm{kR}}_1^3}\right)\left(\frac{l}{2}-z\right)e^{-{\mathrm{jkR}}_1}+\left(\frac{j}{R_2^2}+\frac{1}{{\mathrm{kR}}_2^3}\right)(\frac{l}{2}+\\ z)e^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-\left[\frac{1}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{1}{R_2}{e}^{-{\mathrm{jkR}}_2}\right]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\text{;}\end{array}& \left(9\right)\end{array}$

where

R.sub.1=√{square root over (ρ.sup.2+(z−l/2).sup.2)},  (10a)

R.sub.2=√{square root over (ρ.sup.2(z+l/2).sup.2)};  (10b)

[0034] Example 1, in the free space I.sub.0=1 A, l=2 m, f=10.sup.6 Hz, μ=μ.sub.0=4π×10.sup.−7 H/m,

[00019]$\mathrm{.Math.}={\mathrm{.Math.}}_0\approx \frac{{10}^{-9}}{36\pi }$

F/m and σ=0 S/m are put into the formula (2) to obtain current distribution of a finite-length harmonic linear current source in a free space, as shown in FIG. 2;
according to the formula (6), formula (8) and formula (9) of the exact solution, in the free space, the field generated by the harmonic linear current source is (wherein l=2 m, I.sub.0=1 A and f=10.sup.6 Hz):

[00020]$\begin{array}{cc}\begin{array}{c}{H}_{\phi }=\frac{I_0}{4\pi \rho }[\left(\frac{l\text{/}2-z}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{l\text{/}2+z}{R_2}{e}^{-{\mathrm{jkR}}_2}\right)\cos \left(\frac{\mathrm{kl}}{2}\right)+j(e^{-{\mathrm{jkR}}_1}+\\ e^{-{\mathrm{jkR}}_2})\sin \left(\frac{\mathrm{kl}}{2}\right)]\\ =\frac{1}{4\pi \rho }[\left(\frac{1-z}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{1+z}{R_2}{e}^{-{\mathrm{jkR}}_2}\right)\cos \left(k\right)+j(e^{-{\mathrm{jkR}}_1}+\\ e^{-{\mathrm{jkR}}_2})\sin \left(k\right)]\end{array}& \left(11\right)\\ \begin{array}{c}{E}_{\rho }=\frac{{\mathrm{jI}}_0}{4\pi \rho }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[\left(\frac{1}{{\mathrm{kR}}_1^3}+\frac{j}{R_1^2}\right){\left(\frac{l}{2}-z\right)}^2{e}^{-{\mathrm{jkR}}_1}-\frac{1}{{\mathrm{kR}}_1}{e}^{-{\mathrm{jkR}}_1}-\\ \left(\frac{1}{{\mathrm{kR}}_2^3}+\frac{j}{R_2^2}\right){\left(\frac{l}{2}+z\right)}^2{e}^{-{\mathrm{jkR}}_2}+\frac{1}{{\mathrm{kR}}_2}{e}^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-[\frac{1}{R_1}(\frac{l}{2}-\\ z)e^{-{\mathrm{jkR}}_1}-\frac{1}{R_2}\left(\frac{l}{2}+z\right)e^{-{\mathrm{jkR}}_2}]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\\ \approx \frac{j30}{\rho }\{[\left(\frac{1}{{\mathrm{kR}}_1^3}+\frac{j}{R_1^2}\right){\left(1-z\right)}^2{e}^{-{\mathrm{jkR}}_1}-\frac{1}{{\mathrm{kR}}_1}{e}^{-{\mathrm{jkR}}_1}-(\frac{1}{{\mathrm{kR}}_2^3}+\\ \frac{j}{R_2^2}){\left(1+z\right)}^2{e}^{-{\mathrm{jkR}}_2}+\frac{1}{{\mathrm{kR}}_2}{e}^{-{\mathrm{jkR}}_2}]\cos \left(k\right)-[\frac{1}{R_1}\left(1-z\right)e^{-{\mathrm{jkR}}_1}-\\ \frac{1}{R_2}\left(1+z\right)e^{-{\mathrm{jkR}}_2}]\sin \left(k\right)\}\end{array}& \left(12\right)\\ \begin{array}{c}{E}_z=\frac{{\mathrm{jI}}_0}{4\pi }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[\left(\frac{j}{R_1^2}+\frac{1}{{\mathrm{kR}}_1^3}\right)\left(\frac{l}{2}-z\right)e^{-{\mathrm{jkR}}_1}+\left(\frac{j}{R_2^2}+\frac{1}{{\mathrm{kR}}_2^3}\right)(\frac{l}{2}+\\ z)e^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-\left[\frac{1}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{1}{R_2}{e}^{-{\mathrm{jkR}}_2}\right]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\\ \approx j30\{[\left(\frac{j}{R_1^2}+\frac{1}{{\mathrm{kR}}_1^3}\right)\left(1-z\right)e^{-{\mathrm{jkR}}_1}+\left(\frac{j}{R_2^2}+\frac{1}{{\mathrm{kR}}_2^3}\right)(1+\\ z)e^{-{\mathrm{jkR}}_2}]\cos \left(k\right)-\left[\frac{1}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{1}{R_2}{e}^{-{\mathrm{jkR}}_2}\right]\sin \left(k\right)\}\end{array}& \left(13\right)\end{array}$

[0035] According to the formula (10), it can be known that in the above formula (11), formula (12) and formula (13):

R.sub.1=√{square root over (ρ.sup.2(z−l/2).sup.2)}=√{square root over (ρ.sup.2+(z−1).sup.2)}

R.sub.2=√{square root over (ρ.sup.2+(z+l/2).sup.2)}=√{square root over (ρ.sup.2(z+1).sup.2)}

[0036] According to the formula (1b), it can be known that the wave number k in the above formula (11), formula (12) and formula (13) are:

[00021]$k=\sqrt{2\pi f\mu \left(2\pi f\mathrm{.Math.}-j\sigma \right)}\approx \sqrt{2\pi \times {10}^6\times 4\pi \times {10}^{-7}\left(2\pi \times {10}^6\times \frac{{10}^{-9}}{36\pi }-j0\right)}=\sqrt{8{\pi }^2\times {10}^{-1}\left(\frac{{10}^{-3}}{18}\right)}=\frac{2\pi }{3}\times {10}^{-2}$

[0037] The field point coordinate P(ρ,φ,z) is put into the formula (11), formula (12) and formula (13) to obtain the spatial distribution of the field generated by the finite-length harmonic linear current source in the free space, wherein l=2 m, I.sub.0=1 A and f=10.sup.6 Hz.

[0038] Example 2, in the wet soil environment

I.sub.0=1 A, l=2000 m, f=100 Hz, μ=μ.sub.0=4π×10.sup.−7 H/m,

[00022]$\mathrm{.Math.}=10\times {\mathrm{.Math.}}_0\approx 10\frac{{10}^{-9}}{36\pi }F\text{/}m$

and σ=0.01 S/m are put into the formula (2) to obtain current distribution of a finite-length harmonic linear current source in the wet soil environment, as shown in FIG. 3;
according to the formula (6), formula (8) and formula (9) of the exact solution, in the wet soil environment, the field generated by the harmonic line current source is (wherein l=2000 m, I.sub.0=1 A and f=100 Hz):

[00023]$\begin{array}{cc}\begin{array}{c}{H}_{\phi }=\frac{I_0}{4\pi \rho }[\left(\frac{l\text{/}2-z}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{l\text{/}2+z}{R_2}{e}^{-{\mathrm{jkR}}_2}\right)\cos \left(\frac{\mathrm{kl}}{2}\right)+\\ j\left(e^{-{\mathrm{jkR}}_1}+e^{-{\mathrm{jkR}}_2}\right)\sin \left(\frac{\mathrm{kl}}{2}\right)]\\ =\frac{1}{4\pi \rho }[\left(\frac{1000-z}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{1000+z}{R_2}{e}^{-{\mathrm{jkR}}_2}\right)\cos \left(1000k\right)+\\ j\left(e^{-{\mathrm{jkR}}_1}+e^{-{\mathrm{jkR}}_2}\right)\sin \left(1000k\right)]\\ E_{\rho }=\frac{{\mathrm{jI}}_0}{4\pi \rho }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[\left(\frac{1}{{\mathrm{kR}}_1^3}+\frac{j}{R_1^2}\right){\left(\frac{l}{2}-z\right)}^2{e}^{-{\mathrm{jkR}}_1}-\frac{1}{{\mathrm{kR}}_1}{e}^{-{\mathrm{jkR}}_1}-\\ \left(\frac{1}{{\mathrm{kR}}_2^3}+\frac{j}{R_2^2}\right){\left(\frac{l}{2}+z\right)}^2{e}^{-{\mathrm{jkR}}_2}+\frac{1}{{\mathrm{kR}}_2}{e}^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-\\ \left[\frac{1}{R_1}\left(\frac{l}{2}-z\right)e^{-{\mathrm{jkR}}_1}-\frac{1}{R_2}\left(\frac{l}{2}+z\right)e^{-{\mathrm{jkR}}_2}\right]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\end{array}& \left(14\right)\\ \begin{array}{c}\approx \frac{0.05}{\rho \sqrt{5}}\{[\left(\frac{1}{{\mathrm{kR}}_1^3}+\frac{j}{R_1^2}\right){\left(1000-z\right)}^2{e}^{-{\mathrm{jkR}}_1}-\frac{1}{{\mathrm{kR}}_1}{e}^{-{\mathrm{jkR}}_1}-\\ \left(\frac{1}{{\mathrm{kR}}_2^3}+\frac{j}{R_2^2}\right){\left(1000+z\right)}^2{e}^{-{\mathrm{jkR}}_2}+\frac{1}{{\mathrm{kR}}_2}{e}^{-{\mathrm{jkR}}_2}]\cos \left(1000k\right)-\\ \left[\frac{1}{R_1}{\left(1000-z\right)}^{-{\mathrm{jkR}}_1}-\frac{1}{R_2}\left(1000+z\right)e^{-{\mathrm{jkR}}_2}\right]\sin \left(1000k\right)\}\\ E_z=\frac{{\mathrm{jI}}_0}{4\pi }\sqrt{\frac{\mu }{\mathrm{.Math.}-j\frac{\sigma }{\omega }}}\{[\left(\frac{j}{R_1^2}+\frac{1}{{\mathrm{kR}}_1^3}\right)\left(\frac{l}{2}-z\right)e^{-{\mathrm{jkR}}_1}+\\ \left(\frac{j}{R_2^2}+\frac{1}{{\mathrm{kR}}_2^3}\right)\left(\frac{l}{2}+z\right)e^{-{\mathrm{jkR}}_2}]\cos \left(\frac{\mathrm{kl}}{2}\right)-\\ \left[\frac{1}{R_1}{e}^{-{\mathrm{jkR}}_1}+\frac{1}{R_2}{e}^{-{\mathrm{jkR}}_2}\right]\sin \left(\frac{\mathrm{kl}}{2}\right)\}\end{array}& \left(15\right)\\ \begin{array}{c}\approx \frac{0.05}{\sqrt{5}}\{[\left(\frac{j}{R_1^2}+\frac{1}{{\mathrm{kR}}_1^3}\right)\left(1000-z\right)e^{-{\mathrm{jkR}}_1}+\\ \left(\frac{j}{R_2^2}+\frac{1}{{\mathrm{kR}}_2^3}\right)\left(1000+z\right)e^{-{\mathrm{jkR}}_2}]\cos \left(1000k\right)-[\frac{1}{R_1}{e}^{-{\mathrm{jkR}}_1}+\\ \frac{1}{R_2}{e}^{-{\mathrm{jkR}}_2}]\sin \left(1000k\right)\}\end{array}& \left(16\right)\end{array}$