An Analytical Determination of the Magnetic Field in a System of Finite-Length Ribbon Busbars

Abstract

Using the analytic method based on the Biot–Savart law for the electromagnetic field, the distribution of the magnetic field of a ribbon busbar of finite length was determined. The analytical formulas describing the magnetic field in all areas of the considered ribbon busbars were obtained. The Mathematica program was used to visualize the solutions obtained. The Mathematica programme is a good and convenient tool for analytical measurements using the integration function and conversion of the analytical solutions, for the determination of field quantities, and for the graphical visualisation of the obtained final solutions. This allowed for quick field analysis to be conducted after changes were made in the geometrical or electrical parameters of the systems under examination.

1. Introduction

In the course of teaching the electromagnetic field theory, challenges often arise that require extensive calculations, not only in numerically solving problems [1,2,3,4,5,6,7,8,9] but also in their analytical resolution—involving integration, differentiation, rearrangement, and condensation of expressions, as well as seeking special solutions for terminal values of certain parameters or terminal domains [10,11,12,13,14,15]. The swift visualization of the obtained solutions is equally important, particularly spatial visualization, which enables us to understand the geometrical changes in a given configuration, the values of electrical parameters (current, voltage, external electric and magnetic fields, frequency, etc.), and the electrical parameters of the configuration (conductivity, permittivity, and magnetic permeability, etc.) [10,12,16,17,18,19,20]. Possessing a suitable tool to accomplish these objectives is highly instructive, aiding in memorizing analysed issues and assisting students in understanding and interpreting derived formulas. This enables a deeper insight into the characteristics of the electromagnetic field [13,14].

Tools such as QuickField, Flux, Comsol, Matlab, Agros, and Ansys are specialized instruments that can be used for solving field problems [21,22,23,24,25,26,27]. However, various mathematical programs can be easily applied to solve electric field problems. MatCad and Mathematica are undoubtedly classified among these programs [28,29]. The Mathematica program offers advanced tools for data visualization and computation results, allowing for the presentation of findings in an attractive and clear manner. It is well-regarded among a wide range of users, including scientists, engineers, students, and professionals from various fields.

A precise determination of the magnetic field is necessary in the context of electromagnetic compatibility, which requires thorough determination of magnetic field intensities at industrial frequencies. Magnetic fields can be determined using analytical or analytical–numerical methods employing the vector magnetic potential [30]. From classical electrodynamics formulas, electromagnetic fields can be determined in all areas of busbar conductors, as well as the current density in phase conductors, consequently leading to the flow of phase currents. In works [31,32], authors derived formulas for the self-inductance of a finite-length rectangular busbar conductor using the method of integral equations. Similarly, in works [33,34], authors applied the method of integral equations to determine the impedance of an isolated rectangular busbar conductor at low frequencies, while also comparing the obtained results with those derived from currently used formulas available in the professional literature.

This analysis will demonstrate the process of determining the magnetic field in a busbar with a long rectangular cross-section, a busbar of finite length, and its surroundings using the Mathematica program [35]. The solution obtained will be utilized by the Mathematica program to describe the magnetic field around a ribbon busbar with finite length. The manuscript presents an extension of the author’s work [36]. All the shown formulas enable the calculation of the magnetic field at any point in systems of ribbon busbars, including three-phase systems, under both current symmetry and asymmetry. The derived formulas can be utilized in the design of rectangular busbars [32,33,34] and in superconducting applications [37,38,39]—for calculating the electromagnetic field and current densities in phase conductors—thus resulting in the distribution of phase currents [40,41,42,43,44].

2. The Magnetic Field of a Long Rectangular Busbar

Suppose we have a rectangular busbar with dimensions $a\times b$ (Figure 1) carrying a constant or low-frequency sinusoidal current with the complex RMS value I. Therefore, at each point of the busbar, the current density is constant and equal to

$$\underset{\_}{J}=\frac{\underset{\_}{I}}{a b}$$

(1)

The magnetic field in any point $X(x,y)$, on the inside or outside of the busbar, can be calculated as a superposition of the fields generated by the current at source pointss $Y(x^{\prime },y^{\prime })$ with a distance of r from point $X(x,y)$—Figure 1. For point $Y(x^{\prime },y^{\prime })$, we assume the elementary area as

$$\mathrm{d}s=\mathrm{d}{x}^{\prime } \mathrm{d}{y}^{\prime }$$

(2)

Next, as the current density in the cross-section of the busbar is constant, the current of the busbar can be calculated as

$$\mathrm{d}\underset{\_}{I}=\underset{\_}{J} \mathrm{d}s=\frac{\underset{\_}{I}}{a b} \mathrm{d}{x}^{\prime } \mathrm{d}{y}^{\prime }$$

(3)

According to the Biot–Savart law, there is an elementary magnetic field strength vector in point $X(x,y)$:

$$\mathrm{d}\underset{\_}{\mathit{H}}=\frac{\mathrm{d}\underset{\_}{I}}{2 \pi r} {\mathbf{1}}_z\times {\mathbf{1}}_r=\mathrm{d}{\underset{\_}{H}}_x {\mathbf{1}}_x+\mathrm{d}{\underset{\_}{H}}_y {\mathbf{1}}_y$$

(4)

where ${\mathbf{1}}_x$, ${\mathbf{1}}_y$, and ${\mathbf{1}}_z$ are unit vectors of the Cartesian coordinate system. ${\mathbf{1}}_r$ is the unit vector of vector $\mathit{r}$ connecting point $Y(x^{\prime },y^{\prime })$ and point $X(x,y)$, while $\mathrm{d}{\underset{\_}{H}}_x$ and $\mathrm{d}{\underset{\_}{H}}_y$ are the components of vector $\mathrm{d}\underset{\_}{\mathit{H}}$. Distance can be obtained as follows:

$$r=\sqrt{{(x-x^{\prime })}^2+{(y-y^{\prime })}^2}$$

(5)

Vector module (4):

$$\mathrm{d}H=\frac{\mathrm{d}I}{2 \pi r}=\frac{I}{2 \pi a b} \frac{\mathrm{d}{x}^{\prime } \mathrm{d}{y}^{\prime }}{r}$$

(6)

Its components have the following formulas (Figure 1):

$$\mathrm{d}{\underset{\_}{H}}_x=-\mathrm{d}\underset{\_}{H} \cos \alpha =-\frac{\underset{\_}{I}}{2\pi a b} \frac{y-y^{\prime }}{{(x-x^{\prime })}^2+{(y-y^{\prime })}^2} \mathrm{d}{x}^{\prime } \mathrm{d}{y}^{\prime }$$

(7)

and

$$\mathrm{d}{\underset{\_}{H}}_y=\mathrm{d}\underset{\_}{H} \sin \alpha =\frac{\underset{\_}{I}}{2\pi a b} \frac{x-x^{\prime }}{{(x-x^{\prime })}^2+{(y-y^{\prime })}^2} \mathrm{d}{x}^{\prime } \mathrm{d}{y}^{\prime }$$

(8)

Therefore, we can calculate the ${\underset{\_}{H}}_x$ and ${\underset{\_}{H}}_y$ components of the total magnetic field strength vector $\underset{\_}{\mathit{H}}={\underset{\_}{H}}_x {\mathbf{1}}_x+{\underset{\_}{H}}_y {\mathbf{1}}_y$ using the following integrals:

$${\underset{\_}{H}}_x(x,y)={\displaystyle \underset{-\frac{b}{2}}{\overset{\frac{b}{2}}{\int }}{\displaystyle \underset{-\frac{a}{2}}{\overset{\frac{a}{2}}{\int }}\mathrm{d}{\underset{\_}{H}}_x \mathrm{d}{x}^{\prime } \mathrm{d}{y}^{\prime }}}=-\frac{\underset{\_}{I}}{2\pi a b} {\displaystyle \underset{-\frac{b}{2}}{\overset{\frac{b}{2}}{\int }}{\displaystyle \underset{-\frac{a}{2}}{\overset{\frac{a}{2}}{\int }}\frac{y-y^{\prime }}{{(x-x^{\prime })}^2+{(y-y^{\prime })}^2} \mathrm{d}{x}^{\prime } \mathrm{d}{y}^{\prime }}}$$

(9)

and

$${\underset{\_}{H}}_y(x,y)={\displaystyle \underset{-\frac{b}{2}}{\overset{\frac{b}{2}}{\int }}{\displaystyle \underset{-\frac{a}{2}}{\overset{\frac{a}{2}}{\int }}\mathrm{d}{\underset{\_}{H}}_y \mathrm{d}{x}^{\prime } \mathrm{d}{y}^{\prime }}}=\frac{\underset{\_}{I}}{2\pi a b} {\displaystyle \underset{-\frac{b}{2}}{\overset{\frac{b}{2}}{\int }}{\displaystyle \underset{-\frac{a}{2}}{\overset{\frac{a}{2}}{\int }}\frac{x-x^{\prime }}{{(x-x^{\prime })}^2+{(y-y^{\prime })}^2} \mathrm{d}{x}^{\prime } \mathrm{d}{y}^{\prime }}}$$

(10)

In the outer area of the busbar (i.e., for $x>\frac{a}{2} \cup x<-\frac{a}{2} \cup y>\frac{b}{2} \cup y<-\frac{b}{2}$), the observation point $X(x,y)$ will never coincide with the source point $Y(x^{\prime },y^{\prime })$, and the integrals above are proper.

In the inner area of the busbar (i.e., for $-\frac{a}{2}\le x\le \frac{a}{2}\cap -\frac{b}{2}\le y\le \frac{b}{2}$), the observation point $X(x,y)$ may coincide with the source point $Y(x^{\prime },y^{\prime })$, and the integrals above are improper but convergent. The issue of convergence for these integrals occurs in the second integration. The solution results in a component of magnetic field strength along the Ox axis

$$\begin{array}{l}{\underset{\_}{H}}_x(x,y)=\frac{\underset{\_}{I}}{8 \pi a b}\\ \left\{\begin{array}{l}2 \left(b-2 y\right) \left[\mathrm{arctg} \frac{a-2 x}{b-2 y}+\mathrm{arctg} \frac{a+2 x}{b-2 y}\right]-2 \left(b+2 y\right) \left[\mathrm{arctg} \frac{a-2 x}{b+2 y}+\mathrm{arctg} \frac{a+2 x}{b+2 y}\right]+\\ +(a+2 x) \ln \frac{{(a+2 x)}^2+{(b-2 y)}^2}{{(a+2 x)}^2+{(b+2 y)}^2}+(a-2 x) \ln \frac{{(a-2 x)}^2+{(b-2 y)}^2}{{(a-2 x)}^2+{(b+2 y)}^2}\end{array}\right\}\end{array}$$

(11)

Due to the denominators of functions inverse to trigonometric functions in Formula (11), it is necessary to calculate the component on the upper and lower sides of the busbar cross-section, which results in the following:

$$\begin{array}{l}{\underset{\_}{H}}_x(x,y=\frac{b}{2})=\underset{y\to \frac{b}{2}}{\mathrm{Limit}}{\underset{\_}{H}}_x(x,y)=\\ =\frac{\underset{\_}{I}}{8 \pi a b} \left\{\begin{array}{l}-4 b \left[\mathrm{arctg}\frac{a-2 x}{2 b}+\mathrm{arctg}\frac{a+2 x}{2 b}\right]+\\ +(a+2 x) \ln \frac{{(a+2 x)}^2}{4 b^2+{(a+2 x)}^2}+(a-2 x) \ln \frac{{(a-2 x)}^2}{4 b^2+{(a-2 x)}^2}\end{array}\right\}\end{array}$$

(11a)

and

$$\begin{array}{l}{\underset{\_}{H}}_x(x,y=-\frac{b}{2})=\underset{y\to -\frac{b}{2}}{\mathrm{Limit}}{\underset{\_}{H}}_x(x,y)=\\ =\frac{\underset{\_}{I}}{8 \pi a b} \left\{\begin{array}{l}4 b \left[\mathrm{arctg}\frac{a-2 x}{2 b}+\mathrm{arctg}\frac{a+2 x}{2 b}\right]-\\ -(a+2 x) \ln \frac{{(a+2 x)}^2}{4 b^2+{(a+2 x)}^2}-(a-2 x) \ln \frac{{(a-2 x)}^2}{4 b^2+{(a-2 x)}^2}\end{array}\right\}\end{array}$$

(11b)

Similarly, we can calculate the component of the magnetic field strength along the Oy axis to acquire the following:

$$\begin{array}{l}{\underset{\_}{H}}_y(x,y)=\frac{\underset{\_}{I}}{8 \pi a b}\\ \left\{\begin{array}{l}-2 \left(a-2 x\right) \left[\mathrm{arctg}\frac{b-2 y}{a-2 x}+\mathrm{arctg}\frac{b+2 y}{a-2 x}\right]+2 \left(a+2 x\right) \left[\mathrm{arctg}\frac{b-2 y}{a+2 x}+\mathrm{arctg}\frac{b+2 y}{a+2 x}\right]-\\ \hspace{1em}-(b+2 y) \ln \frac{{(a-2 x)}^2+{(b+2 y)}^2}{{(a+2 x)}^2+{(b+2 y)}^2}-(b-2 y) \ln \frac{{(a-2 x)}^2+{(b-2 y)}^2}{{(a+2 x)}^2+{(b-2 y)}^2}\end{array}\right\}\end{array}$$

(12)

In this case, it is also necessary to calculate the component on the right and left sides of the busbar cross-section, which results in

$$\begin{array}{l}{\underset{\_}{H}}_y(x=\frac{a}{2},y)=\underset{x\to \frac{a}{2}}{\mathrm{Limit}}{\underset{\_}{H}}_y(x,y)=\\ =\frac{\underset{\_}{I}}{8 \pi a b} \left\{\begin{array}{l}4 a \left[\mathrm{arctg}\frac{b-2 y}{2 a}+\mathrm{arctg}\frac{b+2 y}{2 a}\right]-\\ -(b+2 y) \ln \frac{{(b+2 y)}^2}{4 a^2+{(b+2 y)}^2}-(b-2 y) \ln \frac{{(b-2 y)}^2}{4 a^2+{(b-2 y)}^2}\end{array}\right\}\end{array}$$

(12a)

and

$$\begin{array}{l}{\underset{\_}{H}}_y(x=-\frac{a}{2},y)=\underset{x\to -\frac{a}{2}}{\mathrm{Limit}}{\underset{\_}{H}}_y(x,y)=\\ =\frac{\underset{\_}{I}}{8 \pi a b} \left\{\begin{array}{l}-4 a \left[\mathrm{arctg}\frac{b-2 y}{2 a}+\mathrm{arctg}\frac{b+2 y}{2 a}\right]+\\ +(b+2 y) \ln \frac{{(b+2 y)}^2}{4 a^2+{(b+2 y)}^2}+(b-2 y) \ln \frac{{(b-2 y)}^2}{4 a^2+{(b-2 y)}^2}\end{array}\right\}\end{array}$$

(12b)

The formulas specified above allow us to calculate the components of the magnetic field strength in any point $X(x,y)$ inside a rectangular busbar. The total magnetic field module is calculated using the following formula:

$$H(x,y)=\sqrt{H_x^2(x,y)+H_y^2(x,y)}$$

(13)

If we only use Formulas (11) and (12), it should be noted that these formulas also describe the magnetic fields of the outer area of the busbar.

The formulas derived above describe the magnetic field in a rectangular cross-section busbar and are very complex, making it difficult to predict the changes in magnetic field distribution after changing, for example, the cross-section dimension ratio of the busbar. However, any software with a graphic module (e.g., Mathematica) will offer a ready-made, excellent tool for quick visualisation of the solutions [35]. In the considered case of a busbar, the distribution of magnetic field strength module at any point $X(x,y)$ on the inside or outside of the busbar is illustrated in Figure 2 [45], where the field is expressed in relative units, as the following function:

$$h(x,y)=\frac{H(x,y)}{H_0}$$

(14)

where the reference of the magnetic field is

$$H_0=\frac{I}{2 \pi r^{\prime } }$$

(15)

and $r^{\prime }=\sqrt{x^2+y^2}$, as if the magnetic field at a given point was calculated for a current-carrying busbar located at the origin of the coordinate system, i.e., for a thread wire.

The distribution of the module of this field on a cross-section plane of the busbar is illustrated in Figure 3.

From Figure 3, it follows that the magnetic field inside the busbar as well as outside the busbar significantly differs from the magnetic field calculated from the Biot–Savart law (for a thread wire). Considering the dimensions of the busbar with a uniform current density distribution, the magnetic field inside the busbar may be smaller as well as larger than the field calculated at these points from the thread wire. In the external region of the busbar, the magnetic field is greater than the field generated by the thread current, and obviously, as one moves away from the origin of the coordinate system, the field becomes that of the thread current (the relative value equals one—from a large distance, the busbar appears more and more like a thread wire).

Changing the proportions between the dimensions a and b of the busbar changes the magnetic field distribution, which is illustrated in Figure 4.

3. The Magnetic Field of a Long Ribbon Busbar

In the case of a long ribbon busbar (i.e., a busbar with a rectangular cross-section where $b>>a$), the magnetic field generated by a constant or low-frequency sinusoidal current with a complex RMS value $\underset{\_}{I}$ in any point $X(x,y)$ is much easier to calculate than in the case of a rectangular busbar because the calculation requires a single integration. Additionally, it is possible to limit the solution to the wire’s axis of symmetry.

However, if we already have Formulas (11) and (12), we can calculate the magnetic field components by determining the limits of these functions with the cross-section dimension of the busbar $a\to 0$. Therefore, we have the following formulas:

$$H_x^{(t)}(x,y)=\underset{a\to 0}{\mathrm{Limit}}{H}_x(x,y)=\frac{I}{4 \pi b} \ln \frac{4 x^2+{(b-2 y)}^2}{4 x^2+{(b+2 y)}^2}$$

(16)

and

$$H_y^{(t)}(x,y)=\underset{a\to 0}{\mathrm{Limit}}{H}_y(x,y)=\frac{I}{2 \pi b} \left[\mathrm{arctg}\frac{b-2 y}{2 x}+\mathrm{arctg}\frac{b+2 y}{2 x}\right]$$

(17)

In special cases, from the above formulas, we obtain the following:

• Field strength on the Ox axis as

  $$H_{x, Ox}^{(t)}(x)=\underset{y\to 0}{\mathrm{Limit}}{H}_x^{(t)}(x,y)=0$$

  (18)

  and

  $$H_{y, Ox}^{(t)}(x)=\underset{y\to 0}{\mathrm{Limit}}{H}_y^{(t)}(x,y)=\frac{I}{\pi b} \mathrm{arctg}\frac{b}{2 x}$$

  (19)
• Field strength on the Oy axis as

  $$\begin{array}{l}{\underset{\_}{H}}_{x0y}^{(t)}(y)={\underset{\_}{H}}_{x0}^{(t)}(x=0,y)=\underset{x\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_{x0}^{(t)}(x,y)=\\ \frac{\underset{\_}{I}}{4 \pi b}\left\{ \ln \frac{l-\sqrt{{(b-2 y)}^2+l^2}}{l+\sqrt{{(b-2 y)}^2+l^2}}+\ln \frac{l+\sqrt{{(b+2 y)}^2+l^2}}{l-\sqrt{{(b+2 y)}^2+l^2}}\right\}\end{array}$$

  (20)

  and

  $${\underset{\_}{H}}_{y0y}^{(t)}(y)={\underset{\_}{H}}_{y0}^{(t)}(x=0,y)=\underset{x\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_{y0}^{(t)}(x,y)=0$$

  (21)

The distribution of the magnetic field module around the ribbon busbar is illustrated in Figure 5. The field is expressed in relative units, with Function (14), where the reference of the magnetic field corresponds to (15).

4. The Magnetic Field of a Ribbon Busbar of Finite Length

In the case of a ribbon busbar of length l (Figure 6), if we already have formulas from the works [20,42,43,44], we can calculate the magnetic field components by determining the limits of these functions (the details are located in Appendix A) with the cross-section dimension of the busbar $a\to 0$. Therefore, we obtain the following formulas:

• The magnetic field strength component along the Ox axis is

$$\begin{array}{l}{\underset{\_}{H}}_x^{(t)}(x,y,z)=\underset{a\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_x(x,y,z)=\\ \frac{\underset{\_}{I}}{8 \pi b}\left\{\begin{array}{l} 2\ln \frac{2z-l+\sqrt{4 x^2+{(b-2 y)}^2+{(l-2z)}^2}}{2z-l+\sqrt{4 x^2+{(b+2 y)}^2+{(l-2z)}^2}}+\\ \ln \frac{\sqrt{4 x^2+{(b+2 y)}^2+{(l-2z)}^2}(l-2z)+8lz+(l+2z)\sqrt{4 x^2+{(b+2 y)}^2+{(l+2z)}^2}}{\sqrt{4 x^2+{(b-2 y)}^2+{(l-2z)}^2}(l-2z)+8lz+(l+2z)\sqrt{4 x^2+{(b-2 y)}^2+{(l+2z)}^2}}+\\ \ln \frac{-\sqrt{4 x^2+{(b+2 y)}^2+{(l-2z)}^2}(l-2z)+8lz+(l+2z)\sqrt{4 x^2+{(b+2 y)}^2+{(l+2z)}^2}}{-\sqrt{4 x^2+{(b-2 y)}^2+{(l-2z)}^2}(l-2z)+8lz+(l+2z)\sqrt{4 x^2+{(b-2 y)}^2+{(l+2z)}^2}}\end{array}\right\}\end{array}$$

(22)

• The magnetic field strength component along the Oy axis is

$$\begin{array}{l}{\underset{\_}{H}}_y^{(t)}(x,y,z)=\underset{a\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_y(x,y,z)=\\ \frac{\underset{\_}{I}}{4 \pi b}\left\{\begin{array}{l} \mathrm{arctg}\frac{(b-2 y) (l-2z)}{2x\sqrt{4 x^2+{(b-2 y)}^2+{(l-2z)}^2}}+\mathrm{arctg}\frac{(b+2 y) (l-2z)}{2x\sqrt{4 x^2+{(b+2 y)}^2+{(l-2z)}^2}}+\\ \mathrm{arctg}\frac{(b-2 y) (l+2z)}{2x\sqrt{4 x^2+{(b-2 y)}^2+{(l+2z)}^2}}+\mathrm{arctg}\frac{(b+2 y) (l+2z)}{2x\sqrt{4 x^2+{(b+2 y)}^2+{(l+2z)}^2}}\end{array}\right\}\end{array}$$

(23)

In special cases where point $X(x,y,z)$ is on the xOy plane (i.e., for $z=0$), the components can be calculated using the following formulas:

$$\begin{array}{l}{\underset{\_}{H}}_{x0}^{(t)}(x,y)={\underset{\_}{H}}_x^{(t)}(x,y,z=0)=\underset{z\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_x^{(t)}(x,y,z)=\\ \frac{\underset{\_}{I}}{4 \pi b}\left\{ \ln \frac{l-\sqrt{4 x^2+{(b-2 y)}^2+l^2}}{l+\sqrt{4 x^2+{(b-2 y)}^2+l^2}}+\ln \frac{l+\sqrt{4 x^2+{(b+2 y)}^2+l^2}}{l-\sqrt{4 x^2+{(b+2 y)}^2+l^2}}\right\}\end{array}$$

(24)

and

$$\begin{array}{l}{\underset{\_}{H}}_{y0}^{(t)}(x,y)={\underset{\_}{H}}_y^{(t)}(x,y,z=0)=\underset{z\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_y^{(t)}(x,y,z)=\\ \frac{\underset{\_}{I}}{2 \pi b}\left\{ \mathrm{arctg}\frac{l (b-2 y)}{2x\sqrt{4 x^2+{(b-2 y)}^2+l^2}}+\mathrm{arctg}\frac{l (b+2 y)}{2x\sqrt{4 x^2+{(b+2 y)}^2+l^2}}\right\}\end{array}$$

(25)

Moreover, on the Ox axis, these components are

$${\underset{\_}{H}}_{x0x}^{(t)}(x)={\underset{\_}{H}}_{x0}^{(t)}(x,y=0)=\underset{y\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_{x0}^{(t)}(x,y)=0$$

(26)

and

$${\underset{\_}{H}}_{y0x}^{(t)}(x)={\underset{\_}{H}}_{y0}^{(t)}(x,y=0)=\underset{y\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_{y0}^{(t)}(x,y)=\frac{\underset{\_}{I}}{\pi b}\mathrm{arctg}\frac{l b}{2x\sqrt{4 x^2+b^2+l^2}}$$

(27)

If point $X(x,y,z)$ is on the Oy axis, the magnetic field components are

$$\begin{array}{l}{\underset{\_}{H}}_{x0y}^{(t)}(y)={\underset{\_}{H}}_{x0}^{(t)}(x=0,y)=\underset{x\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_{x0}^{(t)}(x,y)=\\ \frac{\underset{\_}{I}}{4 \pi b}\left\{ \ln \frac{l-\sqrt{{(b-2 y)}^2+l^2}}{l+\sqrt{{(b-2 y)}^2+l^2}}+\ln \frac{l+\sqrt{{(b+2 y)}^2+l^2}}{l-\sqrt{{(b+2 y)}^2+l^2}}\right\}\end{array}$$

(28)

and

$${\underset{\_}{H}}_{y0y}^{(t)}(y)={\underset{\_}{H}}_{y0}^{(t)}(x=0,y)=\underset{x\to 0}{\mathrm{Limit}}{\underset{\_}{H}}_{y0}^{(t)}(x,y)=0$$

(29)

The depiction of the magnetic field module distribution for a ribbon busbar of finite length on the xOy plane is presented in Figure 7. The field is represented in relative units according to Formula (14), where the magnetic field strength module is

$$H^{(t)}(x,y,z)=\sqrt{{\left[H_x^{(t)}(x,y,z)\right]}^2+{\left[H_x^{(t)}(x,y,z)\right]}^2}$$

(30)

and the reference of magnetic field is

$$H_0^{(t)}=\frac{I}{2 b}$$

(31)

The impact of the busbar length on the magnetic field distribution is depicted in Figure 8 and Figure 9.

5. Results and Discussion

5.1. The Magnetic Field of a Line with Two Ribbon Busbars of Finite Length

Suppose we have a line made of two ribbon busbars with dimensions $b\times l$ and distance d between the busbars (Figure 10), carrying a constant or low-frequency sinusoidal current with complex RMS values ${\underset{\_}{I}}_1$ and ${\underset{\_}{I}}_2$.

The approach for computing the magnetic field strength at any given point $X(x,y,z)$ remains consistent, as seen in the scenario of a line featuring two rectangular busbars [30,34]. The illustration of the total magnetic field module distribution on the xOy plane for a two-busbar line, where the current direction is identical for each busbar (i.e., where ${\underset{\_}{I}}_1={\underset{\_}{I}}_2=\underset{\_}{I}$), is presented in Figure 11. The field is represented in relative units based on Formula (14) with the magnetic field reference being (31).

The relation between the length of the busbar and total magnetic field distribution is illustrated in Figure 12 and Figure 13. If the current in the busbars flows in opposite directions (i.e., where ${\underset{\_}{I}}_1=-{\underset{\_}{I}}_2=-\underset{\_}{I}$), the distributions of the magnetic field modules are as illustrated in Figure 14, Figure 15, and Figure 16, respectively.

5.2. The Magnetic Field Characteristics of a Three-Phase Line with Three Ribbon Busbars of Finite Length

Consider a three-phase line composed of three ribbon busbars with dimensions $b\times l$ and a distance d between the busbars (refer to Figure 17). These busbars carry a constant or low-frequency sinusoidal current with complex RMS values ${\underset{\_}{I}}_1$, ${\underset{\_}{I}}_2$, and ${\underset{\_}{I}}_3$.

The approach for calculating the magnetic field strength at point $X(x,y,z)$ remains consistent with the method applied in the case of a three-phase line featuring three rectangular busbars. The illustration of the total magnetic field module distribution on the xOy plane for a three-phase three-busbar line, where the current in the busbars is symmetrical (i.e., where ${\underset{\_}{I}}_1=\underset{\_}{I}{ \mathrm{e}}^{{\mathrm{j} 0}^{\mathrm{o}}}$, ${\underset{\_}{I}}_2=\underset{\_}{I}{ \mathrm{e}}^{{-\mathrm{j} 120}^{\mathrm{o}}}$, and ${\underset{\_}{I}}_3=\underset{\_}{I}{ \mathrm{e}}^{{\mathrm{j} 120}^{\mathrm{o}}}$), is presented in Figure 18. The field is quantified in relative units as a function using Formula (14), with the magnetic field reference outlined in Formula (31). The relationship between the length of the busbars and the overall magnetic field distribution in such a line is depicted in Figure 19 and Figure 20.

Current asymmetry changes the magnetic field distribution around the considered. As an example, the following phase currents were assumed: ${\underset{\_}{I}}_1=\underset{\_}{I}{ \mathrm{e}}^{{\mathrm{j} 0}^{\mathrm{o}}}$ and ${\underset{\_}{I}}_2=0.5 \underset{\_}{I}{ \mathrm{e}}^{{-\mathrm{j} 120}^{\mathrm{o}}}$. Therefore, this results in ${\underset{\_}{I}}_3=-{\underset{\_}{I}}_1-{\underset{\_}{I}}_2= 0.866 \underset{\_}{I}{ \mathrm{e}}^{{\mathrm{j} 150}^{\mathrm{o}}}$. The field distributions for this case are illustrated in Figure 21, Figure 22 and Figure 23.

5.3. The Magnetic Field of a Three-Phase Line with Four Ribbon Busbars of Finite Length

Suppose we have a four-wire hight current line with ribbon busbars of dimensions $b\times l$ and distance d between the busbars (Figure 24), with current asymmetry ${\underset{\_}{I}}_1=\underset{\_}{I}{ \mathrm{e}}^{{\mathrm{j} 0}^{\mathrm{o}}}$, ${\underset{\_}{I}}_2=0.5 \underset{\_}{I}{ \mathrm{e}}^{{-\mathrm{j} 120}^{\mathrm{o}}}$, and ${\underset{\_}{I}}_3=\underset{\_}{I}{ \mathrm{e}}^{{\mathrm{j} 120}^{\mathrm{o}}}$, with the current in the neutral rail being ${\underset{\_}{I}}_N={\underset{\_}{I}}_1+{\underset{\_}{I}}_2+{\underset{\_}{I}}_3= 0.5 \underset{\_}{I}{ \mathrm{e}}^{{\mathrm{j} 60}^{\mathrm{o}}}$.

The field distributions for this case are illustrated in Figure 25, Figure 26 and Figure 27. The field is expressed in relative units as a function provided by Formula (14), where the reference of the magnetic field is expressed in Formula (31).

6. Conclusions

The formulas derived above describe the magnetic fields in ribbon busbars and are very complex, making it difficult to predict the changes in magnetic field distribution after altering, for example, the length of the busbar. However, any software with a graphic component (e.g., Mathematica), will offer a ready-made, excellent tool for a quick visualisation of solutions. Therefore, as illustrated in the figures above, it is possible to quickly analyse the magnetic field of a ribbon busbar after changes are made in the geometrical or electric parameters of the examined circuits.

This article derives analytical formulas, which is a huge advantage over numerical solutions. Analytical formulas allow us to obtain precise solutions to problems concerning the magnetic field, depending on their complexity (in this case, for various cases of strip conductors). In contrast, numerical solutions may carry some error, which can vary depending on the numerical methods used and the precision of calculations (their accuracy primarily depends, for example, on the discretization grid and associated computation time). Analytical formulas enable a deeper understanding of the nature of the problem and the mathematical relationships and theories of the electromagnetic field associated with it. They can reveal certain properties or symmetries that may be more challenging to detect with numerical solutions.