Magnetic Fields and Asymptotic Limits of Current Loops

1. Introduction

Ever since the Danish physicist Hans Christian Ørsted discovered [1] that a current carrying wire deflects a magnetic needle kept in its vicinity, philosophers and natural scientists have researched [2] on the connection between magnetic fields as observed near bar magnets and lodestones [3,5], and electric currents. Laplace, Biot, Savart, Ampère, and others experimentally observed [7] that the magnetic field $\mathbf{B}\left(\mathbf{r}\right)$ produced by a steady current density $\mathbf{J}\left({\mathbf{r}}^{\prime }\right)$ distributed over a set ${\mathbf{r}}^{\prime }\in \mathcal{D}$ is given by

$$\begin{array}{c}\hfill \mathbf{B}\left(\mathbf{r}\right)=\frac{{\mu }_0}{4\pi }{\int }_{\mathcal{D}}\frac{\mathbf{J}\times (\mathbf{r}-{\mathbf{r}}^{\prime })}{{∥\mathbf{r}-{\mathbf{r}}^{\prime }∥}^3}d{\mathcal{V}}^{\prime }.\end{array}$$

(1)

This result, commonly known as the Biot–Savart law, was used by Maxwell to formulate the so-called Ampère’s circuital law [8,9], which states that the circulation of the magnetic field $\mathbf{B}$ produced by a steady (time-invariant) current over a closed loop is proportional to the total current flowing through the surface, or, equivalently in differential form, $\nabla \times \mathbf{B}={\mu }_0\mathbf{J},$ where $\mathbf{J}$ is the volume current density and $\nabla \times \mathbf{A}$ for a vector field $\mathbf{A}\left(\mathbf{r}\right)\equiv A_x\left(\mathbf{r}\right)\widehat{\mathbf{x}}+A_y\left(\mathbf{r}\right)\widehat{\mathbf{y}}+A_z\left(\mathbf{r}\right)\widehat{\mathbf{z}}$ is defined as

$$\nabla \times \mathbf{A}:=\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)\widehat{\mathbf{x}}+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)\widehat{\mathbf{y}}+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)\widehat{\mathbf{z}}.$$

With Maxwell’s unification of electromagnetism [10], Ampère’s circuital law was incorporated as part of Maxwell’s equations [4,11,12]. Maxwell’s work also comprehensively laid out the exact connections between sources and fields, as well as electric and magnetic fields [14]. Further unification was achieved with the advent of special relativity [15,16], when it was shown that electric and magnetic fields are in fact components of a rank-2 tensor field, the electromagnetic field [18], through the work of Einstein [16], Lorentz [19], Poincaré [20], and others. During this period, several experiments designed to test various hypotheses around the so-called luminiferous ether [21] also failed and thereby experimentally confirmed the validity of this unification [22,23]. One of the most notable of these experiments was that conducted by Michelson and Morley in 1887 [24,25]. The Michelson–Morley experiment aimed to measure the velocity of the earth relative to the luminiferous ether and involved some of the most precise measurements ever made in physics till then. A state of the art interferometer [26] was used for the measurement, and this remarkably well-designed experiment further contributed in retrospect to the demise of the ether theory [27].

The rest of the paper is organized as follows. Section 2.1 presents the magnetic field and vector potential produced at an arbitrary point by a steady current flowing in a circular loop of finite radius $a.$ Section 2.2 generalizes this result to an arbitrary closed loop [8] on a plane and numerically examines the behavior of the magnetic fields produced by loops of various shapes. Section 3.1 theoretically establishes that the magnetic field and vector potential [6,13] due to a circular current loop approach that of an ideal magnetic dipole as the loop shrinks to a point while keeping the magnetic moment fixed, and Section 3.2 generalizes the convergence result to an arbitrary closed and differentiable loop on a plane. Section 4 concludes the paper.

2. Field of a loop current

In this section, we use the Biot–Savart law (1) to derive the magnetic field $\mathbf{B}$ produced by a plane current loop carrying a steady current $I.$ Without loss of generality, we can take the current loop to be in the $xy$-plane.

2.1. Circular current loop

We first consider the case of a circular loop of radius $a,$ centered at the origin. With our choice of coordinates, the current distribution has azimuthal symmetry, i.e., for any coordinate transformation $\varphi \to \varphi +c$ for any constant $c\in \mathbb{R},$ the form of the current distribution remains unchanged. This implies that the magnetic field at any point $\mathbf{r}$ will be independent of $\varphi$ and will not have any component along the $\widehat{\mathit{\varphi }}$ direction. We can therefore take the point $\mathbf{r}$ to be on the $zx$-plane without loss of generality. The differential magnetic field produced at this point $\mathbf{r}=x\widehat{\mathbf{x}}+z\widehat{\mathbf{z}}$ by a differential element $dl=ad\varphi$ of the current loop located at ${\mathbf{r}}^{\prime }=acos\varphi \widehat{\mathbf{x}}+asin\varphi \widehat{\mathbf{y}}$ is given, using (1), by

$$\begin{array}{cc}\hfill d\mathbf{B}& =\frac{{\mu }_{0}aI}{4\pi }\left(\frac{\widehat{\mathit{\varphi }}\left({\mathbf{r}}^{\prime }\right)\times \left(\mathbf{r}-acos\varphi \widehat{\mathbf{x}}-asin\varphi \widehat{\mathbf{y}}\right)}{{∥\mathbf{r}-acos\varphi \widehat{\mathbf{x}}-asin\varphi \widehat{\mathbf{y}}∥}^3}\right)d\varphi \hfill \\ \hfill & =\frac{{\mu }_{0}aI}{4\pi }\left(\frac{(-sin\varphi \widehat{\mathbf{x}}+cos\varphi \widehat{\mathbf{y}})\times \left((x-acos\varphi )\widehat{\mathbf{x}}-asin\varphi \widehat{\mathbf{y}}+z\widehat{\mathbf{z}}\right)}{{∥(x-acos\varphi )\widehat{\mathbf{x}}-asin\varphi \widehat{\mathbf{y}}+z\widehat{\mathbf{z}}∥}^3}\right)d\varphi \hfill \\ \hfill & =\frac{{\mu }_{0}aI}{4\pi }\left(\frac{zcos\varphi \widehat{\mathbf{x}}+zsin\varphi \widehat{\mathbf{y}}+(a-xcos\varphi )\widehat{\mathbf{z}}}{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{3/2}}\right)d\varphi .\hfill \end{array}$$

(2)

We can integrate (2) to obtain the magnetic field $\mathbf{B}$ at the point $\mathbf{r}.$ As mentioned earlier, due to the azimuthal symmetry of the problem [17], the $\varphi$ component (or since $\mathbf{r}$ is chosen to be in the $zx$-plane, the y component) of $\mathbf{B}$ should be zero. We arrive at the same conclusion from (2) by seeing that

$$\begin{array}{cc}\hfill B_y& =\frac{{\mu }_{0}azI}{4\pi }{\int }_0^{2\pi }\frac{sin\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{3/2}}d\varphi \hfill \\ \hfill & \stackrel{\left(a\right)}{=}\frac{{\mu }_{0}azI}{4\pi }{\int }_{-\pi }^{\pi }\frac{sin\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{3/2}}d\varphi \hfill \\ \hfill & \stackrel{\left(b\right)}{=}0,\hfill \end{array}$$

where $\left(a\right)$ follows since trigonometric functions are periodic with period $2\pi ,$ and $\left(b\right)$ follows since the integrand is odd. We can write the x component of $\mathbf{B}$ as

$$\begin{array}{cc}\hfill B_x& =\frac{{\mu }_{0}azI}{4\pi }{\int }_0^{2\pi }\frac{cos\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{3/2}}d\varphi \hfill \\ \hfill & =\frac{{\mu }_{0}azI}{4\pi }{\int }_{-\pi }^{\pi }\frac{cos\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{3/2}}d\varphi \hfill \\ \hfill & \stackrel{\left(a\right)}{=}\frac{2{\mu }_{0}azI}{4\pi }{\int }_0^{\pi }\frac{cos\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{3/2}}d\varphi \hfill \\ \hfill & =\frac{2{\mu }_{0}azI}{4\pi {\left(x^2+z^2+a^2\right)}^{3/2}}{\int }_0^{\pi }\frac{cos\varphi }{{\left(1-\frac{2ax}{x^2+z^2+a^2}cos\varphi \right)}^{3/2}}d\varphi \hfill \\ \hfill & \stackrel{\left(b\right)}{=}\frac{{\mu }_{0}zm}{4\pi {\left(x^2+z^2+a^2\right)}^{3/2}}·\frac{2}{\pi a}{\int }_0^{\pi }\frac{cos\varphi }{{\left(1-\frac{2ax}{x^2+z^2+a^2}cos\varphi \right)}^{3/2}}d\varphi \hfill \\ \hfill & =\frac{{\mu }_{0}zm}{4\pi {\left(x^2+z^2+a^2\right)}^{3/2}}·\frac{2}{\pi a}{f}^{\left(1\right)}\left(2ax/(x^2+z^2+a^2)\right),\hfill \end{array}$$

(3)

where

$$f^{\left(1\right)}\left(t\right):={\int }_0^{\pi }\frac{cos\varphi }{{\left(1-tcos\varphi \right)}^{3/2}}d\varphi .$$

Here, $\left(a\right)$ follows since the integrand in the previous step is an even function, and in $\left(b\right),$ m is the magnetic dipole moment $\pi a^{2}I.$ Following similar steps, we can integrate the z component of (2) to write

$$\begin{array}{cc}\hfill B_z& =\frac{{\mu }_{0}m}{4\pi {\left(x^2+z^2+a^2\right)}^{3/2}}·\frac{2}{\pi }\left(f^{\left(2\right)}\left(2ax/(x^2+z^2+a^2)\right)-\frac{x}{a}{f}^{\left(1\right)}\left(2ax/(x^2+z^2+a^2)\right)\right),\hfill \end{array}$$

(4)

where

$$f^{\left(2\right)}\left(t\right):={\int }_0^{\pi }\frac{1}{{\left(1-tcos\varphi \right)}^{3/2}}d\varphi .$$

Using (3) and (4) and taking $t:=2ax/(x^2+z^2+a^2)$, we can write the magnetic field $\mathbf{B}\left(\mathbf{r}\right)$ as

$$\begin{array}{cc}\hfill \mathbf{B}\left(\mathbf{r}\right)& =\frac{{\mu }_{0}m}{4\pi {\left(x^2+z^2+a^2\right)}^{3/2}}·\frac{2}{\pi }\left[\frac{z}{a}{f}^{\left(1\right)}\left(t\right)\widehat{\mathbf{x}}+\left(f^{\left(2\right)}\left(t\right)-\frac{x}{a}{f}^{\left(1\right)}\left(t\right)\right)\widehat{\mathbf{z}}\right].\hfill \end{array}$$

(5)

Converting to spherical polar coordinates with $x^2+z^2=r^2,$ $z=rcos\theta ,$ $x=rsin\theta ,$ $\widehat{\mathbf{x}}=sin\theta \widehat{\mathbf{r}}+cos\theta \widehat{\mathit{\theta }},$ and $\widehat{\mathbf{z}}=cos\theta \widehat{\mathbf{r}}-sin\theta \widehat{\mathit{\theta }},$ (5) becomes

$$\mathbf{B}\left(\mathbf{r}\right)=\frac{{\mu }_{0}m}{4\pi {\left(r^2+a^2\right)}^{3/2}}·\frac{2}{\pi }\left[f^{\left(2\right)}\left(t\right)cos\theta \widehat{\mathbf{r}}+\left(\frac{r{f}^{\left(1\right)}\left(t\right)}{a}-f^{\left(2\right)}\left(t\right)sin\theta \right)\widehat{\mathit{\theta }}\right].$$

We thus have the following lemma.

Lemma 1. 

The magnetic field $\mathbf{B}$ at a point $\mathbf{r}$ (with spherical polar coordinates $(r,\theta ,\varphi )$) produced by a circular loop of radius a and magnetic moment m placed on the $xy$-plane with its center at the origin, is given by

$$\begin{array}{c}\hfill \mathbf{B}\left(\mathbf{r}\right)=\frac{{\mu }_{0}m}{4\pi {\left(r^2+a^2\right)}^{3/2}}·\frac{2}{\pi }\left[f^{\left(2\right)}\left(t\right)cos\theta \widehat{\mathbf{r}}+\left(\frac{r{f}^{\left(1\right)}\left(t\right)}{a}-f^{\left(2\right)}\left(t\right)sin\theta \right)\widehat{\mathit{\theta }}\right],\end{array}$$

(6)

where

$$\begin{array}{cc}\hfill f^{\left(1\right)}\left(t\right)& :={\int }_0^{\pi }\frac{cos\varphi }{{\left(1-tcos\varphi \right)}^{3/2}}d\varphi ,\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill f^{\left(2\right)}\left(t\right)& :={\int }_0^{\pi }\frac{1}{{\left(1-tcos\varphi \right)}^{3/2}}d\varphi ,\mathit{and}\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill t& :=\frac{2arsin\theta }{r^2+a^2}.\hfill \end{array}$$

(9)

Remark 1. 

The integrals in eqns (7) and (8) are referred to as elliptic integrals [28]. They have no finite closed form expressions in terms of elementary functions.

Figure 1 illustrates the magnetic field lines on a vertical plane (i.e., containing the z axis) obtained in accordance with Lemma 1. The red line segment represents the diameter of the current loop. The field lines form closed loops and all of them cross the current loop.

Remark 2. 

When the field point $\mathbf{r}$ is on the $xy$-plane (i.e., $\theta =\pi /2$), (6) yields

$$\begin{array}{c}\hfill \mathbf{B}=\frac{{\mu }_0}{4\pi {\left(r^2+a^2\right)}^{3/2}}·\frac{2}{\pi }\left(f^{\left(2\right)}\left(\frac{2ar}{r^2+a^2}\right)-\frac{r}{a}{f}^{\left(1\right)}\left(\frac{2ar}{r^2+a^2}\right)\right)\mathbf{m},\end{array}$$

(10)

which shows that the field on the plane of the loop is always perpendicular to the plane, as one can expect from the form of the Biot–Savart law [39].

Remark 3. 

Lemma 1 has a specialization which is commonly derived in freshman physics classes. For a point $\mathbf{r}$ on the z axis (i.e., $\theta =0$), eqn (6) yields

$$\begin{array}{cc}\hfill \mathbf{B}\left(\mathbf{r}\right)& =\frac{{\mu }_{0}m}{4\pi {\left(r^2+a^2\right)}^{3/2}}·\frac{2}{\pi }\left(f^{\left(2\right)}\left(0\right)\widehat{\mathbf{r}}+\frac{r{f}^{\left(1\right)}\left(0\right)}{a}\widehat{\mathit{\theta }}\right)\hfill \\ \hfill & =\frac{{\mu }_0}{4\pi }\left(\frac{2\mathbf{m}}{{\left(r^2+a^2\right)}^{3/2}}\right).\hfill \end{array}$$

We can also establish the magnetic vector potential for the current loop, as presented in the following lemma.

Lemma 2. 

The magnetic vector potential $\mathbf{A}$ at a point $\mathbf{r}$ (with spherical polar coordinates $(r,\theta ,\varphi )$) produced by a circular loop of radius a and magnetic moment m placed on the $x-y$ plane with its center at the origin, is given by

$$\begin{array}{c}\hfill \mathbf{A}\left(\mathbf{r}\right)=\frac{{\mu }_0}{4\pi {\left(r^2+a^2\right)}^{1/2}}·\frac{2}{\pi asin\theta }{f}^{\left(3\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)\left(\mathbf{m}\times \widehat{\mathbf{r}}\right),\end{array}$$

(11)

where

$$\begin{array}{cc}\hfill f^{\left(3\right)}\left(t\right)& :={\int }_0^{\pi }\frac{cos\varphi }{{\left(1-tcos\varphi \right)}^{1/2}}d\varphi .\hfill \end{array}$$

(12)

Proof. 

The magnetic vector potential $\mathbf{A}\left(\mathbf{r}\right)$ produced by a steady current density $\mathbf{J}\left({\mathbf{r}}^{\prime }\right)$ distributed over a set ${\mathbf{r}}^{\prime }\in \mathcal{D}$ can be written as (see, for example, [33])

$$\begin{array}{c}\hfill \mathbf{A}\left(\mathbf{r}\right)=\frac{{\mu }_0}{4\pi }{\int }_{\mathcal{D}}\frac{\mathbf{J}}{∥\mathbf{r}-{\mathbf{r}}^{\prime }∥}d{\mathcal{V}}^{\prime }.\end{array}$$

(13)

Similar to (2), the differential vector potential at a point $\mathbf{r}=x\widehat{\mathbf{x}}+z\widehat{\mathbf{z}}$ by a differential element $dl=ad\varphi$ of the current loop located at ${\mathbf{r}}^{\prime }=acos\varphi \widehat{\mathbf{x}}+asin\varphi \widehat{\mathbf{y}}$ is given, using (13), by

$$\begin{array}{cc}\hfill d\mathbf{A}& =\frac{{\mu }_{0}aI}{4\pi }\left(\frac{1}{∥\mathbf{r}-acos\varphi \widehat{\mathbf{x}}-asin\varphi \widehat{\mathbf{y}}∥}\right)d\varphi \widehat{\mathit{\varphi }}\hfill \\ \hfill & =\frac{{\mu }_{0}aI}{4\pi }\left(\frac{-sin\varphi \widehat{\mathbf{x}}+cos\varphi \widehat{\mathbf{y}}}{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{1/2}}\right)d\varphi .\hfill \end{array}$$

(14)

We can integrate (14) to calculate the vector potential $\mathbf{A}.$ We have

$$\begin{array}{cc}\hfill A_x& =-\frac{{\mu }_{0}aI}{4\pi }{\int }_0^{2\pi }\frac{sin\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{1/2}}d\varphi \hfill \\ \hfill & \stackrel{\left(a\right)}{=}-\frac{{\mu }_{0}aI}{4\pi }{\int }_{-\pi }^{\pi }\frac{sin\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{1/2}}d\varphi \hfill \\ \hfill & \stackrel{\left(b\right)}{=}0,\hfill \end{array}$$

where $\left(a\right)$ follows since trigonometric functions are periodic with period $2\pi ,$ and $\left(b\right)$ follows since the integrand in the previous step is an odd function. We similarly have

$$\begin{array}{cc}\hfill A_y& =\frac{{\mu }_{0}aI}{4\pi }{\int }_0^{2\pi }\frac{cos\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{1/2}}d\varphi \hfill \\ \hfill & \stackrel{\left(a\right)}{=}\frac{{\mu }_{0}aI}{4\pi }{\int }_{-\pi }^{\pi }\frac{cos\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{1/2}}d\varphi \hfill \\ \hfill & \stackrel{\left(b\right)}{=}\frac{2{\mu }_{0}aI}{4\pi }{\int }_0^{\pi }\frac{cos\varphi }{{\left(x^2+z^2+a^2-2axcos\varphi \right)}^{1/2}}d\varphi \hfill \\ \hfill & =\frac{2{\mu }_{0}aI}{4\pi {\left(x^2+z^2+a^2\right)}^{1/2}}{\int }_0^{\pi }\frac{cos\varphi }{{\left(1-\frac{2ax}{x^2+z^2+a^2}cos\varphi \right)}^{1/2}}d\varphi \hfill \\ \hfill & =\frac{{\mu }_{0}m}{4\pi {\left(x^2+z^2+a^2\right)}^{1/2}}·\frac{2}{\pi a}{f}^{\left(3\right)}\left(\frac{2ax}{x^2+z^2+a^2}\right),\hfill \end{array}$$

(15)

where $f^{\left(3\right)}(·)$ is as defined in (12). Here, $\left(a\right)$ follows since trigonometric functions are periodic with period $2\pi ,$ and $\left(b\right)$ follows since the integrand in the previous step is an even function [42].Converting to spherical polar coordinates with $x^2+z^2=r^2,$ $x=rsin\theta ,$ and $\widehat{\mathbf{y}}=\widehat{\mathit{\varphi }},$ (15) becomes

$$\begin{array}{cc}\hfill \mathbf{A}\left(\mathbf{r}\right)& =\frac{{\mu }_{0}m}{4\pi {\left(r^2+a^2\right)}^{1/2}}·\frac{2}{\pi a}{f}^{\left(3\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)\widehat{\mathit{\varphi }}\hfill \\ \hfill & \stackrel{\left(a\right)}{=}\frac{{\mu }_{0}m}{4\pi {\left(r^2+a^2\right)}^{1/2}}·\frac{2}{\pi asin\theta }{f}^{\left(3\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)\left(\widehat{\mathbf{z}}\times \widehat{\mathbf{r}}\right)\hfill \\ \hfill & =\frac{{\mu }_0}{4\pi {\left(r^2+a^2\right)}^{1/2}}·\frac{2}{\pi asin\theta }{f}^{\left(3\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)\left(\mathbf{m}\times \widehat{\mathbf{r}}\right).\hfill \end{array}$$

This establishes the result. Here, $\left(a\right)$ follows from the observation that

$$\begin{array}{cc}\hfill \widehat{\mathit{\varphi }}& =\widehat{\mathbf{r}}\times \widehat{\mathit{\theta }}\hfill \\ \hfill & =-\frac{1}{sin\theta }\widehat{\mathbf{r}}\times \left(cos\theta \widehat{\mathbf{r}}-sin\theta \widehat{\mathit{\theta }}\right)\hfill \\ \hfill & =\frac{1}{sin\theta }\left(\widehat{\mathbf{z}}\times \widehat{\mathbf{r}}\right).\hfill \end{array}$$

□

In the following subsection, we generalize Lemmas 1 and 2 to general planar loops.

2.2. General current loop on a plane

Consider a current loop on the $xy$-plane described in polar coordinates $(\rho ,\varphi )$ as $\rho =\lambda ·\gamma \left(\varphi /2\pi \right),$ where $\gamma :[0,1]\to {\mathbb{R}}^{+}$ is continuously differentiable almost everywhere in $[0,1]$, bounded below (i.e., ${inf}_{u\in [0,1]}\gamma \left(u\right)>0$), satisfies $\gamma \left(0\right)=\gamma \left(1\right),$ and

$$\begin{array}{c}\hfill {\int }_0^1\gamma {\left(u\right)}^{2}du=\frac{1}{\pi }.\end{array}$$

(16)

Eqn (16) defines the scale of the loop, since the area enclosed by the loop is given by

$$\frac{1}{2}{\int }_0^{2\pi }{r}^{2}d\varphi =\frac{{\lambda }^2}{2}{\int }_0^{2\pi }\gamma {\left(\varphi /2\pi \right)}^{2}d\varphi ={\lambda }^2.$$

Thus, $\mathcal{L}(\lambda ;\gamma ):=\left\{\right(x,y):x=\lambda \gamma (u)cos(2\pi u),y=\lambda \gamma (u)sin(2\pi u),u\in [0,1\left]\right\}$ defines a family of “similar” loops parametrized by $\lambda ,$ where the area enclosed by $\mathcal{L}(\lambda ;\gamma )$ is given by ${\lambda }^2.$ Henceforth, we refer to “the loop $\mathcal{L}(\lambda ;\gamma )$” to refer to a loop in the shape given by $\mathcal{L}(\lambda ;\gamma ).$ The following lemma establishes the magnetic field and vector potential produced by a uniform steady current I flowing in the loop $\mathcal{L}(\lambda ;\gamma ).$

Lemma 3. 

The magnetic vector potential $\mathbf{A}$ at a point $\mathbf{r}$ (with spherical polar coordinates $(r,\theta ,\varphi )$) produced by a loop $\mathcal{L}(\lambda ;\gamma )$ with magnetic moment $\mathbf{m}=m\widehat{\mathbf{z}}$ placed on the $xy$-plane is given by

$$\begin{array}{cc}\hfill \mathbf{A}(\mathbf{r},\lambda ;\gamma )& =\frac{{\mu }_0}{4\pi \lambda }\mathbf{m}\times \hfill \\ \hfill & (\left(\widehat{\mathbf{r}}sin\theta +\widehat{\mathit{\theta }}cos\theta \right){\int }_0^1\frac{2\pi \gamma \left(u\right)cos(2\pi u-\varphi )+{\gamma }^{\prime }\left(u\right)sin(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{1/2}}du\hfill \\ \hfill & +\widehat{\mathit{\varphi }}{\int }_0^1\frac{2\pi \gamma \left(u\right)sin(2\pi u-\varphi )-{\gamma }^{\prime }\left(u\right)cos(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{1/2}}du).\hfill \end{array}$$

(17)

The magnetic field at the point $\mathbf{r}$ is given by

$$\begin{array}{cc}\hfill \mathbf{B}(\mathbf{r},\lambda ;\gamma )& =\frac{{\mu }_{0}m}{4\pi \lambda }(2\pi \lambda cos\theta \widehat{\mathbf{r}}{\int }_0^1\frac{\gamma {\left(u\right)}^2}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du\hfill \\ \hfill & +\widehat{\mathit{\theta }}{\int }_0^1\frac{2\pi r\gamma \left(u\right)cos(2\pi u-\varphi )+r{\gamma }^{\prime }\left(u\right)sin(2\pi u-\varphi )-2\pi \lambda sin\theta \gamma {\left(u\right)}^2}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du\hfill \\ \hfill & +rcos\theta \widehat{\mathit{\varphi }}{\int }_0^1\frac{2\pi \gamma \left(u\right)sin(2\pi u-\varphi )-{\gamma }^{\prime }\left(u\right)cos(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du).\hfill \end{array}$$

(18)

The proof of Lemma 3 follows immediately from the Biot–Savart Law (1) and the vector potential expression (13) very similarly to the proofs of Lemmas 1 and 2, and is omitted.

Remark 4. 

One can easily see that Lemma 3 recovers the results of Lemmas 1 and 2 by setting $\gamma \equiv 1/\sqrt{\pi }$ and $\lambda =a·\sqrt{\pi }.$

Remark 5. 

Similarly to Remark 3, Lemma 3 can be specialized to the case $\theta =0$ (i.e., the field at a point on the z-axis) to obtain

$$\begin{array}{cc}\hfill \mathbf{B}(\mathbf{r},\lambda ;\gamma )& =\frac{{\mu }_0\mathbf{m}}{2}{\int }_0^1\frac{\gamma {\left(u\right)}^2}{{\left({\lambda }^2\gamma {\left(u\right)}^2+r^2\right)}^{3/2}}du\hfill \\ \hfill & +\frac{{\mu }_{0}m}{4\pi \lambda }(\widehat{\mathbf{x}}{\int }_0^1\frac{2\pi r\gamma \left(u\right)cos\left(2\pi u\right)+r{\gamma }^{\prime }\left(u\right)sin\left(2\pi u\right)}{{\left({\lambda }^2\gamma {\left(u\right)}^2+r^2\right)}^{3/2}}du\hfill \\ \hfill & +\widehat{\mathbf{y}}{\int }_0^1\frac{2\pi r\gamma \left(u\right)sin\left(2\pi u\right)-r{\gamma }^{\prime }\left(u\right)cos\left(2\pi u\right)}{{\left({\lambda }^2\gamma {\left(u\right)}^2+r^2\right)}^{3/2}}du).\hfill \end{array}$$

(19)

Remark 6. 

For a point $\mathbf{r}$ on the $xy$-plane, similarly to Remark 2, Lemma 3 reduces to

$$\begin{array}{cc}\hfill \mathbf{B}(\mathbf{r},\lambda ;\gamma )& =\frac{{\mu }_0\mathbf{m}}{4\pi \lambda }{\int }_0^1\frac{2\pi \lambda \gamma {\left(u\right)}^2-2\pi r\gamma \left(u\right)cos(2\pi u-\varphi )-r{\gamma }^{\prime }\left(u\right)sin(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du.\hfill \end{array}$$

(20)

We now study some examples of commonly encountered families of loops $\mathcal{L}(\lambda ;\gamma )$ and examine the magnetic fields and vector potentials produced by steady currents flowing in such loops.

Example 1 

(Polygon). A polygonal loop with N vertices may be defined in terms of the vertex coordinates $\{(r_i,{\varphi }_i),i=1,\dots ,N\},$ where $0\le {\varphi }_i<2\pi$ for each $i,$ and the coordinates are sorted such that ${\varphi }_1<{\varphi }_2<\cdots <{\varphi }_N,$ as

$$\begin{array}{cc}\hfill {\gamma }^{\left(N\right)}\left(u\right)& ={\left(\frac{2}{{\sum }_{j=1}^N{r}_j{r}_{j+1}sin\left({\varphi }_{j+1}-{\varphi }_j\right)}\right)}^{1/2}·\left(\frac{r_i{r}_{i+1}sin\left({\varphi }_{i+1}-{\varphi }_i\right)}{r_{i}sin\left(2\pi u-{\varphi }_i\right)+r_{i+1}sin\left({\varphi }_{i+1}-2\pi u\right)}\right),\hfill \\ \hfill & u\in \left[\frac{{\varphi }_i}{2\pi },\frac{{\varphi }_{i+1}}{2\pi }\right]\hfill \end{array}$$

(21)

for $i=1,2,\dots ,N,$ where we choose the convention ${\varphi }_{N+1}\equiv {\varphi }_1.$ As a special case, if $r_i=a/\left(2sin\left(\pi /N\right)\right)$ for each i and ${\varphi }_i=\left(i-1/2\right)2\pi /N$ for $i=1,2,\dots ,N,$ the polygon becomes regular and (21) reduces to

$$\begin{array}{cc}\hfill {\gamma }_{\mathrm{reg}.}^{\left(N\right)}\left(u\right)& =2sin\left(\frac{\pi }{N}\right){\left(\frac{2}{a^{2}Nsin\left(2\pi /N\right)}\right)}^{1/2}\times \left(\frac{a}{2sin\left(\pi /N\right)}\right)\times \hfill \\ \hfill & \left(\frac{sin\left(2\pi /N\right)}{sin\left(2\pi u-\left(i-1/2\right)2\pi /N\right)+sin\left(\left(i+1/2\right)2\pi /N-2\pi u\right)}\right)\hfill \\ \hfill & ={\left(\frac{1}{Ntan\left(\pi /N\right)}\right)}^{1/2}·\left(\frac{1}{cos\left(2\pi \left(u-i/N\right)\right)}\right)\hfill \end{array}$$

(22)

for $u\in \left[\frac{i-1/2}{N},\frac{i+1/2}{N}mod1\right],$ $i=1,2,\dots ,N.$ Figure 2a geometrically depicts ${\gamma }_{\mathrm{reg}.}^{\left(N\right)}$ for $N=5$ (i.e., a regular pentagon). A common special case of (22) can be obtained by setting $N=4$ (i.e., a square). We have

$$\begin{array}{cc}\hfill {\gamma }_{\mathrm{reg}.}^{\left(4\right)}\left(u\right)& =\frac{1}{2cos\left(2\pi u-i\pi /2\right)}\hfill \\ \hfill & =\left\{\begin{array}{cc}\frac{1}{2sin\left(2\pi u\right)},\hfill & u\in [1/8,3/8],\hfill \\ -\frac{1}{2cos\left(2\pi u\right)},\hfill & u\in [3/8,5/8],\hfill \\ -\frac{1}{2sin\left(2\pi u\right)},\hfill & u\in [5/8,7/8],\hfill \\ \frac{1}{2cos\left(2\pi u\right)},\hfill & u\in [7/8,1]\cup [0,1/8].\hfill \end{array}\right.\hfill \end{array}$$

(23)

A different special case of (21) can be obtained by setting $N=4,$ ${\varphi }_1=\alpha ,$ ${\varphi }_2=\pi -\alpha ,$ ${\varphi }_3=\pi +\alpha ,$ ${\varphi }_4=2\pi -\alpha ,$ and $r_1=r_2=r_3=r_4=a>0,$ for some $\alpha \in (0,\pi /2).$ The reader may recognize this loop as a rectangle, with $\alpha =\pi /4$ corresponding to the square (23). We then have

$$\begin{array}{cc}\hfill {\gamma }_{\mathrm{rect}.}^{\left(\alpha \right)}\left(u\right)& =\left\{\begin{array}{cc}\left(\frac{{\left(tan\alpha \right)}^{1/2}}{2}\right)\frac{1}{sin\left(2\pi u\right)},\hfill & u\in \left[\frac{\alpha }{2\pi },\frac{1}{2}-\frac{\alpha }{2\pi }\right],\hfill \\ -\left(\frac{1}{2{\left(tan\alpha \right)}^{1/2}}\right)\frac{1}{cos\left(2\pi u\right)},\hfill & u\in \left[\frac{1}{2}-\frac{\alpha }{2\pi },\frac{1}{2}+\frac{\alpha }{2\pi }\right],\hfill \\ -\left(\frac{{\left(tan\alpha \right)}^{1/2}}{2}\right)\frac{1}{sin\left(2\pi u\right)},\hfill & u\in \left[\frac{1}{2}+\frac{\alpha }{2\pi },1-\frac{\alpha }{2\pi }\right],\hfill \\ \left(\frac{1}{2{\left(tan\alpha \right)}^{1/2}}\right)\frac{1}{cos\left(2\pi u\right)},\hfill & u\in \left[1-\frac{\alpha }{2\pi },1\right]\cup \left[0,\frac{\alpha }{2\pi }\right].\hfill \end{array}\right.\hfill \end{array}$$

(24)

One can verify that both the square (23) and the rectangle (24) have unit area, by noting that the square has side length $1,$ while the rectangle has side lengths ${\left(\tan \alpha \right)}^{1/2}$ and $1/{\left(\tan \alpha \right)}^{1/2}.$ Figure 2b geometrically depicts ${\gamma }_{\mathrm{rect}.}^{\left(\alpha \right)}$ for $\alpha =\pi /6$ and $\pi /3.$

Example 2 

(Ellipse). An elliptical loop with eccentricity ϵ and major axis of length $2a$ with one focus at the origin and another focus at $(r,\varphi )=(2aϵ,0),$ may be defined as the locus of a point which moves such that the distances of the point from the two foci sums to $2a.$ When $ϵ=0,$ the two foci coincide and the ellipse reduces to a circle of radius $a.$ We have, for any point $(r,\varphi )$ on the ellipse,

$$\begin{array}{cc}\hfill & r+\sqrt{r^2+{\left(2aϵ\right)}^2-4raϵcos\varphi }=2a\hfill \\ \hfill & \Rightarrow r^2+{\left(2aϵ\right)}^2-4raϵcos\varphi ={(2a-r)}^2\hfill \\ \hfill & \Rightarrow 4ra(1-ϵcos\varphi )=4a^2(1-{ϵ}^2)\hfill \\ \hfill & \Rightarrow r=\frac{a\left(1-{ϵ}^2\right)}{1-ϵcos\varphi }.\hfill \end{array}$$

We therefore have

$$\begin{array}{c}\hfill {\gamma }_{\mathrm{ellipse}}^{\left(ϵ\right)}\left(u\right)=\frac{K}{1-ϵcos\left(2\pi u\right)},\end{array}$$

(25)

where K is a constant to be determined from the normalization condition ${\int }_0^1\gamma {\left(u\right)}^{2}du=1/\pi .$ We have

$$\begin{array}{cc}\hfill {\int }_0^1{\gamma }_{\mathrm{ellipse}}^{\left(ϵ\right)}{\left(u\right)}^{2}du& =K^2{\int }_0^1{\left(\frac{1}{1-ϵcos\left(2\pi u\right)}\right)}^{2}du\hfill \\ \hfill & =\frac{K^2}{\pi }{\int }_0^{\pi }\frac{1}{{\left(1-ϵcost\right)}^2}dt\hfill \end{array}$$

(26)

Using the substitution $v:=tan(t/2),$ (26) reduces to

$$\begin{array}{cc}\hfill {\int }_0^1{\gamma }_{\mathrm{ellipse}}^{\left(ϵ\right)}{\left(u\right)}^{2}du& =\frac{2K^2}{\pi }{\int }_0^{\infty }\frac{1+v^2}{{\left(1+v^2-ϵ(1-v^2)\right)}^2}dv\hfill \\ \hfill & =\frac{2K^2}{\pi {(1+ϵ)}^2}{\int }_0^{\infty }\frac{v^2+\frac{1-ϵ}{1+ϵ}+\frac{2ϵ}{1+ϵ}}{{\left(v^2+\frac{1-ϵ}{1+ϵ}\right)}^2}dv\hfill \\ \hfill & =\frac{2K^2}{\pi {(1+ϵ)}^2}\left({\int }_0^{\infty }\frac{1}{v^2+\frac{1-ϵ}{1+ϵ}}dv+\left(\frac{2ϵ}{1+ϵ}\right){\int }_0^{\infty }\frac{1}{{\left(v^2+\frac{1-ϵ}{1+ϵ}\right)}^2}dv\right).\hfill \end{array}$$

Finally, substituting $v:=\sqrt{(1-ϵ)/(1+ϵ)}tan\xi$ lets us write

$$\begin{array}{cc}\hfill {\int }_0^1{\gamma }_{\mathrm{ellipse}}^{\left(ϵ\right)}{\left(u\right)}^{2}du& =\frac{2K^2}{\pi {(1+ϵ)}^2}\left({\left(\frac{1+ϵ}{1-ϵ}\right)}^{1/2}{\int }_0^{\pi /2}d\xi +\left(\frac{2ϵ}{1+ϵ}\right)·{\left(\frac{1+ϵ}{1-ϵ}\right)}^{3/2}{\int }_0^{\pi /2}{cos}^2\xi d\xi \right)\hfill \\ \hfill & =\frac{K^2}{{(1+ϵ)}^2}{\left(\frac{1+ϵ}{1-ϵ}\right)}^{1/2}\left(1+\frac{ϵ}{1-ϵ}\right)\hfill \\ \hfill & =\frac{K^2}{{\left(1-{ϵ}^2\right)}^{3/2}},\hfill \end{array}$$

(27)

which implies that

$$K=\frac{{\left(1-{ϵ}^2\right)}^{3/4}}{{\pi }^{1/2}},$$

letting us update (25) as

$$\begin{array}{c}\hfill {\gamma }_{\mathrm{ellipse}}^{\left(ϵ\right)}\left(u\right)=\frac{{\left(1-{ϵ}^2\right)}^{3/4}}{{\pi }^{1/2}\left(1-ϵcos\left(2\pi u\right)\right)}.\end{array}$$

(28)

Figure 3 depicts 2 such ellipses with eccentricites ${ϵ}_1=1/2$ and ${ϵ}_2=4/5.$ As shown in the figure, the ellipses share a common focus at the origin, while the other focus moves away from the origin as ϵ increases. One can can show that the other focus lies on the x-axis at a distance $2ϵ/\left({\pi }^{1/2}{\left(1-{ϵ}^2\right)}^{1/4}\right)$ from the origin.

Example 3 

(Norm ball). A loop in the form of a p-norm ball for $p\in {\mathbb{R}}^{+}$ may be defined as the set of points $(x,y)$ satisfying ${\left|x\right|}^p+{\left|y\right|}^p=1.$ We have, for any point $(x,y)\equiv (rcos\varphi ,rsin\varphi )$ on the p-norm ball,

$$\begin{array}{cc}\hfill & r^p{|cos\varphi |}^p+r^p{|sin\varphi |}^p=1\hfill \\ \hfill & \Rightarrow r=\frac{1}{{\left({|cos\varphi |}^p+{|sin\varphi |}^p\right)}^{1/p}}.\hfill \end{array}$$

We therefore have

$$\begin{array}{c}\hfill {\gamma }_{\mathrm{norm}}^{\left(p\right)}\left(u\right)=\frac{K_p}{{\left({|cos\left(2\pi u\right)|}^p+{|sin\left(2\pi u\right)|}^p\right)}^{1/p}},\end{array}$$

(29)

where

$$\begin{array}{cc}\hfill K_p& ={\left(\pi {\int }_0^1\frac{du}{{\left({|cos\left(2\pi u\right)|}^p+{|sin\left(2\pi u\right)|}^p\right)}^{2/p}}\right)}^{-1/2}.\hfill \end{array}$$

(30)

Figure 4a plots $K_p$ as a function of $p,$ and Figure 4b depicts ${\gamma }_{\mathrm{norm}}^{\left(p\right)}$ for several values of $p.$

Remark 7. 

Using (30), one can write $K_p$ as

$$\begin{array}{cc}\hfill K_p& ={\left(\pi ·\frac{1}{2\pi }·8·{\int }_0^1{\left(1+x^p\right)}^{-2/p}dx\right)}^{-1/2}\hfill \\ \hfill & =\frac{1}{2}{\left(1+{\int }_0^1\left(\sum _{l=1}^{\infty }{(-1)}^l\frac{x^{pl}}{l!}\prod _{q=0}^{l-1}\left(\frac{2}{p}+q\right)\right)dx\right)}^{-1/2}\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill & =\left\{\begin{array}{cc}1/\sqrt{2},\hfill & p=1,\hfill \\ \frac{1}{2}{\left(1+\sum _{l=1}^{\infty }{(-1)}^l\frac{1}{l!(pl+1)}\prod _{q=0}^{l-1}\left(\frac{2}{p}+q\right)\right)}^{-1/2},\hfill & \mathit{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

(32)

Closed forms of $K_p$ for various values of p can be obtained using the integral form (31) or the series form (32). For example, for $p=1,$ (31) leads to

$$\begin{array}{cc}\hfill K_1& =\frac{1}{2}{\left({\int }_0^1\frac{dx}{{(1+x)}^2}\right)}^{-1/2}\hfill \\ \hfill & =\frac{1}{2}{\left(\frac{1}{2}\right)}^{-1/2}\hfill \\ \hfill & =1/\sqrt{2}.\hfill \end{array}$$

For $p=2,$ (32) yields

$$\begin{array}{cc}\hfill K_2& =\frac{1}{2}{\left(1+\sum _{l=1}^{\infty }{(-1)}^l\frac{1}{l!(2l+1)}\prod _{q=0}^{l-1}\left(q+1\right)\right)}^{-1/2}\hfill \\ \hfill & =\frac{1}{2}{\left(1+\sum _{l=1}^{\infty }\frac{{(-1)}^l}{2l+1}\right)}^{-1/2}\hfill \\ \hfill & \stackrel{\left(a\right)}{=}\frac{1}{2}{\left(\pi /4\right)}^{-1/2}\hfill \\ \hfill & =1/\sqrt{\pi },\hfill \end{array}$$

where $\left(a\right)$ follows from the special case of Gregory’s series [29] for $\pi /4,$ also known as the Leibniz formula for $\pi .$

For $p=4,$ (31) yields

$$\begin{array}{cc}\hfill K_4& =\frac{1}{2}{\left({\int }_0^1\frac{dx}{\sqrt{1+x^4}}\right)}^{-1/2}.\hfill \end{array}$$

The expression $I:={\int }_0^1\left(1/\sqrt{1+x^4}\right)dx$ can be evaluated through the following sequence of steps. First, through the substitution $u:=1/x,$ we have $I={\int }_1^{\infty }\left(1/\sqrt{1+u^4}\right)du,$ whence, we have $I=(1/2){\int }_0^{\infty }\left(1/\sqrt{1+x^4}\right)dx.$ The substitution $u:=1/\left(1+x^4\right)$ leads to

$$\frac{1}{\sqrt{1+x^4}}dx=-\frac{1}{4u^{3/2}{\left(\frac{1}{u}-1\right)}^{3/4}}du=-\frac{1}{4}{u}^{-3/4}{(1-u)}^{-3/4},$$

and therefore, we have

$$\begin{array}{cc}\hfill I& =\frac{1}{2}·\frac{1}{4}{\int }_0^1{u}^{-3/4}{(1-u)}^{-3/4}du\hfill \\ \hfill & \stackrel{\left(a\right)}{=}\frac{1}{8}B\left(1/4,1/4\right)\hfill \\ \hfill & \stackrel{\left(b\right)}{=}\frac{\Gamma {\left(1/4\right)}^2}{8·\sqrt{\pi }},\hfill \end{array}$$

where in $\left(a\right),$ the beta function $B(m,n)$ is defined, for $(m,n)\in {\mathbb{R}}^{+},$ as

$$B(m,n)={\int }_0^1{u}^{m-1}{(1-u)}^{n-1}du={\int }_0^1{u}^{n-1}{(1-u)}^{m-1}du,$$

and in $\left(b\right),$ the gamma function $\Gamma \left(n\right)$ is defined, for $n\in {\mathbb{R}}^{+},$ as

$$\Gamma \left(n\right)={\int }_0^{\infty }{u}^{n-1}exp\left(-u\right)du,$$

with the well-known identity (see, for example, [30])

$$B(m,n)=\Gamma \left(m\right)·\Gamma \left(n\right)/\Gamma (m+n).$$

We therefore have $K_4=2^{1/2}{\pi }^{1/4}/\Gamma (1/4).$

Finally, using (32), we have

$$\underset{p\to \infty }{lim}{K}_p=1/2.$$

All these results agree with the plot shown in Figure 4a.

We now examine the magnetic fields produced by various loops defined in Examples 1, 2, and 3, as predicted by Lemma 3, as a function of the scaling parameter $\lambda .$ To this end, we consider loops $\mathcal{L}(\lambda ;\gamma )$ with magnetic moment $\mathbf{m}=m\widehat{\mathbf{z}}$ and calculate the magnitude of the magnetic field produced at three different fixed points $A_1,$ $A_2,$ and $A_3,$ as demonstrated in Figure 5. In Figure 6, we plot the magnetic field (in $\mu \mathrm{T}$) as a function of the scale $\lambda ,$ for $\lambda \in (0,1/2],$ when the magnetic moment magnitude equals $1\mathrm{A}·{\mathrm{m}}^2$ and the distance of the observation point from the origin is $r=1\mathrm{m}.$

From Figure 6a, Figure 6b, and Figure 6c, we observe that loops of different shapes produce different magnetic fields at the same point for the same $\lambda ,$ as expected; in particular, the fields for ${\gamma }_{\mathrm{ellipse}}^{(0.8)}$ and ${\gamma }_{\mathrm{ellipse}}^{(0.5)}$ show the largest variation with $\lambda ,$ since the position of the center of symmetry of the loop is a function of $\lambda ,$ while all the other classes of loops examined have their center of symmetry at the origin. In spite of the different magnetic fields produced by the different loops for a finite $\lambda ,$ however, the magnetic field magnitudes converge to the same value for every loop as $\lambda$ approaches zero, for each of the points $A_1,$ $A_2,$ and $A_3.$ This is a general result that will be theoretically established in Section 3.2 (Theorem 1). As a special case of Theorem 1, the convergence points in Figure 6a, Figure 6b, and Figure 6c can be inferred to be $(\sqrt{2}/10)\mu \mathrm{T},$ $0.1\mu \mathrm{T},$ and $0.2\mu \mathrm{T},$ respectively.

3. Current loop to ideal magnetic dipole

In this section, we demonstrate that the magnetic field and vector potential due to a plane current loop approach that of an ideal magnetic dipole as the loop shrinks to a point. We first establish the result for a circular loop in Section 3.1 and generalize the result to an arbitrary loop $\mathcal{L}(\lambda ;\gamma )$ in Section 3.2.

3.1. Circular loop

We will show that the magnetic field in Lemma 1 and the vector potential in Lemma 2 approach that of an ideal magnetic dipole as the radius a becomes small. To this end, we first prove the following lemma.

Lemma 4. 

For $f^{\left(1\right)}$ as defined in (7), we have

$$\underset{a\to 0}{lim}\frac{1}{a}{f}^{\left(1\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)=\frac{3\pi sin\theta }{2r}.$$

Proof. 

Since the integrand $cos\varphi /{\left(1-tcos\varphi \right)}^{3/2}$ is bounded on the integration interval $[0,\pi ]$ for $0<t<1,$ we have, by the dominated convergence theorem [30],

$$\begin{array}{cc}\hfill \underset{t\to 0}{lim}{f}^{\left(1\right)}\left(t\right)& =\underset{t\to 0}{lim}{\int }_0^{\pi }\frac{cos\varphi }{{\left(1-tcos\varphi \right)}^{3/2}}d\varphi \hfill \\ \hfill & ={\int }_0^{\pi }\underset{t\to 0}{lim}\frac{cos\varphi }{{\left(1-tcos\varphi \right)}^{3/2}}d\varphi \hfill \\ \hfill & ={\int }_0^{\pi }cos\varphi d\varphi \hfill \\ \hfill & =0.\hfill \end{array}$$

(33)

Further, since the integrand $cos\varphi /{\left(1-tcos\varphi \right)}^{3/2}$ is continuously differentiable in both $\varphi$ and $t,$ we have

$$\begin{array}{cc}\hfill \underset{t\to 0}{lim}{f}^{\left(1\right)\prime }\left(t\right)& \stackrel{\left(a\right)}{=}\frac{3}{2}\underset{t\to 0}{lim}{\int }_0^{\pi }\frac{{cos}^2\varphi }{{\left(1-tcos\varphi \right)}^{5/2}}d\varphi \hfill \\ \hfill & \stackrel{\left(b\right)}{=}\frac{3}{2}{\int }_0^{\pi }\underset{t\to 0}{lim}\frac{{cos}^2\varphi }{{\left(1-tcos\varphi \right)}^{5/2}}d\varphi \hfill \\ \hfill & =\frac{3}{2}{\int }_0^{\pi }{cos}^2\varphi d\varphi \hfill \\ \hfill & =\frac{3\pi }{4}.\hfill \end{array}$$

(34)

Here, $\left(a\right)$ follows from Leibniz’s theorem on differentiation under the integral sign [31] and $\left(b\right)$ again follows by the application of the dominated convergence theorem. Eqns (33), (34) and L’Hôpital’s rule [32] yields

$$\underset{t\to 0}{lim}\frac{f^{\left(1\right)}\left(t\right)}{t}=\underset{t\to 0}{lim}{f}^{\left(1\right)\prime }\left(t\right)=\frac{3\pi }{4}.$$

We then have

$$\begin{array}{cc}\hfill \underset{a\to 0}{lim}\frac{1}{a}{f}^{\left(1\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)& \stackrel{\left(a\right)}{=}\left(\underset{a\to 0}{lim}\frac{2rsin\theta }{r^2+a^2}\right)·\left(\underset{t\to 0}{lim}\frac{f^{\left(1\right)}\left(t\right)}{t}\right)\hfill \\ \hfill & =\frac{2sin\theta }{r}·\frac{3\pi }{4}\hfill \\ \hfill & =\frac{3\pi sin\theta }{2r},\hfill \end{array}$$

where $\left(a\right)$ follows from the substitution $t:=2arsin\theta /\left(r^2+a^2\right)$ in the second factor. This establishes the result. □

We are now ready to examine the behavior of the magnetic field obtained in Lemma 1 as the radius a grows small. We first note that from (8), application of the dominated convergence theorem yields

$$\begin{array}{cc}\hfill \underset{a\to 0}{lim}{f}^{\left(2\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)& ={\int }_0^{\pi }d\varphi =\pi .\hfill \end{array}$$

(35)

We then have, using (6),

$$\begin{array}{cc}\hfill \underset{a\to 0}{lim}\mathbf{B}\left(\mathbf{r}\right)& =\left(\underset{a\to 0}{lim}\frac{{\mu }_{0}m}{4\pi {\left(r^2+a^2\right)}^{3/2}}\right)·\frac{2}{\pi }[cos\theta \left(\underset{a\to 0}{lim}{f}^{\left(2\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)\right)\widehat{\mathbf{r}}\hfill \\ \hfill & +\left(r\left(\underset{a\to 0}{lim}\frac{1}{a}{f}^{\left(1\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)\right)-sin\theta \left(\underset{a\to 0}{lim}{f}^{\left(2\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)\right)\right)\widehat{\mathit{\theta }}]\hfill \\ \hfill & \stackrel{\left(a\right)}{=}\frac{{\mu }_{0}m}{4\pi r^3}·\frac{2}{\pi }\left[\pi cos\theta \widehat{\mathbf{r}}+\left(\frac{3\pi sin\theta }{2}-\pi sin\theta \right)\widehat{\mathit{\theta }}\right]\hfill \\ \hfill \mathrm{(36)}& \hfill & =\frac{{\mu }_{0}m}{4\pi r^3}\left(2cos\theta \widehat{\mathbf{r}}+sin\theta \widehat{\mathit{\theta }}\right)\hfill \hfill \mathrm{(37)}& \hfill & =\frac{{\mu }_0}{4\pi r^3}\left(3\left(\mathbf{m}·\widehat{\mathbf{r}}\right)\widehat{\mathbf{r}}-\mathbf{m}\right).\hfill \end{array}$$

Eqn (36) and (37) are the standard expressions for the field of an ideal magnetic dipole $\mathbf{m}$ placed at the origin (see, for example, [33,34]). Here, $\left(a\right)$ follows from Lemma 4 and (35). We have thus established that a loop of steady current I creates the same field as an ideal magnetic dipole in the limit when the size of the loop goes to zero and the current I goes to $\infty ,$ the magnetic moment $m=\pi a^{2}I$ being kept fixed.

Remark 8. 

The limiting behavior of the vector potential in (11) can also be readily established. We first note that, using similar reasoning as in the proof of Lemma 4, we have

$$\begin{array}{cc}\hfill \underset{a\to 0}{lim}\frac{2}{\pi asin\theta }{f}^{\left(3\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)& =\underset{a\to 0}{lim}\left[\left(\frac{2}{\pi asin\theta }\right)·\left(\frac{2arsin\theta }{r^2+a^2}\right)·\frac{f^{\left(3\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)}{\frac{2arsin\theta }{r^2+a^2}}\right]\hfill \\ \hfill & =\frac{4}{\pi r}{f}^{\left(3\right)\prime }\left(0\right)\hfill \\ \hfill & =\frac{4}{\pi r}{\int }_0^{\pi }{\left[\frac{\partial }{\partial t}\left(\frac{cosu}{{\left(1-tcosu\right)}^{1/2}}\right)\right]}_{t\to 0}du\hfill \\ \hfill & =\frac{2}{\pi r}{\int }_0^{\pi }{cos}^{2}udu\hfill \\ \hfill & =\frac{1}{r}.\hfill \end{array}$$

Using (11), we then have

$$\begin{array}{cc}\hfill \underset{a\to 0}{lim}\mathbf{A}\left(\mathbf{r}\right)& =\left[\underset{a\to 0}{lim}\frac{{\mu }_0}{4\pi {\left(r^2+a^2\right)}^{1/2}}\right]·\left[\underset{a\to 0}{lim}\frac{2}{\pi asin\theta }{f}^{\left(3\right)}\left(\frac{2arsin\theta }{r^2+a^2}\right)\right]\left(\mathbf{m}\times \widehat{\mathbf{r}}\right)\hfill \\ \hfill & =\frac{{\mu }_0}{4\pi r^2}\left(\mathbf{m}\times \widehat{\mathbf{r}}\right),\hfill \end{array}$$

(38)

which is the vector potential produced by a point magnetic dipole $\mathbf{m}$ placed at the origin (see, for example, [36]).

3.2. Arbitrary loop

We now turn to the case of an arbitrary loop $\mathcal{L}(\lambda ;\gamma )$ where $\gamma$ is continuously differentiable almost everywhere in $[0,1]$ and satisfies $\gamma \left(0\right)=\gamma \left(1\right).$ We have the following result.

Theorem 1. 

For $r>0,$ the magnetic field $\mathbf{B}(\mathbf{r},\lambda ;\gamma )$ and the vector potential $\mathbf{A}(\mathbf{r},\lambda ;\gamma )$ as defined in Lemma 3 satisfy

$$\begin{array}{cc}\hfill \underset{\lambda \to 0}{lim}\mathbf{B}(\mathbf{r},\lambda ;\gamma )& =\frac{{\mu }_{0}m}{4\pi r^3}\left(2cos\theta \widehat{\mathbf{r}}+sin\theta \widehat{\mathit{\theta }}\right)=\frac{{\mu }_0}{4\pi r^3}\left(3(\mathbf{m}·\widehat{\mathbf{r}})\widehat{\mathbf{r}}-\mathbf{m}\right),\hfill \\ \hfill \underset{\lambda \to 0}{lim}\mathbf{A}(\mathbf{r},\lambda ;\gamma )& =\frac{{\mu }_0}{4\pi r^2}\left(\mathbf{m}\times \widehat{\mathbf{r}}\right).\hfill \end{array}$$

Proof. 

Using (18), we can write

$$\begin{array}{cc}\hfill \mathbf{B}(\mathbf{r},\lambda ;\gamma )& =\frac{{\mu }_{0}m}{4\pi \lambda }(2\pi \lambda cos\theta ·I_1(\mathbf{r},\lambda ;\gamma )\widehat{\mathbf{r}}+\left(I_2(\mathbf{r},\lambda ;\gamma )-2\pi \lambda sin\theta ·I_1(\mathbf{r},\lambda ;\gamma )\right)\widehat{\mathit{\theta }}\hfill \\ \hfill & +rcos\theta ·I_3(\mathbf{r},\lambda ;\gamma )\widehat{\mathit{\varphi }}),\hfill \end{array}$$

(39)

where

$$\begin{array}{cc}\hfill I_1(\mathbf{r},\lambda ;\gamma )& :={\int }_0^1\frac{\gamma {\left(u\right)}^2}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du,\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill I_2(\mathbf{r},\lambda ;\gamma )& :={\int }_0^1\frac{2\pi r\gamma \left(u\right)cos(2\pi u-\varphi )+r{\gamma }^{\prime }\left(u\right)sin(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du,\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill I_3(\mathbf{r},\lambda ;\gamma )& :={\int }_0^1\frac{2\pi \gamma \left(u\right)sin(2\pi u-\varphi )-{\gamma }^{\prime }\left(u\right)cos(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du.\hfill \end{array}$$

(42)

At a high level, the proof will follow a similar overall approach as in the derivation in Section 3.1. We will first invoke the dominated convergence theorem evaluate the limit of the $\widehat{\mathbf{r}}$ term directly, as well as to show that the quantities $I_2(\mathbf{r},\lambda ;\gamma )$ and $rcos\theta ·I_3(\mathbf{r},\lambda ;\gamma )$ each approach zero as $\lambda \to 0.$ We will then use L’Hôpital’s rule and Leibniz’s theorem on differentiation under the integral sign to evaluate the limits of the $\widehat{\mathit{\theta }}$ and $\widehat{\mathit{\varphi }}$ terms. Let $\mathcal{S}\subseteq [0,1]$ be a set of Lebesgue measure 1 such that $\gamma$ is continuously differentiable everywhere on $\mathcal{S}.$ Then $\gamma$ and ${\gamma }^{\prime }$ are bounded almost everywhere on $\mathcal{S}$ (see, for example, [30]) and therefore, almost everywhere on $[0,1].$ Let $M_1$ and $M_2$ be these upper bounds on $\left|\gamma \left(u\right)\right|$ and $\left|{\gamma }^{\prime }\left(u\right)\right|,$ respectively. Then, for $\lambda <r/2M_1,$ the integrands in (40), (41), and (42) are bounded (uniformly in $\lambda$) by $8M_1^2/r^3,$ $8{\left(4{\pi }^2{M}_1^2+M_2^2\right)}^{1/2}/r^2,$ and $8{\left(4{\pi }^2{M}_1^2+M_2^2\right)}^{1/2}/r^3,$ respectively, each of which is a finite constant and therefore yields a finite integral on $[0,1].$ Thus, using the dominated convergence theorem, we have

$$\begin{array}{cc}\hfill \underset{\lambda \to 0}{lim}{I}_1(\mathbf{r},\lambda ;\gamma )& ={\int }_0^1\underset{\lambda \to 0}{lim}\frac{\gamma {\left(u\right)}^2}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du\hfill \\ \hfill & =\frac{1}{r^3}{\int }_0^1\gamma {\left(u\right)}^{2}du\hfill \\ \hfill & =\frac{1}{\pi r^3}.\hfill \end{array}$$

(43)

Similarly, we have

$$\begin{array}{cc}\hfill \underset{\lambda \to 0}{lim}{I}_2(\mathbf{r},\lambda ;\gamma )& ={\int }_0^1\underset{\lambda \to 0}{lim}\frac{2\pi r\gamma \left(u\right)cos(2\pi u-\varphi )+r{\gamma }^{\prime }\left(u\right)sin(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du\hfill \\ \hfill & =\frac{1}{r^2}{\int }_0^1\left(2\pi \gamma \left(u\right)cos(2\pi u-\varphi )+{\gamma }^{\prime }\left(u\right)sin(2\pi u-\varphi )\right)du\hfill \\ \hfill & =\frac{1}{r^2}{\left[\gamma \left(u\right)sin(2\pi u-\varphi )\right]}_{u=0}^1\hfill \\ \hfill & =0,\hfill \end{array}$$

since $\gamma \left(0\right)=\gamma \left(1\right),$ and

$$\begin{array}{cc}\hfill \underset{\lambda \to 0}{lim}rcos\theta ·I_3(\mathbf{r},\lambda ;\gamma )& =cos\theta ·{\int }_0^1\underset{\lambda \to 0}{lim}\frac{2\pi r\gamma \left(u\right)sin(2\pi u-\varphi )-r{\gamma }^{\prime }\left(u\right)cos(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{3/2}}du\hfill \\ \hfill & =\frac{cos\theta }{r^2}{\int }_0^1\left(2\pi \gamma \left(u\right)sin(2\pi u-\varphi )-{\gamma }^{\prime }\left(u\right)cos(2\pi u-\varphi )\right)du\hfill \\ \hfill & =\frac{cos\theta }{r^2}{\left[-\gamma \left(u\right)cos(2\pi u-\varphi )\right]}_{u=0}^1\hfill \\ \hfill & =0.\hfill \end{array}$$

We then have, using L’Hôpital’s rule,

$$\begin{array}{cc}\hfill & \underset{\lambda \to 0}{lim}\frac{I_2(\mathbf{r},\lambda ;\gamma )}{\lambda }\hfill \\ \hfill & =\underset{\lambda \to 0}{lim}\frac{\partial I_2(\mathbf{r},\lambda ;\gamma )}{\partial \lambda }\hfill \\ \hfill & \stackrel{\left(a\right)}{=}-\frac{3}{2}\underset{\lambda \to 0}{lim}{\int }_0^1(\frac{2\pi r\gamma \left(u\right)cos(2\pi u-\varphi )+r{\gamma }^{\prime }\left(u\right)sin(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{5/2}}\hfill \\ \hfill & \times \left(2\lambda \gamma {\left(u\right)}^2-2r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )\right))du\hfill \\ \hfill & \stackrel{\left(b\right)}{=}\frac{3sin\theta }{r^3}{\int }_0^1\left(2\pi \gamma \left(u\right)cos(2\pi u-\varphi )+{\gamma }^{\prime }\left(u\right)sin(2\pi u-\varphi )\right)\gamma \left(u\right)cos(2\pi u-\varphi )du\hfill \\ \hfill & =\frac{3sin\theta }{2r^3}{\int }_0^1\left(2\pi \gamma {\left(u\right)}^2(1+cos(4\pi u-2\varphi ))+\gamma \left(u\right){\gamma }^{\prime }\left(u\right)sin(4\pi u-2\varphi )\right)du\hfill \\ \hfill & \stackrel{\left(c\right)}{=}\frac{3\pi sin\theta }{r^3}{\int }_0^1\gamma {\left(u\right)}^{2}du+\frac{3sin\theta }{4r^3}{\int }_{-\gamma {\left(0\right)}^{2}sin\left(2\varphi \right)}^{-\gamma {\left(1\right)}^{2}sin\left(2\varphi \right)}dv\hfill \\ \hfill & =\frac{3sin\theta }{r^3},\hfill \end{array}$$

(44)

where $\left(a\right)$ follows from Leibniz’s theorem, since the integrand is continuously differentiable in both $\lambda$ and u, $\left(b\right)$ follows from the dominated convergence theorem (taking the limit inside the integral), and $\left(c\right)$ follows from the substitution $v:=\gamma {\left(u\right)}^{2}sin(4\pi u-2\varphi ).$ Following very similar steps, we have

$$\begin{array}{cc}\hfill & \underset{\lambda \to 0}{lim}\frac{rcos\theta ·I_3(\mathbf{r},\lambda ;\gamma )}{\lambda }\hfill \\ \hfill & =\underset{\lambda \to 0}{lim}rcos\theta \frac{\partial I_3(\mathbf{r},\lambda ;\gamma )}{\partial \lambda }\hfill \\ \hfill & =-\frac{3cos\theta }{2}\underset{\lambda \to 0}{lim}{\int }_0^1(\frac{2\pi r\gamma \left(u\right)sin(2\pi u-\varphi )-r{\gamma }^{\prime }\left(u\right)cos(2\pi u-\varphi )}{{\left({\lambda }^2\gamma {\left(u\right)}^2-2\lambda r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )+r^2\right)}^{5/2}}\hfill \\ \hfill & \times \left(2\lambda \gamma {\left(u\right)}^2-2r\gamma \left(u\right)sin\theta cos(2\pi u-\varphi )\right))du\hfill \\ \hfill & =\frac{3cos\theta sin\theta }{r^3}{\int }_0^1\left(2\pi \gamma \left(u\right)sin(2\pi u-\varphi )-{\gamma }^{\prime }\left(u\right)cos(2\pi u-\varphi )\right)\gamma \left(u\right)cos(2\pi u-\varphi )du\hfill \\ \hfill & =\frac{3cos\theta sin\theta }{2r^3}{\int }_0^1\left(2\pi \gamma {\left(u\right)}^{2}sin(4\pi u-2\varphi )-\gamma \left(u\right){\gamma }^{\prime }\left(u\right)(1+cos(4\pi u-2\varphi ))\right)du\hfill \\ \hfill & =\frac{3cos\theta sin\theta }{4r^3}\left({\int }_{-\gamma {\left(0\right)}^{2}cos\left(2\varphi \right)}^{-\gamma {\left(1\right)}^{2}cos\left(2\varphi \right)}dv-{\int }_{\gamma {\left(0\right)}^2}^{\gamma {\left(1\right)}^2}dz\right)\hfill \\ \hfill & =0.\hfill \end{array}$$

(45)

Finally, using (39), we have

$$\begin{array}{cc}\hfill \underset{\lambda \to 0}{lim}\mathbf{B}(\mathbf{r},\lambda ;\gamma )& =\frac{{\mu }_{0}m}{2}\left(\underset{\lambda \to 0}{lim}{I}_1(\mathbf{r},\lambda ;\gamma )\right)(cos\theta \widehat{\mathbf{r}}-sin\theta \widehat{\mathit{\theta }})+\frac{{\mu }_{0}m}{4\pi }\left(\underset{\lambda \to 0}{lim}\frac{I_2(\mathbf{r},\lambda ;\gamma )}{\lambda }\right)\widehat{\mathit{\theta }}\hfill \\ \hfill & +\frac{{\mu }_{0}m}{4\pi }\left(\underset{\lambda \to 0}{lim}\frac{rcos\theta ·I_3(\mathbf{r},\lambda ;\gamma )}{\lambda }\right)\widehat{\mathit{\varphi }}\hfill \\ \hfill & \stackrel{\left(a\right)}{=}\frac{{\mu }_{0}m}{2\pi r^3}(cos\theta \widehat{\mathbf{r}}-sin\theta \widehat{\mathit{\theta }})+\frac{3{\mu }_{0}m}{4\pi r^3}sin\theta \widehat{\mathit{\theta }}\hfill \\ \hfill & =\frac{{\mu }_{0}m}{4\pi r^3}(2cos\theta \widehat{\mathbf{r}}+sin\theta \widehat{\mathit{\theta }}),\hfill \end{array}$$

which establishes the limit for $\mathbf{B}(\mathbf{r},\lambda ;\gamma ).$ Here, $\left(a\right)$ follows by plugging in (43), (44), and (45). The limit of $\mathbf{A}(\mathbf{r},\lambda ;\gamma )$ can be demonstrated through a similar sequence of steps. □

Remark 9. 

It can be shown that the limiting behavior established in Section 3.1 and Section 3.2 holds for any loop with a piecewise continuous boundary, provided it can be continuously deformed to a point. We refer the reader to [35] for further reading.

Remark 10. 

As remarked in Section 2.2, the limiting magentic field at the points $A_1,$ $A_2,$ and $A_3$ in Figure 5 can be computed as a special case of Theorem 1. To see this, note that using Theorem 1, we have, for $r>0,$

$$\underset{\lambda \to 0}{lim}∥\mathbf{B}(\mathbf{r},\lambda ;\gamma )∥=\frac{{\mu }_{0}m}{4\pi r^3}\sqrt{4{cos}^2\theta +{sin}^2\theta }=\frac{{\mu }_{0}m}{4\pi r^3}\sqrt{3{cos}^2\theta +1}.$$

The values of ${cos}^2\theta$ for $A_1,$ $A_2,$ and $A_3$ are, respectively, $1/3,$ $0,$ and $1,$ which, along with the numerical values $m=1\mathrm{A}·{\mathrm{m}}^2,$ $r=1\mathrm{m},$ and ${\mu }_0/4\pi ={10}^{-7}\mathrm{T}·\mathrm{m}·{\mathrm{A}}^{-1},$ immediately yields the desired limiting values of the magnetic field magnitudes.

4. Discussions

The deep connection between currents and magnetic fields has been explored in great depth and its mysteries gradually revealed through extensive research since Ørsted’s experiment, culminating in Maxwell’s equations [10] and beyond. For a historical perspective of this gradual development, we refer the reader to [37] and references therein.

A popular current area of research specifically related to magnetic fields, is the search for the magnetic monopole [41] and the building of a theoretical framework [43] to study the consequences of the existence of the same [44].