Maxwell–Lodge Effect and Electric Fields of a Solenoid

1. Introduction

The Maxwell–Lodge effect [1,2] refers to the observed electric field outside a long solenoid with a slowly varying current. According to classical electromagnetism, the Coulomb electric field inside and outside the solenoid is zero because there is no net charge. Thus, the observed electric field must be due to induction. For a long solenoid with constant current, the magnetic field inside the solenoid is uniform, while the magnetic field just outside is nearly zero. However, when the current varies slowly, the assumption of a zero magnetic field outside leads to inconsistencies in Maxwell’s equations (as well as in extended electrodynamics) [3,4]. Solutions to Maxwell’s equations under these conditions reveal both an induced electric field and a very small magnetic field outside the solenoid [5]. It is generally accepted that these fields can be treated as quasi-stationary, with negligible electromagnetic radiation.

According to Maxwell’s equations [6], a changing magnetic field generates an electric field, and by the principle of locality in physics, this must occur locally. In the case of a slowly varying current, the magnetic field inside the solenoid changes, while the magnetic field outside remains essentially zero. This makes it difficult for classical electromagnetism to explain the detected electric field outside the solenoid unless one assumes a non-local effect, i.e., the varying magnetic field inside somehow causes the electric field outside [7].

Even though the magnetic field outside is zero, the magnetic vector potential is not [8]. Since the time-varying magnetic vector potential contributes to the electric field, it is proposed that the observed field outside is caused by this variation. While this explanation seems to uphold the locality principle, it assumes that the magnetic vector potential is physically real—a point still debated in classical theory [2]. Many studies [2,8,9,10,11] support the physical reality of magnetic vector potential, though the traditional view treats it as merely a mathematical construct.

A pioneering study explained the induced electromotive force outside a long solenoid using Weber’s force [12], avoiding the controversial reality assumption of a physical vector potential. In this paper, we continue along this path, using the field theory of Weber’s electrodynamics [13] to derive the electric fields of a solenoid. We demonstrate that electron motion in the current produces magnetic and electric fields both inside and outside the solenoid.

2. Electric Fields of Weber’s Electrodynamics

In the electric field theory of Weber’s electrodynamics [13], the force that a charge $Q$ exerts on another charge $q$ can be written in the following form (equation 1) for any given reference frame. Let $Q$ have velocity $v_i$ and acceleration $a_i$​, and $q$ have velocity $u_i$​ and acceleration $w_i$​. The distance between the two charges is $r$, and $n_k$ is the unit vector pointing from charge $Q$ to charge $q$. Here, ${\epsilon }_0$​ is the vacuum permittivity and $c$ is the speed of light.

$$F_k=q\left[E_k+u_i{M}_{ikj}{u}_j+{u_{i}B}_{ik}+{\tilde{E}}_k+{w_{i}L}_{ik}+{\stackrel{̿}{E}}_k\right]$$

(1)

where:

$$\begin{array}{c}{E}_k={\displaystyle \frac{Q{n}_k}{4\pi {\epsilon }_0{r}^2}} \hfill \\ M_{ikj}={\displaystyle \frac{E_k}{c^2}}\left({\delta }_{ij}-{\displaystyle \frac{3}{2}}{n}_i{n}_j\right)\hfill \\ B_{ik}={\displaystyle \frac{E_k}{c^2}}\left(-2{\delta }_{ij}+3n_i{n}_j\right)v_j\hfill \\ {\tilde{E}}_k=v_i{\displaystyle \frac{E_k}{c^2}}\left({\delta }_{ij}-{\displaystyle \frac{3}{2}}{n}_i{n}_j\right)v_j\hfill \\ L_{ik}={\displaystyle \frac{E_{k}r}{c^2}}{n}_i\hfill \\ {\stackrel{̿}{E}}_k={\displaystyle \frac{-E_{k}r}{c^2}}{n}_i{a}_i \hfill \end{array}$$

(2)

In this formulation:

• $E_k$ is the classical Coulomb electric field,
• $B_{ik}$ is equivalent to the magnetic field,
• ${\tilde{E}}_k$ is the electric field due to the charge $Q$ (source) velocity,
• ${\stackrel{̿}{E}}_k$ is the electric field due to the charge $Q$ (source) acceleration,
• $M_{ikj}$ is the electric field acting on the charge $q$ (receiver) velocity,
• $L_{ik}$ is the electric field acting on the charge $q$ (receiver) acceleration.

To obtain the total field from multiple source charges, we sum the contributions of each individual charge. To compute the force on a single receiver charge, we apply Equation (1) using its properties (charge, velocity, and acceleration) and the total electric field resulting from the superposition of all source charges.

3. Long Solenoid with Constant Current

In this section, we analyze a circular electric current in a solenoid centered at the origin, as shown in Figure 1a. Our goal is to calculate the electric fields at point A on the x-axis, as depicted in Figure 1b,c.

Assume that the current is constant. The positive charges in the solenoid are stationary, while the negative charges (electrons) move with a velocity $\stackrel{⃑}{\mathit{v}}$ along the circular path, experiencing an acceleration $\stackrel{⃑}{a}$ that points towards the z-axis (Figure 1b).

Given the coexistence of positive and negative charges in the solenoid, the terms $E_k$, $M_{ikj}$, and $L_{ik}$ will be zero due to the cancellation of the positive and negative charge contributions. Therefore, we focus on calculating the terms $B_{ik}$, ${\tilde{E}}_k$, and ${\stackrel{̿}{E}}_k$ from the moving negative charges in the solenoid.

Let’s begin by describing the current segment $ds=Rd\theta ·dz$, where $\theta$ is the angle around the solenoid’s circular path. The vector $\stackrel{⃑}{r}$ is pointing from the segment $ds$ to point A, with a unit vector $n_i$. The surface density of negative charge is $-\rho$, where $\rho$ is the electron charge density.

The vector $\stackrel{⃑}{r}$, its magnitude and unit vector are given by:

$$\begin{array}{c}\stackrel{⃑}{r}=\left(x-R\cos \theta ,-R\sin \theta ,-z\right)\hfill \\ r=\left|\stackrel{⃑}{r}\right|=\sqrt{{\left(x-R\cos \theta \right)}^2+{\left(R\sin \theta \right)}^2+z^2}\hfill \\ n_i={\displaystyle \frac{1}{r}}\left(x-R\cos \theta ,-R\sin \theta ,-z\right)\hfill \end{array}$$

(3)

Next, the differential Coulomb electric field ${dE}_k$​ due to the current segment is:

$${dE}_k={\displaystyle \frac{ds·\left(-\rho n_k\right)}{4\pi {\epsilon }_0{r}^2}}$$

(4)

The velocity and acceleration components of the moving negative charges (electron) are given by:

$$\begin{array}{c}{v}_j=v\left(-\sin \theta ,\cos \theta ,0\right)\hfill \\ a_i={\displaystyle \frac{v^2}{R}}\left(-\cos \theta ,-\sin \theta ,0\right)\hfill \end{array}$$

(5)

Using the above expressions for $v_j$ and $a_i$, we proceed to calculate the magnetic field and electric field components resulting from these moving electrons. From Equations (2, 3, 4, 5), the differential magnetic field contribution from the current segment is:

$${dB}_{ik}={\displaystyle \frac{ds·\left(-\rho v\right)}{4\pi {\epsilon }_0{r}^5{c}^2}}\left(\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right)$$

(6)

where the components $b_{ij}$ are given by the following expressions:

$$\begin{array}{c}{b}_{11}=2\sin \theta r^{2}x-2\sin \theta r^{2}R\cos \theta +6x^2\sin \theta R\cos \theta -3x\sin \theta R^2{cos}^2\theta -3x^3\sin \theta \hfill \\ b_{12}=-2{sin}^2\theta r^{2}R-3x{sin}^2\theta R^2\cos \theta +3x^2{sin}^2\theta R\hfill \\ b_{13}=-2\sin \theta r^{2}z-3x\sin \theta R\cos \theta z\hfill \\ b_{21}=-2\cos \theta r^{2}x+2{cos}^2\theta r^{2}R+3x^2{sin}^2\theta R-3x{sin}^2\theta R^2\cos \theta \hfill \\ b_{22}=2\sin \theta r^{2}R\cos \theta -3x{sin}^3\theta R^2\hfill \\ b_{23}=2\cos \theta r^{2}z-3x{sin}^2\theta Rz\hfill \\ b_{31}=3x^2\sin \theta z-3x\sin \theta R\cos \theta z\hfill \\ b_{32}=-3x{sin}^2\theta zR\hfill \\ b_{33}=-3x\sin \theta z^2 \hfill \end{array}$$

(7)

We integrate the components along the solenoid, and then we obtain $B_{ik}$:

$$B_{ik}=\left(\begin{array}{ccc}0& B_{12}& 0\\ B_{21}& 0& 0\\ 0& 0& 0\end{array}\right)$$

(8)

In this equation, all other components become zero due to symmetry, except for $B_{12}$ and $B_{21}$, for which we perform numerical integration. The parameters for the numerical integration are predefined as follows: the solenoid radius $R=0.1m$; the solenoid length is 10 m; the electron surface density $\rho =8.5\times {10}^{26}{m}^{-2}$; the electron charge is $1.6\times {10}^{-19}C$; the electron drift velocity is $4\times {10}^{-4}m/s$; the speed of light $c=3\times {10}^{8}m/s$; and the vacuum permittivity ${\epsilon }_0=8.854\times {10}^{-12}F/m$.

The result of the integration is plotted in Figure 2a. Inside the solenoid, the $B_{12}$ and $B_{21}$ components are uniform. Outside the solenoid, the $B_{12}$ and $B_{21}$ components are zero. This outcome is consistent with the magnetic field predicted by classical electromagnetism.

Similarly, from equations (2,3,4,5), we can derive differential electric field ${d\tilde{E}}_k$:

$${d\tilde{E}}_k={\displaystyle \frac{\rho v^2·ds}{8\pi {\epsilon }_0{r}^5{c}^2}}\left(\begin{array}{ccc}{\tilde{e}}_1& {\tilde{e}}_2& {\tilde{e}}_3\end{array}\right)$$

(9)

where:

$$\begin{array}{c}{\tilde{e}}_1=-2r^{2}x+2r^{2}R\cos \theta +3x^3{sin}^2\theta -3x^2{sin}^2\theta R\cos \theta \hfill \\ {\tilde{e}}_2=2\sin \theta r^{2}R-3x^2{sin}^3\theta R\hfill \\ {\tilde{e}}_3=2r^{2}z-3x^2{sin}^2\theta z\hfill \end{array}$$

(10)

We integrate these components along the solenoid to obtain ${\tilde{E}}_k$. Due to symmetry, all components become zero except for ${\tilde{E}}_1$, which is computed through numerical integration.

$${\tilde{E}}_k=\left(\begin{array}{ccc}{\tilde{E}}_1& 0& 0\end{array}\right)$$

(11)

Again, from equation (2,3,4,5), we can derive differential electric field $d{\stackrel{̿}{E}}_k$:

$$d{\stackrel{̿}{E}}_k={\displaystyle \frac{\rho v^2·ds}{4\pi {\epsilon }_0{r}^3{c}^{2}R}}\left(\begin{array}{ccc}{\stackrel{̿}{e}}_1& {\stackrel{̿}{e}}_2& {\stackrel{̿}{e}}_3\end{array}\right)$$

(12)

where:

$$\begin{array}{c}{\stackrel{̿}{e}}_1=-x^2\cos \theta +xR+xR{cos}^2\theta -R^2\cos \theta \hfill \\ {\stackrel{̿}{e}}_2=xR\sin \theta \cos \theta -R^2\sin \theta \hfill \\ {\stackrel{̿}{e}}_3=x\cos \theta z-Rz\hfill \end{array}$$

(13)

We integrate these components along the solenoid to obtain ${\stackrel{̿}{E}}_k$. Due to symmetry, all components become zero except for ${\stackrel{̿}{E}}_1$, which is computed using numerical integration.

$${\stackrel{̿}{E}}_k=\left(\begin{array}{ccc}{\stackrel{̿}{E}}_1& 0& 0\end{array}\right)$$

(14)

The integrated ${\tilde{E}}_1$ and ${\stackrel{̿}{E}}_1$ components are plotted in Figure 2b. Inside the solenoid, ${\tilde{E}}_1$ is zero, while ${\stackrel{̿}{E}}_1$ is non-zero and increases with radius. Outside the solenoid, both ${\tilde{E}}_1$ and ${\stackrel{̿}{E}}_1$ are non-zero; however, their sum is zero.

For a charge $q$ located at point A, with velocity $u_i$ and acceleration $w_i$, the force exerted on it can be calculated using equation (1):

$$F_k=q\left[E_k+u_i{M}_{ikj}{u}_j+{u_{i}B}_{ik}+{\tilde{E}}_k+{w_{i}L}_{ik}+{\stackrel{̿}{E}}_k\right]=q{u_{i}B}_{ik}+{\tilde{E}}_k+{\stackrel{̿}{E}}_k$$

(15)

The force term $q{u_{i}B}_{ik}$ is consistent with the Lorentz force derived from classical magnetic theory. However, the terms ${\tilde{E}}_k$ and ${\stackrel{̿}{E}}_k$ can only be obtained from the field theory of Weber’s electrodynamics.

4. Long Solenoid with Slowly Varying Current

Here, we consider a solenoid with a slowly varying current (Figure 3a). The solenoid has the same geometry and parameters as the one in Section 3 (Figure 1), except that it has an acceleration component in the tangential direction. We will not repeat the analysis of the electric fields due to the current itself. Instead, we will only analyze the electric field contributed by this tangential acceleration component.

$$a_i=a_0\left(-\sin \theta ,\cos \theta ,0\right)$$

(16)

From equation (2,3,4,16), we can calculate differential electric field $d{\stackrel{̿}{E}}_k$ due to this tangential acceleration component:

$$d{\stackrel{̿}{E}}_k={\displaystyle \frac{\rho a_0·ds}{4\pi {\epsilon }_0{r}^3{c}^2}}\left(\begin{array}{ccc}{\stackrel{̿}{e}}_1& {\stackrel{̿}{e}}_2& {\stackrel{̿}{e}}_3\end{array}\right)$$

(17)

where:

$$\begin{array}{c}{\stackrel{̿}{e}}_1=-x^2\sin \theta +xR\sin \theta \cos \theta \hfill \\ {\stackrel{̿}{e}}_2=xR{sin}^2\theta \hfill \\ {\stackrel{̿}{e}}_3=x\sin \theta z\hfill \end{array}$$

(18)

We integrate them along the solenoid, resulting in ${\stackrel{̿}{E}}_k$. Here, all other components become zero due to symmetry except for ${\stackrel{̿}{E}}_2$, for which we perform numerical integration.

$${\stackrel{̿}{E}}_k=\left(\begin{array}{ccc}0& {\stackrel{̿}{E}}_2& 0\end{array}\right)$$

(19)

The parameters we used are the same as those in Section 3, with the addition that $a_0=4\times {10}^{-4}m/s^2$. This represents the maximum acceleration of a sinusoidal current at 1 Hz, assuming a maximum current of $4\times {10}^{-4}m/s$. The integration outcome is plotted in Figure 3b. Inside the solenoid, ${\stackrel{̿}{E}}_2$ increases linearly with radius. Outside the solenoid, ${\stackrel{̿}{E}}_2$ decreases with radius. This outcome is consistent with the electric field derived from classical electromagnetism, assuming an infinitely long solenoid.

5. Short Solenoid with Constant Current

Here, we consider a scenario in which the solenoid is short and the current is constant. The solenoid length is reduced to 0.01 m, and all other parameters are the same as those in Section 3. The calculated $B$ and $E$ components are plotted in Figure 4.

The $B$ components are no longer uniform inside the solenoid. Outside the solenoid, the $B$ components reverse polarity and gradually decrease with radius (Figure 4a). This behavior is consistent with classical electromagnetism. However, classical electromagnetism predicts a zero electric field both inside and outside the solenoid. In contrast, the field theory of Weber’s electrodynamics predicts non-zero $E$ components. Compared with the long solenoid case, the $E$ components are dominated by the velocity-related component ${\tilde{E}}_1$. The acceleration-related component ${\stackrel{̿}{E}}_1$ is relatively small (Figure 4b), and the two components do not cancel each other outside the solenoid.

6. Conclusion

In this paper, we provided a simple and straightforward explanation of the Maxwell-Lodge effect using the electric field theory of Weber’s electrodynamics. The tangential electric field is generated directly by the moving electrons in the solenoid, rather than being induced by the magnetic field or magnetic vector potential. Additionally, we demonstrated the different components of the electric field both inside and outside long and short solenoids, highlighting their similarities and differences compared to classical electromagnetism.