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From Balanis' Antenna Theory: Analysis and Design, 3rd Ed., p. 137:

3.3 The vector potential $\boldsymbol F$ for a magnetic current source $\boldsymbol M$

Although magnetic currents appear to be physically unrealizable, equivalent magnetic currents arise when we use the volume or the surface equivalence theorems. The fields generated by a harmonic magnetic current in a homogeneous region, with $\mathbf J = 0$ but $\mathbf M \neq 0$, must satisfy $\nabla\cdot \mathbf D = 0$. Therefore, $\mathbf E_F$ can be expressed as the curl of the vector potential $\mathbf F$ by $$\mathbf E_F = -\frac1\epsilon \nabla\times\mathbf F$$

I didn't find much information about this "electric vector potential" (I had never heard about it before, actually). I don't understand how the magnetic current density $\mathbf{M}$ is related to the electric field $\mathbf{E}$, and why the conditions imposed ($\mathbf{J}=0$ and $\mathbf{M}\neq0$) lead to $\nabla\cdot\mathbf{D}=0$.

Can someone explain this or tell me where I can find more information about this vector potential?

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4 Answers 4

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Charge conservation demands that $\text{div} \textbf{J} = -\frac{\partial \rho}{\partial t}$ hold always. Thus in regions where there is no electric current, $\textbf{J}=0$, the charge density $\rho(\textbf{r},t)$ must be constant in time, $\rho= \rho(\textbf{r})$ and thus contributes nothing to radiation (which is what Balanis is interested in) and can be neglected. There follows $\text{div}\textbf{D}=0$.

The purpose of introducing the "electric vector potential" is to exploit the mathematical analogy with the physically more meaningful $\textbf{A}$ vector. This is similar to what is done in magnetostatics where the concept of magnetic potential is used in analogy with the conventional electric potential used in electrostatics. Both potentials satisfy Laplace's equations $\text{divgrad}\psi=0$ in regions free of sources, and as long as one stays in simple, singly connected domains the analogy holds.

The use of the $\textbf{F}$ potential follows similar line of thinking. One solves with much pain the electric dipole radiation problem and then to solve for the radiation from a loop one replaces the loop with an equivalent magnetic dipole, a kind of dual to the electric one.

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  • $\begingroup$ Thanks for the answer. My question was only partially answered, though. Could you explain the relationship between $\mathbf E$ and $\mathbf M$? I don't know how to derive any link between them from Maxwell's equations. $\endgroup$
    – Tendero
    Commented Apr 26, 2017 at 17:03
  • $\begingroup$ It would be way too long to go through the full derivation Balanis has in detail, but look at Eqs (3.27) for $\textbf{A}$ and (3.28) for $\textbf{F}$. The integral for $\textbf{A}$ is the result of the electric current $\textbf{J}$, while the integral for $\textbf{F}$ is the result of the magnetic current $\textbf{M}$. They are formally the same but of course $\textbf{J}$ is a true current of real electric charges while $\textbf{M}$ is fictitious. Nevertheless, if you can find the solution for one then you can find its dual approximately that is good enough. $\endgroup$
    – hyportnex
    Commented Apr 26, 2017 at 17:20
  • $\begingroup$ An electric dipole oscillation is the 1st order approximation of a linear dipole antenna of two wires that can be solved with reasonable accuracy using (3.27). To solve the radiation problem for a loop is much more difficult directly but if you take advantage of this analogy and duality represented by (3.28) so that you replace a small loop with an equivalent magnetic dipole of oscillating magnetic charges you get an equally good in practice formula for the radiation formula of the loop. $\endgroup$
    – hyportnex
    Commented Apr 26, 2017 at 17:26
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Short answer: $\nabla \cdot \mathbf{D}=0$ comes from taking the divergence of $\nabla \times \mathbf{H}_F = j \omega \epsilon \mathbf{E}_F$ since we have assumed that $\mathbf{J}=0$ for this case.

We use surface equivalence to replace physical structures with fictitious $\mathbf{J}$ and $\mathbf{M}$ so that we can use the free space Green's function to calculate radiated electric and magnetic fields.


And now to overexplain:

For the case of both electric and magnetic sources, we start with the equations

$\nabla \times \mathbf{E} = -j\omega \mu \mathbf{H} -\mathbf{M}$

$\nabla \times \mathbf{H} = j\omega \epsilon \mathbf{E} + \mathbf{J}$

Using superposition, we can consider the electric and magnetic current sources separately.

$\nabla \times \mathbf{E}_A = -j\omega \mu \mathbf{H}_A$

$\nabla \times \mathbf{H}_A = j\omega \epsilon \mathbf{E}_A + \mathbf{J}$

and

$\nabla \times \mathbf{E}_F = -j\omega \mu \mathbf{H}_F - \mathbf{M}$

$\nabla \times \mathbf{H}_F = j\omega \epsilon \mathbf{E}_F$

and using superposition we have

$\mathbf{E} = \mathbf{E}_A+\mathbf{E}_F$

$\mathbf{H} = \mathbf{H}_A+\mathbf{H}_F$

Solving the electric current case, we use the usual magnetic vector potential $\mathbf{A}$, where $\mathbf{H}_A = \frac{1}{\mu}\nabla \times \mathbf{A}$, and this will yield $\mathbf{E}_A$ and $\mathbf{H}_A$. All of the logic for solving for $\mathbf{E}_F$ and $\mathbf{H}_F$ from the magnetic current source $\mathbf{M}$ is exactly the same, just switch $\mathbf{E}$ and $\mathbf{H}$ (and other constants and signs) and do the same thing.

Take the following divergence

$\nabla \cdot \left( \nabla \times \mathbf{H}_F = j\omega \epsilon \mathbf{E}_F \right)$

to get

$\nabla \cdot \epsilon \mathbf{E}_F=0$

I think this was the part you were missing. Then we can write

$\mathbf{E}_F = -\frac{1}{\epsilon} \nabla \times \mathbf{F}$.

This will allow us to solve for $\mathbf{E}_F$ and $\mathbf{H}_F$.

Leaving some steps out, we'll end up with the vector potentials $\mathbf{A}$ and $\mathbf{F}$ as

$\nabla^2 \mathbf{A} +k^2 \mathbf{A} = -\mu \mathbf{J}$

$\nabla^2 \mathbf{F} +k^2 \mathbf{F} = -\epsilon \mathbf{M}$

This can be seen as 6 separate scalar wave equations, and in free space the solution is well known using Green's functions. Note that using the Lorenz gauge is necessary to obtain this form. Assuming that $\mathbf{J}$ and $\mathbf{M}$ are surface currents on some surface $\Gamma$, then

$\mathbf{A}(\mathbf{r}) = \mu \int_\Gamma g(\mathbf{r},\mathbf{r}') \mathbf{J}(\mathbf{r}') ds'$

$\mathbf{F}(\mathbf{r}) = \epsilon \int_\Gamma g(\mathbf{r},\mathbf{r}') \mathbf{M}(\mathbf{r}') ds'$

where $g(\mathbf{r},\mathbf{r}') = \frac{e^{-jk|\mathbf{r}-\mathbf{r}'|}}{4\pi|\mathbf{r}-\mathbf{r}'|}$

And then once we have $\mathbf{A}(\mathbf{r})$ and $\mathbf{F}(\mathbf{r})$ we can get

$\mathbf{E}(\mathbf{r}) = -j \omega \mathbf{A} - \frac{j}{\omega \mu \epsilon} \nabla (\nabla \cdot \mathbf{A}) - \frac{1}{\epsilon} \nabla \times \mathbf{F}$

$\mathbf{H}(\mathbf{r}) = -j \omega \mathbf{F} - \frac{j}{\omega \mu \epsilon} \nabla (\nabla \cdot \mathbf{F}) + \frac{1}{\mu} \nabla \times \mathbf{A}$

We can also take the limit $|\mathbf{r}| \rightarrow \infty$ to obtain far fields.


As to why we consider fictitious magnetic currents $\mathbf{M}$:

Suppose we have a radiating antenna, and we know the electric and magnetic fields in some close vicinity of the antenna (e.g. using a numerical finite element simulation), but we want to find the radiated fields at some far away point. We can't use simply use the actual electric currents along with the integrals above, since those equations rely on the free space Green's function and the presence of a physical radiating structure means this is not free space.

Given a closed surface $\Gamma$ which contains the antenna, if we know the fields $\mathbf{E}$ and $\mathbf{H}$ on $\Gamma$, then we can consider a equivalent problem in which we (1) introduce fictitious sources $\mathbf{J}=\mathbf{\hat{n}} \times \mathbf{H}$, $\mathbf{M} = -\mathbf{\hat{n}} \times \mathbf{E}$ on $\Gamma$, where $\mathbf{E}$ and $\mathbf{H}$ are the known fields from the original problem, and (2) remove the physical structure and set $\mathbf{E}=\mathbf{H}=0$ inside the surface. Surface equivalence tells us that outside $\Gamma$ the two scenarios will produce the same fields. The equivalent problem has the advantage that it is free space, and therefore we know the Green's function. Therefore we can use the integrals above to calculate the vector potentials and then the electric and magnetic fields at any point outside $\Gamma$.

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  • $\begingroup$ This was exactly the answer I was looking for. Thank you very much! I have one question, however. In this equation: $\nabla \times \mathbf{E} = -j\omega \mu \mathbf{H} -\mathbf{M}$, where does the $\mathbf M$ come from? I understand that the magnetic current has to come up somewhere, but I had never seen it in Maxwell's Equations before. $\endgroup$
    – Tendero
    Commented Apr 26, 2017 at 17:56
  • $\begingroup$ @Tendero $\mathbf{M}$ is entirely fictitious. One way to think about it is that it's a dual quantity to $\mathbf{J}$. I don't have the textbook that you have, but I do have Balanis Advanced Engineering Electromagnetics which has a conversation about duality in electromagnetics at the beginning of Chapter 7. I'm sure Balanis also talks about it in your textbook. The real reason is that you can't do surface equivalence without it. $\endgroup$
    – LedHead
    Commented Apr 26, 2017 at 18:31
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Simply put, if $\nabla\cdot\mathbf D = 0$ in some region (which is simply Gauss's law for the $\mathbf D$ field in a region with no free charges), then $\mathbf D$ is a solenoidal field in that region, and such fields can always be represented in the form $$ \mathbf D = (-)\, \nabla \times \mathbf F. $$ The relationship for $\mathbf E$ follows from the consitutive relationship $\mathbf D = \epsilon\mathbf E$.


As to why this is a useful way to do things, though, I'm not particularly sure - presumably Balanis goes on to talk about applications that would justify that expression. But it's a perfectly legitimate way to do things under those assumptions (no free charges, linear medium).

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  • $\begingroup$ I get that if $\nabla\cdot\mathbf D = 0$, one can express $\mathbf D$ as the curl of some field. What I don't understand is why $\mathbf J =0,\mathbf M\neq 0 \implies \nabla\cdot\mathbf D = 0$, and that was my question. Sorry if I didn't express myself correctly. AFAIK, one can say that $\nabla\cdot\mathbf D = 0$ only if $\rho_{free} = 0$, and that assumption is not mentioned in the derivation I wrote in the question. $\endgroup$
    – Tendero
    Commented Apr 25, 2017 at 16:46
  • $\begingroup$ @Tendero You'll have to ask the author. $\nabla\cdot\mathbf D=0$ is a reasonable assumption but it's independent of the other two. $\endgroup$ Commented Apr 25, 2017 at 16:56
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Magnetic Currents are not necessarily fictitious. In a non-conservative system, it just represents a forcing function to Faraday's law. Like if I rotate a magnet, where the south pole remains stationary, but the north pole moves, then there will be a real net magnetic bound current, also called impressed magnetic currents. This generates an EMF across the rotating magnet. See Some Equivalence Theorems of Electromagnetics and Their Application to Radiation Problems By S. A. SCHELKUNOFF

For a more recent paper see: DOI: 10.1103/PhysRevApplied.15.014007

The electrodynamics of a negative reactance or voltage source, such as a hertzian dipole antenna, require magnetic currents.

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