# How do I calculate electric fields due to currents of magnetic dipoles?

Short version of my question:

Do dipole currents cause fields? I think currents of aligned magnetic dipoles cause an electric field, but I don't know how to calculate this field except in the simplest of cases. I'd like to know how!

Full version of my question:

Suppose I have a wire (or pipe) that carries a steady current of particles, and the wire is shaped like a sine curve in the $X$-$Y$ plane with spatial period $P$ and amplitude $A$. I want to calculate the electric and magnetic fields produced by the current in the wire.

1. If the flowing particles are charged, I can use the Biot-Savart law to write an integral and (at least numerically) calculate the resulting magnetic field. In the limit where $A=0$, I can check my results against a Lorentz-boost calculation, and see that the two methods agree.

2. If the flowing particles are electric dipoles pointing in the $Z$-direction, things are more complicated, but I can still use the Biot-Savart law, adding up the magnetic fields from two equal, opposite, displaced sinusoidal electric currents to get a net magnetic field, albeit one that falls off much more quickly than in the previous case. I can still check my results against the Lorentz boost in the $A=0$ limit, and find that the two methods agree.

3. However, if the flowing particles are magnetic dipoles pointing in the $Z$-direction, I don't know how to calculate their electric fields except in the $A=0$ limit. In this limit, I can use a Lorentz boost argument to show that an electric field exists, but I don't know of any equivalent to the Biot-Savart law for currents of magnetic dipoles. I would be very surprised if the electric field vanished in the $A>0$ case. How do I proceed? Is this covered in elementary textbooks, and I've missed it somehow?

I'm tempted to model this third case in a similar fashion to case 2, as a fictitious pair of equal and opposite "magnetic currents", and use an electric version of Biot-Savart. This produces the correct result in the $A=0$ case, and seems reasonable in the $A>0$ case, but seems to have no basis in any textbook or reference I can find. What am I missing?

Caveats:

1. If at all possible, please base your answer on credible textbooks or peer-reviewed papers. For example, if you think Maxwell's equations is missing a $\vec{v} \times \vec{M}$ term, I'd very much like to see an external link to back it up.

2. This may be obvious from the 'spirit' of the question, but I'm not particularly attached to the sinusoidal shape of the current-carrying wire. Any nontrivial shape of the wire that precludes a Lorentz boost is equally interesting to me, so go ahead and change the wire's shape if it makes the math easier.

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Just a heuristic thought: thinking of magnetic dipoles as "little current loops", for many equal dipoles next to each other the internal currents cancel out and you are left with an edge current. So you get back to your picture of two counter-propagating electric currents infinitesimally separated, only now in the x-y plane. That's for static dipoles. To get a current imagine doing a Lorentz boost and you'll end up with counter-propagating currents of unequal magnitude. – Michael Brown Aug 14 '13 at 15:32
Thinking of them as little current loops is fine, but I don't think this makes the calculation any easier, or any harder. – Andrew Aug 17 '13 at 18:20
@MichaelBrown: You went astray in the last sentence. The Lorentz boost makes the charge density unequal on the two sides, not the current. Here's a more detailed analysis of boosting a current loop: physicsforums.com/showthread.php?t=631446 – Ben Crowell Aug 18 '13 at 17:50
I don't think you've missed anything in elementary textbooks: the situation you describe is highly "contrived" (I don't mean that word to be negative) in the sense that it would be extremely hard to set up in the experimental world: I'm thinking electrons with their spins all aligned in the Z direction. So I'm sure if elementary textbook authors even think of it, they wouldn't put it in a first textbook. The question is an excellent and intriguing thought experiment +1 and I'll certainly think about it, but I think BenCrowell and @MichaelBrown are onto a good ideas so they may well answer. – WetSavannaAnimal aka Rod Vance Aug 19 '13 at 0:04
@BenCrowell Thanks. Now that I actually do the Lorentz boost I see that you get equal and opposite charge densities and currents in the wires (I don't get unequal in magnitude charge densities, only opposite signs). So in the OP's configuration you get orthogonal electric and magnetic dipole fields. A magnetic dipole along the z-axis and an electric dipole in the x-y plane orthogonal to the current. – Michael Brown Aug 19 '13 at 1:09

Since there seems to have been some doubt about whether the electric field of a boosted magnetic dipole is nonzero, this paper by Hnizdo (what a great name!) may be helpful. Section 3 explicitly calculates the field. This is all tangentially related to the Mansuripur paradox as well.

I think dj_mummy's technique works, but it has the disadvantage that it requires you to do a calculation for some finite $l$ and the take the limit $l\rightarrow0$. Here's a different technique, which avoids that.

The vector potential due to a magnetic dipole at rest is $A^\mu=(\phi,\textbf{A})$, with $\phi=0$ and $\textbf{A}=\textbf{m}\times\hat{\textbf{r}}/r^2$. Do a Lorentz boost on this vector, and you get a potential $A^{\mu'}$ for a moving dipole. In fact, you only care about the timelike component of this, so you don't need to compute the rest. Integrate this over all the dipoles (each with its own position and velocity vector), and then take the gradient to get the electric field. Note that although the electric field is $-\nabla\phi-\partial\textbf{A}/\partial t$, the second term can be neglected; the integrated value of $\textbf{A}$ is constant in the $\mu'$ frame, since the dipole current is static in that frame.

Another way to approach this, suggested by Art Brown in a comment below, goes like this. Hnizdo shows that to a sufficient approximation, and ignoring some subtleties related to the definition of multipoles, we can take a magnetic dipole $\textbf{m}$ to have electrical properties, in the lab frame, characterized by an electric dipole moment $\textbf{p}'=\textbf{v}\times \textbf{m}$ (in units with $c=1$). This makes the whole problem look directly analogous to the idea of developing the Biot-Savart law by assembling a collection of magnetic dipoles, so although I haven't worked it out in detail, it sounds like you can get something that's an exact analog of the Biot-Savart law.

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v nice answer. Why not add, from eqn 46 of your ref, that a B-dipole $\boldsymbol{m_0}$ with velocity $\boldsymbol{v}$ acquires an electric dipole moment $\boldsymbol{v \times m_0}/c^2$ (mks, non-relativistic), which can then be integrated over the "circuit"? Very Biot-Savart-like... – Art Brown Aug 19 '13 at 1:06
@ArtBrown: Cool idea! Done. – Ben Crowell Aug 19 '13 at 1:32
The Hnizdo reference is a good one. I'm thinking of writing to him about this; it's a small effect, but I'm surprised it's not mentioned in Jackson. (If it is, I missed it.) – Andrew Aug 21 '13 at 15:24
This reminds me: I have an old question similar in spirit to this one you might be interested in. physics.stackexchange.com/q/6581/2359 – Andrew Aug 21 '13 at 15:25

An easy way I came up with was to take two separate magnetic monopole currents and plug them into Maxwell's modified equations to account for a magnetic monopole.

So based on what you said, here are the given parameters (please bear with my absolute neglect of formatting): The four-velocity field is

V0 = (1-v/c)^-1/2

V1 = v/c * cosα * (1-v/c)^-1/2

V2 = v/c * sinα * (1-v/c)^-1/2

V3 = 0

where α = angle formed by tangent to sine curve with x-axis, it is obviously a function of x

ρ+ = gδ(z-l/2)δ(y-Asin(2πx/P))

ρ- = -gδ(z+l/2)δ(y-Asin(2πx/P))

where g is some proportionality constant and l is the distance between the monopoles. From your question, I assume that all positively charged monopoles flow in the z=l/2 plane and the negative monopoles flow in z=-l/2 plane

jμ = (ρ+ + ρ-)*Vμ

Just plug these values in and you should get your answer. For any arbitrary curve all you need to do is adjust Vμ and ρ+ and ρ-.

You can find Maxwell's modified equations on http://en.wikipedia.org/wiki/Magnetic_monopoles. You could also solve in terms of the Hodge dual of the stress tensor. The equations are given in the same link.

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As Ben Crowell pointed out in his answer, the structure of the dipole is approximated to 2 monopoles separated by a finite distance. One would have to take l to 0 to get closer to the dipole structure you desire. – dj_mummy Aug 18 '13 at 18:38

I think your premise 3) with the Lorentz boost argument isn't correct. Let's consider the $A=0$ case, a current of magnetic dipoles (oriented in $z$-direction) flowing in $x$-direction. Then, the magnetic field does NOT change, and with Faraday's $\nabla \times \ \vec E =-\frac{\partial \vec B}{\partial t}$, $\nabla \times \ \vec E$ vanishes, thus there is no dynamic electric field. But there is no static electric field either, for symmetry reasons (imagine your dipoles pointing in $-z$ direction), and because there isn't an excess of electric charge anywhere.

(Such a symmetry argument may apply to premise 2) as well, so it would be nice to see it worked out in the limiting case when the distance of charges goes to zero)

I think all this applies to the sinusoidal shape as well.

[this answer has been edited, since I do not have comment privileges]

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Perhaps I'm reading your answer wrong. "The magnetic field does change"... did you mean "doesn't"? I agree that there's no dynamic electric field, or anything that changes with time. – Andrew Aug 16 '13 at 13:40
Your symmetry argument, however, would apply equally to case 2 and case 3. Do you think there's a field in case 2? – Andrew Aug 16 '13 at 13:41
@Andrew: If you want Classical Physicist to reply to your comment, you need to +1 his answer . – centralcharge Aug 17 '13 at 10:25
Alright, regarding case 2, flowing electric dipoles. I assume we agree there is an electric field? In the $A=0$ case, in the rest frame of the electric dipoles, the magnetic field is zero. Use a Lorentz boost to calculate the magnetic field in a frame where the line of dipoles is moving (in any direction), and you will find the magnetic field is nonzero in this frame. The symmetry is broken by the motion of the dipoles. – Andrew Aug 17 '13 at 13:54
Sorry about the missing not. In case 2), once you imagine an electric dipole with finite distance of the charges, you do generate a small magnetic field from the differences. But you can't think of a magnetic dipole composed of two monopoles at a finite distance I think. – ClassicalPhysicist Aug 17 '13 at 13:58