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Consider a cylinder of permanently magnetized material, with uniform magnetization pointing along the cylindrical symmetry axis (the $z$-direction). The magnet is rotating about its cylindrical symmetry axis with angular velocity $\omega$. What electric field does the rotating magnet generate?

Backstory: Moving permanent magnets generally generate an electric field, even in cases where $d\vec{M}/d t = 0$. In the case of uniform motion, this electric field is straightforward to determine using a Lorentz boost. I'm interested in cases where the simple Lorentz boost does not work.

EDIT:
As perceived by some of the answers, I am not specifically interested in a cylinder. If your solution is for a ring, a sphere, or pretty much any nontrivial cylindrically symmetric object rotating about its cylindrical symmetry axis, I'm interested, as long as $d\vec{M}/d t = 0$.

Landau and Lifshitz describe a similar, interesting case where the rotating magnet is also a conductor. I'm interested in the case where the rotating object is not a conductor.

Unipolar induction is very interesting, but again, involves a rotating conductor, which I am not asking about.

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If I have time later, I may develop this fully, but in the mean time: In the Lorentz gauge you can determine the propagator (since the vector potential satisfies a d'Alembertian). For any given sources, this can be used to find the solution. For this particular case, intuition says that at distances far from the magnet itself, the dominant contribution will be dipolar --- so you will simply get a rotating magnetic dipole field (treat as a linear superposition of two oscilating ones). See: en.wikipedia.org/wiki/Dipole#Dipole_radiation –  genneth Mar 10 '11 at 9:55
    
You are using the "symmetry" in a wrong way! Even the x- and y- axis respectively are symmetry-axes of that cylinder. –  Georg Mar 10 '11 at 10:46
    
@Georg I changed 'symmetry axis' to 'cylindrical symmetry axis'. –  Andrew Mar 10 '11 at 15:34
    
I'll delete my answer, because You changed the question substantially. BTW a voltage due to induction is never dependend on some conducting object. The voltage is always there, but You wont have a current to measure. –  Georg Mar 10 '11 at 16:06
    
@Georg Which change do you refer to? Specifying a non-conductive object? –  Andrew Mar 10 '11 at 16:27

7 Answers 7

up vote 4 down vote accepted

The electric field is nonzero. For a cylinder of finite length, it's nonvanishing everywhere. In the limiting case of an infinitely long cylinder, the field is only nonvanishing inside the cylinder.

The easiest way to solve this is to use the fact that the electric and magnetic polarizations $(-\textbf{P},\textbf{M})$ transform in exactly the same way as the fields $(\textbf{E},\textbf{B})$ (Hnizdo 2011). Taking the low-velocity limit for convenience, we have $\textbf{P}=\textbf{v}\times\textbf{M}$. This produces a radial polarization with magnitude $P=\omega r M$, corresponding to a constant interior charge density plus a surface charge of the opposite sign. (This agrees with Kostya's answer.) The interior field is clearly nonvanishing. Applying Gauss's law in the limit of an infinitely long cylinder, the exterior field is found to vanish.

Hnizdo and McDonald, "Fields and Moments of a Moving Electric Dipole," 2011, http://www.physics.princeton.edu/~mcdonald/examples/movingdipole.pdf

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In case of infinite cylinder the correct answer is 0. There is no field outside the rotating cylinder.

It was obvious from the very beginning from Gauss law. But I crushed into it "doing it a hard way". Anyway I setteld down all the details of the problem, so let me present my solution:
1. Obtaining the potential inside.
Inside the rotating object we have the Lorentz force that acts on charges (free or bound) inside the medium. The charges redistribute producing the electric field that compensate the force. The electrostatic potential energy, produced by the charge distribution $\rho(r)$ must be equal to the mechanical work against the Lorentz forces:
$$F_r(r) = \rho(r)\frac{B\omega r}{c}\quad\Rightarrow\quad U(r)=\rho(r)\frac{B\omega r^2}{2c}=\rho(r)\,\phi(r)$$ Thus, getting the $\phi(r) = \frac{B\omega r^2}{2c}$ inside the cylinder. Let me stress that $\rho(r)$ can be the density of bound charges, free charges or combined density. The result doesn't depend on the nature of these charges.

2. Obtaining the charge distribution.
Let us first obtain charge density inside the cylinder. For this I'll just substute $\phi(r)$ into the Poisson equation: $$\Delta\phi(r) = -4\pi\rho(r)\quad\Rightarrow\quad \rho(r)=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\frac{B\omega r^2}{2c}\right) = -\frac{B\omega}{2\pi c}$$ Inside the cylinder $\rho$ is constant, producing linearly growing field.

There are also surface charges $\sigma$, responsible for the discontinuity in the electric field. These are obtained from electroneutrality. $$2\pi R\sigma = -\pi R^2\rho \quad\Rightarrow\quad \sigma = \frac{B\omega R}{4\pi c}$$

3. Solving for the potential outside.
Now we have to solve the Laplace equation outside the cylinder. The general solution is: $$\Delta\phi = 0 \quad\Rightarrow\quad \phi(r)=A+B\log r$$ There are two boundary conditions to satisfy: first is the continuity of the potential $$\phi(R) = \frac{B\omega R^2}{2c}$$ and the second is the discontinuity in the field: $$-\phi'(R+0) + \phi'(R-0) = 4\pi\sigma$$ Obtaining: $$ A = \frac{B\omega R^2}{2c} \quad \text{and}\quad B = 0 $$ So the potential is constant outside the cylinder. No field.

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+1 for L&L! However, I'm interested in the case where the spinning object is not a conductor. I've edited the question to clarify this. –  Andrew Mar 10 '11 at 15:36
    
Regarding the statement 'If there are no free charges, then the same result can be obtained considering bound charges and forces acting on them'. In this case, wouldn't the answer depend on the polarizability of the rotating object? –  Andrew Mar 10 '11 at 15:37
    
I appreciate that you returned to the topic! I also appreciate the clear, formal argument. Given your answer to this question, do agree with the answer to this question, describing an infinitely long, straight magnet moving at constant speed? If so, imagine the case where the long, straight magnet (and its trajectory) is very slightly curved. Since this curve will eventually close on itself, forming a rotating object, does the field vanish in this case? –  Andrew Mar 11 '11 at 15:25
    
I'll not start discussion in comments. That is not constructive. Ask another question. –  Kostya Mar 11 '11 at 15:43
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Note that this is for an infinite cylinder. A finite cylinder will produce an electric field outside of the cylinder even though dM/dt = 0. –  Edward Mar 20 '11 at 10:00

The electric field is null: Because of the rotational symmetry assumed there, the magnetic induction $B$ is constant in time, so $\nabla\times\,E = 0$ by Faraday's law. On the other hand no electric charge is present, so $\nabla\cdot E = 0$. This is enough to make $E = 0$.

Moving permanent magnets do generate an electric field, "even in cases where $\frac{dM}{dt}=0$", but the $M$ there is referred to a frame linked with the magnet. The resulting induction $B$, being referred to the laboratory frame, changes with time, hence a nonzero $\frac{dB}{dt}$, and a nonzero $\nabla\times\,E$.

One may object that, in the case of the question, the magnet's frame is moving, too, so a changing B should result. What makes the difference is the rotational symmetry: The field generated by a rotating axisymmetric magnet is independent of its rotation speed, because any given point of the lab "sees" always the same magnetization, hence the same induction too. So $\frac{dB}{dt} = 0$.

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"The resulting induction B, being referred to the laboratory frame, changes with time". Can you expand on this statement? Specifically, can you describe a steady-state situation where a permanent magnet is moving, $d\vec{M}/dt = 0$, and $d\vec{B}/dt$ is nonzero? –  Andrew Mar 10 '11 at 19:12
    
B is a vector field in my answer. So is M. Both are expressed in lab coordinates. If so, M varies with time, even though the magnetic moment M you use (a vector, not a vector field) is constant in time. Distinguishing magnetization (a vector field) and magnetic moment (the integral of the latter) is essential. –  Bossavit Mar 11 '11 at 13:30
    
I edited your answer in order to remove my downvote, since you are absolutely correct. –  Kostya Mar 11 '11 at 14:30
    
Note also that the effect of the magnet rotating wrt a co-axial loop of wire is different to the co-axial loop of wire rotating wrt the magnet. –  John McVirgo Mar 22 '11 at 19:14
    
This answer is incorrect, and contradicts Kostya's correct answer, which showed that the field did not vanish inside the cylinder. The same argument applied to this physics.stackexchange.com/questions/6457/… simpler question would also contradict Lubos Motl's correct answer to that one. On the other hand no electric charge is present This is the mistake. A moving, magnetically polarized material also has an electrical polarization. –  Ben Crowell Aug 21 '13 at 15:59

I disagree with the claim "curlE=0 ... divE=0. This is enough to make E=0". Consider for instance an electric dipole. Outside of the dipole, curlE=0 and divE=0, yet E does not equal 0.

I also think the charge distribution considerations are too limited. For intuitively I expect polarization to show up, but not free charges.

Here's a simple concrete example showing M constant can still have an electric field:

An infinite cylinder, neutral charge, M constant. In the rest frame, outside: E=0,B=0, and inside: E=0,B=const. Now boost to frame moving along the axis, outside: E=0,B=0, and inside B=const, E!=0.

Now consider a cylinder of length L and radius a, and turn it into a nice symmetric ring (so 'outer' radius = L/2 pi, and 'inner' radius = a). Inside the ring, in the limit $a<<L$, we need to get back the infinite cylinder case. So yes, a rotating ring will have non-zero electric field in it. Also, intuitively, before the infinite limit, the ring will have electric field outside it as well.

EDIT(yet again) I need to think about it, but this can probably be made rigorous as such: In the infinite cylinder case, it should be possible to see how M in one frame changes to M and P in another frame. There may be a simple way to use the symmetry of relativity to explain how these mix.

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E has to be symmetrical about the axis in this example, but not for a dipole. –  John McVirgo Mar 22 '11 at 19:16
    
@John A dipole is symmetric about the axis just like a cylinder. The dipole actually has the same symmetry as a finite cylinder. –  Edward Mar 23 '11 at 2:00

The spinning magnet should create an $E$ field analogous to the $B$ field from an electric current. I understand the skepticism and the "source" problem, but a spinning magnet is like a collection of separate magnets spinning around in circles perpendicular to their lengths. I you moved relative to a magnet's pole, that is $\vec v \times B$ and you will find $E$ in the moving frame. It doesn't matter what the source is, $B$ is $B$ to the observer moving through it. Yes it is odd since we have no authentic $\rho$ or $J$ to serve as sources (the magnetic atoms aren't really current loops that can show charge density redistributed by SRT effects, and there is no $dA/dt$ in a constant situation) but the $E$ field should be there. Amazing this is not settled physics. It may mean we should reassess the sourcing equations. See my post at http://tyrannogenius.blogspot.com/2013/11/because-of-relative-motion-of-sources.html.

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Consider another example of similar implications: an "infinite" (or very long) row of side-to-side bar magnets. The rows of N or S poles will produce B field like a long string of charge would an E field. If you move relative to that B, there must be E in your RF. Again, there is neither genuine net charge density nor dA/dt. –  user32133 Nov 3 '13 at 1:28

It would seem that this problem could be re-expressed in a manner that makes the form of the answer intuitively clear.

Consider a cylinder oriented vertically. Its top (disk-shaped) surface is coated with a thin layer of "North" magnetic monopoles. Its bottom surface is similarly coated with "South" magnetic monopoles. The spinning of the cylinder along its vertical axis creates rings of magnetic current due to the resulting circular paths traced out by the magnetic monopoles.

These magnetic currents appear in the Maxwell equation corresponding to Faraday's Law in a manner exactly analogous to the appearance of electric currents in the Maxwell equation corresponding to Ampere's Law. [This term in the Faraday's Law equation is normally zero, because there are no magnetic monopoles, hence no magnetic currents.]

The rings of magnetic current will produce toroidal electric fields (concentrated at the ends of the cylinder). These fields are analogous to the toroidal magnetic fields produced by rings of electric current.

The validity of this answer (that E is not zero, but a toroidal configuration at both ends) depends on whether a magnetic dipole formed from the aforementioned distribution of magnetic monopoles is equivalent to your cylindrical magnet.

Interestingly, although magnetic monopoles could be used to create a magnetic field identical to that of your magnetic (whose field result from electric charges circulating about nuclei), the situations are not equivalent. There are only two ways to generate an electric field -- electric charges or magnetic currents. Spinning the magnet does not spontaneously generate electric charge. Nor, in the absence of mononoples, does it create a magnetic current.

By the way, Gauss' Law does not imply that the electric field is zero. It only implies that the integral of the electric field on a surface enclosing the cylinder is zero. [As pointed out by Edward, for an electric dipole this integral is zero, but the electric field itself is not.]

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The Earth is rotating and has a permanent Magnetic field. It does not exhibits a correspondent Electric field. The answer is 0.

The magnetic field is due to the presence of a relative motion between the charges inside the Earth (electric dipole) and the observer. Thus, it is a derived quantity much like the Coriolis force. Instead of saying magnetic field I think more appropriate to say magnetic force.

by Hans de Vries: The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics He uses only Electrostactic field and the non-simultaneity to derive the Magnetic Field.

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""The Earth is a rotating permanent magnet "" Aha, what is the ferromagnetic material with high coercitive field down in the core? –  Georg Mar 21 '11 at 19:49
    
Surely The Earth is a rotating permanent magnet As it lasts for several billions of years and no one can switch it off is a serious candidate to be 'permanent'. Permanent_magnet WP : Ampère model: "where all magnetization is due to the effect of microscopic, or atomic, circular bound currents, ... throughout the material." Being an electromanet or a ferromagnetic material the resultant effect is all the same and the origin of the magnetic field is the same. The Vries doc is not ok? –  Helder Velez Mar 21 '11 at 20:38
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""Lasts for several billions of years"" ? You should read wiki on earth magnetic field. They say that it changes polarity every 300 000 years. The rest of Your comment is below any comment from me. –  Georg Mar 21 '11 at 21:03
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@Helder Velez; You might want to edit out the claim that the earth is a rotating permanent magnet. Do any kind of literature search and you'll find otherwise. In fact, it takes incredible amounts of power to keep the temporary field we see in place. –  Carl Brannen Mar 22 '11 at 0:49
    
@Carl @Georg Which of the 3 words you dont like in? "rotating permanent magnet" an electromagnet is not a magnet? Permanent is not restricted to ferromagnetic materials. They are formed in the already presence of Earth mag.field, that exists since : ".. only 10 million years after the Earth began to form, ..setting up the formation of Earth's magnetic field." I got downvotes everytime I mentioned the Vries doc. It is annoying because no one is saying that it has an error. –  Helder Velez Mar 22 '11 at 23:59

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