As far as I know, it's possible to create a radially polarised ring magnet, where one pole is on the inside, and the field lines cross the circumference at right angles.

Radially polarised ring magnet

So imagine if I made one which was shaped like a sector of a torus.

Radially polarised ring magnets torus

Then I forced a load of these magnets into a complete torus.

Impossible torus magnet

Clearly this magnet is impossible because there's no way for the field lines to get back into the middle. So what happens to the field in this case? Does it disappear completely? Do the magnets blow up?

  • $\begingroup$ Related: physics.stackexchange.com/q/18115/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 7, 2014 at 5:03
  • $\begingroup$ Not sure if the magnet will blow up, but my brain has after reading this! $\endgroup$
    – uhoh
    Sep 13, 2017 at 10:39

7 Answers 7


The field would disappear completely.

I think the simplest explanation is in terms of the surface currents that account for the field (assuming constant magnetization, which is reasonable for thin slices). For the initial torus magnet (your second image) the magnetic field is generated, in practice, by surface currents on the planar ends of the torus. One runs clockwise as seen on the image - the one nearest the viewer - while the rear planar face has an anticlockwise current on it.

If you now stack two of those together, the surface currents on the faces in contact will cancel out, giving a bigger version of the same thing. The field will then be created just by the surface currents at the ends of the tube.

You then propose adding more magnets in this fashion until the ring is closed. That will bring the two remaining planar surfaces with surface currents into contact, cancelling out those currents. The resulting magnet will have zero surface current and therefore zero magnetic field.

  • $\begingroup$ What happens to the energy of assembly? It takes a lot of energy to assemble that many like poles. $\endgroup$
    – KDN
    Feb 21, 2013 at 18:17
  • $\begingroup$ It takes a lot of energy to make the rings in the first place! You need to put four south poles together and that's also hard. Also note that the surface currents are in anti-Helmholtz configuration, so for thin rings the field will be quadrupole and rather weak. $\endgroup$ Feb 21, 2013 at 19:01
  • $\begingroup$ The practical answer is : demagnetisation. Any permanently magnetic material has some maximum divergence of the field it can sustain. This is what limits the field strength at the poles of a magnet. (BTW the modern "super" magnets are good in this respect, they are not just "stronger") If You stack enough rings in a linear way, the field emerging at the ends of the hole gets too strong and too divergent, the last rings magnetisation will snap, heat will be released. $\endgroup$
    – Georg
    Jul 16, 2013 at 10:05
  • 1
    $\begingroup$ @Emilio: I'm awarding you the bounty, despite the fact that you haven't said anything about where the energy goes, because you have the only answer that I think isn't incorrect. I believe that where the energy is stored is in the cost of field lines going through a permanent magnet. I don't think this will necessarily demagnetize the magnets, as in my experience two bar magnets are not demagnetized if I bring their north poles together. If they're superconducting magnets, I assume they would get demagnetized. $\endgroup$ Jul 24, 2013 at 0:29
  • $\begingroup$ @PeterShor Thanks, though I know there's still unresolved issues. Here's one thought for the energy conservation, though: I think the problem can be reduced to taking to coaxial thin coils, far away from each other and in anti-Helmholtz configuration, and pushing them together until they meet. Then the magnetic energy in the anti-Helmholtz field will disappear with the field. Where does that energy go? $\endgroup$ Jul 24, 2013 at 1:44

I think Emilio Pisanty's answer is good enough. But here is another longer, 'magnetic charge' approach. (

Let's specify the coordinates first (sorry I borrow your picture).

enter image description here

It's obvious that the toroid is symmetrical under rotation along $\hat{\phi}$ direction. Thus we can't have magnetic field along $\hat{\phi}$. Which means it is sufficient for us to find the magnetic field on the $xz$ plane, and we can generalize later by rotating this $xz$ plane.

We have some constrains to consider here due to the shape of torus:

  • $\vec{B}$ is symmetrical under reflection over $\hat{x}$ axis and $\hat{z}$ axis.

  • $\nabla\times B=0$.

  • $\nabla.B=0$

The most general field lines in $xz$ plane that satisfy these conditions roughly looks like this.

enter image description here

Now we only have two possible directions, the one shown in the picture or the opposite of that(or zero everywhere). We can apply magnetic Gauss law here with the Gaussian surface marked with black dotted line(in 3D point of view this black dotted line is rotated along $\phi$ so that the product looks like a mountain).

The only part of the Gaussian surface which has magnetic flux through it is the top part. The remaining area is intentionally shaped to follow the field lines tangentially so that there's no flux through it. Now we only need to determine whether the magnetic charge inside the envelope is positive or negative(Positive charges are shown in blue, and negative charges in red). As we can see in the picture, the inner magnetic field which is represented by yellow lines may bend slightly from radial direction towards the direction which allows the envelope to catch more or less positive charge. So the total charge inside the envelope might be positive or negative. However if the yellow lines bend in a way like that, then we will meet a kink somewhere between the inner and outer field or we may also get a non-zero curl field and these are impossible. So we are only left with a radial inner field. And therefore the total charge inside the envelope is zero and there can't be any flux passing through it. So in conclusion the field outside must vanish everywhere.

  • $\begingroup$ This correctly determines that the only configuration in which a hollow torus can exist with N poles on its external surface and S poles on its internal surface is for there to be no field at all, since it the torus were to be closed, it would create a non-physical magnetic configuration. But that doesn't answer the question "what happens" when this configuration is attempted. It takes a lot of energy to assemble the $N-1$ pieces before the $N$th piece is added; what happens when we attempt to add the $N$th piece? The magnetic field doesn't just quietly go away... $\endgroup$
    – KDN
    Feb 21, 2013 at 18:22
  • $\begingroup$ I don't see why it will create a non-physical magnetic configuration. Can you specify what rule might be violated ? I think it's okay for the magnetic field outside to disappear entirely, it doesn't mean that the interaction energy will also go away with it. Higher potential energy(depends on negative of the cross term of magnetic energy density) doesn't always mean higher total field energy, because only part of the field energy is responsible for interactions(force). The final configuration will have a very high potential energy and tends to blow up, but there's no singularity involved here. $\endgroup$
    – Emitabsorb
    Feb 21, 2013 at 21:31
  • $\begingroup$ Also we don't need so much energy to insert the last piece of magnet, the field is already quite weak before the insertion. Can the downvoter please explain? $\endgroup$
    – Emitabsorb
    Feb 21, 2013 at 21:50
  • $\begingroup$ It would be non-physical to have a ring of S poles inside with no path to connect to the N poles. By adding magnet segments to one another, the total magnetic flux passing from the inner tube of the torus (the S pole) to the outer surface (the N pole) is increasing, but the path through which it can do this remains the same size (the diameter of the inner tube). This means that the flux density at the exits to the partial torus increases as new segments are added. If the last segment could be added, it would cut off all path for the field lines. $\endgroup$
    – KDN
    Feb 22, 2013 at 0:40
  • $\begingroup$ In the attempt to add the last segment, all of the magnetic flux from the S poles has to squeeze through the opening left, and as the last segment is inserted this hole is diminished. Eventually, the magnetic flux density will be so high that the system must break. $\endgroup$
    – KDN
    Feb 22, 2013 at 0:42

I am writing a new answer so that psx community can judge my idea regarding the energy issue independently.

The work done in order to assemble the pieces together will simply be converted to the magnetic potential energy of the system. When the pieces are put together, the system will have a tendency to blow apart without involving any singularity and demagnetization is unnecessary(the magnetic field outside the "meat" will decreases as more pieces are put together). Even though the magnetic field vanishes everywhere except in the "meat", it does not mean that there is no magnetic potential energy.

The total magnetic field in all space says nothing about the amount of extractable potential energy of the system. Mathematically the total magnetic field energy of a system is

$E=\frac{μ_0}{2} \int B_{tot}^2 dV$

Where $B_{tot}=Σ B_i$ , $B_i$ is the magnetic field due to a small portion of magnet.

However not all of this energy is extractable, the extractable potential energy of the system is the negative* of sum of the cross terms

$U=-\frac{μ_0}{2} \int (Σ_{i≠j} B_i B_j) dV$

We know that it is mathematically possible to have $B_{tot}=0$ everywhere while having $Σ_{i≠j} B_i B_j\neq0$. So it is possible that $U\neq0$ in such case, which removes the paradox.

We can also write the magnetic potential energy in terms of magnetic moments

$U=-∫\vec{B_{ext}} . \vec{dm}$

Where $\vec {dm}$ is infinitesimal magnetic dipole element, and $\vec{B_{ext}}$ is the total magnetic field at the dipole excluding the magnetic field due the dipole itself.

Though we assume that the magnetization is permanent, meaning that $\vec{dm}$ is unchanged, there is nothing that prevents $\vec{B_{ext}}$ from changing. If we draw the field lines, we can see that it is more likely that $\vec{B_{ext}}$ on a single piece of magnetic dipole increases as more and more segments are put together. We can also see that $\vec{B_{ext}}$ tends to be anti-parallel to $\vec {dm}$. Therefore the total potential energy will be positive and its value increases as more and more pieces are put together. Which suggests that the segments tend to blow apart when put together. Due to the geometry of the system it is hard to give a rigorous proofs of my last few statements. However, no matter how complicated the system is, we can still be sure that there are many rooms for the energy to hide. Thus the conservation of energy is still safe.

*Note that the negative sign here is not obvious. In the case of electric charges we would pick the positive of the sum, instead of its negative. The explanation of the sign can be found in my blog http://wp.me/p1AI8o-2P; It is also explained briefly in Feynman Lectures on Physics Vol.2 ch.15.

  • $\begingroup$ I don't see it. If we assume the potential energy in magnetic configurations is stored in the magnetic field, this can't work. When the pieces are together, you start with a configuration having no magnetic field. Now, let the pieces push themselves apart, doing work. You obtain a configuration with a non-zero magnetic field, which clearly has more potential energy. $\endgroup$ Jul 19, 2013 at 12:00
  • $\begingroup$ @PeterShor the total magnetic field does not say anything about the potential energy of the system. The potential energy is not given by $E=\frac{\mu_0}{2}\int B_{tot}^2 dV$, but it is given by the cross terms $-\frac{μ_0}{2} \int (Σ_{i≠j} B_i B_j) dV$. Any other terms in $E$ are infinitely large and independent of configurations. When the magnetic field is zero, we only have $B_{tot}=Σ B_i$. But it does not mean that every single $B_i$ is zero, thus there is no reason to say that $-\frac{μ_0}{2} \int (Σ_{i≠j} B_i B_j) dV$ is zero. $\endgroup$
    – Emitabsorb
    Jul 19, 2013 at 18:13
  • $\begingroup$ So we cannot say that a configuration with zero total magnetic field everywhere has lower potential energy than configuration with non-zero magnetic field. And when we integrate $-\frac{μ_0}{2} \int (Σ_{i≠j} B_i B_j) dV$, we have to cover the region of magnetization(the "meat"). The magnetic fields in this region are very likely to change with different configurations and thus is the potential energy contribution. So we have two layers of shield that can protect the conservation of energy from being violated. $\endgroup$
    – Emitabsorb
    Jul 19, 2013 at 18:22
  • $\begingroup$ But $\frac{\mu_0}{2} \int B^2_{tot} dV = \frac{\mu_0}{2} \sum_i \int B_i^2 + \frac{\mu_0}{2} \sum_{i \ne j} \int B_i B_j dV$. So the total magnetic field and your formula for potential energy differ only by a constant. $\endgroup$ Jul 19, 2013 at 19:46
  • $\begingroup$ @PeterShor yes, but there's no problem with it.So $E=Constant-U$ and $dE=-dU$, if we remove a segment $dE$ is positive so $dU$ is negative.Thus the potential energy decreases as it should,and there's no problem.It's all because of the (-) sign. But of course we've to include the field energy in the "meat" of the magnet to get the total energy. I was trying to say that the extractable potential energy is non-zero, so when the pieces blow apart they do not get kinetic out of nothing. Btw I made a mistake when I wrote "Any other terms in $E$ are infinitely large", that is only for point charges. $\endgroup$
    – Emitabsorb
    Jul 19, 2013 at 20:54

An object becomes magnetized not only by aligning all relevant molecules into a lattice-type structure, but also by flipping what are called molecular domains. These larger level magnetic regions must also be aligned, usually in rapid succession, during the application of high electric current in the magnetizing process. Aligning the North rings side by side will become increasingly difficult as anyone who has tried to manually build a Halbach Array knows. Assuming that the final piece is in place, (and this requires an external frame), the magnetic field will only be zero theoretically because that assumes a perfectly structured lattice at the molecular level and it assumes that the larger magnetic domains will not flip under the strain. The real world imperfections will find the system's weak spots and create holes in which South will flow through, much like Swiss cheese. This principle is also evident on large surface disk magnets, neodymium included, as their Gauss readings are not uniform and show substantial variation. In some cases, if the molecular lattice is weak enough, physical chunks of magnetic material will blow off, adding to the holes. In summary, trying to physically shut down a magnetic field of an object at the macro level is not the same as uniformly demagnetizing that object at the microscopic level. -- James R. Williamson, Jr.


I come to the same conclusion as Emilio Pisanty directly, but by using superposition.

A simpler model with the same features as the toroidal one is to assemble prismoidal magnets to form a sphere, say with the North poles outside and the South poles inside. Even simpler, work in a lower dimension, in a plane, and assemble trapezoidal magnets in a circle, again, oriented with the same pole outside.

The following picture represents the magnet, and the field lines. A dot marks the origin of each field line.

Spherical magnet, field lines

Then, by superposition, the magnetic field cancels itself. One way to think about this is that, because of the spherical symmetry, the field is identical to a spherical field sourced in the center, so that the North and South poles coincide.

Peter Shore asked about energy conservation. I guess a lot of energy is needed to assemble together the magnets, and it can be recovered if we let the magnets forming the sphere free, since they continue to repel each other.


My intuition tells me that if the radius of the ring is large enough, a radial magnetic field should be possible. A ring having a small radius would cause a cancellation of the field. Say for example the ring was 10 ft in diameter. Surely such a diameter would permit a radial field. Magnetic fields will only cancel each other if they are sufficiently close to each other.

A sphere on the other hand is a bit of a weird case. since there is no "escape" for the internal pole, it would seem logical to say there would be perfect cancellation of the field. Unless nodes form creating a field that varies at regular intervals.

I don't see why a spherically radial field wouldn't be possible if the magnets aroung the sphere were evenly spaced.


If all of the segments necessary to form such a toroid could be put in place, it would form a larger, and more powerful, version of one of the segments except that it would be hollow instead of solid. This is possible due to the fact that a magnetic field is capable of penetrating, and passing through, the material of which the segments are made, and also of realigning the particles/atoms of the material. The field, and the alignment of particles/atoms in the material, would simply shift in order to acheive a field as uniformly balanced as possible (one with as close to uniform field pressure as possible).


Not the answer you're looking for? Browse other questions tagged or ask your own question.