# How would I go about detecting monopoles?

A question needed for a "solid" sci-fi author: How to detect a strong magnetic monopole? (yes, I know no such thing is to be found on Earth).

Think of basic construction details, principles of operation and necessary components of a device capable of detecting/recognizing a macroscopic object emitting magnetic field of equivalent of order ~0.1-10 Tesla near its surface, but with only one pole, reliably distinguishing it from normal (2-pole) magnets, preferably at a distance.

Preferably a robust method, not involving extremely advanced technology. Detect the presence, possibly distance (or field strength) and direction.

I know of SQUIDs, but these concentrate on extreme sensitivity. I'm thinking of something less sensitive but more robust (like, no need for the monopole to fall through the loop) and still able to recognize a monopole against a magnet.

Also, how would such a macroscopic object behave practically? Such a "one-pole magnet" about the size and strength of a refrigerator magnets - how would it behave around ferromagnetics, normal magnets and so on?

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Consider the motion of a magnetic monopole in a completely symmetric Maxwell system, where $$\nabla\cdot {\vec B}~=~4\pi\rho_{mag},$$ and $$\nabla\times{\vec E}~=~4\pi{\vec J}_{mag}~-~\frac{\partial{\vec B}}{\partial t}$$ The first equation is then a Gauss’ law for magnetic monopole charge, and the second is a magnetic current form of the Maxwell-Faraday equation. For the occurrence of a magnetic monopole flying through space this will act as a transient current. The last term on the right hand side is a displacement monopole current in this case. The left hand side will by Stokes’ law $\int\nabla\times{\vec E}\cdot da~=$ $\int{\vec E}\times d{\vec l}$, produce an electric current in a loop. So the right hand side could be measured by the torque this magnetic field induces on an ordinary magnetic dipole. The right hand side measured in a solenoid. If the left hand side and the last right hand side term do not equal each other in the standard form of the Maxwell equation with ${\vec J}_{mag}~=~0$, this would be a signal for the detection of a magnetic monopole.

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No need to go all around :) –  anna v Mar 26 '11 at 16:24

Blas Cabrera designed and built a magnetic monopole detector. Here's how he did it (more or less):

http://www.slac.stanford.edu/cgi-wrap/getdoc/ssi82-025.pdf

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This still requires the monopole to pass through the loop. If it's a macroscopic object, that's not quite a practical approach (though "detect all magnets, then determine whether given magnet is a monopole" could be a workable solution.) –  SF. Mar 25 '11 at 10:06
Yeah, it requires the monopole to go through the loop. That's a great way of distinguishing monopoles from dipoles. You can make the loop as big as you want, so I don't see how it would be an issue for a "macroscopic object". Unless you're using the word "macroscopic" in a different sense than I know it. –  Anonymous Coward Mar 27 '11 at 23:09

In this I am replying to the question stated about macroscopic monopoles, as you describe them.

A magnetic monopole would attract magnetized matter, falling in strength over distance by 1/r^2.

You would know it is a monopole if you go with your normal everyday compass and your spaceship all around it and see that the compass is pointing always north, or always south, all around. This could be done quite a distance away if it is a monopole in the range of Tesla, as you seem to ask.

As long as the dimensions are macroscopic, as simple compass will tell you if is a monopole using it to map it all around. A macroscopic monopole would be attracted to the opposite side pole of a dipole magnet and to ferromagnetic materials just as the dipole magnets attract them. There are commercial instruments for measuring magnetic fields.

You can see I am a science fiction fan. My favorite is Terry Pratchett, where he talks of magnetism as "the love of iron" :).

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I wouldn't call Discworld SciFi, but still I like your quote –  Tobias Kienzler Mar 25 '11 at 9:44
@Tobias Kienzler He is utilizing the infinite probable universes and creates his own physics and science. fun. There is even a book called "the science of discworld" which actually was my first acquaintance with the author. –  anna v Mar 25 '11 at 9:48
I loved that one. Maybe we can agree on calling it Science Fantasy? I mean, there are dragons and wizzards involved after all :-7 –  Tobias Kienzler Mar 25 '11 at 10:47
Shouldn't the attraction of a magnetic monopole for a dipole would go down faster than $1/d^2$? You have to differentiate, so it'd go down like $1/d^3$. –  Peter Shor Mar 26 '11 at 1:25
@Peter Shor the monopole itself has a 1/r^2 field. The compass is a dipole but in this case it reacts to the monopole field strength, I think, and it is quantitative. Now the magnetometers should just measure magnetic fields, with whatever method they have inside :). –  anna v Mar 26 '11 at 16:22

Chapter 6.11 of Jackson's Electrodynamics mentions that Dirac's quantization argument fixes the strength of a magnetic monopole, were one to exist. Jackson goes on to say:

...Their coupling strength is enormous, making their extraction from matter with dc magnetic fields and their subsequent detection very simple in principle. For instance, the energy loss in matter by a relativistic Dirac monopole is approximately the same as that of a relativistic heavy nucleus with Z=137n/2. It can presumably be distinguished from such a nucleus if it is brought to rest because it will not show an increase in ionization at the end of its range...

The method suggested wouldn't work at a distance, but it seems to cover the heart of your question.

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@Deepak Any long, thin magnet magnetized along its length is a 'flux tube' connecting two 'monopoles'. No, it would not exhibit the same thermodynamic properties. Yes, I would be able to 'isolate' the north and south ends (whatever that means), and have them move independently (you could use a flexible magnet, just like they did). Of course you would observe flux tubes- if you didn't, $\nabla \cdot \vec{B} \neq 0$, and that would be remarkable. Read the link in my last comment! –  Andrew Mar 25 '11 at 12:11