# Can the poles of a magnet have varying intensity?

In re-reading Is it possible to separate the poles of a magnet? (amongst others) the question mentioned in the title here just occurred to me.

It may not be possible, at our current levels of technology & science, to expect/create a monopole.

Can a magnet, instead, be asymmetric?

i.e. Could the poles of a magnet have varying intensity? If yes, would such asymmetry be preliminary evidence that a pragmatic monopole (one pole very weak compared to the other) may exist?

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Depending on how you define a magnet. See example for the asymmetric magnetic field here: physics.stackexchange.com/questions/47185/… –  hwlau Dec 28 '12 at 16:40

No, it isn't. While it is possible to create geometrically asymmetric magnetic fields, the net magnetic flux through any bounded surface must be zero. While you could, for instance, construct a magnetic where the N pole had much lower magnetic field intensity than the S pole, the N pole face would have to be much larger than the S face, and the total magnetic flux through the N and S poles would still be "symmetric". The total magnetic flux would be the same through both faces.

If an asymmetric magnetic could exist, it could be represented mathematically as the sum of a dipole (symmetric) magnet and a monopole. But, since monopoles are forbidden (even on a microscopic scale), no such asymmetric magnetic field configuration may exist.

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Yes, it can. Apart from a magnet, it is possible to create an asymmetric magnetic field with an asymmetric coil. Still, it does not create a monopole.

Electric charges can also have different densities $\rho_+(\vec{x})\ne\rho_-(\vec{x})\;$ and create an asymmetric electric field, so what?

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Unquestionably the answer is yes. One would only have to look at the largest magnet in the solar system (the sun) to confirm this. If you have any doubt please go to the following site and learn http://wso.stanford.edu/Polar.html

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Are you referring to the oscillation of North and South, which do not cancel out, for example in this graph? And have scientist also made the same conclusion? –  fibonatic Dec 7 '14 at 14:53
This is something approaching a link-only answer. –  HDE 226868 Dec 7 '14 at 14:58