Hole in the disc shaped magnet

Let's suppose we have a magnetic disc. One half is magnetic north pole and the other half is magnetic south pole. Now suppose we drill a hole in the disc. Does anybody has an idea how the magnetic poles are arranged on the edge of the hole? Is the edge north, south or on the upper half is south and lower half is north ? (Th hole must be in the middle to make a ring)

I placed an image of the disc, the hole and polarities of the parts. I want to find out if we place some sensors in positions 1,2,3,4,5,6 and 7 what polarites the sensor will detect ? An if we cut the yellow line of this material at positions 5-2 and 2-5 what will be the polarities ?

I drew an image of the disc but I have no enough reputations to place images so for now I will try to explain the details on my question.

Postions: 1 is center of the disc 2 are inner edge (cricle) at 3 and 9 o'clock 3 is at 12 o'clock 4 is at 6 o'clock 5 are outer edge of the disc at 3 and 9 o'clock 6 is on the outer edge between 7 and 8 o'clock 7 is on the outer edge between 1 and 2 o'clock.

In the future when I earned at least 10 reputation I will post the image.

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1
You have to define the "halving" first, to make Your question reasonable. Is it along a diameter of the disk, or along the middle of the two face planes ? – Georg Nov 9 '11 at 10:45

1 Answer

If the external upper edge is North, then the internal (near hole) upper edge is South, and vice versa. (And the same thing holds for the other edge.) So along an axis, you have North-South-North-South in this order rather than North-North-South-South.

You may see it if you realize that the magnetic disk may be constructed out of little magnetic blocks; you may 3D-pixelate the disk to have a better idea. For the little blocks, it's still true that they have the opposite poles (North vs South) on the opposite sides so if the external edge is North, the internal edge on the same side has to be South, and so on.

If the magnetic field $\vec B$ is outgoing from one side of the block, it must be incoming into the opposite one. So the conclusion above is right, at least when the hole is large enough i.e. if the disk with hole is a thin enough annulus. But I believe the sign to stay the same even if the hole is tiny relatively to the disk radius.

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Thank you for your explanation. I supposed the same answer but I don't understand what happens in the middle between the upper and lower part of the disc. Let's suppose we draw a line between N and S what happens at this junction line of two halfs. – Patrik Nov 10 '11 at 13:43
In practice that is a question of how that ring was magnetized! Depending on magnetic properties and field strengths wanted or not wanted in that hole, one can achieve nearly any form of field. – Georg Nov 11 '11 at 20:08
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