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Suppose we have a cylindrical wire of radius a carrying a current I. If the current is uniformly distributed over the surface of the wire, the magnetic field inside is zero.

We can prove that easily using Ampere's law. However, when I try to picture it in my head I can't see why.

I imagine that the cylinder is made up of infinite straight wires , each of them contributing to the total magnetic field by $B=\frac{μ_0I}{2\pi a}$ . If we had two straight wires it's obvious that exactly between them the B fields are cancelled out.

With that in mind it's easy to see that in the middle of the cylinder every B component is cancelled by its symmetrical one. What about the rest points? It seems that one side is more powerful.

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This diagram might give you the insight you are looking for:

enter image description here

I visualize two segments of current (into the plane of the image) on opposite ends of an arbitrary (off axis) point, where both cover the same solid angle. The total amount of current increases with distance (for the same incremental angle, you "see" more of the surface); the actual contribution is the same (as the field goes as $\frac{1}{r}$). Thus the contribution of the two current segments are equal and opposite. You can repeat this for any angle, and see that "everything cancels out".

I added a little construction so you can see the angle between the current segment and the direct line to the point of interest is the same for both segments; this confirms that the amount of current "seen" on the two sides within the same angle is indeed proportional to the distance, so the contributions are equal and opposite.

Do you think you could extend this with formal mathematical analysis?

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Just for picturing it...meaning this is not exact: If you are not in the middle, where it is clear, but close to one side you may see it like this: the length over which you see a wire approximately straight is sort of given by the angle. Hence, the amount of wire with approximately opposite direction on the other side increases with the distance to the other side. This factor would compensate for the field decrease by reciprocal distance.

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  • $\begingroup$ See my diagram... our answers crossed in cyberspace! $\endgroup$ – Floris May 31 '17 at 14:53
  • $\begingroup$ @Floris ...yes....they did...yours is more detailed of course. $\endgroup$ – mikuszefski May 31 '17 at 14:54

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