# Cylinder , charge on surface, why is B inside zero?

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.

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".