Why Is the Magnetic Field Uniform in a Non-Coaxial Cylindrical Cavity?

[c299792458]

Homework Statement

We are given an infinitely long cylinder of radius b with an empty cylinder (not coaxial) cut out of it, of radius a. The system carries a steady current (direction along the cylinders) of size I. I am trying to find the magnetic field at a point in the hollow. I am told that the answer is that the magnetic field is uniform throughout the cavity. and is proportional to d\over b^2-a^2 where d is the distance between the centers of the cylinders.

The Attempt at a Solution

I have found by using Ampere's law that the magnetic field at a point at distance r from the axis in a cylinder of radius R carrying a steady current, I, is given by \mu_0 I r\over 2\pi R^2. So I thought I would use superposition. But what I get is {\mu_0 I \sqrt{(x-d)^2+y^2}\over 2\pi b^2}-{\mu_0 I \sqrt{(x)^2+y^2}\over 2\pi a^2}. However this is not the given answer!

 

[M Quack]

You are on the right track, but you have to superpose the magnetic field vectors.

 

[c299792458]

@M Quack: Thank you. I don't know how to change these into vectors, could you please kindly give me another nudge? Thanks again.

 

[M Quack]

The magnetic field generated by a long wire goes right around the wire. So it is perpendicular to the raidal vector.

If the wire is along (0,0,z) and your point at (x,y,z), you know that B_z=0 and that
B is perpendicular to (x,y,0). What vector has these properties?