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I realize that this sounds like a physics question, but what I am stuck on is a mathematical issue, so I hope you won't mind me posting this question here.

I have a cylinder given by the equation $(x-l)^2+y^2\leq b$. There is a hollowed-out cylinder in it given by $x^2+y^2. We are given that $l\in(0,b-a)$.

In the non-hollow region of the tube flows a current $I$. I wish to find the magnetic field in the hollow where the answer says that the magnetic field is uniform and its magnitude is proportional to $l\over (b^2+a^2)$.

I have previously found that the magnitudes of the magnetic fields at a point $(x,y)$ in the hollow are ${\mu_0I\over 2\pi b^2}\sqrt{(x+d)^2+y^2}$ and ${\mu_0I\over 2\pi a^2}\sqrt{x^2+y^2}$ respectively. And they are in the directions perpendicular to the vectors $(x+d,y)$ and $(x,y)$ respectively , and by the right hand rule, I believe they should be in the directions ${1\over \sqrt{(x+d)^2+y^2}}(y,-x-d)$ and ${1\over \sqrt{x^2+y^2}}(-y,x)$ respectively. It is clear from symmetry that the $\hat{x}$-component should be $0$, but I can't make it so. Also I can't get the right magnitude either. I think there is something wrong with the vectors I have found, but I really don't know what is wrong. Please help me :( Thank you.

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    @joriki: Thank you! :)2012-03-01

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If I understand your question, then you're imagining having plugged the hole with a material that has the same current density as the original configuration, and then subtracting the magnetic field of the plug later.

However, then you shouldn't be using the same current for the two cylinders. The given $I$ is the net current, which is originally distributed over a cross section of $\pi(b^2-a^2)$. This should not change when we plug the hole and calculate on cylinders with cross sections $\pi b^2$ and $\pi a^2$. So the current to use in the solid-cylinder formulas should be $\frac{b^2}{a^2-b^2}I$ and $\frac{a^2}{a^2-b^2}I$, respectively -- which happens to cancel out the cross sections in the denominators nicely.

Once that is done with, don't bother to separate the fields into magnitude and direction. The square roots will look ugly and just cancel out anyway. Just compute directly with vector-valued quantities proportional to $(-y,x)$ and $(-y,x+l)$ in Cartesian coordinates.

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    Thank you so so so much!! You have no idea how long I have stared at this thing. Thanks again!!2012-03-01