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Take a long cylinder of radius $ a$. It has a long cylindrical hole of radius $b$ parallel to the cylinder axis. The distance between the two axes is $d$. If the cylinder has a uniform current density of $J$, calculate the magnetic field at the centre of the cable, and at the centre of the hole.

I understand this to be a problem involving superposition. I have managed to calculate the magnetic field in the hole to be $ \frac{\mu_0 Jd}{2} $

I am confused how to get the magnetic field at the centre of the cylinder. I thought to calculate the magnetic field of the whole cylinder without a hole, and the subtract from that my answer for the magnetic field in the hole i.e.:

$B_{net} = B_{cylinder} - B_{hole} $

but this does not give me the answer i expect.

Any help is appreciated. Thank you

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Consider the hole as just another cylinder with opposite current density of the same absolute value which compensates the current density field of the large cylinder within the hole.

When you have understood this concept apply superposition of the B-fields.

One hint: The answer is maybe much simplier than you expect and this is directly related to the cylinder axes and to the B-field of a symmetric circular-cylindrical conductor in dependence of the radius (as distance from the axis).

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  • $\begingroup$ This is what I had attempted before.I have calculated the magnetic field in the hole as stated in the question, but am confused about the field in the main cylinder. Is the field in the main cylinder (without a hole) given by $ \mu_0 Ja / 2$? $\endgroup$ – user1887919 Feb 24 '14 at 16:32
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    $\begingroup$ No, the field inside of the main cylinder is not constant. Consider a circular path around the axis with radius $\rho<a$. How large is the current enclosed by this path? How long is the path? This gives you the value of $H$ and $B$. $\endgroup$ – Tobias Feb 24 '14 at 17:59
  • $\begingroup$ Current enclosed is given by the product of the current density and the area so $I_{enc} = JA=J\pi \rho^2 $ Path length is just given by $2\pi\rho $? So then apply Ampere's law? $\endgroup$ – user1887919 Feb 24 '14 at 18:41
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    $\begingroup$ Yes, so we get $\vec{B}=\frac{1}{2}\mu_0 J\rho\,\vec{e}_\alpha$ inside the single conductor. Only at the boundary you get $\vec{B}=\frac{1}{2}\mu_0 J a\,\vec{e}_\alpha$. Ah yes, and what is the value at $\rho=0$ for a single conductor (i.e. on the axis)? $\endgroup$ – Tobias Feb 24 '14 at 18:47
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    $\begingroup$ On the axis the field is zero because of continuity of $B$. This is all for the single circular cylindrical conductor. Now, you can combine the fields of two such things. Be careful, the field of the hole may be an outer field if $d<b$. $\endgroup$ – Tobias Feb 24 '14 at 21:28

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