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While studying the shielded wires, I noticed that the magnetic field of the inner conductor can penetrate the shield conductor (can be calculated in the region 3). However, the boundary condition of the magnetic field at the surface (between dielectric and perfect conductor) of a perfect conductor impose that the tangential component of $H$($H_{t}=J_{s}$ surface current density) and the normal component is $H_{n}= 0$. Inside the perfect conductor, we have $H_{t}= 0$ and $H_{n}= 0$

So why do we superpose the magnetic field of the inner conductor and the outside conductor when calculating the magnetic field in the region 3? Why do we still have $H_{i}$(inner cond) in the region 3?

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  • $\begingroup$ Because we haven't found magnetic monopoles yet. That seems flip, perhaps, but how would you propose to block the magnetic field, using Maxwell's equations? Materials like mu-metal re-direct the magnetic field lines to minimize them in a defined region. $\endgroup$ – Jon Custer Oct 25 '16 at 13:19
  • $\begingroup$ Based on the boundary conditions ( dielectric- perfect conductor) there is no magnetic field inside the perfect conductor, basically it blocks the magnetic field. $\endgroup$ – aymene chafik Oct 25 '16 at 13:29
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In your shielded wire the current on the outside conductor is equal but opposite to the current on the inner conductor. The magnetic field in the outside region 3 is determined by the total current on the wire plus shield according to the Maxwell equation $$ ∇ ×\vec H=\vec j+\epsilon ∂ \vec E/∂t$$ which gives for the tangential magnetic field strength H on a concentric circle in region 3 with radius $r$ $$ 2\pi r H= I$$ where I is the total current (convective and displacement) passing through the enclosed area of the circle which is zero. Therefore the magnetic field outside the outer conductor is zero. If you only consider the magnetic field due to the current on the inner conductor you would get a finite value. Therefore, you have to include both the current on the inner conductor and the opposite current on the outer conductor in your magnetic field determination in region 3 giving there $\vec H=0$.

Note: In the coaxial cable with perfect inner and outer conductors, the boundary condition $J_s=H_t$ at the inner radius $R_B$ produces a total surface current of magnitude $$I_s=2πR_B J_s=2πR_B H_t=2πR_BI/(2πR_B)=I$$ where $I$ is the total positive current on the inner conductor. The direction of $I_s$ is opposite to $I$ so that the total current encircled by a path in the outer metal and beyond $I_{tot}=I+Is=0$. Thus in the outer metal and in the outside region 3 the total magnetic field $H$ is zero. The shielding in the outer metal and in region 3 of the magnetic field due to the current $I$ on the inner conductor is caused by the compensating magnetic field produced by the surface current $I_S$ at the inner surface of the outer conductor which is equal and opposite to the current $I$ on the inner conductor.

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  • $\begingroup$ I totally agree with you regarding the maxwell-amper law. However there is a contradiction, if we consider just the current of the inner conductor, we will get a finite value outside the outer conductor (using amper law). But according to the boundaries conditions, the outer conductor is going to block the magnetic field generated by the current of the inner conductor, thus the no magnetic field outside. $\endgroup$ – aymene chafik Oct 26 '16 at 10:29
  • $\begingroup$ @aymene chafik - In the coaxial cable with perfect inner and outer conductors, the boundary condition $J_s=H_t$ at the inner radius $r=R_b$ produces a total surface current of magnitude $$ I_s=2\pi R_b J_s= 2\pi R_b H_t= 2\pi R_b I /(2 \pi R_b)$ =I$$ where I is the total positive current on the inner conductor. The direction of $I_s$ is opposite to $I$ so that the total current encircled by a path in the metal $I_{tot}=I+I_s=0$ . Thus in the outer metal and in the outside region 3 the total H is zero. The H shielding in the metal is due to the H produced by the surface current at $R_B$. $\endgroup$ – freecharly Oct 26 '16 at 18:13
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A perfect conductor will completely shield the magnetic field. This is the cause of the famous Meissner effect where we see magnets floating over superconductors.

But ordinary conductors will only partially shield magnetic fields if they are constantly changing magnetic fields.

Thanks for the feedback on my answer:

There are a couple of tangled parts to this question, so I think the confusion lies with making assumptions about the original problem statement. So I am going to make assumptions about those assumptions.

This typical problem of a current carrying wire in a current carrying conduit is analogous to find the electric field with a point charge at the center and a sphere of charge. The point charge has an electric field we get from Coulomb's law, the charged shell contributes a zero field inside (all the fields from the charges on the spherical shell cancel in the interior) but has a field outside that's identical if all the charge was concentrated at the center.

In the problem of a wire inside a hollow cylinder, the magnetic field in region 1 is just the magnetic field (using the usual long straight wire equation) from the wire in the center. The cylinder makes no contribution to the magnetic field in the interior. On the outside, the contribution of the field from the cylinder is also just like a long straight wire. The result is, we just add the two magnetic fields for region 3.

So I think the confusion is about how the cylinder doesn't contribute in the interior. This doesn't mean the conduit is cancelling the field, it just means the magnetic field is just due to the wire in that region. Just like the case of the spherical charged shell.

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  • $\begingroup$ thank you for your response. Ordinary conductors will let the magnetic field penetrate until the skin depth ,if the thikness of the condcutor shield is greater than the skin depth so the magnetic field will not exist in the other region. Its weird because according to the literature, it exist a mutual inductance between the generator wire and the receptor wire. $\endgroup$ – aymene chafik Oct 25 '16 at 14:33
  • $\begingroup$ There is indeed a difference between a perfect conductor and a super conductor. A perfect conductor would freeze the present magnetic flux which was present when it entered this state. So the inside of a perfect conductor might preserve a magnetic field inside. $\endgroup$ – user_na Oct 25 '16 at 16:09
  • $\begingroup$ This doesn't answer the actual question (last sentence) of this post. $\endgroup$ – freecharly Oct 25 '16 at 17:17

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