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Assuming that you have a infinite long conducting cylindrical wire: Is it possible to let the magnetic field be constant inside the cavity and why?

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    $\begingroup$ To me, this reads like a homework type question, so I think you should give your own thoughts and references in your post. Something like this hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html $\endgroup$ – user198207 Jun 24 '18 at 11:45
  • $\begingroup$ Where is the return current going ? $\endgroup$ – my2cts Jun 24 '18 at 12:30
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    $\begingroup$ If we assume infinitely long conducting pipe will there be any role for the return current? $\endgroup$ – AP1 Jun 24 '18 at 12:32
  • $\begingroup$ This is an interesting question, hope I understood it the right way. Do you found an answer by yourself? $\endgroup$ – abu_bua Jun 25 '18 at 2:21
  • $\begingroup$ Does your problem have any practical relevance? $\endgroup$ – abu_bua Jun 25 '18 at 14:22
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Using Ampere's law

$$\nabla \times \mathbf H = \mathbf{J} + \frac{\partial \mathbf D}{\partial t}$$

and assuming to have steady-state condition ($\partial \over \partial t$ = 0)

yields to

$$\nabla \times \mathbf H = \mathbf{J} \quad .$$

It is common to make use of the vector potential $\mathbf{A}$, defined by $\mathbf{B} = \nabla \times \mathbf{A}$ :

$$ \nabla \times \mathbf H = \nabla \times \Bigg( \cfrac{1}{\mu} \nabla \times \mathbf A \Bigg) = \cfrac{1}{\mu} \nabla \big( \nabla \cdot \mathbf A \big) - \cfrac{1}{\mu} \big( \nabla \cdot \nabla \big) \mathbf A =\mathbf J $$

Using the Coloumb gauge $\nabla \cdot \mathbf A = 0$ and $\mathbf B = \mu \mathbf H$ yields

$$\Delta \mathbf A = -\mu \mathbf J \quad .$$

Since you have an cylindrical wire with a cavity inside it make sense to use cylindrical coordinates to solve this equation. If you assume that you have an infinite long wire and symmetry the derivatives $\frac{\partial}{\partial \phi} = \frac{\partial}{\partial z} = 0 $ and the Poison's equation reduces to a very simple differential equation:

$$\Delta_{\mathrm{cyl}} \mathbf{A(r,\phi,z)} = \cfrac{1}{r} \Bigg[ \cfrac{\partial}{\partial r} \Bigg( r \cfrac{\partial \mathbf{A_z}}{\partial r} \Bigg) \Bigg] \mathbf{e_z} = - \mu \mathbf{J}\quad .$$

You have only to solve for the z component:

$$ \cfrac{1}{r} \Bigg[ \cfrac{\partial}{\partial r} \Bigg( r \cfrac{\partial A_z}{\partial r} \Bigg) \Bigg] = -\mu J_z \qquad \qquad \mathrm (1). $$




Case-to-case Analysis

Now you have to distinguish between the inner non-conducting area, and the outer conducting area!

1. Inner (cavity) case

If we integrate this differential equation to get a solution for the cavity we can set $J_z = 0$. Now by integrating once by $\mathrm{d}r$ you get

$$\cfrac{\partial A_z}{\partial r} = \cfrac{c}{r} \quad ,$$

where $c$ is an integration constant. The solution can be easily found to be

$$A_r = c_1, \quad A_{\phi}= c_2, \quad A_z = c_2 \mathrm{ln}(r) + c_3 \quad , $$

where $c_2$ and $c_3$ are integration constants.

To get the magnetic field you have to take the curl of $\mathbf A$, which is in our case with $\frac{\partial}{\partial \phi} = \frac{\partial}{\partial z} = 0 $:

$$\mathbf B = \nabla_{\mathrm{cyl}} \times \mathbf A = - \cfrac{\partial A_z}{\partial r} \mathbf e_{\phi} \qquad \qquad \mathrm{(2)} .$$

Hence the solution including the integration constant determined by $\mathbf B_0$, and setting $c_4 = -c_2$, is

$$ \mathbf B_i = \mathbf B_{0} + \cfrac{c_4}{r} \mathbf e_{\phi}$$ or since you need it later using $\mathbf H = \frac{1}{\mu} \mathbf B$ $$ \mathbf H_i = \mathbf H_{i0} + \cfrac{c_5}{r} \mathbf e_{\phi} \qquad \qquad \mathrm{(3)} .$$ The index indicates that it is the inner (cavity) area.

2. Conducting Ring

Now you have to find a solution for the conducting ring of the cylinder. So using equation (1) and integrating once yields

$$r \cfrac{\partial A_z}{\partial r} = -\mu \int r \cdot J_z\; \mathrm{d}r + c_6 \quad , $$ or

$$\cfrac{\partial A_z}{\partial r} = -\cfrac{\mu}{r} \int r \cdot J_z\; \mathrm{d}r + \cfrac{c_6}{r} \quad . $$

Using equation (2) again and taking the last equation

$$\mathbf B_o = \nabla_{\mathrm{cyl}} \times \mathbf A = - \cfrac{\partial A_z}{\partial r} \mathbf e_{\phi} \\ \quad \quad = \cfrac{\mu}{r} \int r \cdot J_z\; \mathrm{d}r \; \mathbf e_{\phi} - \cfrac{c_6}{r} \; \mathbf e_{\phi} + \mathbf B_{o0} \quad ,$$

and since you will need $\mathbf H_o$ later: $$\mathbf H_o = \cfrac{1}{r} \int r \cdot J_z\; \mathrm{d}r \; \mathbf e_{\phi} + \cfrac{c_7}{r} \; \mathbf e_{\phi} + \mathbf H_{o0} \qquad \qquad \mathrm{(4)}.$$

The index $o$ denotes the outer (conducting) area.


3. Solving for an inner magnetic field to be constant

Now you can solve your problem. The figure below shows the tangential components of the magnetic field at a conducting boundary surface, as in your problem. (Figure taken from Griffith's Introduction to Electrodynamics 4e , pp.250) Amperian loop with tangential components

Therefore the boundary condition at the surface is:

$$\mathbf{H_{o\ \mathrm{tangential}}} - \mathbf{H_{i\ \mathrm{tangential}}} = \mathbf K \times \mathbf{\hat{n}}$$

where $\mathbf K$ denotes the surface current.

In other words, at the radius $R$, where the cavity and the conducting medium touch each other, the condition above has to be meet.
In cylindrical coordinates and for a cylinder as in your case that means:

$$H_{o}(R) \; \mathbf e_{\phi} = H_{i}(R) \; \mathbf e_{\phi} \quad .$$

You can take now equation (4) and (3) and set it equal as described above:

$$ \cfrac{1}{r} \int r \cdot J_z\; \mathrm{d}r \; \mathbf e_{\phi} + \cfrac{c_7}{r} \; \mathbf e_{\phi} + \mathbf H_{o0} \; \mathbf e_{\phi} = \mathbf H_{i0} \; \mathbf e_{\phi}+ \cfrac{c_5}{r} \mathbf e_{\phi}\qquad.$$

Multiplying this equation with $r$ on both sides and differentiating both sides by $\frac{\mathrm{d}}{\mathrm{d}r}$ yields to

$$r \cdot J_z = \mathbf H_{i0} - \mathbf H_{o0} \cdot \mathbf e_{\phi}\\ r \cdot J_z = K(R) \quad, $$

therefore the current density has to be

$$\bbox[5px,border:2px solid red]{ J_z = \cfrac{1}{r}J_{R} }$$


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