Magnetic field decay in cylinder coaxial with solenoid

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SUMMARY

The discussion focuses on the magnetic field decay in a coaxial cylinder inserted into a long solenoid carrying a time-varying current. The governing equation is given by $$ \nabla^2 \vec{B} = \sigma \mu \mu_0 \frac{\partial \vec{B}}{\partial t} + \mu \mu_0 \epsilon \epsilon_0 \frac{\partial^2 \vec{B}}{\partial^2 t} $$, with the solution for the magnetic field expressed as $$ B_z = n I_0 \mu \mu_0 e^{i(\omega t - x/d)} e^{-x/d} $$, indicating a traveling wave with decaying amplitude. The induced electric field is derived as $$ E = \frac{1}{\sigma d} n I_0 e^{i\omega t} e^{-(a-r)/d} \sin{r/d} $$, which reflects the influence of the cylinder's conductivity and the solenoid's magnetic field.

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Alettix
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Homework Statement


For a medium of conductivity ##\sigma##:
$$ \nabla^2 \vec{B} = \sigma \mu \mu_0 \frac{\partial \vec{B}}{\partial t} + \mu \mu_0 \epsilon \epsilon_0 \frac{\partial^2 \vec{B}}{\partial^2 t} $$
A long solenoid with ##r=b## has n turns per unit length of superconducting wire anc carries ##I = I_0 e^{i\omega t} ##. A lonf cylinder of radius a (a<b), permeability ##\mu## and conductivity ##\sigma## is inserted coaxially in the cylinder.

a) If a>>d where ## d = \sqrt{2/\omega \sigma \mu \mu_0}##. Hence describe how the magnetic field decays with distance in from the cylinder.
b) show that the induced E field is:
$$ E = \frac{1}{\sigma d} n I_0 e^{i\omega t } e^{-(a-r)/d} \sin{r/d}$$

Homework Equations


Field in infinite solenoid: ## B = \mu_0 n I##

The Attempt at a Solution


I am unfortunately stuck from the very beginning of this part of the problem. In a long solenoid the field is taken to be entirely along the axis and being uniform. Moreover, ## \nabla \cdot \vec{B} = 0## requires the radial field at the cylinder to be zero in any case and we have cylindrical symmetry. So I suppose what we are looking for is the radial variation of the z component of the field.

I understand that the question is somehow getting to the skin depth, but because we have a second order time derivative this doesn't work out as nicely as with the standard derivation for a perpendicularly incoming electric field.

I have tried to set an ansatz ##B_z = B_0 e^{i(\omega t - kr)} ## and plug into the equation given. Using cylindrical polars this leaves me with $$k^2 + ik/r = \mu \mu_0 (\epsilon \epsilon_0 \omega^2 - \omega \sigma i) $$ which unfortunately doesn't have any nice solution with the skindepth, and worse is dependent on r.

Where do I go wrong and how do I tackle the problem?
Many thanks in advance!

UPDATE: So making some approximations of neglecting curvature and high conductivity, I ended up with: $$ B_z = n I_0 \mu \mu_0 e^{i(\omega t - x/d)} e^{-x/d} $$
ie a traveling wave with quickly decaying amplitude, as expected. (x = a-r)

Is this result correct?

Applying ##\nabla \times \vec{H} \approx \sigma \vec{E} ## it doesn't give me the correct result for the second part. I simply end up with: $$ E =- \frac{1}{\sigma d} n I_0 e^{i(\omega t + x/d) } e^{-(a-r)/d} (1+i)$$
 
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Alettix said:
b) show that the induced E field is:
$$ E = \frac{1}{\sigma d} n I_0 e^{i\omega t } e^{-(a-r)/d} \sin{r/d}$$
Should the argument of the sine function be ##(a-r)/d## ?

I have tried to set an ansatz ##B_z = B_0 e^{i(\omega t - kr)} ##
...
...
I ended up with: $$ B_z = n I_0 \mu \mu_0 e^{i(\omega t - x/d)} e^{-x/d} $$
ie a traveling wave with quickly decaying amplitude, as expected. (x = a-r)
Try to obtain the most general solution. You obtained a specific solution representing waves traveling inward (in the positive x-direction) with a decaying factor ##e^{-x/d}##. Is there also a solution representing a wave propagating outward (in the negative x-direction) that contains the same factor ##e^{-x/d}##?

What is the boundary condition for E at the surface of the cylinder? (Hint: superconducting wires)
 

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