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Calculate the angle between the E-field and Current vectors in an anisotropic conductive material

Karl Karlsson

New member
Sep 13, 2020
2
In a certain anisotropic conductive material, the relationship between the current density \(\displaystyle \vec j\) and the electric field \(\displaystyle \vec E\) is given by: \(\displaystyle \vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)\) where \(\displaystyle \vec n\) is a constant unit vector.



i) Calculate the angle between the vectors \(\displaystyle \vec j\) and \(\displaystyle \vec E\) if the angle between \(\displaystyle \vec E\) and \(\displaystyle \vec n\) is α



ii) Now assume that \(\displaystyle \vec n=\vec e_3\) and define a coordinate transformation ξ = x, η = y, ζ = γz where γ is a constant. For what value of γ does the conductivity tensor component take the form \(\displaystyle \sigma_{ab} = \bar \sigmaδ_{ab}\) and what is the value of the constant \(\displaystyle \bar\sigma\) in the new coordinate system?





My attempt:



I don't really know if I get it into the simplest possible form but i guess one way of solving i) would be:



\(\displaystyle \vec E\cdot\vec j = |\vec E|\cdot|\vec j|\cdot cos(\phi)= \sigma_0\vec E^{2} + \sigma_1\vec n\cdot \vec E(\vec n\cdot\vec E) \implies \phi =arccos(\frac {\sigma_0|\vec E^{2}| + \sigma_1\cdot cos(α)\cdot|\vec E|\cdot cos(α)|\cdot|\vec E|} {|\vec E|\cdot|\vec j|})\)



Is this the best way to solve this?



On ii) i am completely lost. What do the coordinate transformations mean? x, y and z are not even in the given expression \(\displaystyle \vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)\). I have already found a matrix \(\displaystyle \sigma\) that transforms \(\displaystyle \vec E\) to \(\displaystyle \vec j\). Do they want me to find eigenvectors and eigenvalues? Why?



Thanks in advance!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
In a certain anisotropic conductive material, the relationship between the current density \(\displaystyle \vec j\) and the electric field \(\displaystyle \vec E\) is given by: \(\displaystyle \vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)\) where \(\displaystyle \vec n\) is a constant unit vector.

i) Calculate the angle between the vectors \(\displaystyle \vec j\) and \(\displaystyle \vec E\) if the angle between \(\displaystyle \vec E\) and \(\displaystyle \vec n\) is α
My attempt:
\(\displaystyle \vec E\cdot\vec j = |\vec E|\cdot|\vec j|\cdot cos(\phi)= \sigma_0\vec E^{2} + \sigma_1\vec n\cdot \vec E(\vec n\cdot\vec E) \implies \phi =arccos(\frac {\sigma_0|\vec E^{2}| + \sigma_1\cdot cos(α)\cdot|\vec E|\cdot cos(α)|\cdot|\vec E|} {|\vec E|\cdot|\vec j|})\)
That is correct.
We can still simplify the expression a bit though.

ii) Now assume that \(\displaystyle \vec n=\vec e_3\) and define a coordinate transformation ξ = x, η = y, ζ = γz where γ is a constant. For what value of γ does the conductivity tensor component take the form \(\displaystyle \sigma_{ab} = \bar \sigmaδ_{ab}\) and what is the value of the constant \(\displaystyle \bar\sigma\) in the new coordinate system?

On ii) i am completely lost. What do the coordinate transformations mean? x, y and z are not even in the given expression \(\displaystyle \vec j = \sigma_0\vec E + \sigma_1\vec n(\vec n\cdot\vec E)\). I have already found a matrix \(\displaystyle \sigma\) that transforms \(\displaystyle \vec E\) to \(\displaystyle \vec j\). Do they want me to find eigenvectors and eigenvalues? Why?
The formula for the conductivity tensor is $j_i = \sigma_{ij} E_j$.
With $\vec n=\vec e_3=\delta_{i3}\vec e_i$ and the given equation, we get $j_i = \sigma_{ij} E_j = \sigma_0 E_i + \sigma_1 \delta_{i3}(\vec e_3 \cdot \vec E)
=\sigma_0 E_i + \sigma_1 \delta_{i3}E_3$, so $\vec j = \sigma_0 E_1 \vec e_1 + \sigma_0 E_2 \vec e_2 + (\sigma_0 + \sigma_1) E_3 \vec e_3$.
In the new coordinate system, this must become $j_a^* = \bar\sigma \delta_{ab}E_b^*=\bar\sigma E_a^*$. That is $\vec{j}=\bar\sigma E_1 \vec e_1 + \bar\sigma E_2 \vec e_2 + \bar\sigma E_3 \gamma \vec e_3$.
It follows that $\bar\sigma=\sigma_0$ and $\gamma = \frac{\sigma_0}{\sigma_0+\sigma_1}$.