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I am trying to calculate the Efield from a varying Bfield. The Bfield is from a current loop.
[tex]\vec{B}(\vec{r},t)=\vec{B_{0}}(\vec{r})Sin(\omega t)[/tex]
where [tex]\vec{B_{0}}=[/tex] http://www.netdenizen.com/emagnet/offaxis/iloopoffaxis.htm"
Now since B is varying with time there should be an electric field. Since B is harmonic E should also vary harmonically but with a phase delay. I am assuming that omega is small enough and the length scales are small enough that this can be treated quasistatic. If I assume E takes this form...
[tex]\vec{E}(\vec{r},t)=\vec{E_{0}}(\vec{r})Cos(\omega t)[/tex]
then faraday's equation gives
[tex]\nabla \times \vec{E_{0}}=\omega \vec{B_{0}}[/tex]
I tried to solve for E this way but I am not sure if it is possible to do analytically since it involves elliptic integrals. I attempted to integrate using Mathematica but that didn't work either. I can go into more detail on this approach but I do not feel like it will work.
The other way I attempted was to us ampere's equation. This would give [tex]\nabla \times \vec{B_{0}}=\frac{\omega}{c^{2}} \vec{E_{0}}[/tex]
This seemed like it would be nice and easy but it didn't work out either. Here is what I did.
[tex]\nabla \times \vec{B_{0}}=\frac{1}{r}(\frac{\partial B_{z}}{\partial \theta}r\frac{\partial B_{\theta}}{\partial z})\hat{e_{r}} +
(\frac{\partial B_{r}}{\partial z}  \frac{\partial B_{z}}{\partial r})\hat{e_{\theta}} +
\frac{1}{r}(\frac{\partial}{\partial r} (r B_{\theta})  \frac{\partial B_{r}}{\partial \theta} )\hat{e_{z}}[/tex]
well we already know [tex]\vec{B_{\theta}}=0[/tex] and that there is no dependence on [tex]\theta[/tex] for [tex]\vec{B_{r}}[/tex] and [tex]\vec{B_{z}}[/tex] so
[tex]\nabla \times \vec{B_{0}}= (\frac{\partial B_{r}}{\partial z}  \frac{\partial B_{z}}{\partial r})\hat{e_{\theta}}[/tex]
I first tried to take these derivative by hand but after I had worked it all out (it gets kind of long) I decided to let Mathematica do it and see what it gave. I attached the notebook. Well Mathematica give 0 as the derivative. Right off the bat that does not seem right since looking that the vector field it is obvious that there should be a theta component in the curl. Also this would suggest that there is no Efield from a changing Bfield which isn't right either.
I do not really want to finish working through the derivatives by hand since they are messy and this point I will have no idea if I am right since if they cancel then it agrees with mathematica but seems unphysical and if they don't cancel then it would probably but to messy to know if it is right.
Any ideas on how to determine the Efield in this case? I would be exterminly happy also if someone was to just point to the literature where this has already been treated. I had a hard time searching for it since I don't know the proper keywords to use.
This problem is a part of the research that I am working on. I will be using these fields in a particle tracing code. It has really had me stuck for a while.
Thank you very much for any help or suggestions. I am willing to provide any clarifications that are necessary.
[tex]\vec{B}(\vec{r},t)=\vec{B_{0}}(\vec{r})Sin(\omega t)[/tex]
where [tex]\vec{B_{0}}=[/tex] http://www.netdenizen.com/emagnet/offaxis/iloopoffaxis.htm"
Now since B is varying with time there should be an electric field. Since B is harmonic E should also vary harmonically but with a phase delay. I am assuming that omega is small enough and the length scales are small enough that this can be treated quasistatic. If I assume E takes this form...
[tex]\vec{E}(\vec{r},t)=\vec{E_{0}}(\vec{r})Cos(\omega t)[/tex]
then faraday's equation gives
[tex]\nabla \times \vec{E_{0}}=\omega \vec{B_{0}}[/tex]
I tried to solve for E this way but I am not sure if it is possible to do analytically since it involves elliptic integrals. I attempted to integrate using Mathematica but that didn't work either. I can go into more detail on this approach but I do not feel like it will work.
The other way I attempted was to us ampere's equation. This would give [tex]\nabla \times \vec{B_{0}}=\frac{\omega}{c^{2}} \vec{E_{0}}[/tex]
This seemed like it would be nice and easy but it didn't work out either. Here is what I did.
[tex]\nabla \times \vec{B_{0}}=\frac{1}{r}(\frac{\partial B_{z}}{\partial \theta}r\frac{\partial B_{\theta}}{\partial z})\hat{e_{r}} +
(\frac{\partial B_{r}}{\partial z}  \frac{\partial B_{z}}{\partial r})\hat{e_{\theta}} +
\frac{1}{r}(\frac{\partial}{\partial r} (r B_{\theta})  \frac{\partial B_{r}}{\partial \theta} )\hat{e_{z}}[/tex]
well we already know [tex]\vec{B_{\theta}}=0[/tex] and that there is no dependence on [tex]\theta[/tex] for [tex]\vec{B_{r}}[/tex] and [tex]\vec{B_{z}}[/tex] so
[tex]\nabla \times \vec{B_{0}}= (\frac{\partial B_{r}}{\partial z}  \frac{\partial B_{z}}{\partial r})\hat{e_{\theta}}[/tex]
I first tried to take these derivative by hand but after I had worked it all out (it gets kind of long) I decided to let Mathematica do it and see what it gave. I attached the notebook. Well Mathematica give 0 as the derivative. Right off the bat that does not seem right since looking that the vector field it is obvious that there should be a theta component in the curl. Also this would suggest that there is no Efield from a changing Bfield which isn't right either.
I do not really want to finish working through the derivatives by hand since they are messy and this point I will have no idea if I am right since if they cancel then it agrees with mathematica but seems unphysical and if they don't cancel then it would probably but to messy to know if it is right.
Any ideas on how to determine the Efield in this case? I would be exterminly happy also if someone was to just point to the literature where this has already been treated. I had a hard time searching for it since I don't know the proper keywords to use.
This problem is a part of the research that I am working on. I will be using these fields in a particle tracing code. It has really had me stuck for a while.
Thank you very much for any help or suggestions. I am willing to provide any clarifications that are necessary.
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