Faraday's and Ampere's circuital law

robert25pl

For electric field [tex]E=E_{0}e^{-\alpha}^{z}cos(\omega t)a_{x}[/tex] in free space (J=0), find B that satisfies Faraday’s law in differential form and then determine if the pair of E and B satisfy Ampere’s circuital law in differential form.

[tex]\nabla \times E = -\frac{\partial B}{\partial t}[/tex]

Can someone give me hint for next step. Thanks

dextercioby

Compute the curl.You should have written it (for ease and rigor)

[tex] \vec{E}(z,t)=E_{0}e^{-\alpha z} \cos \omega t \ \vec{i} [/itex]

After that,u need to integrate wrt time the negative of the curl you had just computed.

Daniel.

robert25pl

From computed curl:

[tex]\frac{\partial E\vec{i}}{\partial z} = -\frac{\partial B\vec{y}}{\partial t}[/tex]

And for B I got:

[tex]B =-E_{0}\alpha\omega e^{-\alpha z} \sin \omega t \ \vec{j} [/tex]

The answer for B in the book is different so I think I'm doing something wrong

dextercioby

You might.Here's the curl

[tex]\nabla\times\vec{E}=\left|\begin{array}{ccc}\vec{i}& \vec{j} &\vec{k}\\\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ E_{0}e^{-\alpha z}\cos \omega t & 0 & 0 \end{array} \right| [/tex]

Okay?

Daniel.

robert25pl

This is what I got for curl
[tex]-E_{0}\alpha e^{-\alpha z} \cos \omega t \ \vec{j} [/tex]
then using Faraday's law
[tex]\nabla \times E = -\frac{\partial B}{\partial t}[/tex]

[tex]B=\frac{E_{0}}{\omega}\alpha e^{-\alpha z} \sin \omega t \ \vec{j} [/tex]

I fixed what was wrong, but I still have problem with vector j. Book says it should be z.

dextercioby

It can't,unless there was something different to start with...

Your work is correct.

Daniel.