Faraday's and Ampere's circuital law

[robert25pl]

For electric field E=E_{0}e^{-\alpha}^{z}cos(\omega t)a_{x} 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.

\nabla \times E = -\frac{\partial B}{\partial t}

Can someone give me hint for next step. Thanks

 

[dextercioby]

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

\vec{E}(z,t)=E_{0}e^{-\alpha z} \cos \omega t \ \vec{i} [/itex] <br /> <br /> After that,u need to integrate wrt time the negative of the curl you had just computed.<br /> <br /> Daniel.

 

[robert25pl]

From computed curl:

\frac{\partial E\vec{i}}{\partial z} = -\frac{\partial B\vec{y}}{\partial t}

And for B I got:

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

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

 

[dextercioby]

You might.Here's the curl

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

Okay?

Daniel.

 

[robert25pl]

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

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

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.