## Faraday's and Ampere's circuital law

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
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Compute the curl.You should have written it (for ease and rigor) $$\vec{E}(z,t)=E_{0}e^{-\alpha z} \cos \omega t \ \vec{i} [/itex] After that,u need to integrate wrt time the negative of the curl you had just computed. Daniel.  From computed curl: [tex]\frac{\partial E\vec{i}}{\partial z} = -\frac{\partial B\vec{y}}{\partial t}$$ 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

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## Faraday's and Ampere's circuital law

You might.Here's the curl

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

Okay?

Daniel.
 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.