# Faraday's and Ampere's circuital law

1. Apr 2, 2005

### 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

2. Apr 2, 2005

### dextercioby

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

$$\vec{E}(z,t)=E_{0}e^{-\alpha z} \cos \omega t \ \vec{i} [/itex] After that,u need to integrate wrt time the negative of the curl you had just computed. Daniel. 3. Apr 2, 2005 ### robert25pl From computed curl: [tex]\frac{\partial E\vec{i}}{\partial z} = -\frac{\partial B\vec{y}}{\partial t}$$

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

4. Apr 4, 2005

### dextercioby

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.

5. Apr 5, 2005

### robert25pl

This is what I got for curl
$$-E_{0}\alpha e^{-\alpha z} \cos \omega t \ \vec{j}$$
$$\nabla \times E = -\frac{\partial B}{\partial t}$$

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

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

6. Apr 5, 2005