# Homework Help: Find expression for electric field from magnetic field

1. Nov 30, 2015

### charlief

1. The problem statement, all variables and given/known data
In a region of space, the magnetic field depends on the co-ordinate $z$ and is given by $$\mathbf{B} = \hat{\jmath} B_0 \cos \left(kz - \omega t \right)$$ where $k$ is the wave number, $\omega$ is the angular frequency, and $B_0$ is a constant.
The Electric Field in Cartesian coordinates is $\mathbf{E} = \hat{\imath} E_x + \hat{\jmath} E_y + \hat{k} E_z$. Given that $E_y = E_z = 0$ and $E_z = \omega B_0/k$ at $z = t = 0$, determine an expression for $E_x$.

2. Relevant equations

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

3. The attempt at a solution
Using the cross product rule, I changed $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$ to $-\frac{\partial \mathbf{B}}{\partial t} \times \nabla = \mathbf{E}$. Calculated $-\frac{\partial \mathbf{B}}{\partial t} = -\hat{\jmath} B_0\omega\sin(kz - \omega t)$. Then working out the cross product I got $\mathbf{E} = \hat{\imath} \frac{\partial}{\partial z}[-B_0\omega\sin(kz - \omega t)] = -\hat{\imath} B_0 k \omega \cos(k z - \omega t)$. So inputting $z = t = 0$ clearly gives $B_0 \omega k$ instead of $\omega B_0 / k$.

I cannot see where in my method I have gone wrong and I am not sure this method is correct?
Thank you so much

2. Nov 30, 2015

### AhmirMalik

Why do you say that $$E_{z}=0$$ and then $$E_{z}=(\omega B_{0})/k$$. You can't say that the z component of electric field is zero and non zero at the same time... clear that confusion then I'll solve your problem.

3. Nov 30, 2015

### charlief

Apologies it was a typo, I meant "$E_y = E_z = 0$ and $E_x = (\omega B_0)/k$"

4. Nov 30, 2015

### AhmirMalik

Oh.. then you just have to use this;

$$\vec{\nabla}\times \vec{B}=\mu_{0}\vec{J}+\mu_{0}\epsilon_{0}\frac{\partial \vec{E}}{\partial t}$$

Since you dont have any currents, so first term on the right hand side is zero, so you are left with;

$$\vec{\nabla}\times \vec{B}=\mu_{0}\epsilon_{0}\frac{\partial \vec{E}}{\partial t}$$

which is easy to solve.

5. Nov 30, 2015

### charlief

Thank you very much!!