Electromagnetism - finding electric field from magnetic field I'm trying to solve the following question, but to no avail.

Let $B=\left(0,B_0\left[1+\sin(kx+\omega t)\right],0\right)$ be the magnetic field of some electromagnetic plane wave.

*

*Find the wave direction vector.

*Find the electric field.

*Compute the Poynting vector and the energy density.


Part $1$:
We have  $$B_y=B_0+B_0\sin(kx+\omega t)=B_0+B_0\cos\left(\frac{\pi}{2}-kx-\omega t\right)=B_0+B_0\mathbf{Re}\left[e^{j\left(\frac{\pi}{2}-kx-\omega t\right)}\right]$$ hence, $\vec{k}=\left(k,0,0\right)\Rightarrow \hat{k}=(1,0,0)$.
Now, my problem is with part $2$. I tried to use Maxwell equations $\displaystyle \nabla\times B=\mu_0 J+\frac{1}{c^2}\frac{\partial E}{\partial t}$ and $\displaystyle \nabla\times E=-\frac{\partial B}{\partial t}$. In the first equation, I don't $J$ and in the second one, I don't know how to solve for $E$.
Part $3$ is simple after solving part $2$.
How should one solve part $2$? any help would be appreciated.
Thanks!
 A: I'll just give you hints on each question.


*

*I assume you are using 3D cartesian coordinates. Two components appear to be zero. What can you deduce on the vector's direction ? Is this related to the direction of propagation ?

*You indeed have two Maxwell equations that involve both electric and magnetic field. You thus have to choose one of them. Derivate something with respect to time is a simple operation, while taking the rotational is a much more complicated one. You already know the expression of the magnetic field, so I would apply the complicated operation to this field rather than to the unknown electric field.
In what medium are you calculating the field ? Is there conducting charges to support an electric current $J$ ?
A: It sometimes helps to sketch the waveform of the magnetic field at a given time.  
Perhaps $t=0$ in this example?  

The direction of the wave that you are given can readily be identified from what is in the bracket.
In this example both terms have a plus sign in front of them $(+kx+\omega t)$ so it is a wave travelling in the negative $x$ direction which would also be the case if it was $(-kx-\omega t)$.
If the signs were different $(+kx-\omega t)$ or $(-kx+\omega t)$ then it would be a wave travelling in the positive $x$ direction.  
A way of deciding is to see that when $x=0$ and $t=0$ then $B=B_o$
Now if the time advances a little for what value of $x$ will the magnetic field be $B_o$?  
Looking at the equation $B_y=B_0 ( 1+\sin(kx+\omega t))$ we need $kx + \omega t$ to be zero.  
$\omega t$ is positive so $kx$ must be negative which means that $x$ is negative.  
The wave is travelling in the negative $x$ direction and that must be the direction of the Poynting vector $\vec S$.  
This will immediately enable you to find the direction of the electric field at any point as $\vec S = \dfrac {\vec E \times \vec B}{\mu_o}$
To find the electric field it is probably best to use $\nabla \times \vec B = \dfrac {1}{\mu_o \epsilon_o } \dfrac {\partial \vec E}{\partial t}$ with there being no $\vec J $ term as there are no charges moving around.
From the diagram you can see that the only no zero term on the left hand side is $\dfrac{\partial B_y}{\partial x} \hat Z$.
