Find the corresponding magnetic field given an electric field of light Given the following electric field:
$\vec E(r,t) = Re\{[E_o\hat{x} + iE_o\hat{y}]\exp[\ 2\pi i \ ((2 \cdot  10^4 \mathrm{cm}^{-1})\ z\ -\ (5 \cdot 10^{14} \mathrm{s}^{-1})\ t) ]$}
I wish to find an expression for the magnetic field in a medium of refractive index $n > 1$. What I know so far:
$\vec\nabla\cdot\vec B = 0 \qquad \vec\nabla \times \vec B=\epsilon\mu\frac{\partial \vec E}{\partial t}+\mu \vec J$
Because $\vec J=0$ and charge density is 0 (electromagnetic wave), this will be $\vec E=\vec B\times \vec v$. However, I'm drawing a blank on how to finish solving the problem. Can anybody point me in the right direction?
EDIT: Forgot to mention that $\mu_r = 1$.
 A: I think it's easier if you use the third Maxwell's equation, i.e. the Faraday law: 
$$ \nabla \times \vec E = -\frac{\partial \vec B}{\partial t} $$
After you take the curl, you integrate the result over time. If you wish to find the $\vec H$-field you then use the following relationship: $\vec H = \frac{\vec B}{\mu}$.
A: A rather general rule to search for solutions to such differential equations is to "guess" a solution and plug it into the equation (it is easier to use the equation suggested by rnels12). Assume that
$$\vec{B} = Re \left\{ \left[B_{0x} \hat{x} + B_{0y} \hat{y}\right]\exp\left[2\pi i(\lambda^{-1}z-ft)\right] \right\}$$
Why this assumption? $\vec{B}$ must (not always, but certainly in this case) describe a plane wave just like $\vec{E}$.
Strip the real-part operator first for $\vec{E}$ and $\vec{B}$ to avoid unnecessary confusion. You can place it back once you find the complex-valued solution. Also, hint: $\lambda^{-1} f^{-1} = \sqrt{\varepsilon\mu}$.
If we assume that $\varepsilon$ and $\mu$ are tensors, $\vec{B}$ could have a more complicated form (e.g. not orthogonal to $\vec{E}$). Further complications, such as not being in phase if we allow magnetoelectric activity. It would be cool if someone has the time to cover these cases in another answer.
