The Poynting vector $\mathbf{S}$ represents the flow of energy in an EM field. Specifically, if $u$ is the energy density of the field, the Poynting vector satisfies the continuity equation for it:
$$\frac{\partial u}{\partial t}+\nabla\cdot\mathbf{S}=0$$ in vacuum. (This is Poynting's theorem.)
In your particular problem, $E$ and $B$ are perpendicular and their cross product is proportional to the product of their amplitudes. Thus
$$S_z={c\over\mu_0}B^2.$$
You then have to use your knowledge of $B$ to work out $S$.
Ok will attempt an answer to this one, although i state that the language of differential geometry and differential forms (especially in relativistic 4-D space) is not sth i use every-day. However i totally agree that a physical picture is important (especially in physics :))
Well the process of "zero-ing" a derivative (or differential form) is not only for extremum values (minimum/maximum) It is also related to "exactness"
So what does this mean in less "symbolic" and more "physical" terms.
Since the differential and the exactness/closedness relates to a kernel of a transformation, in simple terms this means that there is a "stable-point of (dynamic) balance" at that area (nothing is "lost" or "added").
When sources are added, the form is no longer "exact". Sources are not zero, so the kernel is not zero and there are interactions.
If you want a "more geometric" picture (eg in terms of "rotation"), this is related to the relativistic concept of "time" (which i disagree at some points).
Nevertheless one can still assume a geometric picture like this:
The A potential (which is "un-observable") is used as a "phase". This A potential can be seen as a configuration which has a stable form but can change position.
Think of it as a DNA molecule which is formed into a loop. The information on the DNA molecule is not changed but its relative permutation/rotation/position CAN change and the difderential form just states that this should be irrelevant ("gauge invariance")
Best Answer
Yes, you are basicaly right.
The Poynting vector gives you the momentum of the the EM wave. At the quantum level, it is an operator of the form (see page 7 of http://www.physics.usu.edu/torre/3700_Spring_2013/What_is_a_photon.pdf) : $$\hat{\mathbf{P}} =\sum_\mathbf{k} \mathbf{k}\, \hat a^\dagger_\mathbf{k} \hat a_\mathbf{k}$$ for a given polarisation (here $\hbar=1$).
This, if the system has just one photon with momentum $\mathbf{q}$, the average value $\mathbf{P}$ of the Poynting vector will be $\mathbf{P}=\mathbf{q}$ has you would expect.