How to find the amplitude of magnetic field of this plane wave

Properties of an electromagnetic wave in vacuum:

1. ${bf E}$, ${bf B}$ and ${bf k}$ are mutually perpendicular. The mutual orientation is such that ${bf E}times {bf B}$ is in the same direction as ${bf k}$.

2. ${bf k}$ defines the direction of travel of the wave.

3. The amplitude of the B-field is $c^{-1}$ that of the E-field, where $c = omega /k$ (you really ought to see what ${bf k}$ is from this). If the wave is not in vacuum, but a non-conducting medium, $c$ is replaced by $c/n$, where $n$ is the refractive index.

4. E-field and B-field are in phase.

So here, $k =2pi/lambda = 6283$ m$^{-1}$. The vector direction is along the z-axis.

The magnitude of $k$ is not required here since in air we can assume the wave speed is still $c$ and thus the H-field will have a magnitude of $E_0/mu_0 c$ (SI units). The direction must be perpendicular to the E-field and perpendicular to $vec{k}$.

Now to find this, you could use Faraday's law and do the curl of the E-field and then integrate with respect to time; or you could just see that two possible vectors that are in the xy plane and perpendicular to $hat{x} + hat{y}$ are $hat{x} - hat{y}$ or $-hat{x} + hat{y}$. Which is it? Well the tiebreaker is that $vec{E} times vec{H}$ (the Poynting vector) must be in the direction of wave motion. Looking at the two possibilities, we see it is the first one. Thus $$vec{H}_0 = frac{A}{mu_0 c} (hat{x} - hat{y})$$

I wonder why nobody mentioned the obvious: divide by the impedance of free space. The E field is in volts per meter, the H field is in amps per meter, and the ration of volts to amps is 377 ohms.