# Why do electromagnetic waves have the magnetic and electric field intensities in the same phase?

The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
\begin{align}
\nabla \times \mathbf E & = -\frac{\partial\mathbf B}{\partial t} \\
\nabla \times \mathbf B & = \frac{1}{c^2} \frac{\partial\mathbf E}{\partial t},
\end{align}
where the notation $\nabla \times{\cdot}$ is a spatial derivative (the curl). This means that *both* sides have derivatives, and if you're applying them to a function like $\cos(kx-\omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.

E and B are in phase for a running plane wave, but are 90 degrees out of phase for a standing wave. This can be easily seen by considering the vector potential, $A(t, x) $. Using $E = \partial_t A$ and $B=\partial_x A$. For $A=sin(\omega t - kx) $ you find that E and B are in phase. For $A=sin(\omega t) sin(kx) $, a standing wave, E and B are out of phase.