What would Maxwell's equations
\begin{align} \boldsymbol{\nabla} \times \boldsymbol{E} &= \frac{\partial \, \boldsymbol{B}}{\partial \, t}, &\boldsymbol{\nabla} \times \boldsymbol{B} &= -\,\frac{\partial \, \boldsymbol{E}}{\partial \, t}. \end{align}
have to look like to correspond to the following sketch and the two wave equations?
The sketch shows two wave functions with $\boldsymbol{B}$ shifted by a quarter of a wavelenght.
\begin{align}\boldsymbol{E}(t, x) &= \boldsymbol{a} \sin{(k \, x - \omega \, t + \phi)} \\[3pt] \boldsymbol{B}(t, x) &= \boldsymbol{b} \cos{(k \, x - \omega \, t + \phi)} \end{align}
The question came to mind after reading this question about Non-standard representation of the free electromagnetic plane wave.
$\underline{\text{Modifying Maxwell's equations}}$
Consider the $\color{red}{\text{modifications}}$ to Faraday's law and Ampere's law $^\dagger$
$$ \nabla \times E = \color{red}{-i}\partial_t B \qquad ; \qquad \nabla \times B = -\color{red}{i}\partial_t E $$
The wave equations for $E$ and $B$ follow as usual by taking the curl of each equation and appealing to Gauss' laws $\nabla \cdot E=0$ and $\nabla \cdot B=0$
$$ \nabla^2E=\partial_{tt}E \qquad ; \qquad \nabla^2B=\partial_{tt}B $$
Which admit plane wave solutions. Consider the particular solutions
$$ E(z,t)=E_0e^{i(kz-\omega t)} \hat{x} \\ B(z,t)=B_0e^{i(kz-\omega t)} \hat{y} \tag{1} $$
Using "Faraday's law" we find the relation between $E_0$ and $B_0$
$$ B_0=\frac{ik}{\omega}E_0 $$
From "Ampere's law" we learn that $|\omega|=|k|$. Noting that $i=e^{i\pi/2}$ (and with $\omega$ and $k$ real) we have
$$ B(z,t)=E_0 e^{i(kz-\omega t+\pi/2)} \hat{y} $$
So $B$ is exactly $\pi/2$ out of phase with $E$, as necessary.
$\dagger$ It is amusing that these look very nearly like a Wick rotation.
$\underline{\text{Without modifying Maxwell's equations}}$
In a conductor, with a current density given by $J=\sigma E$, we can again take the curl of Faraday's law and Ampere's law to get damped wave equations for $E$ and $B$
$$ \nabla^2E=\sigma \partial_t E+\partial_{tt}E \qquad ; \qquad \nabla^2B=\sigma \partial_t B+\partial_{tt}B $$
These admit solutions of the form (1) except that $k \in \mathbb{C}$ such that
$$ k^2=\omega^2+i \sigma \omega $$
Substituting into Faraday's law yields
$$ B_0=\frac{k}{\omega}E_0=\frac{|k|e^{i\phi}}{\omega}E_0 $$
After a little algebra, we find that the phase difference $\phi$ between $E$ and $B$ is given by
$$ \phi=\frac{1}{2}\tan^{-1}\left( \frac{\sigma}{\omega} \right) $$
This is nice, but the solutions with complex $k$ are not free waves, they are damped exponentials, for instance
$$ E(z,t)=E_0e^{-k_I z}e^{i(k_R z-\omega t)} $$
Where $k_R$ and $k_I$ are the real and imaginary parts of $k$.