Wave equations in free space ($q=0, \vec J=0$)
$$ \vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t} \quad \vec \nabla \times \vec B = \mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t} \quad \vec \nabla \cdot \vec E=0 \quad \vec \nabla \cdot \vec B = 0 $$
We can combine these equations
$$ \begin{aligned} \vec \nabla \times \vec E&=-\frac{\partial \vec B}{\partial t} \\\Rightarrow \qquad \vec \nabla \times (\vec \nabla \times \vec E)&=-\frac{\partial}{\partial t} \underbrace{\vec \nabla \times \vec B}{\mu_0 \epsilon_0 \frac{\partial \vec E}{\partial t}} \\ \Rightarrow \qquad \vec \nabla (\underbrace{\vec \nabla \cdot \vec E}{0})-\nabla^2 \vec E&=-\mu_0 \epsilon_0 \frac{\partial}{\partial t} \left ( \frac{\partial}{\partial t} \vec E\right ) \\ \Rightarrow \qquad \nabla^2 \vec E&=\mu_0\epsilon_0 \frac{\partial ^2 \vec E}{\partial t^2}
\end{aligned} $$
where we can write $c=1/\sqrt{\mu_0 \epsilon_0}$ and we get
$$ \nabla^2 \vec E= \frac{1}{c^2}\frac{\partial ^2 E}{\partial t^2} $$
Similarly for the magnetic field we get
$$ \nabla^2 \vec B=\frac{1}{c^2}\frac{\partial^2 \vec B}{\partial t^2} $$
$$ \vec E=\vec E_0 e^{i(\vec k \cdot \vec r-\omega t)} $$
Similarly $\vec B=\vec B_0 e^{i(\vec k \cdot \vec r-\omega t)}$, here $\vec k$ is the wavevector
$$ \vec k=(k_x,k_y,k_z)=\left ( \frac{2\pi}{\lambda_x},\frac{2\pi}{\lambda_y},\frac{2\pi}{\lambda_z} \right ) $$
💼 Case: consider a general magnetic field:
$$ \vec B=(B_x,B_y,B_z)=e^{-i\omega t} (B_{0x}e^{i\vec k \cdot \vec r},B_{0y}e^{i\vec k \cdot \vec r},B_{0z}e^{i\vec k \cdot \vec r}) $$
Using maxwell’s equation
$$ \begin{aligned} \vec \nabla \cdot \vec B &=0 \\ \frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial x}+\frac{\partial B_z}{\partial x}&=0 \\ ik_xB_x+ik_yB_y+ik_zB_z&=0 \\ \vec k \cdot \vec B&=0 \end{aligned} $$
💎 Conclusion: $\vec B \perp \vec k$ always, similarly in free space $\vec E \perp \vec k$ ( $\vec k \cdot \vec E=0$)
Using another maxwell relation
$$ \begin{aligned} \vec \nabla \times \vec E&=-\frac{\partial \vec B}{\partial t} \\ (ik_x,ik_y,ik_z)\times \vec E&=i\omega \vec B \\ \vec k \times \vec E&=\omega \vec B \\ \hat k \times \vec E&=c\vec B \end{aligned} $$
🗒️ Note: this only applies in free space
Poynting vector:
$$ \vec S= \frac{1}{\mu_0} \vec E \times \vec B $$
💼 Case: for plane wave $\vec E=\vec E_0 \cos(\vec k \cdot \vec r-\omega t)$, $\vec B=\vec B_0 \cos(\vec k \cdot \vec r-\omega t)$
$$ \vec S= \frac{\vec E_0 \times \vec B_0}{\mu_0 } \cos^2(\vec k \cdot \vec r-\omega t) $$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/e4e92d3c-f07f-4404-bfe4-728672b7acc0/Irradiance.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/e4e92d3c-f07f-4404-bfe4-728672b7acc0/Irradiance.png" width="40px" /> Irradiance ( or intensity): $I$ is defined as the energy in a wave crossing unit area per unit volume, averaged over many $1/f$ intervals. It is proportional to the square of the electric field amplitude
</aside>
In the above case
$$ I=\left < |\vec S| \right >=\frac{|\vec E_0 \times \vec B_0|}{2\mu_0}=\frac{c \epsilon_0 E^2_0}{2} $$
