Are the electric and magnetic fields of an EM wave also perpendicular in the near-field or the intermediate zone? Are the electric and magnetic fields of an electromagnetic wave perpendicular to each other even in the near-field zone or in the intermediate zone (when the radiation zone approximation is not valid)?
The example given by ProfRob is nice but I would like to know whether E and B fields are always perpendicular for any arbitrary radiation source when the radiation zone approximation is not valid.
 A: Do some degree, the answer depends on exactly how you define "electromagnetic wave" and "radiation source".
Jackson (Third Edition) works out the exact expressions for the ${\bf E}$ and ${\bf B}$ fields from a radiating source in eqs. 9.18, making the sole approximation that the source is small compared to the wavelength of radiation and distance from the source under consideration. (The distance under consideration can be either larger or smaller than the wavelength, so this encompasses both the near and far zones.) He finds that the ${\bf E}$ and ${\bf B}$ fields are indeed always exactly perpendicular (although in the near zone, the ${\bf E}$ field does have a radial component, unlike in the far zone where it becomes transverse).
In full generality, where the wavelength, the spatial extent of the source, and the distance under consideration are all arbitrary, the ${\bf E}$ and ${\bf B}$ fields don't need to be perpendicular. This is essentially just a completely general solution to Maxwell's equations, and the Lorentz-covariant scalar field ${\bf E} \cdot {\bf B}$ certainly isn't required to be zero in general. The fact that you mention the "near-field and intermediate zones" in your question suggests that you have some specific assumptions in mind about the nature of the radiation sources, wavelengths, and/or spatial region under consideration, but we can't answer your question without more details about what those assumptions are.
A: I will assume you mean the fields generated by an oscillating electric dipole.
The E- and B-fields are
$$ {\bf E} = \frac{j}{2\pi \epsilon_0} kp_0 \cos\theta\left(1 - \frac{j}{kr}\right) \frac{e^{j(\omega t -kr)}}{r^2}\ {\bf \hat{r}} - \frac{k^2}{4\pi \epsilon_0} p_0 \sin \theta \left(1 - \frac{j}{kr} - \frac{1}{k^2 r^2}\right) \frac{e^{j(\omega t -kr)}}{r}\ {\bf \hat{\theta}}$$
$${\bf B} = -\frac{k^2}{4\pi\epsilon_0 c} p_0 \sin \theta\left(1 - \frac{j}{kr}\right)\frac{e^{j(\omega t -kr)}}{r}\ {\bf \hat{\phi}}\ ,$$
where $k$ is the magnitude of the wave vector, $p_0$ is the electric dipole moment amplitude and $\omega$ is the angular frequency.
As you can see, the scalar product of these two fields is always zero and the E- and B-fields are perpendicular, whatever the value of $r$ and $\theta$.
However, you want to know about more general fields. Well we could show that they are not always perpendicular by considering a trivial counterexample.
Take the dipole mentioned above and surround it with a small oscillating current loop - an oscillating magnetic dipole, with fields
$$ {\bf E} = \frac{\mu_0}{4\pi} k^2 m_0 c\sin\theta \left(1  + \frac{j}{kr}\right) \frac{e^{j(\omega t -kr)}}{r}\, {\bf \hat{\phi}} $$
$${\bf B} =   -j\frac{\mu_0}{2\pi} k m_0\cos\theta\left(1 +\frac{j}{kr} \right)           \frac{e^{j(\omega t -kr)}}{r^2}\, {\bf \hat{r}} -  \frac{\mu_0}{4\pi} k^2 m_0\sin\theta \left(1+ \frac{j}{kr} -\frac{1}{k^2r^2}  \right)\frac{e^{j(\omega t -kr)}}{r}\, {\bf \hat{\theta}}    $$
Since the solutions of Maxwell's equations superpose, then we can take the scalar product of the total fields to be
$${\bf E}\cdot{\bf B} = \frac{\mu_0  }{4\pi^2 \epsilon_0} k^2 p_0 m_0 \cos^2\theta \left(1 +\frac{1}{k^2 r^2}\right) \frac{e^{2j(\omega t-kr)}}{r^4} + \frac{\mu_0  }{16\pi^2 \epsilon_0} k^4 p_0 m_0 \sin^2\theta \left(\left(1-\frac{1}{k^2r^2}\right)^2 +\frac{1}{k^2 r^2}\right) \frac{e^{2j(\omega t-kr)}}{r^2} - \frac{\mu_0  }{16\pi^2 \epsilon_0} k^4 p_0 m_0 \sin^2\theta \left(1 +\frac{1}{k^2 r^2}\right) \frac{e^{2j(\omega t-kr)}}{r^2}$$
In the limit where $kr \gg 1$ then indeed the scalar product approaches zero. This is the radiation field limit where the waves approximate to plane waves and the E- and B-fields must be perpendicular. However for smaller $r$ values, this isn't true. The scalar product depends on $r$ and $\theta$ and is generally non-zero.
Thus for any arbitrary distribution of charge and current sources, the E- and B-fields will not be perpendicular. However, if you go to the radiation field limit and at a distance much larger than the source size, then the superposed fields from a single source will approximate to plane waves with perpendicular E- and B-fields.
NB. It is trivial to show that the E- and B-fields of plane waves arriving from different directions from different sources are not necessarily perpendicular.
A: Imagine the following setup. We have a capacitor whose $E$ field is oriented in the $z$ direction. If you put AC voltage on that capacitor with frequency $\omega$ it will radiate (it's like an electric dipole). Now put a solenoid concentric with that capacitor (either inside the capacitor or around it or something) so that its $B$ field is in the $z$ direction. If you put an AC voltage on that solenoid with frequenecy $\omega$ it will also radiate (it's like a magnetic dipole).
If you now drive the capacitor and the solenoid simultaneously you will have some weird radiator with some weird radiation pattern in the far field, but in the near field, we can see that $E$ and $B$ are parallel.
A: If we are only considering  the radiative effects of the EM field, we can prove that in general they are always perpendicular to eachother. In general this is not true for the total EM field, but for the radiative components it is always true!
This can be proved by analysing the dot products between the radiative components of jefimenkos equations, for the electric and magnetic field.
The radiative components of jefimenkos equations are -
$\vec{E} = \frac{1}{4\pi\epsilon_0}\int-\frac{1}{|\vec{r}-\vec{r'}|}\frac{1}{c^2} \frac{\partial \vec{J_{tr}}}{\partial t} d^3r'$
$\vec{B} =\frac{-\mu_{0}}{4\pi}\int \frac{\vec{r}-\vec{r'}}{|\vec{r}-\vec{r'}|^3}×\frac{1}{c} \frac{\partial \vec{J_{tr}}}{\partial t} d^3r'$
From here it is seen that $\vec{E} \cdot \vec{B} = 0$ as the difference in the integrand is a cross product, so by definition, the contribution to the integral at each point in space must be perpendicular
(I believe the logic involving analysing the integrants itself to determine the outcome of the dot product is correct, please tell me if I am wrong though)
