Possibly you are talking about the difference between the "far field" and "near field" solutions for the simple oscillating electric dipole.
Often when dealing with such a system, if we are looking at the field more than a few wavelengths away from the dipole (or more formally, $kr \gg 1$ or $r \gg \lambda/2\pi$) then the solution looks like a spherically expanding electromagnetic wave; the E-field and B-field are in phase, mutually perpendicular and at right angles to the outward propagation direction. For a dipole moment aligned with the z-axis, the E-field is polarised in the $\theta$ (poloidal) direction and B-field is in the $\phi$ (toroidal) direction.
But if $r \leq \lambda$ then the solution is more complicated. The E-field has both a $\theta$ and a radial component. The B-field is just toroidal, but contains two terms with differing radial dependencies.
In these extra terms for the nearby fields, the E-field becomes much more dominant (in transverse electromagnetic waves it is normally $c$ times bigger). Furthermore it is out of phase with the B-field by $\pi/2$. You can see this from the equations below - when $r$ is small, the first term in $B_{\phi}$ dominates the B-field, whereas it is the first (even stronger) terms in the $E_r$ and $E_{\theta}$ components that dominate the E-field.
These are different in magnitude from the B-field by a factor that includes $i$ and hence are out of phase by $\pi/2$.
Perhaps this is what you mean?
The Maths:
The solutions for the E- and B-field from a simple oscillating dipole are
$$E_{r} = \frac{p_0 \cos\theta}{4\pi \epsilon_0} \frac{k^2 \exp(ikr)}{r}\left[ \frac{2}{k^2r^2} - \frac{2i}{kr} \right]$$
$$E_{\theta} = \frac{p_0 \sin\theta}{4\pi \epsilon_0} \frac{k^2 \exp(ikr)}{r}\left[ \frac{1}{k^2r^2} - \frac{i}{kr} -1 \right]$$
$$B_{\phi} = \frac{p_0 \sin\theta}{4\pi \epsilon_0} \frac{k^2 \exp(ikr)}{r}\left[ - \frac{i}{kr} -1 \right] \left(\frac{\epsilon_0}{\mu_0} \right)^{1/2}$$