Electrons have an electric field. By the influence of an external electric field, the electron is attracted or repelled. Electrons also possess a magnetic dipole moment. This moment is coupled to the rotation of the electron (the magnetic dipole moments of an electron and a positron are opposed with the same direction of rotation). When electrons moving through a magnetic field the magnetic dipoles and therefore also the axes of rotation of the electrons are aligned. Now arises a gyroscopic effect and the electron goes under emission of a photon out of alignment again. Detailed see here.
The acceleration of electrons can be done not only by electric fields. Under the influence of electromagnetic fields, electrons are also accelerated. With a positive acceleration the energy of photons is transmitted parcial to the electron. In case of negative acceleration more energy is emitted as received. As described above, a magnetic field through which electron moves - not parallel to the magnetic field lines! - influential on the electron too. The electron emit photons and its kinetic energy get lost.
1) A stationary charge that has always been stationary is associated with an electric field and only an electric field. The electric field points towards the charge and every point that us equally far away has an equally strong field and the fields gets four times as weak if you go twice as far away.
2) A uniformly moving charge that has always been stationary is associated with a magnetic field and also an electric field. And the electric field for the moving charge is different than the electric field for the stationary charge in #1. For instance it points towards the line the charge is moving on, and always to a point where the charge used to be but no long is (in fact you can imagine that at every the particle was at a certain place that a sphere expands from that place and time at the speed of light and on that sphere the electric field points towards where that particle was even though it isn't there any more) and points that are currently equally far from the charge do not have equally strong fields.
But wait, there is more! When you see a pure electric field, a friend moving relative to you sees a combination of electric and magnetic fields. And the electric and magnetic fields you see for the moving charge at some point are literally just the electric and magnetic field you'd see when a friend moving with the charge saw a pure electric field pointing radially from/to the charge. So really your motion has just made the electric field from the proton's frame look like an electric field.
You can combine electric and magnetic fields into a single unified Faraday field with six independent components that different people break into different pieces to see electric and magnetic fields.
3) If you are accelerating at a place p at a moment t then there is again an expanding shell moving outwards from that place p expanding at the speed of light, c. And on this shell there are additional electric and magnetic fields, different than any we have seen before.
These will be in addition to those due to the location and velocity of the charge. They can point in different directions than before. And even though they are different, they share some properties for instance the new magnetic field is still orthogonal to the the new electric field. But other things are different, for instance they don't get weak as quickly you could go out a million times as far out and the new fields from the acceleration only get a million times smaller on average so really far away that's what you are likely to notice. The fields caused by the acceleration. On the other hand if the acceleration was a tiny acceleration you might have to go very far before that is the stronger field.
There is a sense where (squared) field strength is conserved and so just moved around. And the reason the acceleration fields get weaker less quickly is because the flow of field strength is orthogonal to both the electric and magnetic fields. In fact that is why you need magnetic fields to allow electric fields to change, you need magnetic fields to flow the field strength. And so the fields associated with the acceleration are trying to send the energy outwards to the future location of the shell since the new (acceleration based) fields are both tangent to the surface of the shell. Magnetic fields due to a single charge are always tangent to the expanding shell but for the acceleration fields both are tangent, so if it was just up to them all the field strength would flow outwards to the next instants slightly bigger shell.
This contrasts with the static fields which have the electric field pointing orthogonally to the shell and since the magnetic field is orthogonal to the electric field and tangent to the shell, the magnetic field do the the position of the charge is zero (it is both orthogonal to and tangent to the shell). So on its own the static fields wouldn't flow any field strength.
But they do interact since the field strength flows orthogonal to the total fields. Plus there are the fields due to velocity, which on some points of the shell help and in some places hurt. When they keep field strength from travelling outwards they could be redistributing it across the shell
And the fields due to the velocity are sometimes helping and sometimes hurting because the part in the direction of the acceleration has a part that (like the static field) doesn't contribute to flow of field strength. While the part of the velocity orthogonal to the acceleration actually just contributes to another acceleration field but one that in some parts if the shell makes the other one stronger and in other parts makes it weaker.
Finally. It is important to note that the acceleration fields flow the field strength in a way to make there be more fields on the later shells that are like themselves, so when the other fields cancel each other out the acceleration fields succeed at going to the next shell in a self propagating way.
So when the acceleration fields have their way they stay at full strength and have progeny like themselves but when the other fields have their way they don't flow the field strength out to the next part of the shell.
So in a battle like that it isn't surprising that in those later shells it is almost all acceleration fields.
So these new fields that are associated with acceleration are not just the same fields that look different because a different person is looking. They are different fields. Because they point tangent to the shell, so flow their field strength outwards to the next shell.
Best Answer
The electric $\:\mathbf{E}\:$ and magnetic $\:\mathbf{B}\:$ parts of the electromagnetic field produced by a moving charge $\:q\:$ are :
SOURCE 1
From Jackson's 'Classical Electrodynamics', 3rd Edition, equations (14.14) and (14.13)
\begin{align} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf{E}(\mathbf{x},t) & = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{n})^3 R^2}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\boldsymbol{\times}\left[(\mathbf{n}-\boldsymbol{\beta})\boldsymbol{\times} \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{14.14}\\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf{B}(\mathbf{x},t) & = \left[\mathbf{n}\boldsymbol{\times}\mathbf{E}\right]_{\mathrm{ret}} \tag{14.13} \end{align} where \begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{01a}\\ \dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{01b}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'} \tag{01c} \end{align}
SOURCE 2
From Feynman's Lectures, Volume II 'Mainly Electromagnetism and Matter', New Millennium Edition, equations (21.1) \begin{align} \mathbf{E} & = \frac{q}{4 \pi \epsilon_0} \left[ \frac{ \mathbf{e}_{r'}}{r'^2} + \frac{r'}{c} \frac{d}{dt} \left(\frac{\mathbf{e}_{r'} }{r'^2}\right) + \frac{1}{c^2} \frac{d^2}{dt^2} \mathbf{e}_{r'} \right] \tag{21.1-Feynman}\\ c\mathbf{B} & = \mathbf{e}_{r'}\boldsymbol{\times}\mathbf{E} \nonumber \end{align}
Without looking at the details it seems from the second equations, those giving $\:\mathbf{B}\:$ from $\:\mathbf{E}\:$, that the former is produced by the latter. But this is not the case : the electromagnetic field is an entity and its separation to the electric and magnetic parts depends upon the observer's inertial frame of reference.
Note that these equations are derived from the so called Liénard-Wiechert potentials which in turn are produced from the Maxwell's equations in empty space
\begin{align} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{E} & = -\frac{\partial \mathbf{B}}{\partial t} \tag{02a}\\ \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} & = \mu_{0}\mathbf{j}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{02b}\\ \nabla \boldsymbol{\cdot} \mathbf{E} & = \frac{\rho}{\epsilon_{0}} \tag{02c}\\ \nabla \boldsymbol{\cdot}\mathbf{B}& = 0 \tag{02d} \end{align} with electric charge and electric charge current densities \begin{align} \rho(\mathbf{x},t) & = q\cdot \delta\left(\mathbf{x}-\mathbf{x}_{q}\right) \tag{03a}\\ \mathbf{j}(\mathbf{x},t) & = q\cdot \delta\left(\mathbf{x}-\mathbf{x}_{q}\right)\cdot\dfrac{\mathrm d\mathbf{x}_{q}}{\mathrm d t} \tag{03b} \end{align}
The fact that the electromagnetic field is an entity is more clear if we take a look at its transformation. So let $\:\mathrm{S}\:$ and $\:\mathrm{S}'\:$ two inertial frames of reference and the frame $\:\mathrm{S}'\:$ moves uniformly with velocity $\:\mathbf{v}=\upsilon\mathbf{n}, \upsilon \in \left(-c,+c\right)\:$ with respect to $\:\mathrm{S}$, see Figure in the bottom. Then from $\:\mathrm{S}\:$ to $\:\mathrm{S}'$ \begin{align} \mathbf{E}' & =\gamma \mathbf{E}\!-\!\left(\gamma\!-\!1\right)\left(\mathbf{E}\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}+\,\dfrac{\gamma}{c}\left(\mathbf{v}\boldsymbol{\times}c\mathbf{B}\right) \tag{04a}\\ c \mathbf{B}' & = \gamma c\mathbf{B}\!-\!\left(\gamma\!-\!1\right)\left(c\mathbf{B}\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}\!-\!\dfrac{\gamma}{c}\left(\mathbf{v}\boldsymbol{\times}\mathbf{E}\right) \tag{04b} \end{align} From equations (04) the six scalar components bind together more strongly in the so-called electromagnetic field tensor \begin{equation} F=F^{\mu\nu} \begin{bmatrix} \hphantom{-}0 & -E_1 & -E_2 & -E_3 \hphantom{-}\vphantom{\dfrac12}\\ \hphantom{-}E_1 & \hphantom{-} 0 & -cB_3 & \hphantom{-} cB_2 \hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-}E_2 & \hphantom{-} cB_3 & \hphantom{-} 0 & -cB_1 \hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-}E_3 & -cB_2 & \hphantom{-} cB_1 & \hphantom{-} 0 \hphantom{-}\vphantom{\dfrac12} \end{bmatrix} \tag{05} \end{equation} transformed under the Lorentz Transformation $\:\Lambda\:$ as follows \begin{equation} F'=\Lambda F \Lambda \tag{06} \end{equation} or by tensors as \begin{equation} F'^{\alpha\beta}=\Lambda^{\alpha}_{\mu} F^{\mu\nu}\Lambda^{\beta}_{\nu} \tag{07} \end{equation}