Is it correct way of saying that a moving electric field causes magnetic field? $$\mathbf B=\mathbf V \times \mathbf E \mu _0 \epsilon _0$$
$\mathbf V$ is the velocity vector of moving electric field. Rest of the parameters follow usual notations for those.
 A: First of all, fields don't move because fields are everywhere in space. For example temperature is a field that has a value at every point in space and all the time, but when the sources moves, the field changes. 
So, I suppose you mean by "moving electric field" that the electric field change with respect to time in a way that the values become shifted by $\mathbf{v} dt$ for a time interval $dt$. Electric and magnetic fields can change over time according to some situations, like changing sources such as moving charges or alternating currents. However, electric and magnetic fields do not change each other. I will explain how you get the relation in your question at the end of the answer but first I need to clarify some important points.
Electromagnetic Potentials
As a matter of fact, there is no cause-and-effect relation between electric and magnetic fields. Rather they both caused by more fundamental fields, called electromagnetic potentials, one is scalar and denoted by $\phi (\mathbf{r},t)$, and the other is vector and denoted by $\mathbf{A}(\mathbf{r},t)$. Their derivatives define how the electric and magnetic field will correlate, as follows:
\begin{align}
\tag{1.a}
\mathbf{E} & \equiv - \mathbf{\nabla} \phi - \frac{\partial \mathbf{A}}{\partial t} \\
\tag{1.b}
\mathbf{B} & \equiv \mathbf{\nabla} \times \mathbf{A}
\end{align}
So, counting the degrees of freedom, $\mathbf{E}$ and $\mathbf{B}$ have 6 in total but they can be derived from 4. The reason why Maxwell, Faraday and Ampére didn't start from them is because they were perceived by forces.
Maxwell's Equations for Potential Fields
The electric and magnetic fields correlate according to the potentials, and those potentials change with respect to the Maxwell's equations which become as follows, in this potential formulation:
\begin{align}
\tag{2.a}
\Box \varphi + \frac{\partial}{\partial t} \left ( \mathbf{\nabla} \cdot \mathbf{A} + \frac{1}{c^2} \frac{\partial \varphi}{\partial t}\right ) &= - \frac{\rho}{\varepsilon_0} 
\\
\tag{2.b}
\Box  \mathbf A - \mathbf \nabla \left ( \mathbf \nabla \cdot \mathbf A + \frac{1}{c^2} \frac{\partial \varphi}{\partial t} \right ) &= - \mu_0 \mathbf J
\end{align}
where $\Box = \nabla^2- \frac{1}{c^2} \frac{\partial^2}{\partial t^2}$ is the d'Alembert operator, and $\rho(\mathbf r, t)$ and $\mathbf{J} (\mathbf{r}, t)$ are charge density and current density, respectively. The expression in paranthesis can be eliminated by shifting those potentials, $\mathbf{A} \rightarrow \mathbf{A} - \mathbf{\nabla} \chi$ and  $\varphi \rightarrow \varphi + \partial_t \chi$, with respect to a continuous scalar function, $\chi(\mathbf r, t)$, such that
\begin{align}
\tag{Lorenz gauge}
\mathbf {\nabla} \cdot \mathbf{A} + \frac{1}{c^2} \frac{\partial \varphi}{\partial t} = 0
\end{align}
however, due to their definition in (1), electric and magnetic fields will remain unchanged under this shift, i.e. $\mathbf E \rightarrow \mathbf E$ and $\mathbf B \rightarrow \mathbf B$. Therefore,
\begin{align}
\tag{3.a}
\Box \varphi &= - \frac{\rho}{\varepsilon_0} 
\\
\tag{3.b}
\Box  \mathbf A  &= - \mu_0 \mathbf{J}
\end{align}
Correlation of E&M Fields Without Sources
So, your question reduces to how these fields correlate in this way even there is no source to change the potentials. The answer becomes obvious when you write Maxwell's equations of (3) without sources, that is:
\begin{align}
\tag{4}
\Box \varphi = 0 
&& and &&
\Box  \mathbf A  = 0
\end{align}
These are just wave equations. So, if you have an antenna producing a changing $\varphi$ and $\mathbf A$, which is basically a rod that moves charges back and forth, then at really far distance, where the effects of the sources become negligible, there would still be a wave traveling to that distance. This kind of wave is called electromagnetic wave since it has both electric and magnetic components. So, EM wave is an energy transferred by the electromagnetic fields because of some source that is no longer effecting it.
Moving Sources
One can ask then how moving charges produce magnetic field, if the field isn't moving? The answer is right there: moving charge means a nonzero current density which produces vector potential according to (3.b), which produces magnetic fields due to (1.b).
However, one can also obtain this result by just switching from an observer A who sees the charge is at rest to an observer B that sees the charge is moving at a velocity, $\mathbf{v}$. Observer A experiences $ \Box \varphi = -\rho / \epsilon_0$ and, say, $\mathbf{A} =0$. In order to obtain what Observer B experiences, one needs to use Lorentz transformations, because Electrodynamics is a Lorentz invariant theory. So, one needs to construct the kinematics of classical electrodynamics on the theory of Special Relativity.
Hence, Lorentz transformations of the potential fields are as follows:
\begin{align}
\varphi' &= \gamma (\varphi - \mathbf{v} \cdot \mathbf{A} ) \\
\mathbf {A}'_{||} &= \gamma (\mathbf{A} _{||} - \mathbf{v} \varphi / c^2 ) \\
\mathbf {A}'_\perp & = \mathbf{A} _\perp
\end{align}
where $\mathbf{A}_{||}$ and $\mathbf {A}_\perp$ are parallel and perpendicular components of the vector potential, and $\gamma = 1/\sqrt{1-v^2/c^2}$ is the Lorentz factor.
Therefore, Observer B will have the following field values in terms of Observer A's fields according to Lorentz transformations:
\begin{align}
\varphi^B &= \gamma \varphi^A \\
\mathbf{A}^B &= -\gamma \mathbf{v} \varphi^A / c^2
\end{align}
which gives a nonzero magnetic field as
\begin{align}
\mathbf{B}^B &= \nabla \times \mathbf{A}^B\\
&= - \frac{1}{c^2}\left( \varphi^B  \nabla \times \mathbf{v} - \mathbf{v} \times \nabla \varphi^B  \right) \\
&= \frac{1}{c^2} \mathbf{v} \times \mathbf{E}^B
\end{align}
where $\mathbf v$ is constant (or at least curl-free since we only switch to inertial observers).
