Why do electrons in currents not produce any Electric Field? Why a current carrying wire in which current i(t)  is a function of time does not produces any electric field ?
Athough i(t) happens due to flow of free electrons in conductor then why these electrons in motion in a  conductor doesn't produce any Electric Field ?
 A: 
Why a current carrying wire in which current i(t) is a function of time does not produces any electric field ?

Who says that it does not produce an electric field?
Current in a wire will induce a magnetic field.
Time-varying current in a wire will induce a time-varying magnetic field.
A time-varying magnetic field will induce an electric field.
Those are the basic principles upon which inductors work, and all wires have inductance to some extent.

Addendum:

Why we neglect electric field due to current carrying wire in most of the cases ;atleast what i have encountered so far ?

A good question.
The short answer is that in a wire, charges re-arrange themselves so that electric fields are displaced, they appear in places other than where one might naively expect.
The longer answer is something like this.
Charge is conserved, so the divergence of the (conduction) current density $\vec{J}$ gives the rate of change of charge density ($\rho$).
$$\nabla \cdot \vec{J} = \frac{\partial \rho}{\partial t}$$
That is, if more (conduction) current flows into a region than flows out, then charges will accumulate in that region.
This gives us the property that if currents in a closed loop circuit are unequal in different places along the circuit, then charges will redistributed themselves.  Conversely, if the charge densities are constant with time, then the current must be uniform throughout the closed loop circuit.
The accumulation of charge in a region will in turn affect the flow of (conduction) current into and out of that region. As charge accumulates in a region, it will "discourage" the flow of conduction current into that region, and "encourage" the flow of conduction current out of that region. This will tend to establish an equilibrium condition in which the (conduction) current into a region is equal to the (conduction) current out of that region.
In equilibrium, charges have re-arranged themselves so that the (conduction) current through an unbranched circuit is the same everywhere in that circuit.
Now we consider where these charges have arranged themselves.
By the "microscopic Ohm's law", the (conduction) current density and $\vec{E}$ field are related by the conductivity of a conductor $\sigma$.
$$\vec{J} = \sigma \vec{E}$$
If we take the divergence of both sides we have
$$\nabla \cdot \vec{J} = \nabla \cdot ( \sigma \vec{E})$$
We have already seen that in equilibrium
$$\frac{\partial \rho}{\partial t} = \nabla \cdot \vec{J} = 0$$
So, it must also be the case that
$$\nabla \cdot ( \sigma \vec{E}) = 0$$
However, there is a vector calculus identity that says
$$\nabla \cdot (\sigma\vec{E}) = \sigma(\nabla \cdot \vec{E}) + (\nabla \sigma) \cdot \vec{E}$$
So, it must be the case that
$$\sigma(\nabla \cdot \vec{E}) + (\nabla \sigma) \cdot \vec{E} = 0$$
By Gauss's law the divergence of the $\vec{E}$ field is proportional to the charge density $\rho$.
$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$
so in equilibrium
$$\sigma\frac{\rho}{\epsilon_0} + (\nabla \sigma) \cdot \vec{E} = 0$$
Now this says that wherever the $\nabla \sigma$ (the gradient of the conductivity) is 0, at equilibrium it must be that
$$\sigma\frac{\rho}{\epsilon_0}=0$$
So we at last arrive at the conclusion that
Wherever the gradient of the conductivity is 0, and assuming the conductivity itself is not zero, when equilibrium is established the charge density $\rho$ is zero. In other words, charges can only aggregate where there are changes in the conductivity of a circuit. [For example, where the leads of a resistor are attached to the body of the resistor]. Charges cannot aggregate in a uniformly conductive material.
So, returning to the original question of

Why we neglect electric field due to current carrying wire in most of the cases

it is because when an electric field is applied to a wire, charges will re-arrange themselves along the boundaries where conductivity changes, or in areas where there is a gradient in conductivity. This rearrangement of charges will induce a new electric field, which will be largely cancel the original field where the conductivity is high, and make a field appear where conductivity is restricted. Thus, the original electric field is displaced and appears, not where naively expected.
A: 
why these electrons in motion in a conductor doesn't produce any Electric Field ?

From  Jefimenko’s equations we have $$ E(t,\vec r)= \frac{1}{4\pi \epsilon_0} \int \left( \frac{\vec r - \vec r’}{|\vec r - \vec r’|^3}\rho(t_r,\vec r’) + \frac{\vec r - \vec r’}{c|\vec r - \vec r’|^2} \frac{\partial}{\partial t}\rho(t_r,\vec r’) -\frac{1}{c^2|\vec r - \vec r’|}\frac{\partial}{\partial t} \vec J(t_r,\vec r’) \right) d\vec r’$$
So if there is no E field then it is because all the terms above are zero or negligible. In particular, the only term which is related to current is proportional to $$\frac{1}{c^2}\frac{\partial}{\partial t}\vec J$$ which is 0 for a DC circuit and due to the very small factor $1/c^2$ is often negligible even when it is not exactly 0.
A: In a powered uniform conductor, shaped into one or more (not necessarily circular) loops, the current (and current density) is constant.  That requires a uniform E field in the wire and a constant gradient in the charge density.  (There are more free electrons near the negative terminal of the power supply.) So, the electrons which make up the current, do contribute to the maintenance of the uniform field in the wire (and also contribute to a field which exists outside of the wire).
