Does the electric field ($E$ caused by induction) of a moving conductor in a magnetic field drop when connected to a curcuit? We know that when a conductor moves in a homogenous magnetic field that is perpendicular to itself due to the amount of electric charge gathered on one end of the conductor( as a result of the lorentz force) an electric field is caused by induction, with a certain electromotive force that is equal to the voltage difference between the two ends of the conductor. However, if the conductor had resistance, when connected to a circuit, the voltage between the two ends of the conductor changes due to current flowing and is in fact equal to $V= emf - I*r $,  where $r$ the resistance of the conductor. If all this is true, then due to the overall voltage of the conductor dropping, shouldn't the electric field (E) inside also get reduced?
 A: 
due to the amount of electric charge gathered on one end of the conductor( as a result of the lorentz force) an electric field is caused by induction, with a certain electromotive force that is equal to the voltage difference between the two ends of the conductor.

No, the electromotive force is not due to electric charge gathered on ends of the conductor. The motional EMF is due to motion of the conductor in magnetic field, and is present whenever the conductor moves in magnetic field, even when there no charge separation on the ends of the conductor yet.
Let the conductor be a rod moving in a plane perpendicular to magnetic field.
When the motion starts from rest, in the first instant, there is no charge separated yet, and the motional EMF
$$
\mathscr{E}_m = (\mathbf v\times \mathbf B) \cdot \mathbf L
$$
is the only force that acts on current in the rod. According to generalized Ohm's law:
$$
rI = (\mathbf v\times \mathbf B) \cdot \mathbf L
$$
where $r$ is resistance of the rod.
If the rod is isolated (not part of a circuit), then this current stops almost immediately, because it causes charge concentration on surface of the rod; this charge creates electric field $\mathbf E$ that counteracts the effect of EMF on current, and the above equation has to be modified into
$$
rI = (\mathbf v\times \mathbf B) \cdot \mathbf L + \mathbf E\cdot \mathbf L.
$$
In equilibrium state (when the rod has constant velocity and current has stopped) this electric field cancels out the force per unit charge due to motional EMF on current. Thus in equilibrium, electric field inside the rod is
$$
\mathbf E  = -\mathbf v\times \mathbf B
$$
and thus difference of potential of the rod ends has magnitude
$$
|u| = |\mathbf E \cdot \mathbf L |= |(\mathbf v\times \mathbf B)\cdot \mathbf L|.
$$
If the rod ends are connected to a circuit (e.g. via rails), then such large charge concentration is prevented, and electric field won't be strong enough to cancel the EMF. Current won't stop as fast; it will flow as long as the rod is moving. However, there is also a force on the rod body slowing it down. So unless there is a force pulling the rod and maintaining the velocity, the rod that is moving on its own will slow down and eventually stop, and motional EMF will decrease to zero.

the voltage between the two ends of the conductor changes due to current flowing and is in fact equal to $V=emf−I∗r$, where $r$ is the resistance of the conductor. If all this is true, then due to the overall voltage of the conductor dropping, shouldn't the electric field (E) inside also get reduced?

Yes, magnitude of electric field inside is decreased when current is allowed to flow, and thus also difference of potential on the rod will have lower magnitude. The exact value depends on the rest of the circuit. If the rest is just the rails and connecting wire with net ohmic resistance $R$ and we neglect inductance and capacitance of the circuit, then Kirchhoff's second circuital law says
$$
(R + r)I = \mathscr{E}_m,
$$
and thus drop of potential on the rod in direction of current is
$$
u = -R I 
$$
$$
u = rI - \mathscr{E}_m.
$$
The drop is negative, thus the rod behaves as battery: mobile charged particles increase their potential when they get displaced from one end to another in direction of current. Also we can see magnitude of the drop is lower than $\mathscr{E}_m$, due to the term $rI$.
