If you are not well-acquainted with special relativity, there is no way to truly explain this phenomenon. The best one could do is give you rules steeped in esoteric ideas like "electromagnetic field" and "Lorentz invariance." Of course, this is not what you're after, and rightly so, since physics should never be about accepting rules handed down from on high without justification.
The fact is, magnetism is nothing more than electrostatics combined with special relativity. Unfortunately, you won't find many books explaining this - either the authors mistakenly believe Maxwell's equations have no justification and must be accepted on faith, or they are too mired in their own esoteric notation to pause to consider what it is they are saying. The only book I know of that treats the topic correctly is Purcell's Electricity and Magnetism, which was recently re-released in a third edition. (The second edition works just fine if you can find a copy.)
A brief, heuristic outline of the idea is as follows. Suppose there is a line of positive charges moving along the $z$-axis in the positive direction - a current. Consider a positive charge $q$ located at $(x,y,z) = (1,0,0)$, moving in the negative $z$-direction. We can see that there will be some electrostatic force on $q$ due to all those charges.
But let's try something crazy - let's slip into $q$'s frame of reference. After all, the laws of physics had better hold for all points of view. Clearly the charges constituting the current will be moving faster in this frame. But that doesn't do much, since after all the Coulomb force clearly doesn't care about the velocity of the charges, only on their separation. But special relativity tells us something else. It says the current charges will appear closer together. If they were spaced apart by intervals $\Delta z$ in the original frame, then in this new frame they will have a spacing $\Delta z \sqrt{1-v^2/c^2}$, where $v$ is $q$'s speed in the original frame. This is the famous length contraction predicted by special relativity.
If the current charges appear closer together, then clearly $q$ will feel a larger electrostatic force from the $z$-axis as a whole. It will experience an additional force in the positive $x$-direction, away from the axis, over and above what we would have predicted from just sitting in the lab frame. Basically, Coulomb's law is the only force law acting on a charge, but only the charge's rest frame is valid for using this law to determine what force the charge feels.
Rather than constantly transforming back and forth between frames, we invent the magnetic field as a mathematical device that accomplishes the same thing. If defined properly, it will entirely account for this anomalous force seemingly experienced by the charge when we are observing it not in its own rest frame. In the example I just went through, the right-hand rule tells you we should ascribe a magnetic field to the current circling around the $z$-axis such that it is pointing in the positive $y$-direction at the location of $q$. The velocity of the charge is in the negative $z$-direction, and so $q \vec{v} \times \vec{B}$ points in the positive $x$-direction, just as we learned from changing reference frames.
For the magnetic field, the currents are one source of the magnetic, but this problem is more linked to the source of the current in the wire. For a conductor with finite conductivity, an electric field is needed in order to drive a current in the wire.
If we assume your wire is straight, this required field is uniform. One way to realize this field is by taking two oppositely charged particles and send them to infinity while increasing the magnitude of their charge to maintain the correct magnitude of electric field. In this limit, you will obtain a uniform electric field through all space.
Now, put your conductor in place along the axis between the voltage sources--a current will flow. In the DC case, this gives rise to the magnetic field outside of the wire. As for the electric field, a conductor is a material with electrons that can move easily in response to electric fields and their tendency is to shield out the electric field to obtain force balance. Because the electrons can't just escape the conductor, they can only shield the field inside the conductor and not outside the conductor. With this model, we see that the electric field is entirely set up by the source and placing the conductor in the field really just establishes a current. Note here that if you bend the wire or put it at an angle relative to the field, surface charges will form because you now have a field component normal to the surface.
For the limit of an ideal conductor, no electric field is needed to begin with to drive the current and so there isn't one outside the wire.
For the AC case, solving for the fields becomes wildly complicated very fast as now the electric field driving particle currents has both a voltage source and a time-varying magnetic source through the magnetic vector potential. The essential physics is the same, though, as the source will establish the fields (in zeroth order), and the addition of the conductor really just defines the path for particle currents to travel. In the next order, the current feeds back and produces electromagnetic fields in addition to the source(s) and will affect the current at other locations in the circuit.
I guess a short answer to your question is that there are always fields outside of the current-carrying wire and the electric field outside disappears only in the ideal conductor limit. Conductors generally do not require very strong fields to drive currents anyway so that the electric field outside is usually negligible, but don't neglect it for very large potentials in small circuits.
Best Answer
Yes. The electron is moving (in our reference frame), so now there is a magnetic field (in our reference frame), but nothing happens to the electric field.
The electrons in the conductor produce an electric field outside the conductor; however, realistically, there will be just as many protons in the conductor as electrons, and hence the net electric field outside the conductor is zero.
The electric field is still there (in some sense), but its zero, because the electrons and protons in the conductor cancel each other out, so we don't care about it. (Actually, I believe that if you take into account relatavistic effects, which is probably silly not to do in the context of electrodynamics, there will be a nonzero electric field). That being said, if for some reason there were stream of moving electrons with no protons, then we would observe both a (nonzero) magnetic and electric field.
Special relativity says this can't happen :). In any case, if we play dumb for a moment, the only thing that would change is the current, and hence the strength of the magnetic field.