Moving charge always produces a magnetic field. If you have a non-zero current then you have non-zero moving charge and a magnetic field will be produced.
You can achieve essentially no magnetic field though by using two wires right next to each other each carrying current in the opposite directions. As long as the wires are very close and the amount of current they carry is very close the magnetic fields they produce will nearly cancel. This is why a clamp meter can't measure current around two conductors carrying current in opposite directions.
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.
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Best Answer
In a circuit involving potential drop (so, not purely a current wo/voltage,) the e-field around conductors is perpendicular and radial, and the e-field around resistors is radial with some tilt.
For conductors with arbitrarily small resistance, the flux-lines of e-field appear in the space outside the conductor, and are connecting the surface-charge with charges found upon other, distant parts of the circuit (e.g. parallel wires having opposite charge.) Here's the oversimplified visual version:
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Simple circuit:
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. Notice that the fields are those of a 2-wire waveguide or transmission line? Exactly right. The same physics applies at Zero Hz DC, and also applies at 60Hz AC, and also at radio frequencies.