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.
If a source charge exerts force on the test charge, it should also feel the same force in opposite direction. So, won't it mean source charge is experiencing force due to its own electric field?
There will be force on the source charge, but due to electric field of the test charge. If the test charge was not present, there would be no force on the source charge.
Best Answer
an $E$ field is simply a region in which an $E$ charge experiences a force. therefore when two charges are sufficiently close by, they experience a mutual force. every charge is surrounded by its own $E$ field, in the same way every magnet is surrounded by its own magnetic field (or $B$ field) and every massive particle is surrounded by its own gravitational field. none of these need to be moving to set up a field.
when two charges interact they necessarily alter the shape of the $E$ field, especially between them. however the amount of alteration depends on the magnitude of the charge involved. therefore a negligible third charge placed in between the first two charges will experience a force in a straight line connecting them. but a third charge of equal magnitude will bend the $E$ field.
a charge entering an $E$ field experiences a force known as the Coulomb force. since charges have mass, the observation is that the charge accelerates along the $E$ field line, however that line may be shaped as per (2).
a moving charge (i.e. current) generates a $B$ field around that charge