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.
First, just as a reminder, the force of a particle due to a magnetic field is $$\vec{F_B}=q\vec{v}\times\vec{B}$$. As you mentioned the force due to the magnetic field is always perpendicular to the velocity but it is also perpendicular to the magnetic field so the differential work is $$dw_B=\vec{F_B}\cdot d\vec{s}=q(\vec{v}\times\vec{B})\cdot \vec{v}dt=0 $$ so we have verified that the work is indeed zero due to the magnetic field. $\vec{F_B}$ cannot change the speed of the particle so it cannot change it's kinetic energy, but it can change the direction of $\vec{v}$. The total force on the particle is $$\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$$ so the total work on a particle is $$dw=q\vec{E}\cdot\vec{v}dt$$ which is non zero.
Now lets consider the wire which is really just a group of moving charges. positive particles inside the wire are pushed upward in the wire due to the magnetic force and negative particles are pushed downward (where the up-down directions are orthogonal to the wire velocity v and magnetic field B). The separation of charges induces an E field which produces a force on the respective particles in the opposite direction. In equilibrium the two forces balance and we have $$qvB=qE$$ and the emf across length l is $$\mathcal{E}=El=Bvl$$. When you are talking about an EMF in the wire you should consider the wire in a magnetic field closed by a loop outside the field (see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c4 , In your problem the circuit is not closed so the wire goes nowhere) The induced emf can be seen as the work done per unit charge so it is really the E field that changes the velocity of the particles (changes their kinetic energy)perpendicular to the B field.
Best Answer
Magnetic forces do not perform work because they are always perpendicular to the motion of the charged particle they act on. However, it is possible to transform the energy stored in a magnetic field into the bulk motion of a conductor carrying a current. I'll give one heuristic.
Consider two long wires running parallel. Give them currents running in the same direction, and hold the wires still for a time that's long compared to the distance between them divided by the speed of light.
Now there is a magnetic field from the first wire at the location of the second and vice versa. The wires attract each other. Once released, they experience an acceleration towards each other.
However, this analysis only holds up until the first moment the wires are released. After that, they are moving towards each other. This motion of the wires entails a motion of the charge they carry - a current. We will need to account for this current.
Say the wires run horizontally on your computer monitor, and the current is to the right. Then the top wire is pulled down, and its current is now mostly to the right and a little bit down. The magnetic field it feels is pointing towards you, out of the monitor. Thus, the force on this top wire points mostly down, but it also points a little bit backwards.
Similarly, the force on the bottom wire is mostly up, but also a little bit backwards. The backwards components of the magnetic forces act to decrease the currents in the wires. If the currents in the wire decrease, the magnetic field gets weaker. In this way energy from the magnetic fields gets converted into the bulk motion of the wires carrying the currents.
The magnetic forces did not actually do any of the work here, though. In the absence of any wire, the charged particles in the magnetic field would like to move in circles. Instead, electrical forces between the particles and the wire they're trapped in forced the particles to stay inside the wire instead. These electrical interactions between the charged particles and the rest of the wire did the work that accelerated the wires towards each other.