Final answer:
The wire with higher conductivity will have a lower electric field when carrying the same current due to the direct relationship between conductivity and current density, and the inverse relationship with resistivity. Drift velocity is lower in the better conductor because of the greater availability of charge carriers.
Step-by-step explanation:
The question pertains to the conductivity of two different wires connected in series and the relationship between their electric fields and conductivity. When two wires of the same cross-sectional area are connected in series and carry the same current, the wire with higher conductivity will have a lower electric field. Conductivity (σ) is inversely related to resistivity (ρ) and directly related to the current density (J) through the relation J = σE. Assuming the current (I) is the same through both wires, and their cross-sectional area (A) is also the same, the wire with higher conductivity will exhibit a lower electric field because current density is constant (J = I/A). The equation for drift velocity v = J/(nq) further reveals that in a better conductor, with more free charge carriers (n), the drift velocity would be lower for the same current density due to the increased number of charges available to carry the current. The difference in resistivity of the two wires can be deduced given the resistivity of one wire and the resistance of both using the relation R = ρL/A, where L is the length of the wire.