Drift speed inversely proportional to wire cross-sectional area

Steps:

• Current I is the same in both wires since they are joined.
• Average drift speed $v_d = \frac{I}{n e A}$, where $n$ (electron density) and $e$ (charge) are identical for copper.
• Cross-sectional area $A_P = \pi (d/2)^2 = \frac{\pi d^2}{4}$; $A_Q = \pi (2d/2)^2 = \pi d^2$.
• Thus, $A_Q = 4 A_P$, so $\frac{v_{dP}}{v_{dQ}} = \frac{A_Q}{A_P} = 4$.

Why D is correct:

• $v_d \propto \frac{1}{A}$; with $A_Q = 4 A_P$, $v_{dP} = 4 v_{dQ}$ by the drift speed formula.

Why the others are wrong:

• A: Assumes $v_d \propto A$, reversing the inverse relationship.
• B: Assumes $v_d \propto \frac{1}{d}$, ignoring area scales with $d^2$.
• C: Assumes $v_d \propto \frac{1}{\sqrt{A}}$, confusing with linear dimension.

Final answer: D