Question Analysis: The wire is connected in series, meaning the current $I$ flowing through all three parts (AC, CD, and DB) must be the same due to the principle of conservation of charge.
Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]
The relationship between current $I$ and drift velocity $v_d$ is given by the formula:
$$I = n e A v_d$$Where $n$ is the electron density, $e$ is the elementary charge, and $A$ is the cross-sectional area of the wire ($A = \pi r^2$). Since the wire is made of the same metal, $n$ and $e$ are constant.
Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]
Since $I$ is constant for all three parts, we have:
$$v_d = \frac{I}{n e A} \implies v_d \propto \frac{1}{A} \implies v_d \propto \frac{1}{r^2}$$Given the radii are $r_1 = r$, $r_2 = r/3$, and $r_3 = r/2$, we calculate the drift velocities:
- For part AC: $v_1 \propto \frac{1}{r^2}$
- For part CD: $v_2 \propto \frac{1}{(r/3)^2} = \frac{9}{r^2} = 9v_1$
- For part DB: $v_3 \propto \frac{1}{(r/2)^2} = \frac{4}{r^2} = 4v_1$
Comparing the values: $9v_1 > 4v_1 > v_1$, which implies $v_2 > v_3 > v_1$.
Step 3 (Final Concluded Answer): [0.25 Mark]
The correct option is (B) $v_2 > v_3 > v_1$.
💡 EL10 Concept Insight: Think of the wire like a garden hose: if you squeeze the hose to make it thinner, the water has to speed up to get the same amount of water through the narrow opening in the same amount of time. Since the middle section is the thinnest, the electrons must move the fastest there!