Question

Two copper wires having their radii in the ratio of 3 : 2 are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.

Hint:

In series connection, drift velocity is inversely proportional to cross-sectional area ($v_d \propto \frac{1}{A} \propto \frac{1}{r^2}$). Thinner wire has higher drift velocity!

Answer 1

Concept:
• When components are connected in series, the same current $I$ flows through each of them.

• Current expression: $I = n e A v_d = n e (\pi r^2) v_d$.

Step 1: Given Data
Ratio of radii of two copper wires: $\frac{r_1}{r_2} = \frac{3}{2}$.
Since both wires are made of copper, electron number density $n$ is identical for both wires.
Since wires are connected in series across battery: $I_1 = I_2 = I$.

Step 2: Relating Drift Velocity to Radius
From current formula $I = n e A v_d$:
\[ I = n e (\pi r^2) v_d \]
Since $I$, $n$, $e$, and $\pi$ are constants:
\[ r_1^2 v_{d1} = r_2^2 v_{d2} \]
\[ \frac{v_{d1}}{v_{d2}} = \left( \frac{r_2}{r_1} \right)^2 \]

Step 3: Calculation
Substitute $\frac{r_1}{r_2} = \frac{3}{2} \implies \frac{r_2}{r_1} = \frac{2}{3}$:
\[ \frac{v_{d1}}{v_{d2}} = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \]

Step 4: Conclusion
The ratio of drift velocities of electrons in the two copper wires is $4 : 9$.