Question

The lengths of two wires made of the same material are in the ratio 2:3 and their radii are in the ratio 1:2. If the two wires are connected in parallel to a battery, then the ratio of the drift velocities of free electrons in the two wires is

• A. 2:1
• B. 3:1
• C. 3:2
• D. 3:4

Hint:

In a parallel circuit, the voltage (V) is the same across components. The electric field E inside a uniform wire is \(E=V/L\). Since drift velocity is proportional to the electric field (\(v_d \propto E\)), it follows that \(v_d \propto V/L\). For a parallel connection, this simplifies to \(v_d \propto 1/L\).

Answer 1

The Correct Option is C

The relationship between drift velocity (\(v_d\)), current (I), and cross-sectional area (A) is \( I = n e A v_d \), where n is the number density of electrons and e is the electron charge.
From this, the drift velocity is \( v_d = \frac{I}{neA} \).
The resistance of a wire is given by \( R = \rho \frac{L}{A} \), where \(\rho\) is the resistivity, L is the length, and A is the area.
The area is \( A = \pi r^2 \). So, \( R = \rho \frac{L}{\pi r^2} \).
When the wires are connected in parallel to a battery, the potential difference V across both wires is the same.
The current in each wire is given by Ohm's law: \( I = \frac{V}{R} \).
Substitute the expression for I into the drift velocity formula:
\( v_d = \frac{V/R}{neA} = \frac{V}{neAR} = \frac{V}{ne(\pi r^2)(\rho L / \pi r^2)} = \frac{V}{ne\rho L} \).
This shows that for a given material (n, e, \(\rho\) are constant) and a given voltage V, the drift velocity is inversely proportional to the length of the wire: \( v_d \propto \frac{1}{L} \).
We are given the ratio of the lengths: \( L_1 : L_2 = 2:3 \).
The ratio of the drift velocities will be the inverse of the ratio of the lengths.
\( \frac{v_{d1}}{v_{d2}} = \frac{L_2}{L_1} = \frac{3}{2} \).
So, the ratio of the drift velocities is 3:2.
Note that the radii of the wires do not affect the drift velocity in this specific parallel connection scenario.