Question

How is $v_d$ affected when the length of the conductor is doubled, keeping the voltage applied across the conductor constant ?

Hint:

Always check what is kept constant in the problem. If the problem had said "keeping the current constant", then since $I = neAv_d$, changing length wouldn't affect $v_d$ as long as $A$ is unchanged! But here, Voltage is constant, which changes the Electric Field.

Answer 1

Concept:
• Drift velocity ($v_d$) is directly driven by the electric field ($E$) established inside the conductor.
• The relationship is $v_d = \frac{e E}{m} \tau$, where $\tau$ is the average relaxation time.
• The electric field is related to the applied voltage ($V$) and the length of the conductor ($L$) by the equation $E = \frac{V}{L}$.

Step 1: Express drift velocity in terms of voltage and length
Start with the fundamental equation for drift velocity:
\[ v_d = \frac{e E \tau}{m} \]
Substitute the expression for the uniform electric field $E = \frac{V}{L}$:
\[ v_d = \frac{e \left(\frac{V}{L}\right) \tau}{m} \]
\[ v_d = \frac{e V \tau}{m L} \]
This equation shows that for a constant applied voltage $V$ and a given material at a constant temperature (so $\tau$ is constant), the drift velocity is inversely proportional to the length of the conductor.
\[ v_d \propto \frac{1}{L} \]

Step 2: Analyze the effect of doubling the length
Let the initial length be $L$ and the initial drift velocity be $v_{d1}$.
When the length is doubled, the new length is $L' = 2L$.
The applied voltage $V$ remains constant.
The new drift velocity $v_{d2}$ will be:
\[ v_{d2} = \frac{e V \tau}{m (2L)} \]
\[ v_{d2} = \frac{1}{2} \left( \frac{e V \tau}{m L} \right) \]
\[ v_{d2} = \frac{v_{d1}}{2} \]

Step 3: Conclusion
When the length of the conductor is doubled while keeping the applied voltage constant, the internal electric field is halved. Consequently, the drift velocity of the electrons is also halved.