A conductor of length 'L' is connected to a battery of emf 'E'. The drift velocity of electron in the conductor is '\( v_{d} \)'. The drift velocity of electron in the conductor on stretching it four times, to length 4 L keeping emf 'E' constant will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.25 v_{d} \)
The drift velocity is set by the electric field inside the conductor and the relaxation time \( \tau \) of the electrons.
\( \displaystyle v_{d} = \frac{e E_{f} \tau}{m} \)
With a battery of emf \( E \) across a conductor of length \( L \), the field inside it is \( E_{f} = E/L \). So
\( \displaystyle v_{d} = \frac{e E \tau}{m L} \;\Rightarrow\; v_{d} \propto \frac{1}{L} \)
Stretching the wire does not change the material, so \( \tau \) is the same. With the same emf across the length \( 4L \), the field is one quarter of its old value.
\( \displaystyle v_{d}' = \frac{v_{d}}{4} = 0.25\,v_{d} \)