Question

Derive the expression for drift velocity of carriers in a conductor. A conductor of length L is connected to a d.c. . The source of emf ‘E’. If the length of the conductor is doubled, keeping ‘E’ constant, explain how its drift velocity would be affected.

Answer 1

Hint: Drift velocity is the average velocity with which the electrons get drifted towards the positive terminal of the conductor in an electric field.

Complete answer:
At any instant of time, the velocity acquired by the electrons with thermal velocity \[{{u}_{1}}\]and acceleration \[a\] is:
\[{{v}_{1}}={{u}_{1}}+a{{\tau }_{1}}\]
Average velocity \[{{v}_{d}}\] of all the electrons in the conductor would be the sum of averages of each term. The sum of averages of the thermal velocities of \[n\] electrons in the conductor is 0.
Hence,
\[{{v}_{d}}=0+a{{\tau }_{{}}}\]
Here
\[\tau =\dfrac{{{\tau }_{1}}+{{\tau }_{2}}+.....+{{\tau }_{n}}}{n}\]
The acceleration of an electron placed in an electric field is,
\[a=\dfrac{F}{m}=-\dfrac{eE}{M}\]
Substituting this in the formula of \[{{v}_{d}}\], we get
\[{{v}_{d}}=-\dfrac{eE}{M}\tau \]
The negative sign shows that the drift velocity is in the opposite direction of the applied electric field.
For second part of the question, we know that the drift velocity is inversely proportional to the distance of the conductor
\[{{v}_{d}}\propto \dfrac{1}{l}\]
Therefore the correct answer is when the length of the conductor is doubled the drift velocity becomes one by two(½ of l).

Note:
Drift velocity is of the order of \[{{10}^{-4}}m{{s}^{-1}}\] and the average relaxation time is of the order of \[{{10}^{-14}}s\].