Question

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0 \times {10^{ - 7}}{m^2}$ carrying a current of $1.5A$ . Assume the density of conduction electrons to be $9 \times {10^{28}}{m^{ - 3}}$ .

Answer 1

Hint: The current due to the drift movement of electrons inside an electrically stressed conductor is known as drift current. The drift velocity electrons is mathematically expressed as: $u = \mu E$
      where $E$ is external electric field
                   $u$ is drift velocity
                   $\mu $ is mobility of electron

Formula used:
${v_d} = \dfrac{I}{{neA}}$
where ${v_d}$ : drift velocity
              $I$ : current through the wire
              $n$ : density of conduction electrons
              $e$ : charge of electron
             $A$ : cross-sectional area

Complete step by step answer:
The free electron gets a directional flow when external electric force is applied, initially electrons inside a conductor were in random movement but in the presence of an external electric field, the free electrons move to higher potential by random movement. The current developed due to this drift movement of electrons under the influence of an external field is called drift current and the velocity is termed as drift velocity.

Given data from the above question are:
Cross sectional area $A = 1.0 \times {10^{ - 7}}{m^2}$
Density of conduction electrons, $n = 9 \times {10^{28}}{m^{ - 3}}$
Current in the copper wire $I = 1.5A$
The drift velocity is calculated as follows:
${v_d} = \dfrac{I}{{neA}}$
Substituting the given values, we get the drift velocity as
${v_d} = \dfrac{{1.5}}{{9 \times {{10}^{28}} \times e \times 1.0 \times {{10}^{ - 7}}}}$
Where $e = 1.6 \times {10^{ - 19}}C$ is the charge of electrons.
$ \Rightarrow {v_d} = \dfrac{{1.5}}{{9 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 1.0 \times {{10}^{ - 7}}}}$
$ \Rightarrow {v_d} = 0.00104m/s$
$ \Rightarrow {v_d} = 1.04 \times {10^{ - 3}}m/s$

Therefore the required drift velocity is $0.00104{m/s}$ or $1.04 \times {10^{ - 3}}{m/s}$. Drift velocity is usually small as this velocity is due to a shift in current which accelerates the electron to travel with a small velocity.

Note: Always ensure that all quantities are in SI units during the calculations. Drift velocity is the average velocity attained due to electric field in a charged particle. The S.I unit of drift velocity is $m/s$. The drift velocity of electrons is small in comparison to the average electron speed associated with its internal energy.