Question

Estimate the average drift of conduction electrons in a copper wire of cross sectional area \[2.5 \times {10^{ - 7}}{m^2}\] carrying a current of \[1.8A\]. Assume the density of conduction electrons to be \[9 \times {10^{28}}{m^{ - 3}}\].

Answer 1

Hint: The drift velocity of a charged particle is the average velocity of the charged particle, such as electrons, within the material, due to an electric field. If an electron is propagating randomly at the Fermi velocity, average velocity will be equal to zero.

Complete step by step solution:
Let’s define all the data given in the question;
The cross sectional area of the copper wire= \[2.5 \times {10^{ - 7}}{m^2}\]
Current carrying= \[1.8A\]
Density of the conduction electron= \[9 \times {10^{28}}{m^{ - 3}}\]
We know, current in terms of drift speed and the area of cross section is given by;
$I = neA{v_d}$
Here, $n$ is the number of conduction electrons per unit volume.
$e$ is the charge of electron
${v_d}$ is the drift speed
$A$ is the area of cross section of the conducting copper wire
From the above equation, we are arranging the drift velocity on the left hand side of the equation and all the other terms on the right hand side of the equation we get,
Drift velocity, ${v_d} = \dfrac{I}{{neA}}$
Applying the known values from the question to the above equation we get,
Current, $I = 2.7A$
Number of conduction electrons per unit volume, $n = 9 \times {10^{28}}{m^{ - 3}}$
Charge of electron, $e = 1.6 \times {10^{ - 19}}$
Are of cross section,\[A = 2.5 \times {10^{ - 7}}{m^2}\]
We get
${v_d} = \dfrac{{2.7}}{{9 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 2.5 \times {{10}^{ - 7}}}}$
$ \Rightarrow {v_d} = 7.5 \times {10^{ - 4}}m/s$

That is the average drift of conduction electrons in a copper wire, ${v_d} = 7.5 \times {10^{ - 4}}m/s$.

Note: Free electrons gain velocity when a potential difference is applied across a conductor, and the direction will be opposite to the electric field between successive collisions, thus acquiring a velocity component in that direction in addition to its random thermal velocity. As a result, there exists a definite small drift velocity of electrons, and this drift velocity will be superimposed on the random motion of free electrons and causing a net flow of electrons in the direction opposite to the field.