Question

If a current flowing through a copper wire of $1\,mm$ diameter is $1.1\,A$ Calculate drift velocity assuming that each copper atom contributes $1$ free electron (density of $Cu = 9\,gmc{m^{ - 3}}$ and atomic weight $= 63$ )

Answer 1

In order to solve this question we need to understand drift velocity. As the name suggests drift velocity is average velocity of electrons with which they drift. Initially when the wire is not connected with battery then free electrons move randomly in with no prescribed direction but when wire is connected to potential difference then an electric field set up across wire forcing electrons to move in opposite direction of applied electric field, so electrons drift with some velocity known as drift velocity.

Complete step by step answer:
Drift velocity of wire is directly proportional to current and inversely proportional to number density of electrons, cross sectional area of wire and electronic charge.So it is mathematically expressed as,
${v_d} = \dfrac{I}{{neA}}$
Here, $I$ is current, $n$ is number density of electrons, $e$ is electronic charge and $A$ is cross sectional area

Given, $I = 1.1A$, Diameter $d = 1\,mm = {10^{ - 3}}m$.
So cross sectional area is, $A = \dfrac{{\pi {d^2}}}{4}$
Putting value we get, $A = \dfrac{{\pi \times ({{10}^{ - 6}})}}{4}$
$A = 0.78 \times {10^{ - 6}}{m^2}$
Electronic charge is, $e = 1.6 \times {10^{ - 19}}C$
Since Number of copper atom is,
${N_A} = 6.093 \times {10^{23}}$
Here, ${N_A}$ is Avogadro number
Since each atom contributes one electron, so total number of electrons is,
${N_1} = 6.093 \times {10^{23}}$

Mass of the copper atom is, $m = 63$ and density is $\rho = 9\,gmc{m^{ - 3}}$.
So using density formula, $\rho = \dfrac{m}{V}$
$V = \dfrac{m}{\rho }$
Putting values we get, $V = \dfrac{{63}}{9}c{m^3}$
$V = 7c{m^3}$
$\Rightarrow V = 7 \times {10^{ - 6}}{m^3}$
So number density is given by, $n = \dfrac{{{N_1}}}{V}$
Putting values we get,
$n = \dfrac{{6.093 \times {{10}^{23}}}}{{7 \times {{10}^{ - 6}}}}$
$\Rightarrow n = 0.87 \times {10^{29}}$
Using values of $n,e,A\& I$ in ${v_d}$ us get,
${v_d} = \dfrac{{1.1}}{{(0.87 \times {{10}^{29}} \times 1.6 \times {{10}^{ - 19}} \times 0.78 \times {{10}^{ - 6}})}}$
$\therefore {v_d} = 0.1\,mm\,{\sec ^{ - 1}}$

Note: It should be remembered that drift velocity is due to electrons in semiconductor too but there electrons behave as minority charge carrier in p type semiconductor and holes are in minority in n type semiconductor so they drift in presence of electric field or when diode is forward biased. Also drift velocity is directly proportional to Electric field applied and proportionality constant is mobility.