Question

A copper wire of 3 $m{m^2}$ cross-sectional area carries a current of 5 ampere. The magnitude of the drift velocity for the electrons in the wire is:
(Assume copper to be monovalent, ${M_{Cu}} = 63.5kg/kmol$ and density of copper $ = 8920kg/{m^3}$)
$
  {\text{A}}{\text{. 0}}{\text{.24}}m/s \\
  {\text{B}}{\text{. 0}}{\text{.12}}m/s \\
  {\text{C}}{\text{. 2}}{\text{.4}}m/s \\
  {\text{D}}{\text{. 0}}{\text{.06}}m/s \\
 $

Answer 1

Hint: The current through a conductor in terms of the drift velocity of electrons is equal to the product of drift velocity, charge on the electrons, number density of charge carriers and the cross-sectional area. We have required quantities from which we can calculate the drift velocity using this expression.

Formula used: The current through a conducting wire can be defined in terms of drift velocity of electrons by the following expression:
$I = {V_d}enA$ …(i)
Here I is the magnitude of current flowing through a given conductor, ${V_d}$ denotes the drift velocity of the electrons in the given metal, e is the charge on an electron which is given as
$e = 1.6 \times {10^{ - 19}}C$
A represents the cross-sectional area of the given conductor and n is the number density of the electrons in the conductor.

Complete step by step answer:
We are given a copper wire which has cross-sectional area given as
$A = 3m{m^2} = 3 \times {10^{ - 6}}{m^2}$
A current of 5 amperes is flowing through this wire. Therefore, we have
$I = 5A$
We are given the molecular weight and density of copper as well from which we can calculate the number density of electrons in copper, which is equal to number of electrons in a mole divided by the volume of copper which is equal to mass of copper divided by its density.
$n = \dfrac{{\rho {N_A}}}{{{M_{Cu}}}}$
Here ${N_A}$ represents the Avogadro's number whose value is given as
${N_A} = 6.023 \times {10^{23}}$
Now we can calculate the magnitude of the drift velocity for the electrons in the wire by using the equation (i) in the following way.
${V_d} = \dfrac{I}{{enA}} = \dfrac{{I \times {M_{Cu}}}}{{e\rho {N_A}A}}$
Now we will insert all the known values in this expression, we get
$
  {V_d} = \dfrac{{5 \times 63.5}}{{1.6 \times {{10}^{ - 19}} \times 8920 \times 6.023 \times {{10}^{23}} \times 3 \times {{10}^{ - 6}}}} \\
   = 0.12m/s \\
$

So, the correct answer is “Option B”.

Note: 1. The drift velocity of electrons signifies the velocity with which the electrons are pushed towards the positive terminal of the battery when a potential difference is applied on the conductor.
2. The student can also insert all the units in the expression for drift velocity in order to check if we obtain the units of velocity.