Question

An electric current of 16 A exists in a metal wire of cross-section ${{10}^{-6}}{{m}^{2}}$ and length 1 m. Assuming one free electron per atom. The drift speed of free electrons in the wire will be:
(Density of metal = \[5\times {{10}^{3}}kg{{m}^{-3}}\], atomic weight = 60)
A. $5\times {{10}^{-3}}m{{s}^{-1}}$
B. $2\times {{10}^{-3}}m{{s}^{-1}}$
C. $4\times {{10}^{-3}}m{{s}^{-1}}$
D. $7.5\times {{10}^{-3}}m{{s}^{-1}}$

Answer 1

Hint: First we will find density of free electrons in the given material using the density and the molecular mass of the material which have been provided. Then we will use it and other quantities like the charge on the electron and the cross-sectional area of the wire to find the drift velocity of the electrons in the wire using the expression for current in the wire which gives the dependence of current on the drift velocity.
Formula used:
Current, $I=nqA{{v}_{d}}$

Complete answer:
The density of the metal is given as $5\times {{10}^{3}}kg{{m}^{-3}}=5\times {{10}^{6}}g{{m}^{-3}}$. Each of the atoms of the metal are given to have an atomic weight of 60 amu. If we take the ratio of density and the mass of each atom, we will get the number density of free electrons in the material.
$n=\dfrac{5\times {{10}^{6}}}{\left( \dfrac{60}{6.022\times {{10}^{23}}} \right)}=\dfrac{5\times {{10}^{6}}\times 6.022\times {{10}^{23}}}{60}=5.018\times {{10}^{28}}e{{m}^{-3}}$
We get $6.022\times {{10}^{23}}$electrons per unit volume of the material. Now according to the formula, we will get the following expression for drift velocity
$I=nqA{{v}_{d}}\Rightarrow {{v}_{d}}=\dfrac{I}{nqA}$
Putting in the values for the formula we get the drift velocity as
${{v}_{d}}=\dfrac{I}{nqA}=\dfrac{16}{5.018\times {{10}^{28}}\times 1.6\times {{10}^{-19}}\times {{10}^{-6}}}=2\times {{10}^{-3}}m{{s}^{-1}}$
We get the drift velocity of free electron in the given material that would give a current of 16 Ampere to be $2\times {{10}^{-3}}m{{s}^{-1}}$.

So, the correct answer is “Option B”.

Note:
Although the drift velocity of electrons is very low, the speed with which the current spreads through the wire is very fast and is almost instantaneous. Also take care that here we took the number density of free electrons in the material as take the ratio of density and the mass of each atom, we have to take the product of the number of free electrons provided by one atom with that ratio. As the number of free electrons provided by one atom is given to be one, we have foregone that in this question.