Question

A current of $1.34\text{A}$ exists in a copper wire of cross-section $1{\text{mm}}^2$. Assuming that each copper atom contributes one free electron, the drift speed of the free electrons in the wire will be:
$($Density of copper is $8990{\text{kg/m}}^3$, atomic mass is $63.50$ and charge on electron is $1.6\times {10}^{-19}\text{C})$

• A. $0.3\text{mm/s}$
• B. $0.5\text{mm/s}$
• C. $0.1\text{mm/s}$
• D. $0.05\text{mm/s}$

Answer 1

The correct option is C $0.1\text{mm/s}$

The density of copper is $8990{\text{kg/m}}^3$.

Mass of $1{\text{m}}^3$ volume of the copper is $8990\text{kg}$
$i.e.m=8990\times {10}^3\text{gm}$

Number of moles of copper in $1{\text{m}}^3$ is

$n^{\prime }=\frac{8990\times {10}^3\text{g}}{63.5\text{g/mole}}=1.4\times {10}^5\text{moles}$

since each mole will contain $N_A$ atoms,
where $N_A=6.023\times {10}^{23}$

The number of atoms in $1{\text{m}}^3$ copper will be,

$n=1.4\times {10}^5\times 6.023\times {10}^{23}$

$\Rightarrow n\approx 8.4\times {10}^{28}$

Using current drift velocity relation,
$I=neA{v}_d$

$\Rightarrow v_d=\frac{I}{neA}$

$\Rightarrow v_d=\frac{1.34}{8.4\times {10}^{28}\times 1.6\times {10}^{-19}\times (1\times {\left({10}^{-3}\right)}^2)}$

$\Rightarrow v_d=\frac{1.34}{8.4\times 1.6\times {10}^3}=0.997\times {10}^{-4}$

$\therefore v_d\approx {10}^{-4}\text{m/s}=0.1\text{mm/s}$

Hence, option $\left(c\right)$ is correct.

$$\begin{array}{c}\text{Why this question ?}\\ \text{Tip :In the question it has been given that}\\ \text{1 electron is contributed per atom, this gives the}\\ \text{hint to find the number of atoms in given}\\ \text{volume of copper using mole-concept.}\end{array}$$