Question

Current is flowing with a current density $J=480amp/{cm}^{2}$ in a copper wire. Assuming that each copper atom contributes one free electron and gives that Avogadro number $=6.0\times{10}^{23}$ atoms/mole density of copper $=9.0 g/{cm}^{3}$. Atomic weight of copper = 64 g/mole. Electric charge $=1.6\times{10}^{-19}$ coulomb. The drift velocity is:
A. 1mm/s
B. 2mm/s
C. 0.5mm/s
D. 0.36mm/s

Answer 1

Hint: The ultimate goal is to find the drift velocity, that can be found by the formula ${v}_{d}=\dfrac{J}{ne}$ meters per second, where J is the current density, n is the charge-carrier number density and e is the charge.
Here, we need to find the number of charge carriers $n$ and the drift velocity $J$ that in turn we can find using the formulae $J = \dfrac{I}{A}$ and $n={N}\times \dfrac{1}{m} \times \gamma$ respectively.

Formulae used:
1. Current density $J = \dfrac{I}{A}$ amperes per square meter, where I is current and A is the cross-sectional area.
2. Drift velocity ${v}_{d}=\dfrac{J}{ne}$ meters per second, where J is the current density, n is the charge-carrier number density and e is the charge.
3. Number of charge-carriers, $n={N}\times \dfrac{1}{m} \times \gamma$, where N is the Avogadro number, m is the atomic weight of element and $\gamma$ is the density of element.

Complete step-by-step answer:
We have been given that current density $J=480amp/{cm}^{2}$, Avogadro number $N=6.0\times{10}^{23}$ atoms/mole, density of copper $\gamma=9.0 g/{cm}^{3}$, atomic weight of copper m = 64 g/mole, electronic charge $e=1.6\times{10}^{-19}$ coulomb.
Now, we will find the number of charge carriers with the help of formula $n={N}\times \dfrac{1}{m} \times \gamma……(i)$
Also the current density can be written as $J = \dfrac{I}{A}……(ii)$
And now, if we use the equation (i) and (ii) in the drift velocity formula which is ${v}_{d}=\dfrac{J}{ne}$
We will get, ${v}_{d}=\dfrac{Jm}{Ne\gamma} \\
\implies{v}_{d}=\dfrac{480\times64}{6.0\times{10}^{23}\times1.6\times{10}^{-19}\times9}\\
\implies{v}_{d}=\dfrac{30720}{86.4\times{10}^{4}}\\
\implies{v}_{d}=355.5\times{10}^{-4} cm/s=0.36mm/s$
Hence, option d is the correct answer.

Additional Information:
Current can be defined as the flow of electrically charged particles in the electron-deficient atoms. The amount of flow of this current per unit cross-section area is called the current density and is expressed in amperes per square meter. These electrons attain an average velocity in a material when subjected to an electric field, that is called the drift velocity.

Note: One may get confused with the number of charge carriers and the Avogadro number and may get the wrong answer due to that. So, we must know that n is the number of charge-carriers while N is the Avogadro number.