
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
503.4k+ views
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.
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
The probability that a leap year will have only 52 class 12 maths CBSE

Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE

And such too is the grandeur of the dooms We have imagined class 12 english CBSE

What does the god that failed refer to class 12 english CBSE

Which country did Danny Casey play for class 12 english CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE
