J & K CET Engineering J and K - CET Engineering Solved Paper-2003

  • question_answer
    Current is flowing with a current density  \[J=480\,A/c{{m}^{2}}\]in a copper wire. Assuming that each copper atom contributes one free electron and given that Avogadro number \[=6.0\times {{10}^{23}}\text{ }atoms/mol\] Density of copper \[=9.0\text{ }g/c{{m}^{3}}\] Atomic weight of copper \[=64\text{ }g/mol\] Electronic charge \[=1.6\times {{10}^{-19}}C\] The drift velocity of electrons is

    A)  \[1\text{ }mm/s\]

    B)  \[\text{2 }mm/s\]

    C)  \[0.5\text{ }mm/s\]     

    D)  \[0.36\text{ }mm/s\]

    Correct Answer: D

    Solution :

    Drift velocity is the average velocity that a particle such as an electron, attains due to an electric field. It is given by \[{{v}_{d}}=\frac{I}{nqA}\] where I is current, n the number of electrons, A the area, q the charge. Given, \[\frac{I}{A}=\frac{480A}{c{{m}^{2}}},\,\,\,q=1.6\times {{10}^{-19}}C\] \[n=\frac{6\times {{10}^{23}}\times 9}{64}\] \[\therefore \] \[{{v}_{d}}=480\times \frac{64}{6\times {{10}^{23}}\times 9\times 1.6\times {{10}^{-19}}}\] \[\Rightarrow \] \[{{v}_{d}}=\frac{480\times 64}{6\times 9\times 1.6\times 10000}cm/s\] \[\Rightarrow \] \[{{v}_{d}}=\frac{32}{900}cm/s\] \[=\frac{32\times 10}{900}mm/s\] \[=0.36mm/s\] \[\Rightarrow \] \[{{v}_{d}}=0.36mm/s\]


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