CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    A conductor wire having \[{{10}^{29}}\] free electrons 1 \[{{m}^{3}}\] carries a current of 20A. If the cross-section of the wire is \[1\text{ }m{{m}^{2}}\], then the drift velocity of electrons will be \[(e=1.6\times {{10}^{-19}})\]

    A)  \[1.25\times {{10}^{-4}}m{{s}^{-1}}\]    

    B)  \[1.25\times {{10}^{-3}}m{{s}^{-1}}\]

    C)  \[1.25\times {{10}^{-5}}m{{s}^{-1}}\]       

    D)  \[6.25\times {{10}^{-3}}m{{s}^{-1}}\]

    Correct Answer: B

    Solution :

    Current, \[I=nAe{{v}_{d}}\] \[\therefore \] Drift velocity \[{{v}_{d}}=\frac{I}{nAe}\]                                 \[=\frac{20}{{{10}^{29}}\times {{10}^{-6}}\times 1.6\times {{10}^{-19}}}\]                 \[{{v}_{d}}=1.25\times {{10}^{-3}}m{{s}^{-1}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner