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Punjab Medical Punjab - MET Solved Paper-2001

  • question_answer
    There is a current of \[40\] ampere in a wire of \[{{10}^{-6}}{{m}^{2}}\] area of cross-section. If the number of free electron per \[{{m}^{3}}\] is\[{{10}^{29}}\], then the drift velocity will be

    A) \[1.25\times {{10}^{3}}\,\,m/s\]               

    B) \[2.50\times {{10}^{-3}}\,\,m/s\]

    C) \[25.0\times {{10}^{-3}}\,\,m/s\]             

    D) \[250\times {{10}^{-3}}\,\,m/s\]

    Correct Answer: B

    Solution :

    The relation between drift velocity \[{{v}_{d}}\] and current \[i\] are related as given by \[i=A\,\,ne\,\,{{v}_{d}}\],            So,          \[{{v}_{d}}=\frac{i}{A\,\,ne}\]                 \[{{v}_{d}}=\frac{40}{{{10}^{-6}}\times 1.6\times {{10}^{-19}}\times {{10}^{29}}}\]                 \[=\frac{40}{16}\times {{10}^{-3}}\,\,m/s\]                 \[=2.5\times {{10}^{-3}}\,\,m/s\]


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