A) \[8.1\times {{10}^{4}}s\]
B) \[2.7\times {{10}^{4}}s\]
C) \[9\times {{10}^{3}}s\]
D) \[3\times {{10}^{3}}s\]
Correct Answer: B
Solution :
(b) \[2.7\times {{10}^{4}}s\] Here, number density of free electrons, \[n=8.5\times {{10}^{28}}{{m}^{-3}}\] Area of cross-section of a wire, \[A=2.0\times {{10}^{-6}}{{m}^{2}}\] Length of wire, \[l=3.0\text{ }m\] Current, \[l~=3.0\text{ }A\] The drift velocity of an electron is \[{{v}_{d}}=\frac{l}{neA}\] ..(1) The time taken by the electron to drift from one end to other end of the wire is \[t=\frac{l}{{{v}_{d}}}=\frac{\ln eA}{l}\] [Using eq.(1)] \[=\frac{(3.0m)(8.5\times {{10}^{28}}{{m}^{-3}})(1.6\times {{10}^{-19}}C)(2.0\times {{10}^{-6}}{{m}^{2}})}{(3.0A)}\]\[=2.7\times {{10}^{4}}s\]
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