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The number of density of free electron in a copper conductor estimated at $8.5 \times 10^{20}m^{-3}$. How long does an electron take a drift from one end of a wire $3.0 \; m$ long to its other end ? The area of cross - section of the wire is $2 \times 10^{-6}\;m^2$ and it is carrying of $3.0 \;A$

$\begin{array}{1 1} 7\;hours\;and\;33\;minutes \\ 6\;hours\;and\;33\;minutes \\ 8\;hours\;and\;30\;minutes \\ 5\;hours\;and\;30\;minutes \end{array}$

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A)
Solution :
Calculate the drift velocity $V_d= \large\frac{I}{neA}$
Given $n= 85 \times 10^{28}/m^3$
$A= 2 \times 10^{-6}\;m^{2}$
$I= 3A$
$e= 1.6 \times 10^{-7} \;C$
Time taken by electron to drift from one end to continue of the wire,
$t=\large\frac{\text{length of the wire}}{\text{drift velocity}}$
$\quad= \large\frac{l}{V_d}$
Use the relation ,
$I= ne AV_d$
=> $V_d= \large\frac{I}{n_eA}$
$t=\large\frac{l n_e A}{I}$
substituting the values we get,
$t= \large\frac{ln_eA}{I}$
$\quad= \large\frac{3 \times 8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}{3}$
$\therefore t= 2.72 \times 10^4\;S$
$\qquad=$7 hours and minute
The time taken by an electron to drift from one end to another end is 7 hours and 33 mins