A copper wire of cross sectional area 10^-6m² and length 10 cm carriers a current of 1 A. If electron density of copper is 8.5 x 10²⁸m^-3 the average drift speed of an electron is

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$13 .6 \times 10^{- 3} {ms}^{- 1}$

$\frac{1}{13 .6} \times 10^{- 3} {ms}^{- 1}$

$13 .5 \times 10^{- 4} {ms}^{- 1}$

$\frac{1}{13 .6} \times 10^{- 4} {ms}^{- 1}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{13 .6} \times 10^{- 3} {ms}^{- 1} \\ A = 10^{- 6} m^{2} \\ 1 = 0 .1 m i = 1 A \\ n = 8 .5 \times 10^{28} m^{- 3} \end{array} We have i = {neAv}_{d} \Rightarrow v_{d} = \frac{i}{neA}$

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