A current of 16A is made to pass through a conductor in which the number density of free electrons is $4 \times 10^{28} m^{- 3}$ and is area of cross section is $10^{- 5} m^{2}$ . The average velocity of free electrons in the conductor is

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$1.6 \times 10^{- 4} {ms}^{- 1}$

$3.2 \times 10^{- 4} {ms}^{- 1}$

$6.4 \times 10^{- 4} {ms}^{- 1}$

$2.5 \times 10^{- 4} {ms}^{- 1}$

answer is B.

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Detailed Solution

$v_{d} = \frac{i}{neA}$

$= \frac{16}{4 \times 10^{28} \times 1.6 \times 10^{- 19} \times 10^{- 5}}$

$= \frac{16}{4 \times 1.6 \times 10^{4}}$

$= \frac{16}{4 \times 16 \times 10^{3}}$

$= \frac{1}{4 \times 10^{3}}$

$= 2.5 \times 10^{- 4} m / s$

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