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Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross-sectional area ( $5 m m^{2}$ ) is $v .$ If the electron density of copper is $9 \times 10^{28} / m^{3}$ the value of $v$ in $m m / s$ is close to (charge of electron to be $= 1.6 \times 10^{- 19} C$ )

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0.02

0.2

answer is D.

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Detailed Solution

As $l = n e A v_{d}$
$v_{d} = 1 / n e A = 1.5 / (9 \times 10^{28} \times 1.6 \times 10^{- 19} \times 5 \times 10^{- 6})$
$v = 0.02 \times 10^{- 3} m / s = 0.02 m m / s$

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