Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm² is v . If the electron density in copper is $9 \times 10^{28} / m^{3}$ the value of v in mm/s is close to (take 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

$\begin{array}{l} I = n_{e} {Av}_{d} \\ V_{d} = \frac{I}{n_{e} A} = \frac{1 .5}{9 \times 10^{28} \times 1 .6 \times 10^{- 19} \times 5 \times 10^{- 6}} = 0 .02 \times 10^{- 3} m / s \\ \Rightarrow V_{d} = 0 .02 mm / s \end{array}$

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