Q. 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 = $1.6 \times 10^{-19}C$)

Solution:

$I = neAv_{d} $
$\Rightarrow v_{d} = \frac{I}{neA} = \frac{1.5}{9 \times10^{28} \times1.6 \times 10^{-19} \times 5\times 10^{-6}} $
= 0.02 m/s

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