An electron takes 40 times 10^3 s to drift f | vedclass

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(D) The drift velocity $v_d$ is given by the ratio of length $L$ to the time $t$ taken to drift across the wire:$v_d = \frac{L}{t} = \frac{2 \ m}{40 \

Vedclass · Accepted Answer

$5 	imes 10^{28} \ m^{-3}$

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(D) The drift velocity $v_d$ is given by the ratio of length $L$ to the time $t$ taken to drift across the wire:$v_d = \frac{L}{t} = \frac{2 \ m}{40 	imes 10^3 \ s} = 0.5 	imes 10^{-4} \ m/s = 5 	imes 10^{-5} \ m/s$.The current $I$ in a conductor is related to the number density $n$ of free electrons by the formula:$I = n e A v_d$where $e = 1.6 	imes 10^{-19} \ C$ is the charge of an electron and $A$ is the cross-sectional area.Given $A = 4 \ mm^2 = 4 	imes 10^{-6} \ m^2$ and $I = 1.6 \ A$,we rearrange for $n$:$n = \frac{I}{e A v_d}$$n = \frac{1.6}{(1.6 	imes 10^{-19}) 	imes (4 	imes 10^{-6}) 	imes (5 	imes 10^{-5})}$$n = \frac{1.6}{1.6 	imes 10^{-19} 	imes 20 	imes 10^{-11}}$$n = \frac{1}{20 	imes 10^{-30}} = \frac{1}{2} 	imes 10^{29} = 5 	imes 10^{28} \ m^{-3}$.

An electron takes $40 \times 10^3 \ s$ to drift from one end of a metal wire of length $2 \ m$ to its other end. The area of cross-section of the wire is $4 \ mm^2$ and it is carrying a current of $1.6 \ A$. The number density of free electrons in the metal wire is

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