The number density of free electrons in a cop | vedclass

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Question

(B) The relationship between current $I$ and drift velocity $v_d$ is given by $I = n A v_d e$,where $n$ is the number density,$A$ is the cross-section

Vedclass · Accepted Answer

$5.4 	imes 10^4 \ s$

Vedclass · Answer

(B) The relationship between current $I$ and drift velocity $v_d$ is given by $I = n A v_d e$,where $n$ is the number density,$A$ is the cross-sectional area,and $e$ is the elementary charge.Since $v_d = \frac{l}{t}$,where $l$ is the length of the wire and $t$ is the time taken,we can write:$I = n A \left( \frac{l}{t} ight) e$Rearranging for $t$:$t = \frac{n A l e}{I}$Given values:$n = 8.5 	imes 10^{28} \ m^{-3}$$A = 1.0 	imes 10^{-6} \ m^2$$l = 6 \ m$$e = 1.6 	imes 10^{-19} \ C$$I = 1.5 \ A$Substituting these values:$t = \frac{8.5 	imes 10^{28} 	imes 1.0 	imes 10^{-6} 	imes 6 	imes 1.6 	imes 10^{-19}}{1.5}$$t = \frac{81.6 	imes 10^3}{1.5} = 54.4 	imes 10^3 \ s$$t = 5.44 	imes 10^4 \ s$Thus,the correct option is $B$.

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