There are 0.8 times 10^{23} free electrons / | vedclass

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(A) The formula for drift velocity is $I = neAv_d$, where $I$ is the current, $n$ is the number density of electrons, $e$ is the elementary charge, $A

Vedclass · Accepted Answer

$1.56 	imes 10^{-5} \, m/s$

Vedclass · Answer

(A) The formula for drift velocity is $I = neAv_d$, where $I$ is the current, $n$ is the number density of electrons, $e$ is the elementary charge, $A$ is the cross-sectional area, and $v_d$ is the drift velocity.Given values:$I = 0.2 \, A$$n = 0.8 	imes 10^{23} \, 	ext{electrons}/cm^3 = 0.8 	imes 10^{29} \, 	ext{electrons}/m^3$$e = 1.6 	imes 10^{-19} \, C$$A = 0.01 \, cm^2 = 10^{-6} \, m^2$Rearranging for $v_d$:$v_d = \frac{I}{neA}$$v_d = \frac{0.2}{(0.8 	imes 10^{29}) 	imes (1.6 	imes 10^{-19}) 	imes (10^{-6})}$$v_d = \frac{0.2}{1.28 	imes 10^4}$$v_d = 0.156 	imes 10^{-4} \, m/s = 1.56 	imes 10^{-5} \, m/s$.

There are $0.8 \times 10^{23}$ free electrons $/ cm^3$ in copper. If a $0.2 \, A$ current is flowing in a copper wire, then the drift velocity of electrons will be, given the cross-sectional area of the wire is $0.01 \, cm^2$.

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