Question

The number density of free electrons in a copper conductor is 8.5 × 10 to the power 28 m-3 . The area of cross section of the wire is 2.0 × 10-6m2 and it is carrying a current of 3.0 A.Calculate drift velocity.(e = 1.6*10 to the power minus 19) *

Answer 1

Explanation:

Number density of electrons, n = 8.5 x 1028 m– 3 Current carried by the wire, I = 3.0 A Area of cross-section of the wire, A = 2.0 x 10–6 m2 ... Charge on electron, e = 1.6 x 10–19 C

Answer 2

Explanation:

$\Large{\green{\underline{\underline{\sf{\blue{Solution:}}}}}}$

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$\hookrightarrow$ Number density of free electrons in a copper conductor, $\sf n\:=\:8.5\times 10^{28}m^{-3}$

$\hookrightarrow$ Length of the copper wire, $\sf l\:=\:3.0\,m$

$\hookrightarrow$ Area of cross-section of the wire, $\sf A\:=\:2.0\times 10^{-6}m^2$

$\hookrightarrow$ Current carried by the wire, $\sf I\:=\:3.0\,A$

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Current carried by the wire is given by the relation,

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$\longrightarrow$ $\boxed{\tt{\red{I\:=\:nA\,e\,V_d}}}$

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Where,

$\mapsto$ $\sf e\:=\:1.6\times 10^{-19}C$ = Electric charge

$\mapsto$ $\sf V_d\:=\: \dfrac{Length\:of\:the\:wire\,(l)}{Time\:taken\:to\:cover\,l(t)}$

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$\leadsto$ $\sf I\:=\:nA\,e \dfrac{l}{t}$

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$\leadsto$ $\sf t\:=\: \dfrac{nA\,e\,l}{I}$

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$\sf t\:=\: \dfrac{\cancel{3}\times 8.5\times 10^{28}\times 2\times 10^{-6}\times 1.6\times 10^{-19}}{\cancel{3}}$

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$\leadsto$ $\sf t\:=\:2.7\times 10^4s$

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$\leadsto$ Hence, the time taken by an electron to drift from one end of the wire to the other is $\sf{\orange{\:2.7\times 10^4s}}$