A current of 1.0 Aexist in copper wire of crossection 1.0mm^2 assuming

1.	A current of 1.0 Aexist in copper wire of crossection 1.0mm^2 assuming one free electron per atom moleculethe drift spred of the electrons in the wire thedensity of copper is 9000 km/m^3
Answer» Answer : $\sf 7.3 \times 10^{-5}\;m/s.$ EXPLANATION : GIVEN that : A current of 1.0 A exist in copper wire of CROSSECTION 1.0mm^2. We know that : $\boxed{\sf Current\; related\; to\; drift\; velocity\; = <klux>N</klux>\times e \times A\times vd}}$ Note : n is the number density of atoms. e is the charge on electron. A is area of cross section. Gram atomic weight of Copper atom 63.5. Number density of atoms : $\sf \implies n = \dfrac{6.0\times 10^{23} \times 9000}{6.35\times 10^{-3}} = 8.5 \times 10^{28}m^{-3}\\ \\\implies \|vd\| = \dfrac{I}{n \times e \times A} = \dfrac{1}{8.5 \times 10^{28} \times 1.062 \times 6^{-19} \times 10^{-6}} = 7.3 \times 10^{-5}\;m/s.$ $\rule{300}{1.5}$ ★ Extra Information : ★ Current : The flow of electricity CALLED current. The SI unit of current is Ampere (A). Current is denoted by the symbol I. $\star\;\boxed{\tt I = \dfrac{V}{R} }}$ ★ Speed : Speed is the rate that how fast or slow something is. Speed is measurment that how somthing fast\slow move from here to there. $\star\;\boxed{\tt Speed = \dfrac{Time}{Distance}}$

A current of 1.0 Aexist in copper wire of crossection 1.0mm^2 assuming one free electron per atom moleculethe drift spred of the electrons in the wire thedensity of copper is 9000 km/m^3

Answer»

Answer :

$\sf 7.3 \times 10^{-5}\;m/s.$

EXPLANATION :

GIVEN that :

A current of 1.0 A exist in copper wire of CROSSECTION 1.0mm^2.

We know that :

$\boxed{\sf Current\; related\; to\; drift\; velocity\; = <klux>N</klux>\times e \times A\times vd}}$

Note :

n is the number density of atoms.
e is the charge on electron.
A is area of cross section.

Gram atomic weight of Copper atom 63.5.

Number density of atoms :

$\sf \implies n = \dfrac{6.0\times 10^{23} \times 9000}{6.35\times 10^{-3}} = 8.5 \times 10^{28}m^{-3}\\ \\\implies |vd| = \dfrac{I}{n \times e \times A} = \dfrac{1}{8.5 \times 10^{28} \times 1.062 \times 6^{-19} \times 10^{-6}} = 7.3 \times 10^{-5}\;m/s.$

$\rule{300}{1.5}$

★ Extra Information :

★ Current :

The flow of electricity CALLED current. The SI unit of current is Ampere (A). Current is denoted by the symbol I.

$\star\;\boxed{\tt I = \dfrac{V}{R} }}$

★ Speed :

Speed is the rate that how fast or slow something is. Speed is measurment that how somthing fast\slow move from here to there.

$\star\;\boxed{\tt Speed = \dfrac{Time}{Distance}}$