Question

A current of 1.0 A
exist in copper wire of crossection 1.0mm^2 assuming one free electron per atom molecule
the drift spred of the electrons in the wire the
density of copper is 9000 km/m^3​

Answer 1

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 :

$\boxed{\sf Current\; related\; to\; drift\; velocity\; = n\times e \times A\times vd}}$

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.

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

Answer 2

Answer:

:Solution :

Refer this photos .

For showing full answer click on photo .