Question

A copper wire of radius 1.0 mm carries a current
of 10 A. Calculate the drift velocity of the electrons.
Metallic copper has one conduction electron per
atom, the atomic mass of copper is 64 u and density
of copper is 8900 kg m-3. Given lu= 1.66 x 10-27
kg.​

Answer 1

Answer:

DRIFT VELOCITY ( Vd ) = 4 × 10^-28 m / s.

Explanation:

★ CURRENT ELECTRICITY ★

★ GIVEN ;

» Current ( i ) = 10 Ampere

» Copper Conductor = 1 Free Electron

» Radius of Wire ( R ) = 1 mm

» Atomic Mass of Copper ( MM ) = 64 amu

» Density of COPPER ( p ) = 8900 Kg / m³

» 1 amu = 1.66 × 10^-27 Kg

★ No. of FREE ELECTRONS PER UNIT VOLUME ( n ) = ???

★ DRIFT VELOCITY ( Vd ) = ???

________[ BY USING FORMULA ]_______

[ $no. \: of \: free \: electrons \: (n) \: =( \frac{avogadro \: no. \: \times \: density}{atomic \: weight \: of \: metal \: } )$ ]

★ n = [ 6 × 10²³ × 8900 ] / 64 amu

» n = [ 6 × 10²³ × 8900 ] / 64 × 1.66 × 10^-27

» n = [ 534 × 10^25 ] / 106.25 × 10^-27

★» No. of Free Electrons = 5 × 10^52 «★

______________________________________

♒ SUBSTITUTION ABOVE VALUE IN GIVEN ; " FORMULA " -------- ♣

[ $current \: (i) \: = no. \: of \: free \: electrons \: (n) \: \times \: e \: \times \: area \:(a) \times \: drift \: velocity \: (vd)$ ]

★ i = n × e × π r² × Vd

» i = 5 × 10^52 × 1.6 × 10^-19 × π ×

» i = 5 × 10^52 × 1.6 × 10^-19 × π × ( 10-³ )² × Vd

» 10 = 8 × π × 10^³³ × 10-^6 × Vd

» 10 = 25.12 × 10^27 × Vd

» Vd = [ 10 / 25 × 10^27 ]

» Vd = [ 10 × 10^-27 / 25 ]

» Vd = [ 100 × 10^-28 / 25 ]

★ ⏩ VD = 4 × 10^-28 m/s ⏪ ★

________________________________________

ANSWER :- DRIFT VELOCITY OF ELECTRONS HAVING CURRENT 10 AMPERE IS ;

♐⬜ VD = 4 ❌ 10^-28 m/s ♐

Answer 2

Answer:

Answer:

DRIFT VELOCITY ( Vd ) = 4 × 10^-28 m / s.

Explanation:

★ CURRENT ELECTRICITY ★

★ GIVEN ;

» Current ( i ) = 10 Ampere

» Copper Conductor = 1 Free Electron

» Radius of Wire ( R ) = 1 mm

» Atomic Mass of Copper ( MM ) = 64 amu

» Density of COPPER ( p ) = 8900 Kg / m³

» 1 amu = 1.66 × 10^-27 Kg

★ No. of FREE ELECTRONS PER UNIT VOLUME ( n ) = ???

★ DRIFT VELOCITY ( Vd ) = ???

________[ BY USING FORMULA ]_______

[  ]

★ n = [ 6 × 10²³ × 8900 ] / 64 amu

» n = [ 6 × 10²³ × 8900 ] / 64 × 1.66 × 10^-27

» n = [ 534 × 10^25 ] / 106.25 × 10^-27

★» No. of Free Electrons = 5 × 10^52 «★

______________________________________

SUBSTITUTION ABOVE VALUE IN GIVEN ; " FORMULA "

★ i = n × e × π r² × Vd

» i = 5 × 10^52 × 1.6 × 10^-19 × π ×

» i = 5 × 10^52 × 1.6 × 10^-19 × π × ( 10-³ )² × Vd

» 10 = 8 × π × 10^³³ × 10-^6 × Vd

» 10 = 25.12 × 10^27 × Vd

» Vd = [ 10 / 25 × 10^27 ]

» Vd = [ 10 × 10^-27 / 25 ]

» Vd = [ 100 × 10^-28 / 25 ]

VD = 4 × 10^-28 m/s