Question

The magnitude of current density in a
copper wire is 500 A/cm². If the number
of free electrons per cm3 of copper is
8.47x10^22 calculate the drift velocity of
the electrons through the copper wire
(charge on an e = 1.6x10-19 C)​

Answer 1

Answer:

CURRENT ELECTRICITY

Explanation:

★ DRIFT VELOCITY ( Vd) ★

Given ;

★ Current Density ( j ) = 500 A/cm³

★ N ( no. of free electrons ) = 8.47 × 10²²

★ e ( charge of electron ) = 1.6 × 10^-19 C

«» BY USING FORMULA ;

$current \: density \: (j) \: = n \times e \: \times vd$

«★» j = N × e × Vd

» 500 = 8.47 × 10²² × 1.6 × 10^-19 × Vd

» 500 = 8470 × 10^19 × 1.6 × 10^-19 × Vd

» 500 = 847 × 16 × Vd

» Vd = 500 / 13,552

«★★» Vd = 0.0368 m/s

ANSWER :- DRIFT VELOCITY OF ELECTRONS OF COPPER WIRE IS 0.0368 m/s.

Answer 2

» The magnitude of current density in a

copper wire is 500 A/cm².

Here ..

• Current density (J) = 500 A/cm²

» The number of free electrons per cm³ of copper is 8.47x10²².

• Number of free electrons (n) = 8.47 × 10²²

• Charge on electron (e) = 1.6 × ${10}^{-19}$

_______________ [ GIVEN ]

• We have to find the drift velocity $(v_{d})$.

_____________________________

Current density : It is defined as the ratio of magnitude of current per unit area of cross-section.

i.e.

J = $\dfrac{I}{A}$

J = $\dfrac{v _{d}enA }{A}$

=> J = $v_{d}en$

Put the known values in above formula.

→ 500 = $v_{d}$ × 1.6 × ${10}^{-19}$ × 8.47 × 10²².

→ 500 = $v_{d}$ × 13.552 × 10³

→ $v_{d}$ = $\dfrac{500}{13.552}$ × 10-³

→ $v_{d}$ = 36.89492 × 10-³

→ $v_{d}$ = 0.03689 m/s

_____________________________

Drift velocity is 0.03689 m/s.

__________ [ ANSWER ]

_____________________________