Question

In a copper wire that has a diameter of 1.63 mm, the drift velocity is 7.08x10^-4 m/s. Assume that there is one free electron for conduction per copper atom. Copper has an atomic mass of 1.055x10^-25 kg and a density of about 9 g/cm^3.

Required:
a. What is the current in the wire?
b. What is the current density?

Answer 1

(a). The current in the wire is 20.16 A.

(b). The current density is 9.66 A/mm².

Step-by-step explanation:

Given that,

Diameter = 1.63 mm

Drift velocity
`v_(d)=7.08*10^(-4)\ m/s`

Atomic mass of copper
`m=1.055*10^(-25)\ kg`

Atomic weight of copper =63.5 g

Density = 9 g/cm³

We need to calculate the number density of electron

Using formula of number density

`n=(N*\rho)/(w)`

Where, w= atomic weight

Put the value into the formula

`n=(6.023*10^(23)*9*10^(6))/(63.5)`

`n=0.853*10^(29)\ electron/m^3`

(a).We need to calculate the current in the wire

Using formula of drift velocity

`v_(d)=(I)/(nqA)`

`I=v_(d)nqA`

Where, A = cross section area

q = charge of electron

n = number density of electron

`v_(d)` = drift velocity

Put the value into the formula

`I=7.08*10^(-4)*0.853*10^(29)*1.6*10^(-19)*\pi*((1.63*10^(-3))/(2))^2`

`I=20.16\ A`

(b). We need to calculate the current density

Using formula of current density

`J=nqv_(d)`

Put the value into the formula

`J=0.853*10^(29)*1.6*10^(-19)*7.08*10^(-4)`

`J=9662784\ A/m^2`

`J=9.66\ A/mm^2`

Hence, (a). The current in the wire is 20.16 A.

(b). The current density is 9.66 A/mm².