Question

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electrons in a copper wire of radius 0.625 mm carrying a current of 3 A?

Answer 1

The drift velocity of electrons in a copper wire is
`1.756*10^(-4)\ m/s`

Step-by-step explanation:

Given that,

Density
`\rho=8.93\ g/cm^3`

Mass
`M=63.5\ g/mol`

Radius = 0.625 mm

Current = 3 A

We need to calculate the drift velocity

Using formula of drift velocity

`v_(d)=(J)/(ne)`

Where, n = number of electron

j = current density

We need to calculate the current density

Using formula of current density

`J=(I)/(\pi r^2)`

`J=(3)/(3.14*(0.625*10^(-3))^2)`

`J=2.45*10^(6)\ A/m^2`

Now, we calculate the number of electron

Using formula of number of electron

`n=(\rho)/(M)N_(A)`

`n=(8.93*10^(6))/(63.5)*6.2*10^(23)`

`n=8.719*10^(28)\ electron/m^3`

Now put the value of n and current density into the formula of drift velocity

`v_(d)=(2.45*10^(6))/(8.719*10^(28)*1.6*10^(-19))`

`v_(d)=1.756*10^(-4)\ m/s`

Hence, The drift velocity of electrons in a copper wire is
`1.756*10^(-4)\ m/s`

Answer 2

`V_d` = 1.75 × 10⁻⁴ m/s

Step-by-step explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire,
`A = (\pi d^2)/(4)` =
`A = (\pi 0.625^2)/(4)`

Now,

The current density, J is given as

`J=(I)/(A)=(3)/( (\pi 0.625^2)/(4))`= 2444619.925 A/mm²

now, the electron density,
`n = (\rho)/(M)N_A`

where,

`N_A`=Avogadro's Number

`n = (8.93)/(63.5)(6.2* 10^(23))=8.719* 10^(28)\ electrons/m^3`

Now,

the drift velocity,
`V_d`

`V_d=(J)/(ne)`

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

`V_d=(2444619.925)/(8.719* 10^(28)* (1.6* 10^(-19))e)` = 1.75 × 10⁻⁴ m/s