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

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

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, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$= 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$=Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

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

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

Answer 2

Answer:

The drift velocity of electrons in a copper wire is $1.756\times10^{-4}\ m/s$

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

$v_{d}=\dfrac{J}{ne}$

Where, n = number of electron

j = current density

We need to calculate the current density

Using formula of current density

$J=\dfrac{I}{\pi r^2}$

$J=\dfrac{3}{3.14\times(0.625\times10^{-3})^2}$

$J=2.45\times10^{6}\ A/m^2$

Now, we calculate the number of electron

Using formula of number of electron

$n=\dfrac{\rho}{M}N_{A}$

$n=\dfrac{8.93\times10^{6}}{63.5}\times6.2\times10^{23}$

$n=8.719\times10^{28}\ electron/m^3$

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

$v_{d}=\dfrac{2.45\times10^{6}}{8.719\times10^{28}\times1.6\times10^{-19}}$

$v_{d}=1.756\times10^{-4}\ m/s$

Hence, The drift velocity of electrons in a copper wire is $1.756\times10^{-4}\ m/s$