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irina [24]
3 years ago
10

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 electro

ns in a copper wire of radius 0.625 mm carrying a current of 3 A?
Physics
2 answers:
viktelen [127]3 years ago
4 0

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

Kruka [31]3 years ago
4 0

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

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