Question

The mass density of copper is 8.95×10^3 kg m^−3 . If one charge carrier is contributed
by each copper atom, calculate the number density of charge carriers in copper. If
5A current is flowing in a copper wire of cross-sectional area 4×10^−6 m^2 , calculate
the drift velocity of electrons.

Answer 1

Answer on Question #74045 Physics / Electric Circuits

The mass density of copper is $\rho = 8.95 \times 10^{3} \, \mathrm{kg} \, \mathrm{m}^{-3}$. If one charge carrier is contributed by each copper atom, calculate the number density of charge carriers in copper. If $I = 5 \, \mathrm{A}$ current is flowing in a copper wire of cross-sectional area $A = 4 \times 10^{-6} \, \mathrm{m}^{2}$, calculate the drift velocity of electrons.

Solution:

The number density of charge carriers

$$n = \frac{\rho}{m_0}$$

where $m_0 = \frac{\mu}{N_A} = \frac{0.063}{6.02 \times 10^{23}} = 1.05 \times 10^{-25} \, \mathrm{kg}$ is the mass of copper atom.

So

$$n = \frac{8.95 \times 10^{3}}{1.05 \times 10^{-25}} = 8.55 \times 10^{28} \, \mathrm{1/m^3}$$

The electrical current

$$I = envA$$

So, the drift velocity of electrons

$$v = \frac{I}{enA} = \frac{5}{1.6 \times 10^{-19} \times 8.55 \times 10^{28} \times 4 \times 10^{-6}} = 9.14 \times 10^{-5} \, \mathrm{m/s}$$

Answers:

$$n = 8.55 \times 10^{28} \, \mathrm{1/m^3}$$ $$v = 9.14 \times 10^{-5} \, \mathrm{m/s}$$

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