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Question : (a) The 12-gauge copper wire in a typical residential buildi...

(a) The 12-gauge copper wire in a typical residential building has a cross-sectional area of 3.31 × 10^—6 m^2. If it carries a current of 10.0 A, each copper atom contributes one free electron to the current. The density of copper is 8.95 g/cm^3. v_{d} =\frac{I}{n q A} =\frac{10.0 \mathrm{C} / \mathrm{s}}{\left(8.49 \times 10^{28} \mathrm{~m}^{-3}\right)\left(1.60 \times 10^{-19} \mathrm{C}\right)\left(3.31 \times 10^{-6} \mathrm{~m}^{2}\right)} =2.22 \times 10^{-4} \mathrm{~m} / \mathrm{s}  and using  the classical model of electron conduction, estimate the average time between collisions for electrons in household copper wiring

 

 

 

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From Equation 27.17, we see that

\tau=\frac{m_{e}}{n q^{2} \rho}

where \rho=1.7 \times 10^{-8} \Omega \cdot \mathrm{m} for copper and the carrier density is n=8.49 \times 10^{28} electrons / \mathrm{m}^{3} for the wire described in Example 27.1. Substitution of these values into the expression above gives

\tau=\frac{\left(9.11 \times 10^{-31} \mathrm{~kg}\right)}{\left(8.49 \times 10^{28} \mathrm{~m}^{-3}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}\left(1.7 \times 10^{-8} \Omega \cdot \mathrm{m}\right)} =2.5 \times 10^{-14} \mathrm{~s}

 

(b) Assuming that the average speed for free electrons in copper is 1.6 \times 10^{6} \mathrm{~m} / \mathrm{s} and using the result from part (a), calculate the mean free path for electrons in copper.

\ell =\bar{v} \tau=\left(1.6 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)\left(2.5 \times 10^{-14} \mathrm{~s}\right) =4.0 \times 10^{-8} \mathrm{~m}

which is equivalent to 40 \mathrm{nm} (compared with atomic spacings of about 0.2 \mathrm{nm} ). Thus, although the time between collisions is very short, an electron in the wire travels about 200 atomic spacings between collisions.