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In silver, the number density of free electrons is n=5.8×n=5.8 \times 102810^{28} per cubic meter. (a)(a) Using the value of the resistivity given in Following table, calculate the average time between collisions of one of these electrons. (b)(b) Assuming that the electric field in a current-carrying silver wire is 8.0 V/m8.0 \mathrm{~V} / \mathrm{m}, calculate the average drift velocity of an electron.

 MATERIAL ρα Silver 1.6×108Ωm3.8×103/C Copper 1.7×1083.9×103 Aluminum 2.8×1083.9×103 Brass 7×1082×103 Nickel 7.8×1086×103 Iron 10×1085×103 Steel 11×1084×103 Constantan 49×1081×105 Nichrome 100×1084×104 At a temperature of 20C\begin{aligned} &\begin{array}{lcc} \text { MATERIAL } & \rho & \alpha \\ \hline \text { Silver } & 1.6 \times 10^{-8} \Omega \cdot \mathrm{m} & 3.8 \times 10^{-3} /{ }^{\circ} \mathrm{C} \\ \text { Copper } & 1.7 \times 10^{-8} & 3.9 \times 10^{-3} \\ \text { Aluminum } & 2.8 \times 10^{-8} & 3.9 \times 10^{-3} \\ \text { Brass } & \approx 7 \times 10^{-8} & 2 \times 10^{-3} \\ \text { Nickel } & 7.8 \times 10^{-8} & 6 \times 10^{-3} \\ \text { Iron } & 10 \times 10^{-8} & 5 \times 10^{-3} \\ \text { Steel } & \approx 11 \times 10^{-8} & 4 \times 10^{-3} \\ \text { Constantan } & 49 \times 10^{-8} & 1 \times 10^{-5} \\ \text { Nichrome } & 100 \times 10^{-8} & 4 \times 10^{-4} \end{array}\\ &\text { At a temperature of } 20^{\circ} \mathrm{C} \text {. } \end{aligned}

Question

The temperature coefficient of resistivity α\alpha is defined as the fractional increase of resistivity per degree Celsius, α=(1/ρ)dρ/dT\alpha=(1 / \rho) d \rho / d T. According to the following table, what is the value of α\alpha for copper?

 MATERIAL ρα Silver 1.6×108Ωm3.8×103/C Copper 1.7×1083.9×103 Aluminum 2.8×1083.9×103 Brass 7×1082×103 Nickel 7.8×1086×103 Iron 10×1085×103 Steel 11×1084×103 Constantan 49×1081×105 Nichrome 100×1084×104 At a temperature of 20C\begin{aligned} &\begin{array}{lcc} \text { MATERIAL } & \rho & \alpha \\ \hline \text { Silver } & 1.6 \times 10^{-8} \Omega \cdot \mathrm{m} & 3.8 \times 10^{-3} /{ }^{\circ} \mathrm{C} \\ \text { Copper } & 1.7 \times 10^{-8} & 3.9 \times 10^{-3} \\ \text { Aluminum } & 2.8 \times 10^{-8} & 3.9 \times 10^{-3} \\ \text { Brass } & \approx 7 \times 10^{-8} & 2 \times 10^{-3} \\ \text { Nickel } & 7.8 \times 10^{-8} & 6 \times 10^{-3} \\ \text { Iron } & 10 \times 10^{-8} & 5 \times 10^{-3} \\ \text { Steel } & \approx 11 \times 10^{-8} & 4 \times 10^{-3} \\ \text { Constantan } & 49 \times 10^{-8} & 1 \times 10^{-5} \\ \text { Nichrome } & 100 \times 10^{-8} & 4 \times 10^{-4} \end{array}\\ &\text { At a temperature of } 20^{\circ} \mathrm{C} \text {. } \end{aligned}

Solution

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We have the definition for temperature coefficient α=(1ρ)dρdT\alpha=\left(\dfrac{1}{\rho}\right)\dfrac{d\rho}{dT}

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