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$\begin{array}{l}{\text { A } 1.50 \text { -m cylindrical rod of diameter } 0.500 \mathrm{cm} \text { is con- }} \\ {\text { nected to a power supply that maintains a constant potential differ- }} \\ {\text { ence of } 15.0 \mathrm{V} \text { across its ends, while an ammeter measures the }} \\ {\text { current through it. You observe that at room temperature }\left(20.0^{\circ} \mathrm{C}\right)} \\ {\text { the ammeter reads } 18.5 \mathrm{A}, \text { while at } 92.0^{\circ} \mathrm{C} \text { it reads } 17.2 \mathrm{A} \text { . You can }} \\ {\text { ignore any thermal expansion of the rod. Find (a) the resistivity at }} \\ {20.0^{\circ} \mathrm{C} \text { and (b) the temperature coefficient of resistivity at } 20^{\circ} \mathrm{C} \text { for }} \\ {\text { the material of the rod. }}\end{array}$

* $R=\frac{V}{I}$$,\quad R=\frac{\rho L}{A}$$,\quad R=R_{0}\left[1+\alpha\left(T- T_{0}\right)\right]$

(a) $R=\frac{V}{I}=\frac{15}{18.5}=0.811 \Omega$

$R=\frac{\rho L}{A} \longrightarrow$$\rho=\frac{R A}{L}=$$\frac{R \pi\left(\frac{0}{2}\right)^{2}}{L}=\frac{0.811 * \pi*\left(\frac{0.005}{2}\right)^{2}}{1.5}$

$=1.06 * 10^{-5} \Omega \cdot m$

(b) $At \ 92^{\circ} c \rightarrow R=\frac{V}{I}=\frac{15}{17.2}$$=0.872 \Omega$

$R=R_{0}$$\left[1+\alpha\left(T-T_{0}\right)\right] \rightarrow$$0.872=(0.811)[1+\alpha(92-20)]$

$∴ \alpha=0.0105\left(c^{\circ}\right)^{-1}$

$\begin{array}{l}{\text { A copper wire has a square cross section } 2.3 \mathrm{mm} \text { on a }} \\ {\text { side. The wire is } 4.0 \mathrm{m} \text { long and carries a current of } 3.6 \mathrm{A} \text { . The }} \\ {\text { density of free electrons is } 8.5 \times 10^{28} / \mathrm{m}^{3} \text { . Find the magnitudes of }} \\ {\text { (a) the current density in the wire and (b) the electric field in the }} \\ {\text { wire. (c) How much time is required for an electron to travel the }} \\ {\text { length of the wire? }}\end{array}$

* $E = \rho.{J}$$,\quad J=\frac{I}{A}$$,\quad I=n |q| V_{d} A$

$\rho=1.72*10^{-8} \Omega \cdot m$

$n=8.5 * 10^{28} / \mathrm{m}^{3}$

(a) $J=\frac{I}{A}$$=\frac{3.6}{\left(2.3*10^{-3}\right)^{2}}=6.81*10^{5} \mathrm{A/m}^{2}$

(b) $E=\rho. J=1.72*10^{-8} * 6.81 * 10^{5}=0.012 \mathrm{V} / \mathrm{m}$

(c) $t=\frac{L}{V_{d}}=$$\frac{\ln |q| A}{I}=$$\frac{4*8.5 * 10^{28} * 1.6 *10^{-19} *\left(2.3710^{-3}\right)^{2}}{3.6}$

$∴ t =8 * 10^{4} \mathrm{5}= 1323 min=2.2 hrs!$

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