Question 24.2: One end of the copper wire in example 1 is welded to one end......
One end of the copper wire in example 1 is welded to one end of an aluminum wire with a 4 mm diameter. The composite wire carries a steady current equal to that of Example 24.1 (i.e. I = 1 A). (a) What is the current density in each wire? (b) What is the value of the drift speed vd in the aluminum? [Aluminum has one free electron per atom and density 2.7 × 10³ kg/m³]
(a) Except near the junction, the current density in a copper wire of radius r_{Cu} = 1 mm and aluminum wire of radius r_{A1} = 2 mm are:
J_{\mathrm{Cu}}={\frac{I}{A_{\mathrm{Cu}}}}={\frac{I}{\pi r_{\mathrm{Cu}}^{2}}}={\frac{1\,\mathrm{A}}{\pi\times(10^{-3}\,\mathrm{m})^{2}}}=3.18\times10^{5}\,\mathrm{A/m^{2}}
J_{\mathrm{Al}}={\frac{I}{A_{\mathrm{Al}}}}={\frac{I}{\pi\,r_{\mathrm{Al}}^{2}}}={\frac{1\,\mathrm{A}}{\pi\,\times\,(2\times10^{-3}\,{\mathrm{m}})^{2}}}=7.96\times10^{4}\,{\mathrm{A/m}}^{2}
(b) From the periodic table of elements, see Appendix C, the molar mass of aluminum is M(Al) = 26.98 kg/kmol. As in Example 24.1, we find:
n=\frac{N_{A} \ \rho}{M}\;=\;\frac{(6.022\times10^{26}\,\mathrm{atoms/kmol})(2.7\times10^{3}\,\mathrm{kg/m^{3}})}{26.98\,\mathrm{kg/kmol}}= 6.03 × 10^{28} atoms/m³
v_{\mathrm{d}}={\frac{I}{n e{\ A}}}={\frac{1\,{\mathrm{C/s}}}{(6.03\times10^{28}\,\mathrm{electrons/m^{3}})(1.6\times10^{-19}\,{\mathrm{C})}(\pi\times(2\times10^{-3}\,{\mathrm{m}})^{2})}}= 8.25 × 10^{−6} m/s = 2.97 cm/h
