# hw51 - Problem 8. Each atom in aluminum contributes about...

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Problem 8. Each atom in aluminum contributes about 3.5 conduction electrons. What is the drift speed in a 0.21-cm-diameter aluminum wire carrying 20 A? Solution As in Example 27-1, the drift speed of electrons in the wire is v d = I = ( 1 4 π d 2 ) ne , where n is the number density of conduction electrons. Using data from the text’s inside back cover, we find n = (3.5 electrons/ion)(2702 kg/m 3 ) ÷ ( 26 . 98 u/ion × 1 . 66 × 10 27 kg/u ) = 2 . 11 × 10 29 electrons/m 3 , for aluminum. Thus, v d = 20 A = 1 4 ( 0 . 21 cm ) 2 × (2.11 × 10 m 3 )(1.6 × 10 19 C) = 0.171 mm/s. Problem 10. The filament of the light bulb in Example 27-6 has a diameter of 0.050 mm. What is the current density in the filament? Compare with the current density in a 12-gauge wire (diameter 0.21 cm) supplying current to the light bulb. The current density in the filament (Equation 27-3a, with numbers from Example 27-6) is J = (0.833 A) = (0.025 mm) 2 = 4.24 × 10 8 A/m 2 , in the direction of the current. A 12-gauge wire, carrying the same current, would have a current density smaller by a factor of the square of the ratio of the diameters, ( 0 . 05 mm = 2 . 1 mm ) 2 = 5 . 67 × 10 4 , or J = 2 . 41 × 10 5 2 . Problem 12. A piece of copper wire joins a piece of aluminum wire whose diameter is twice that of the copper. The same current flows in both wires. The density of conduction electrons in copper is

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## hw51 - Problem 8. Each atom in aluminum contributes about...

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