Ch 28 Physics for Scientists and Engineers

# Ch 28 Physics for Scientists and Engineers - CURRENT AND...

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CURRENT AND CONDUCTIVITY 28.1. Solve: The wire's cross-sectional area is A = mz = ~(1.0 x 10" m)' = 3.1415 x lo4 m2, and the electron current through this wire is 2.0 X loi9 s-l . Using Table 28.1 for the electron density of iron and Equation 28.3, the drift velocity is i 2.0~101~ S-l 'd =-= = 7.5 x 10-5m/s = 75 pm/s nA (8.5 x m")(3.1415 x 10" m') Assess: The drift speed of electrons in metals is small. 28.2. Solve: We estimate a distance of 5 ft from the wall switch to the ceiling and then a distance of 8 ft to the center of the room. This yields a total length of approximately L = 4.0 m that an electron will travel in time t. The drift speed of electrons in copper is vd = 1 .O x IO4 m / s . Thus, m =4.0x104 s = 11 hours L 1.0~10~ m/s t=-= 28.3. Solve: Given a drift speed of 100 pds, a time of 11 hours to cover m is physically reasonable. Using Equation 28.3 and Table 28.1, the electron current is i = nAv, = (5.9 x 10' m-3)~(0.5 x lo-' m)'(S.O x 10" m / s) = 2.3 x 10" S-' The time for 1 mole of electrons to pass through a cross section of the wire is The drift speed is small, and Avogadro's number is large. A time of the order of 3 days is reasonable. 28.4. Solve: Equation 28.2 is Ne = nAvdAt. Using Table 28.1 for the electron density, we get ~ 4( 1 .O 10l6) = 9.26 lo4 m = 0.926 mm n(5.8 x 10" m-3)(8.0 x lo4 m / s)(320 10" s) 28.5. Solve: Using Equation 28.2, = 6.0 x m-j Ne*- 1.44 x 1014 - Av,Ar (4.0 x 10" m2)(2.0 / s)(3.0x 10" s) From Table 28.1, the metal is aluminum. 28.6. Solve: For L = 1.0 cm = 1.0 x lo-* m , the surface area of the wire is A=(2m)L=nDL=n(1.0x10-3m)(1.0x 10~zm)=(1.0x10~5m')~ 28-1

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28-2 Chapter 28 The surface charge density of the wire is Q (lo00 cm-' x 1 cm)1.60 x C = 5.1 x lo-'' C / m2 77=x= (1.0 x 10-~ m2)z 28.7. Solve: moles, the number of atoms is (a) Each gold atom has one conduction electron. Using Avogadro's number and n as the number of The density of gold is p = 19,300 kg/m3, the atomic mass is MA = 197 g mol-', r = 0.5 x 6.02 x loz3 mol-'. Substituting these values, we get N = 4.63 x 10" electrons. (b) If all the electrons are transferred a charge of (4.63 x 1021)(-1.60 X deliver a charge of -32 nC, however, the electrons within a length 1 have to be delivered. Thus, m, L = 0.1 m, and NA = C) = -740.8 C will be delivered. To -32 x 10-~ c 1= (10 cm) = 4.32 x lo-'' cm = 4.32 x m -740.8 C 28.8. Solve: Model: Use the conduction model to relate the drift speed to the electric field strength. From Equation 28.7, the electric field is mvd - (9.11~1O-~'kg)(2.0~10~ m/s) eZ (1.60 x C)(5.0 x s) = 0.023 N / C E=- - 28.9. Model: We will use the model of conduction to relate the electric field strength to the mean free time between collisions. Solve: From Equation 28.8, the electric field is mi (9. I I x kg)(5.0 x 1019 8) =0.31 N/C E=-- - nez4 (8.5 x 10% m-3)(1.6x C)(4.2 x s)a(0.9 x m) 28.10. Visualize: + - +~~~~ ~~1- - + c----, - - + + Solve: (a) The charge density is not uniform along the wire. If it was, there would be no electric field inside the wire. The charge density is most positive near the positive terminal of the battery. It gradually decreases until becoming neutral halfway around the circuit. It then becomes increasingly negative, and is most negative near the negative terminal of the battery.
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Ch 28 Physics for Scientists and Engineers - CURRENT AND...

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