31 1 Solve The wire s cross sectional area is A 2 r 1 0 10 m 3 2 3 1415 10 m 6 2 and the electron current through this wire is i 2 0 10 s 19 1 Using Table 31 1 for the electron density of iron and Equation 31 3 the drift velocity is eN t v d i nA 8 5 10 m 28 3 1415 10 m 6 2 1 19 2 0 10 s 3 7 5 10 5 m s 75 m s Assess The drift speed of electrons in metals is small 31 2 Solve Using Equation 31 3 and Table 31 1 the electron current is i nAv d 5 9 10 m 28 3 2 0 5 10 m 5 0 10 m s 5 3 2 3 10 s 18 1 The time for 1 mole of electrons to pass through a cross section of the wire is N A t 1 mole i 23 6 02 10 18 2 3 10 s 1 2 62 10 s 3 0 days 5 Assess The drift speed is small and Avogadro s number is large A time of the order of 3 days is reasonable 31 3 Solve Equation 31 2 is N e nAv d Using Table 31 1 for the electron density we get t A 2 D 4 N nv d e t D N 4 e t nv d 5 8 10 m 8 0 10 m s 320 10 s 28 3 6 16 4 1 0 10 4 9 3 10 m 0 93 mm 4 31 4 Solve The number of electrons crossing a cross sectional area of a wire is the electron current i 3 2 i nAv d 6 0 10 m 28 3 1 6 10 m 2 4 2 0 10 m s 2 4 10 s 19 1 The electron density n for aluminum is taken from Table 31 1 The number of electrons passing through the cross section in one day is Ne i t 2 4 1019 s 1 365 days 24 hr day 3600 s hr 7 6 1026 electrons Assess The large electron density compensates for the small drift velocity to deliver a huge number of electrons in current 31 5 Solve Using Equation 31 2 n N Av d e t 1 44 10 4 4 00 10 m 2 00 10 m s 3 0 10 s 6 2 6 14 6 0 10 m 28 3 From Table 31 1 the metal is aluminum 31 6 Model Use the conduction model to relate the drift speed to the electric field strength Solve From Equation 31 7 the electric field is E mv d e 31 9 11 10 kg 2 0 10 m s 1 60 10 C 5 0 10 s 4 19 14 0 023 N C 31 7 Solve For L 1 0 cm 1 0 10 m 2 the surface area of the wire is A 2 r L DL 1 0 10 3 m 1 0 10 2 m 1 0 10 5 m2 The surface charge density of the wire is Q A 1000 cm 1 cm 1 60 10 19 C 1 1 0 10 m 5 2 5 1 10 12 C m 2 31 8 Solve of moles the number of atoms is a Each gold atom has one conduction electron Using Avogadro s number and n as the number N nN V M The density of gold is 19 300 kg m3 the atomic mass is MA 197 g mol 1 r 0 50 10 3 m L 0 10 m and NA 6 02 1023 mol 1 Substituting these values we get N 4 6 1021 electrons b If all the electrons are transferred a charge of 4 6 1021 1 60 10 19 C 740 C will be delivered To deliver a charge of 32 nC however the electrons within a length l have to be delivered Thus m M N N N A A A A A A A 2 r L M l 9 32 10 C 740 C 10 cm 4 3 10 10 cm 4 3 10 12 m 31 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 31 8 the electric field is 31 1 E mi ne A 8 5 10 m 1 6 10 28 3 19 9 11 10 19 C 4 2 10 kg 5 0 10 s s 15 0 9 10 m 3 2 0 31 N C 31 10 Solve a The electron current is i nA 17 1 3 5 10 s 5 0 10 m 4 3 2 6 0 10 m 28 i nAv d v d 7 43 10 m s 6 The electron drift speed is 7 4 10 6 m s The electron density for aluminum is taken from Table 31 1 b The electron drift velocity is related to the electric field in the wire by v d E e m mv d eE 31 9 11 10 kg 7 43 10 m s 1 6 10 C 2 0 10 V m 3 6 19 2 1 10 14 s Assess This is about the right order of magnitude based on the examples in the text 31 11 Visualize The current density J in a wire as given by Equation 31 13 does not depend on the thickness of the wire Solve a The current in the wire is I wire J A wire wire 5 4 5 10 A m 2 1 5 10 m 3 0 795 A 1 2 2 Because current is continuous Iwire Ifilament Thus Ifilament 0 795 0 80 A b The current density in the filament is J filament I filament A filament 0 795 A 1 2 0 12 10 m 3 2 7 7 0 10 A m 2 31 12 Solve a The current density is b The electron current or number of electrons per second is J I A I R 2 0 85 A 0 00025 m 2 1 2 1 73 10 A m 7 2 N e t I e 0 85 A 19 1 60 10 0 85 C s 19 C 1 60 10 C 5 3 10 s 18 1 31 13 Solve a From Equation 31 13 and Table 31 1 the current density in the gold wire is J nev d 3 5 9 10 m 1 60 10 28 19 C 3 0 10 m s 4 2 83 10 A m 6 2 The current density is b The current is 6 2 8 10 A m 2 I JA 2 83 10 A m 6 2 0 25 10 m 3 0 56 A 2 31 14 Solve From Equation 31 13 the current in the wire is JA 6 I 2 5 7 50 10 A m 2 5 10 m 75 10 m 0 141 mA 6 …
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