chapter17 - Current and Resistance 91 13. One way of...

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PROBLEM SOLUTIONS 17.1 The charge that moves past the cross section is ∆∆ QIt = ( ) , and the number of electrons is n Q e It e == ( ) = × ( ) ( ) 80 0 10 10 0 60 0 3 .. . C s min s min 1.60 10 C electrons 19 ( ) × 300 10 20 . The negatively charged electrons move in the direction opposite to the conventional current ± ow. 17.2 (a) ²rom Example 17.2 in the textbook, the density of charge carriers (electrons) in a copper wire is n 846 10 28 . electrons m 3 . With Ar q e π 2 and , the drift speed of electrons in this wire is v d I nqA I ne r ( ) = × ( ) 2 370 10 1 60 . . Cs 8.46 m 28 3 × ( ) × ( ) −− 10 1 25 10 557 10 19 3 2 5 C m ms . . (b) The drift speed is smaller because more ele ctrons are being conducted . To create the same current, therefore, the drift speed need not be as great.
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92 Chapter 17 17.3 The period of the electron in its orbit is Tr = 2 π v , and the current represented by the orbiting electron is I Q t e T e r == = = × ( ) × ( ) v 2 2 19 10 1 60 10 61 9 .. ms C 2 5 29 10 105 10 105 11 3 . × ( ) = m Cs mA 17.4 If N is the number of protons, each with charge e , that hit the target in time t , the average current in the beam is IQ t N e t ∆∆ , giving N It e = ( ) = × ( )( ) × 125 10 23 0 160 10 6 19 s Cp . . roton protons 180 10 16 . 17.5 (a) The carrier density is determined by the physical characteristics of the wire, not the current in the wire. Hence, n is unaffected . (b) The drift velocity of the electrons is v d I nqA = . Thus, the drift velocity is doubled when the current is doubled. 17.6 The mass of a single gold atom is m M N A atom 23 gmo l 6.02 10 atoms mol 1 × 197 327 .0 g 1 0 k g 22 25 −− . The number of atoms deposited, and hence the number of ions moving to the negative electrode, is n m m × × atom 25 kg 10 kg 325 10 993 10 3 2 . . . 1 Thus, the current in the cell is I Q t ne t = × ( ) × ( ) 9 93 10 1 60 10 21 19 C 2.78 h ( )( ) 3 600 0 159 159 s 1 h A m A . 17.7 The drift speed of electrons in the line is v d I nqA I ne d ( ) 2 4 , or v d = ( ) × ( ) × ( ) 4 1 000 85 10 002 28 19 A m C 3 . 0 23 2 m 10 m s 4 ( ) . The time to travel the length of the 200-km line is then t L d × ×× v 200 10 234 1 3 m yr 3.156 10 s 47 . = 27 yr
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Current and Resistance 93 17.8 Assuming that, on average, each aluminum atom contributes three electrons, the density of charge carriers is three times the number of atoms per cubic meter. This is n density mass per atom M N N M A A = == 3 33 ρρ , or n = × ( ) ( )( ) 3 6 02 10 2 7 10 1 23 6 .. m o l g c m c m m 3 26 98 18 10 29 . . gmo l m 3 The drift speed of the electrons in the wire is then v d I neA × ( ) × ( ) 50 160 10 40 19 . Cs 1.8 10 m C 29 3 × ( ) 10 6 m 2 4.3 10 m s 5 17.9 (a) Using the periodic table on the inside back cover of the textbook, we F nd M ±e 3 g mol g mol kg g ( )( ) = 55 85 55 85 1 10 55 8 . 51 0 3 × kg mol (b) ±rom Table 9.3, the density of iron is ρ ±e 3 kg m 786 10 3 . , so the molar density is molar density kg m ±e ±e ±e 3 ( ) × M 55 85 3 .
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chapter17 - Current and Resistance 91 13. One way of...

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