Chapter_25_Solutions - 25.1 Q = It =(3.6 A(3(3600 s = 3.89...

Info iconThis preview shows pages 1–18. Sign up to view the full content.

View Full Document Right Arrow Icon
25.1: Q = It = (3.6 A)(3)(3600s) = 3.89 × 10 4 C.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
25.2: a) Current is given by I = Q t = 420C 80(60 s) = 8.75 × 10 2 A. b) I = nqv d A v d = I nqA = 8.75 × 10 2 A (5.8 × 10 28 )(1.6 × 10 19 C)( π (1.3 × 10 3 m) 2 ) = 1.78 × 10 6 ms .
Background image of page 2
25.3: a) v d = I nqA = 4.85 A (8.5 × 10 28 )(1.6 × 10 19 C)( π 4 )(2.05 × 10 3 m) 2 ) = 1.08 × 10 4 ms travel time = d v d = 0.71 m 1.08 × 10 4 = 6574 s = 110 min b) If the diameter is now 4.12 mm, the time can be calculated using the formula above or comparing the ratio of the areas, and yields a time of 26542 s = 442 min. c) The drift velocity depends on the diameter of the wire as an inverse square relationship.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
25.4: The cross-sectional area of the wire is A = π r 2 = (2.06 × 10 3 m) 2 = 1.333 × 10 5 m 2 . The current density is J = I A = 8.00A 1.333 × 10 5 m 2 = 6.00 × 10 5 Am 2 We have v d = J ne; Therefore n = J v d e = 6.00 × 10 5 2 (5.40 × 10 5 ms )(1.60 × 10 19 C electron) = 6.94 × 10 28 electrons m 3
Background image of page 4
25.5: J = nq v d ,soJ v d is constant. J 1 v d1 = J 2 v d2 , v = v (J 2 J 1 ) = v (I 2 I 1 ) = (1.20 × 10 4 ms )(6.00 1.20) = 6.00 × 10 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
25.6: The atomic weight of copper is 63.55 g mole, and its density is 8.96 g cm 3 . The number of copper atoms in 1 .00 m 3 is thus (8.96 g cm 3 )(1.00 × 10 6 cm 3 m 3 )(6.023 × 10 23 atoms mole ) 63.55 g mole = 8.49 × 10 28 atoms m 3 Since there are the same number of free electrons m 3 as there are atoms of copper m 3 (see Ex. 25.1), The number of free electrons per copper atom is one.
Background image of page 6
25.7: Consider 1 m 3 of silver. density = 10.5 × 10 3 kg m 3 ,som = 10.5 × 10 3 kg M = 107.868 × 10 3 kg mol ,son = mM = 9.734 × 10 4 mol and N = nN A = 5.86 × 10 28 atoms m 3 If there is one free electron per m 3 , there are 5.86 × 10 28 free electrons m 3 . This agrees with the value given in Exercise 25.2.
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
25.8: a) Q total = (n Cl + n Na )e = (3.92 × 10 16 + 2.68 × 10 16 )(1.60 × 10 19 C) = 0.0106C I = Q total t = 0.0106 C 1.00 s = 0.0106A = 10.6 mA. b) Current flows, by convention, in the direction of positive charge. Thus, current flows with Na toward the negative electrode. +
Background image of page 8
25.9: a) Q = Idt = 0 8 (55 0.65 t 2 )dt 0 8 = 55t 0 8 + 0.65 3 t 3 0 8 = 329 C. b) The same charge would flow in 10 seconds if there was a constant current of: I = Qt = (329 C) (8 s) = 41.1 A.
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
25.10: a) J = I A = 3.6 A (2.3 × 10 3 m) 2 = 6.81 × 10 5 A/m 2 . b) E = ρ J = (1.72 × 10 8 Ω⋅ m)(6.81 × 10 5 A/m 2 ) = 0.012 V m . c) Time to travel the wire’s length: t = l v d = l nqA I = (4.0 m)(8.5 × 10 28 m 3 )(1.6 × 10 19 C)(2.3 × 10 3 2 3.6 A = 8.0 × 10 4 s = 1333 min 22 hrs!
Background image of page 10
25.11: R = ρ L A = (1.72 × 10 8 Ω⋅ m)(24.0 m) ( π 4 )(2.05 × 10 3 m) 2 = 0.125 .
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
25.12: R = ρ L A L = RA = (1.00 )( π 4 )(0.462 × 10 3 m) 2 1.72 × 10 8 Ω⋅ m = 9.75 m.
Background image of page 12
25.13: a) tungsten: E = ρ J = I A = (5.25 × 10 8 m 3 )(0.820 A) ( π 4 )(3.26 × 10 3 m) 2 = 5.16 × 10 3 Vm . b) aluminum: E = J = I A = (2.75 × 10 8 m 3 )(0.820 A) ( 4 )(3.26 × 10 3 2 = 2.70 × 10 3 .
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
25.14: R Al = R Cu ρ Al L A Al = Cu L A Cu π d Al 2 4 Al = d Cu 2 4 Cu d Cu = d Al Cu Al d Al = (3.26 mm) 1.72 × 10 8 Ω⋅ m 2.75 × 10 8 m = 2.6 mm.
Background image of page 14
25.15: Find the volume of one of the wires: R = ρ L A so A = L R and volume = AL = L 2 R = (1.72 × 10 8 Ω⋅ m)(3.50 m) 2 0.125 = 1.686 × 10 6 m 3 m = (density)V = (8.9 × 10 3 kg m 3 )(1.686 × 10 6 m 3 ) = 15 g
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
25.16: r 1 = 3.5 cm 2 = 1.75 cm r 2 = 3.25 mm 2 = 1.625 mm R = ρ l A l = 2 π r 1 (per coil) × 75 coils A = r 2 2 = RA l = R r 2 2 (2 r 1 )75 = Rr 2 2 150r 1 = (1.74 )(1.625 × 10 3 m) 2 150(1.75 × 10 2 = 1.75 × 10 6 Ω⋅ m
Background image of page 16
25.17: a) From Example 25.1, an 18-gauge wire has A = 8.17 × 10 3 cm 2 I = JA = (1.0 × 10 5 A/cm 2 )(8.17 × 10 3
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 18
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/01/2008 for the course PHYSICS 240 taught by Professor Davewinn during the Fall '08 term at University of Michigan.

Page1 / 84

Chapter_25_Solutions - 25.1 Q = It =(3.6 A(3(3600 s = 3.89...

This preview shows document pages 1 - 18. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online