CHAPTER 18
ELECTRICAL PROPERTIES
PROBLEM SOLUTIONS
18.1 This problem calls for us to compute the electrical conductivity and resistance of a silicon specimen. (a) We use Equations (18.3) and (18.4) for the conductivity, as
1 Il = VA Il V d 2
2
=
SOURCES OF MAGNETIC FIELD
28
28.1.
! ^ EXECUTE: (a) r = ( 0.500 m ) i , r = 0.500 m ! ! ^ v r = vr^ i = -vrk j ^
! IDENTIFY and SET UP: Use Eq.(28.2) to calculate B at each point. ! ! ! ! ! qv r 0 qv r ^ r ^ B= 0 = , since r = . 4 r 2 4 r 3
MAGNETIC FIELD AND MAGNETIC FORCES
27
27.1.
! IDENTIFY and SET UP: Apply Eq.(27.2) to calculate F . Use the cross products of unit vectors from Section 1.10. ! ^ j EXECUTE: v = ( +4.19 104 m/s ) i + ( -3.85 104 m/s ) ^ ! ^ (a) B = (1.40 T ) i !
PARTICLE PHYSICS AND COSMOLOGY
44
44.1.
(a) IDENTIFY and SET UP: Use Eq.(37.36) to calculate the kinetic energy K. 1 EXECUTE: K = mc 2 - 1 = 0.1547 mc 2 2 2 1- v / c
m = 9.109 10 -31 kg, so K = 1.27 10-14 J (b) IDENTIFY and SET UP: The tota
NUCLEAR PHYSICS
43
43.1.
(a) (b) (c)
28 14 85 37
Si has 14 protons and 14 neutrons. Rb has 37 protons and 48 neutrons. Tl has 81 protons and 124 neutrons.
205 81
43.2.
(a) Using R = (1.2 fm)A1 3 , the radii are roughly 3.6 fm, 5.3 fm, and 7.1
ATOMIC STRUCTURE
41
L = l (l + 1) . Lz = ml . l = 0, 1, 2,., n - 1. ml = 0, 1, 2,., l . cos = Lz / L .
41.1.
IDENTIFY and SET UP:
EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2 , Lz = ,0, - . l = 2 : L = 6 , Lz = 2 , ,0, - , -2 . (b) In ea
QUANTUM MECHANICS
40
n2h 2 . 8mL2
40.1.
IDENTIFY and SET UP: The energy levels for a particle in a box are given by En = EXECUTE: (a) The lowest level is for n = 1, and E1 =
(1)(6.626 10-34 J s) 2 = 1.2 10-67 J. 8(0.20 kg)(1.5 m) 2
1 2E 2(1.2
CHAPTER 3
THE STRUCTURE OF CRYSTALLINE SOLIDS
PROBLEM SOLUTIONS
3.1 Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the
CHAPTER 9
PHASE DIAGRAMS
PROBLEM SOLUTIONS
9.1 Three variables that determine the microstructure of an alloy are 1) the alloying elements present, 2) the concentrations of these alloying elements, and 3) the heat treatment of the alloy.
9.2 In or
ELECTROMAGNETIC INDUCTION
29
29.1.
29.2.
IDENTIFY: Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the coil. SET UP: The flux through a coil of N turns
INDUCTANCE
30
Apply Eq.(30.4). di (a) E2 = M 1 = (3.25 10-4 H)(830 A/s) = 0.270 V; yes, it is constant. dt
30.1.
IDENTIFY and SET UP: EXECUTE: (b) E1 = M
di2 ; M is a property of the pair of coils so is the same as in part (a). Thus E1 = 0.270 V.
CHAPTER 13
APPLICATIONS AND PROCESSING OF CERAMICS
PROBLEM SOLUTIONS
13.1 The two desirable characteristics of glasses are optical transparency and ease of fabrication.
13.2 (a) Devitrification is the process whereby a glass material is caused to
CHAPTER 16
COMPOSITES
PROBLEM SOLUTIONS
16.1
The major difference in strengthening mechanism between large-particle and dispersionstrengthened particle-reinforced composites is that for large-particle the particle-matrix interactions are not trea
CHAPTER 20
MAGNETIC PROPERTIES
PROBLEM SOLUTIONS
20.1 (a) We may calculate the magnetic field strength generated by this coil using Equation (20.1) as
NI l
H =
=
(200 turns)(10 A) = 10,000 A - turns/m 0.2 m
(b) In a vacuum, the flux density is
CHAPTER 19
THERMAL PROPERTIES
PROBLEM SOLUTIONS
19.1 The energy, E, required to raise the temperature of a given mass of material, m, is the product of the specific heat, the mass of material, and the temperature change, T as
E = cpm(T)
The T is
CHAPTER 22
ECONOMIC, ENVIRONMENTAL, AND SOCIETAL ISSUES IN MATERIALS SCIENCE AND ENGINEERING
PROBLEM SOLUTION
22.D1W The three materials that are used for beverage containers are glass, aluminum, and the polymer polyethylene terephthalate (designa
CHAPTER 21
OPTICAL PROPERTIES
PROBLEM SOLUTIONS
21.1 Similarities between photons and phonons are: 1) Both may be described as being wave-like in nature. 2) The energy for both is quantized. Differences between photons and phonons are: 1) Phonons
CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
PROBLEM SOLUTIONS
15.1 From Figure 15.3, the elastic modulus is the slope in the elastic linear region of the 20 C curve, which is
(stress) 30 MPa 0 MPa = = 3.3 GPa (strain) 9 x
CHAPTER 14
POLYMER STRUCTURES
PROBLEM SOLUTIONS
14.1
Polymorphism is when two or more crystal structures are possible for a material of given composition. Isomerism is when two or more polymer molecules or mer units have the same composition, but
CHAPTER 17
CORROSION AND DEGRADATION OF MATERIALS
PROBLEM SOLUTIONS
17.1 (a) Oxidation is the process by which an atom gives up an electron (or electrons) to become a cation. Reduction is the process by which an atom acquires an extra electron (or
CHAPTER 12
STRUCTURES AND PROPERTIES OF CERAMICS
PROBLEM SOLUTIONS
12.1
The two characteristics of component ions that determine the crystal structure are:
1) the
magnitude of the electrical charge on each ion; and 2) the relative sizes of the
CHAPTER 8
FAILURE
PROBLEM SOLUTIONS
8.1 Several situations in which the possibility of failure is part of the design of a component or product are as follows: (1) the pull tab on the top of aluminum beverage cans; (2) aluminum utility/light poles
CHAPTER 7
DISLOCATIONS AND STRENGTHENING MECHANISMS
PROBLEM SOLUTIONS
7.1 The dislocation density is just the total dislocation length per unit volume of material (in this case per 3 5 -2 cubic millimeters). Thus, the total length in 1000 mm of ma
CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
2.1 (a) When two or more atoms of an element have different atomic masses, each is termed an isotope. (b) The atomic weights of the elements ordinarily are not integers because:
CHAPTER 5
DIFFUSION
PROBLEM SOLUTIONS
5.1 Self-diffusion is atomic migration in pure metals-i.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal.
5.2 Self-diffusion m
CHAPTER 4
IMPERFECTIONS IN SOLIDS
PROBLEM SOLUTIONS
4.1 In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation (4.1). As stated in the problem, Q = 0.55 eV/atom. Thus, v
N
N
V = exp
Q
V
kT
=
DIRECT-CURRENT CIRCUITS
26
26.1.
26.2.
26.3.
IDENTIFY: The newly-formed wire is a combination of series and parallel resistors. SET UP: Each of the three linear segments has resistance R/3. The circle is two R/6 resistors in parallel. EXECUTE: T
CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE
25
25.1.
25.2.
IDENTIFY: I = Q / t . SET UP: 1.0 h = 3600 s EXECUTE: Q = It = (3.6 A)(3.0)(3600 s) = 3.89 104 C. EVALUATE: Compared to typical charges of objects in electrostatics, this is a huge amou
CAPACITANCE AND DIELECTRICS
24
24.1.
24.2.
24.3.
Q Vab SET UP: 1 F = 10 -6 F EXECUTE: Q = CVab = (7.28 10 -6 F)(25.0 V) = 1.82 10 -4 C = 182 C EVALUATE: One plate has charge + Q and the other has charge -Q . Q PA and V = Ed . IDENTIFY and SE