1、材料科学基础课后习题答案12 Solutions for Chapter 121. Find: Engineering applications of attractive magnetic force. Solution: Applications include: 1) separating iron (ferrous alloys) from other components in a solid waste stream, 2) As an attachment for cranes used to pick up cars in a junkyard, 3) closing devi
2、ces on cabinets and refrigerators, 4) magnetic storage racks for holding tools, and 5) containment fields for fusion reactors.2. Find: Engineering applications of mutual repulsion of like magnetic poles. Solution: Applications include: 1) magnetically levitated trains (and other levitation applicati
3、ons), 2) magnetic bearings, and 3) magnetic stirring devices.3. FIND: Which classes of materials do you expect to show significant magnetic behavior? GIVEN: Magnetic properties result from the motion of electrons. SOLUTION: Metals have high electronic mobility, so we expect metals to have interestin
4、g magnetic properties. When we realize that electronic spins in moderate-to-high atomic number materials are also important, then metals again come to mind as well as some ceramics. Conductive polymers may or may not have interesting magnetic properties: we just cannot say yet.4. FIND: Does a compas
5、s needle point due North? SOLUTION: It certainly does not. It points to the north magnetic pole, which is not the same as the North pole. COMMENTS: Compass measurements need to be corrected in order to provide true readings. Talk to your local surveyor or satellite dish installer.5. Find: Similariti
6、es between (B,H, and ) and (I,V, and ). Data: Equation 12.3-2: B = oH Equation 10.2-1: V = IR Equation 10.2-2: = (A/L)R Equation 10.2-3: = (1/) Solution: Combining equations 10.2-1 through 10.2-3 yields V = I(L/A)(1/) which can be rearranged to yield (I/A) = (V/L). Comparison of this equation with e
7、quation 12.3-2 shows that the magnetic field strength, H, with units (A/m), is comparable to the electric field strength V/L, with units (V/m). Both are measures of the external driving force. The magnetic induction B, with units (T = V-s/m2), is analogous to the current density I/A, with units (A/m
8、2). Both are measures of the response of the material to the external field. Finally, o (and in other materials) and are the materials properties that describe the ability of a specific material to respond to the external field.6. Find: Relationship between electron motion and magnetic properties. S
9、olution: A. The orbital motion of an electron around the atomic nucleus always results in a diamagnetic contribution to the response of a material to an external magnetic field. B. The spin motion of an electron may result in paramagnetic behavior if there are unpaired electrons within unfilled inne
10、r electron shells.7. Find: Difference between magnetic induction and magnetic field strength. Solution: The magnetic field strength (H) is the driving force while the magnetic induction (B) is the system response. That is, in most cases H is the independent and B the dependent variable in a magnetic
11、 experiment.8. Find: Units for and characteristics of magnetic permeability, relative permeability, and susceptibility. Solution: A. Magnetic permeability, has units of (T-m/A)or using the definition of a tesla (T = V-s/m2). has units of (V-s)/(A-m) = (-s)/m). B. Relative permeability is the ratio r
12、 = /o so it is a dimensionless quantity. C. Susceptibility is also a dimensionless quantity. Because r and are dimensionless, they are the values that are more likely to be found in handbooks. This is because they always have the same value regardless of the system of units being employed.9. FIND: E
13、stimate s for air, water, diamond, alumina, Pyrex glass, and epoxy. Compare to accurate values. DATA: Use the data in Tables 11.3-1 () and 11.5-1 (n) to compute the values. Compare to numbers computed from equation 12.3-6 (r = 1 + ) and Table 12.5-1 (). SOLUTION: We begin by using equation 12.3-8, w
14、hich can be used to provide approximate numbers:n = r r = n / .Then we use 12.3-6 to calculate r from values of . The following table gives the raw numbers and the calculated values: n r = n / r = 1 + air 1.0 1.0 1.0 0 1.0water 1.33 80.4 0.0165 -12.9 x 10-6 1.0diamond 2.43 6.0 0.4 -5.9 x 10-6 1.0alu
15、mina 1.76 9.0 0.2 -37 x 10-6 1.0Pyrex glass 1.47 5.0 0.3 0 1.0epoxy 1.58 3.5 0.45 0 1.0Note that I used of 0 for air, Pyrex glass, and epoxy. They are diamagnetic, so the values are essentially zero, as in the other cases shown. COMMENTS: This is not very good agreement!10. Find: The magnetic field
16、strength to create an induction equal to that of the earth in Mg. Given: (Mg) = 13.1x10-6 Data: (earth) = 6x10-5 T Equation 12.3-5: = o(1+)H o = 4x10-7 (T-m/A) Solution: Solving equation 12.3-5 for the field strength H gives: H = /o(1+). Substituting the values in the problem statement yields: H = (
17、6x10-5T)/(4x10-7 T-m/A)(1+13.1x10-6) H = 47.7 A/m11. Find: Relative permeability and permeability of Mg. Given: (Mg) = 13.1x10-6 Data: Eqn. 12.3-5: = o(1+) Eqn. 12.3-6: r = /o = (1+) o = 4x10-7(T-m/A) Solution: A. r = (1+) = 1 + 13.1x10-6 = 1.0000131 B. = o(1+) = (4x10-7 T-m/A)(1.0000131) = 1.257x10
18、-6 (T-m/A)12. Find: Magnetization and inductance of Mg in a field of 105 A/m. Given: = 13.1x10-6 Data: Eqn. 12.3-1: M = H Eqn. 12.3-3: = o(H + M) o = 4x10-7(T-m/A) Solution: Using eqn. 12.3-1, the magnetization is M = 13.1x10-6(105 A/m) = 1.31 A/m. Using eqn. 12.3-3, the inductance is = o(H + M) = (
19、4x10-7 T-m/A)(105 + 1.31 A/m) = 0.1257 T13. FIND: How can you pick out a steel with 1? GIVEN: Most stainless steels have this property. SOLUTION: Hold a magnet up to the metals. A non-magnetic steel is this type of stainless. Al is easy to detect because of its lowdensity. Brass is a shade of yellow
20、. COMMENTS: Simple and effective14. FIND: Would a giant ferromagnetic single crystal increase in length when placed in a magnetic field? SOLUTION: While I have never seen an equation that describes the growth or shrinkage of a material placed in a magnetic filed, I expect there might be a changes in
21、 dimensions as the dipoles align. The effect would be very small. Since a large single crystal is not huge, the change in length might be difficult to measure.15. Find : Relationship between vectors , H, and M in dia- and paramagnetic materials. Solution: The external magnetic field vector, H, and t
22、he induction vector, B, always point in the same direction. The magnetization vector, M, also points in the same direction as H in paramagnetic materials but is antiparallel to H in diamagnetic materials.16. Find: Why dia- and paramagnetic find few intersecting magnetic applications. Solution: Both
23、dia- and paramagnetic materials have susceptibility values very near zero. Therefore, their response to a magnetic field is nearly indistinguishable from that of a vacuum (or air) and their magnetic properties are not very interesting.17. Find: Explain why NaCl, LiF, Si, Ge, Cu, and Ag are all diama
24、gnetic. Solution: NaCl and LiF are ionic compounds. After electron transfer from the cation to the anion, all ions have filled electron shells. Thus, there are no unpaired electrons and both compounds are diamagnetic. Si and Ge are Group IV elements. Both elements crystallize in the diamond cubic st
25、ructure by forming four covalent bonds. As such, their electron shells are filled, there are no unpaired electrons, and Si and Ge are diamagnetic. Cu and Ag are metals and have some unpaired electrons. Their unpaired electrons, however, are in the valence shell and do not contribute to paramagnetic
26、behavior. Since there are no unpaired electrons in inner electron shells, Cu and Ag are diamagnetic.18. Find: Show electron configuration for Sr through Cd and predict magnetic behavior. Data: Electron configurations in Appendix B. Solution:ELEMENT ELECTRON CONFIGURATION # OF UNPAIRED MAGNETIC (INNE
27、R) e-s CLASS Sr Kr5s2 0 DIAY Kr4d15s2 1 PARAZr Kr4d25s2 2 PARANb Kr4d45s1 4 PARAMo Kr4d55s1 5 PARATc Kr4d55s2 5 PARARu Kr4d75s1 3 PARARh Kr4d85s1 2 PARA Pd Kr4d10 0 DIAAg Kr4d105s1 0 DIACd Kr4d105s2 0 DIA19. Find: Can tin cans be separated from a waste stream using a magnet? Data: From Table 12.5-1,
28、 Sn is either dia- or paramagnetic depending on the crystal structure. Solution: Sn cant be separated from a waste stream using a magnet. Sn cans, however, are made of Fe with a Sn lining, and the ferromagnetic Fe is easily separated using a magnet.20. Find: Contribution of unpaired valence electron
29、s to paramagnetic behavior.Solution: Within a solid the valence electrons of individual atoms are either shared or transferred. In either case, there are no unpaired valence electrons when the solid is considered as a whole. Thus, there is no paramagnetic contribution from the valence electrons.21.
30、Find: Why are most oxides diamagnetic and are there any exceptions to the trend? Data: From Table 12.5-1, we see that only a few oxides such as CeO2, CsO2, and Cr2O3 are paramagnetic. Solution: Most oxides are either ionic, covalent, or a mixture of the two. As such they will exhibit filled valence
31、shells and no unpaired electrons. Thus, they are likely to be diamagnetic. If the oxide contains a transition metal with an unfilled inner electron shell, it may exhibit paramagnetic characteristics.22. Find: Mechanisms by which an external field increases the magnetization of a ferromagnetic material. Solution: In a ferromagnetic material an external filed can increase M either by domain growth or by dipole r
