1816 the electrical resistivity of a beryllium alloy

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18–16 The electrical resistivity of a beryllium alloy containing 5 at% of an alloying element is found to be 50 10 6 ohm cm at 400 C. Determine the contributions to resistivity due to temperature and due to impurities by finding the expected resis- tivity of pure beryllium at 400 C, the resistivity due to impurities, and the defect resistivity coefficient. What would be the electrical resistivity if the beryllium contained 10 at% of the alloying element at 200 C? Solution: From the data in Table 18–3, the resistivity at 400 C should be: Consequently the resistance due to impurities is: Since there are 5 at% impurities present, x 0.05, and the defect resistivity coefficient is: The resistivity at 200 C in an alloy containing 10 at% impurities is: 18–17 Is Equation 18–7 valid for the copper-zinc system? If so, calculate the defect resis- tivity coefficient for zinc in copper. (See Figure 18–11.) Solution: The conductivity and resistivity of pure copper are: For 10 wt% Zn in copper: From Figure 18–11(a), the conductivity of the Cu–10% Zn alloy at zero deformation is about 44% that of pure copper, or ¢ r 0.38 10 5 0.167 10 5 0.213 10 5 r 1 s 0.38 10 5 s 1 5.98 10 5 21 0.44 2 2.63 10 5 x Zn 1 1 x Zn 2 1 0.0975 21 1 0.0975 2 0.088 x Zn 1 10 65.38 2 1 10 65.38 2 1 90 63.54 2 0.0975 s 5.98 10 5 or r 1 s 0.167 10 5 ohm # cm 21.5 10 6 16.1 10 6 37.6 10 6 ohm # cm 10 6 1 0.1 21 1 0.1 2 1 4 10 6 2 3 1 1 0.025 21 200 25 2 4 178.9 r 200 r r d b 8.5 10 6 1 0.05 21 1 0.05 2 178.9 10 6 ohm # cm r d b x 1 1 x 2 or b r d x 1 1 x 2 r d 8.5 10 6 ohm # cm 50 10 6 41.5 10 6 r d r r t r d r t 1 4 10 6 2 3 1 1 0.025 21 400 25 2 4 41.5 10 6 # CHAPTER 18 Electronic Materials 199
200 The Science and Engineering of Materials Instructor’s Solution Manual The following table includes the calculations for other compositions: wt% Zn x Zn x Zn (1 x Zn ) % s s r r 0 0 0 101 5.98 10 5 0.167 10 5 0 10 0.0975 0.088 44 2.63 10 5 0.380 10 5 0.213 10 5 15 0.146 0.125 37 2.21 10 5 0.452 10 5 0.285 10 5 20 0.196 0.158 33 1.97 10 5 0.508 10 5 0.341 10 5 30 0.294 0.208 28 1.67 10 5 0.599 10 5 0.432 10 5 These data are plotted. The slope of the graph is “ b ”: 1.8 10 5 ohm # cm b 0.4 10 5 0.2 10 5 0.19 0.08 18–19 GaV 3 is to operate as a superconductor in liquid helium (at 4 K). The T c is 16.8 K and H o is 350,000 oersted. What is the maximum magnetic field that can be applied to the material? Solution: 18–20 Nb 3 Sn and GaV 3 are candidates for a superconductive application when the mag- netic field is 150,000 oersted. Which would require the lower temperature in order to be superconductive? Solution: 18–21 A filament of Nb 3 Sn 0.05 mm in diameter operates in a magnetic field of 1000 oer- sted at 4 K. What is the maximum current that can be applied to the filament in order for the material to behave as a superconductor? Solution: From Figure 18–12, the maximum current density for Nb 3 Sn in a field of 1000 oersted is about 2 10 6 A/cm 2 . I JA 1 2 10 6 A /cm 2 21 4 21 0.005 cm 2 2 39.3 A T 12.7 K For GaV 3 : 150,000 350,000 3 1 1 T 16.8 2 2 4 T 11.42 K For Nb 3 Sn: 150,000 250,000 3 1 1 T 18.05 2 2 4 150,000 H o 3 1 1 T T c 2 2 4 H c H o 3 1 1 T T c 2 2 4 350,000 3 1 1 4 16.8 2 2 4 330,159 oersted T c 16.8 K H o 350,000 oersted 0.2 0.4 r = 10 5 0.1 0.2 x (1 x )
18–22 Assume that most of the electrical charge transferred in MgO is caused by the diffu- sion of Mg 2 ions. Determine the mobility and electrical conductivity of MgO at 25 C and at 1500 C. (See Table 5–1.)

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