Chap 20 Solns-6E - CHAPTER 20 MAGNETIC PROPERTIES PROBLEM...

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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 H = N I l = (200 turns)(10 A) 0.2 m = 10,000 A- turns/m (b) In a vacuum, the flux density is determined from Equation (20.3). Thus, B o µ o H = 1.257 x 10 -6 H/m () (10,000 A - turns/m) = 1.257 x 10 -2 tesla (c) When a bar of titanium is positioned within the coil, we must use an expression that is a combination of Equations (20.5) and (20.6) in order to compute the flux density given the magnetic susceptibility. Inasmuch as χ m = 1.81 x 10 -4 (Table 20.2), then B = µ o H + µ o M = µ o µ o χ m µ o H1 + χ m ( ) = 1.257 x 10 -6 H/m (10,000 A - turns/m) 1 + 1.81 x 10 -4 ( ) 1.257 x 10 -2 tesla which is essentially the same result as part (b). This is to say that the influence of the titanium bar within the coil makes an imperceptible difference in the magnitude of the B field. (d) The magnetization is computed from Equation (20.6): χ m H = 1.81 x 10 -4 (10,000 A - turns/m) = 1.81 A/m 1
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20.2 (a) This portion of the problem asks that we compute the flux density in a coil of wire 0.1 m long, having 15 turns, that carries a current of 1.0 A, and that is situated in a vacuum. Utilizing Equations (20.1) and (20.3), and solving for B o yields B o = µ o H = µ o N I l 1.257 x 10 6 H/m () (15 turns)(1.0 A) 0.1 m = 1.89 x 10 -4 tesla (b) Now we are to compute the flux density with a bar of the iron-silicon alloy, the B-H behavior for which is shown in Figure 20.24. It is necessary to determine the value of H using Equation (20.1) as N I l (15 turns)(1.0 A) = 150 A - turns/m Using the curve in Figure 20.24, B = 1.65 tesla at H = 150 A-turns/m. (c) Finally, we are to assume that a bar of Mo is situated within the coil, and to calculate the current that is necessary to produce the same B field as when the iron-silicon alloy in part (b) was used. Molybdenum is a paramagnetic material having a χ m of 1.19 x 10 -4 (Table 20.2). Combining Equations (20.2), (20.4), and (20.7) we get B µ B µ o 1 m ( ) And when Mo is positioned within the coil, then 1.65 tesla 1.257 x 10 6 1 + 1.19 x 10 4 = 1.312 x 10 6 A- turns/m Now, the current may be determined using Equation (20.1); I Hl N 1.312 x 10 6 A - turns /m (0.1 m) 15 turns = 8750 A 20.3 This problem asks us to show that χ m and µ r are related according to χ m = µ r - 1. We begin with Equation (20.5) and substitute for M using Equation (20.6). Thus, 2
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B = µ o H + µ o M = µ o µ o χ m H But B is also defined in Equation (20.2) as µ H = µ o µ o χ m which leads to µ = µ o 1 + χ m ( ) If we divide both sides of this expression by µ o µ µ o µ r = 1 + χ m or, upon rearrangement χ m µ r 1 20.4 For this problem, we want to convert the volume susceptibility of silver (i.e., -2.38 x 10 -5 ) into other systems of units. For the mass susceptibility χ m (kg) = χ m ρ (kg /m 3 ) 2.38 x 10 5 10.49 x 10 3 kg/m 3 2.27 x 10 -9 For the atomic susceptibility χ m (a) = χ m (kg) x atomic weight (in kg) [ ] 2.27 x 10 -9 () (0.10787 kg/mol) = 2.45 x 10 -10 For the cgs-emu susceptibilities, 3
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χ m ' = χ m 4 π 2.38 x 10 5 4 π = 1.89 x 10 -6 χ m ' (g) = χ m ' ρ (g/cm 3 ) 1.89 x 10 6 10.49 g/ cm 3 1.80 x 10 -7 χ m ' (a) = χ m ' (g) x atomic weight (in g) [] 1.80 x 10 -7 () (107.87 g/mol) = 1.94 x 10 -5 20.5 (a) The two sources of magnetic moments for electrons are the electron's orbital motion around the nucleus, and also, its spin.
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This homework help was uploaded on 02/07/2008 for the course MATENG MAT201 taught by Professor Na during the Spring '08 term at Wisconsin Milwaukee.

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Chap 20 Solns-6E - CHAPTER 20 MAGNETIC PROPERTIES PROBLEM...

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