Homework 6 Solutions

# 4 m 1885 m 133 10 6 m 2 16 we use rl a 0150

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Unformatted text preview: where r is the radius of the coil, then L = (250)2πr = (250)(2π)(0.12 m) = 188.5 m. If rw is the radius of the wire itself, then its cross-sectional area is 2 A = π rw = π (0.65 × 10–3 m)2 = 1.33 × 10–6 m2. 1043 According to Table 26-1, the resistivity of copper is ρ = 1.69 ×10−8 Ω ⋅ m . Thus, R= ρL A . c169 × 10 = −8 hb g = 2.4 Ω. Ω ⋅ m 188.5 m 133 × 10 . −6 m 2 16. We use R/L = ρ/A = 0.150 Ω/km. (a) For copper J = i/A = (60.0 A)(0.150 Ω/km)/(1.69 × 10–8 Ω · m) = 5.32 × 105 A/m2. (b) We denote the mass densities as ρm. For copper, (m/L)c = (ρmA)c = (8960 kg/m3) (1.69 × 10–8 Ω · m)/(0.150 Ω/km) = 1.01 kg/m. (c) For aluminum J = (60.0 A)(0.150 Ω/km)/(2.75 × 10–8 Ω · m) = 3.27 × 105 A/m2. (d) The mass density of aluminum is (m/L)a = (ρmA)a = (2700 kg/m3)(2.75 × 10–8 Ω · m)/(0.150 Ω/km) = 0.495 kg/m. 17. We find the conductivity of Nichrome (the reciprocal of its resistivity) as follows: σ= 1 ρ = b gb g . 10 m 4.0 A L L Li = = = = 2.0 × 106 / Ω ⋅ m. RA V / i A VA 2.0 V 10 × 10−6 m2 . bg b gc h 18. (a) i = V/R = 23.0 V/15.0 × 10–3 Ω = 1.53 × 103 A. (b) The cross-sectional area is A = π r 2 = 1 π D 2 . Thus, the magnitude of the current 4 density vector is 4 (1.53 ×10−3 A ) i 4i = = 5.41×107 A/m 2 . J= = A π D 2 π ( 6.00 ×10−3 m )2 (c) The resistivity is RA (15.0 ×10−3 Ω)π (6.00 ×10−3 m) 2 ρ= = = 10.6 × 10−8 Ω ⋅ m. L 4(4.00 m) (d) The material is platinum. 19. The resistance of the wire is given by R = ρL / A , where ρ is the resistivity of the material, L is the length of the wire, and A is its cross-sectional area. In this case, CHAPTER 26 1044 A = π r 2 = π ( 0.50 ×10−3 m ) = 7.85 ×10−7 m 2 . 2 Thus, −3 −7 2 RA ( 50 ×10 Ω ) ( 7.85 ×10 m ) = = 2.0 ×10−8 Ω⋅ m. ρ= L 2.0m 20. The thickness (diameter) of the wire is denoted by D. We use R ∝ L/A (Eq. 26-16) and note that A = 1 π D 2 ∝ D 2 . The resistance of the second wire is given by 4 F A IF L I F D I F...
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## This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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