Homework 6 Solutions

26 13 to the three sections of the resistive strip we

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Unformatted text preview: n leads to r1 = 1.55 mm. (b) The 4r22 = 1.5r32 relation leads to r2 = 1.22 mm. 25. Since the mass density of the material does not change, the volume remains the same. If L0 is the original length, L is the new length, A0 is the original cross-sectional area, and A is the new cross-sectional area, then L0A0 = LA and A = L0A0/L = L0A0/3L0 = A0/3. The new resistance is ρL ρ 3 L0 ρL = = 9 0 = 9 R0 , R= A A0 / 3 A0 where R0 is the original resistance. Thus, R = 9(6.0 Ω) = 54 Ω. 26. The absolute values of the slopes (for the straight-line segments shown in the graph of Fig. 26-25(b)) are equal to the respective electric field magnitudes. Thus, applying Eq. 26-5 and Eq. 26-13 to the three sections of the resistive strip, we have i J1 = A = σ1 E1 = σ1 (0.50 × 103 V/m) i J2 = A = σ2 E2 = σ2 (4.0 × 103 V/m) i J3 = A = σ3 E3 = σ3 (1.0 × 103 V/m) . We note that the current densities are the same since the values of i and A are the same (see the problem statement) in the three sections, so J1 = J2 = J3 . (a) Thus we see that σ1 = 2σ3 = 2 (3.00 × 107(Ω · m)−1 ) = 6.00 × 107 (Ω · m)−1. 1046 CHAPTER 26 (b) Similarly, σ2 = σ3/4 = (3.00 × 107(Ω · m)−1 )/4 = 7.50 × 106 (Ω · m)−1 . 27. The resistance of conductor A is given by RA = ρL 2 prA , where rA is the radius of the conductor. If ro is the outside diameter of conductor B and ri is its inside diameter, then its cross-sectional area is π(ro2 – ri2), and its resistance is RB = The ratio is b ρL π ( ro2 − ri 2 ) gb . 2 g 1.0 mm − 0.50 mm R A ro2 − ri 2 = = 2 2 RB rA 0.50 mm b g 2 = 3.0. 28. The cross-sectional area is A = πr2 = π(0.002 m)2. The resistivity from Table 26-1 is ρ = 1.69 × 10−8 Ω · m. Thus, with L = 3 m, Ohm’s Law leads to V = iR = iρL/A, or 12 × 10−6 V = i (1.69 × 10−8 Ω · m)(3.0 m)/ π(0.002 m)2 which yields i = 0.00297 A or roughly 3.0 mA. 29. First we find the resistance of the copper wire to be R= ρL A (1.69 ×10 = −8 Ω ⋅ m ) ( 0.020 m ) π (2.0 ×10 m) −3 2 = 2.69 × 10−5 Ω . With potential difference V = 3.00 nV ,...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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