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Unformatted text preview: nguyen (jmn727) – oldhomework 14 – Turner – (59070) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 6 . 7 g wire has a density of 3 g / cm 3 and a resistivity of 4 . 3 × 10 − 8 Ω · m. The wire has a resistance of 108 Ω. How long is the wire? Correct answer: 74 . 8953 m. Explanation: Let : m = 6 . 7 g = 0 . 0067 kg , ρ = 3 g / cm 3 = 3000 kg / m 3 , r = 4 . 3 × 10 − 8 Ω · m , and R = 108 Ω . ρ is density, and r is the resistivity of the wire, so m = ρ V V = m ρ = A ℓ A = m ρ ℓ Thus the resistance is R = r ℓ A = r ℓ m ρ ℓ = r ρ ℓ 2 m ℓ 2 = R m ρ r ℓ = radicalBigg R m ρ r = radicalBigg (108 Ω) (0 . 0067 kg) (3000 kg / m 3 ) (4 . 3 × 10 − 8 Ω · m) = 74 . 8953 m . 002 (part 2 of 2) 10.0 points The wire is made up of atoms with va lence 3 and molecular mass 18 kg, where u = 1 . 6605 × 10 − 27 kg is the mass of a hy drogen atom. What is the drift speed of the electrons when there is a voltage drop of 172 . 8 V across the wire? Correct answer: 0 . 000414185 m / s. Explanation: Let : n con = 3 , m μ = 18 kg , V = 172 . 8 V , N A = 6 . 02 × 10 23 , ρ = 3 g / cm 3 = 3000 kg / m 3 , ℓ = 74 . 8953 m , and q e = 1 . 602 × 10 − 19 C , where m μ is atomic mass, n con is the num ber of conduction electrons/atom, and N A is Avogadro’s number. Since the current I is given by I = n q A v d , where n is the density of charge carriers, q the charge on an electron and v d the drift speed, we have v d = I n q e A = V n q e A R . (1) The density of charge carriers is found from knowing how many electrons there are per atom and then the total number of atoms in the wire. The mass of one atom is m μ n A and ρ = m V , so n = n con ρ parenleftbigg m μ N A parenrightbigg = n con ρ N A m μ = (3) (3000 kg / m 3 ) (6 . 02 × 10 23 ) 18 kg = 3 . 01 × 10 26 m − 3 . Since the volume is the crosssectional area multiplied by the length this tells us that A ℓ = m ρ nguyen (jmn727) – oldhomework 14 – Turner – (59070) 2 A = m ρ ℓ = 18 kg (3000 kg / m 3 ) (74 . 8953 m) = 8 . 01119 × 10 − 5 m 2 , v d = V n q e A R (1) = 172 . 8 V 3 . 01 × 10 26 m − 3 × 1 1 . 602 × 10 − 19 C × 1 (8 . 01119 × 10 − 5 m 2 )(108 Ω) = . 000414185 m / s ....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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