Current and current density-solutions

Current and current density-solutions - gilvin (jg47854) -...

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? t = 4.43 mC gilvin (jg47854) - 10: Current and current density - meyers - (21235) 1 This print-out should have 11 questions. Multiple-choice questions may continue on 003 10.0 points the next column or page - find all choices before answering. This homework is due Tuesday, March 12, at midnight Tucson time. 001 10.0 points A total charge of 4.43 mC passes through a cross-sectional area of a wire in 0.882 s. What is the current in the wire? Correct answer: 5.02268 mA. Explanation: Let : Δ Q = 4.43 mC and Δ t = 0.882 s . I = Δ Q = 5.02268 mA . Calculate the average drift speed of electrons traveling through a copper wire with a cross- sectional area of 40 mm 2 when carrying a current of 50 A (values similar to those for the electric wire to your study lamp). Assume one electron per atom of copper contributes to the current. The atomic mass of copper is 63.5 g/mol and its density is 8.93 g/cm 3 . Avogadro’s number N A is 6.02 × 10 23 . Correct answer: 9.22817 × 10 −5 m/s. Explanation: Let : N = 1 , M = 63.5 g/mol , ρ = 8.93 g/cm 3 , A = 40 mm 2 0.882 s 002 10.0 points The drift velocity of free electrons in a cop- per wire is 7 mm/s, resistivity is 1.63 × 10 −8 Ω · m, and the free electron density is 8.44 × 10 28 electrons/m 3 . Calculate the electric field in the conductor. Correct answer: 1.54273 N/C. Explanation: = 4 × 10 −5 m 2 , I = 50 A , and q e = 1.6 × 10 −19 C/electron . We first calculate n, the number of current- carrying electrons per unit volume in copper. Assuming one free conduction electron per atom, n = N A ρ M , where N A is Avogodro ’s number and ρ and M are the density and the atomic weight of copper, respectively ( ) Let : v d = 7 mm/s = 0.007 m/s , n ≡ n = 8.44 × 10 28 electrons/m 3 , N A ρ 1 electron
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This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

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Current and current density-solutions - gilvin (jg47854) -...

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