SOL28 - Thevalingam Donald Homework 28 Due midnight Inst...

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Thevalingam, Donald – Homework 28 – Due: May 31 2007, midnight – Inst: Eslami 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points The current in a wire decreases with time according to the relationship I = (2 . 61 mA) × e - at where a = 0 . 13328 s - 1 . Determine the total charge that passes through the wire From t = 0 to the time the current has diminished to zero. Correct answer: 0 . 0195828 C. Explanation: I = dq dt q = Z t t =0 I dt = Z t =0 (0 . 00261 A) e - 0 . 13328 s - 1 t dt = (0 . 00261 A ) e - 0 . 13328 s - 1 t - 0 . 13328 s - 1 f f f f f 0 = 0 . 0195828 C . keywords: 002 (part 1 oF 1) 10 points A copper wire carries 1 amp oF electric cur- rent. What kind oF charge does the electron ±ow create in the wire? 1. No charge. correct 2. Negative 3. Positive Explanation: The electrons are balanced out by an equal number oF protons whether or not they are moving. keywords: 003 (part 1 oF 1) 10 points In a ±uorescent tube oF diameter 4 . 4 cm, 1 . 1 × 10 18 1 / s electrons and 1 × 10 17 1 / s positive ions (with a charge oF + e ) ±ow through a cross-sectional each second. What is the current in the tube? Correct answer: 0 . 192 A. Explanation: Let : n electron = 1 . 1 × 10 18 1 / s , n ion = 1 × 10 17 1 / s , q electron = 1 . 6 × 10 - 19 C , and q ion = 1 . 6 × 10 - 19 C . Note that, while the positive and negative charges ±ow in opposite directions, the total current is their sum: I = I electron + I ion = n electron q electron + n ion q ion = (1 . 1 × 10 18 1 / s) (1 . 6 × 10 - 19 C) + (1 × 10 17 1 / s) (1 . 6 × 10 - 19 C) = 0 . 176 A + 0 . 016 A = 0 . 192 A . keywords: 004 (part 1 oF 5) 10 points A current oF 25 A exists in a copper (Cu) wire which has a diameter oF 1 mm. Each atom oF copper has one conduction band, and the average thermal speed r k T m oF an electron is 100000 m / s . The mass den- sity oF Cu is 8 . 92 g / cm 3 , its molar mass is 63 . 5 g / mol, and Avogadro’s number is 6 . 02214 × 10 23 atoms / mole. The electron mass is 9 . 10939 × 10 - 31 kg, and the resistivity oF copper is 1 . 7 × 10 - 8 Ω · m. What is the current density? Correct answer: 3 . 1831 × 10 7 A / m 2 . Explanation:
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Thevalingam, Donald – Homework 28 – Due: May 31 2007, midnight – Inst: Eslami 2 Let : I = 25 A and r = 0 . 5 mm = 0 . 0005 m . The current density is given by J I A = I π r 2 = (25 A) π (0 . 0005 m) 2 = 3 . 1831 × 10 7 A / m 2 . 005 (part 2 of 5) 10 points What is the density of conduction electrons band of copper? Correct answer: 8 . 45944 × 10 28 m - 3 . Explanation: Let : N A = 6 . 02214 × 10 23 atoms / mole , ρ m = 8 . 92 g / cm 3 = 8920 kg / m 3 , and M = 63 . 5 g / mol = 0 . 0635 kg / mol . The density of the valence electrons is n e = (# of valence electrons) volume = (# of kg) m 3 (# of moles) kg (# of atoms) mol · (1 conduction electron) atom = N A ρ m M = 6 . 02214 × 10 23 atoms / mole 0 . 0635 kg / mol × 8920 kg / m 3 = 8 . 45944 × 10 28 m - 3 . 006 (part 3 of 5) 10 points What is the drift velocity of the electrons? Correct answer: 0 . 00234854 m / s. Explanation: Let : q e = 1 . 60218 × 10 - 19 C . Current density is J I A = n e q e v d v d = J n e q e = 3 . 1831 × 10 7 A / m 2 8 . 45944 × 10 28 m - 3 × 1 1 . 60218 × 10 - 19 C = 0 . 00234854 m / s .
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SOL28 - Thevalingam Donald Homework 28 Due midnight Inst...

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