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hw3 solutions

# hw3 solutions - homework 05 – BOBBITT JAMES – Due 4:00...

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Unformatted text preview: homework 05 – BOBBITT, JAMES – Due: Feb 18 2008, 4:00 am 1 Question 1, chap 27, sect 1. part 1 of 2 10 points The current I = a t 2- b t + c in a section of a conductor depends on time. What quantity of charge moves across the section of the conductor from t = 0 to t = t 1 ? 1. q = a 3 t 3 1- b 2 t 2 1 + c 2. q = a t 3 1- b t 2 + c t 1 3. q = a t 3 1- b 2 t 2 1 + c t 1 4. q = a 3 t 3 1- b 2 t 2 1 + c t 1 correct 5. q = a t 2 1- b t 1 + c Explanation: The unit of current is Coulomb per second I = d q dt or dq = I dt . To find the total charge in coulombs that passes through the conductor, one must inte- grate the current over the time interval. q = integraldisplay t 1 dq = integraldisplay t 1 I dt = integraldisplay t 1 ( a t 2- b t + c ) dt = bracketleftbigg a 3 t 3 1- b 2 t 2 1 + c t bracketrightbiggvextendsingle vextendsingle vextendsingle vextendsingle t 1 = a 3 t 3 1- b 2 t 2 1 + c t 1 . Question 2, chap 27, sect 1. part 2 of 2 10 points If I is in A, and a = 2 C / s 3 , b = 3 C / s 2 , and c = 7 C / s, what quantity of charge moves across the section of the conductor from t 1 = 2 s to t 2 = 4 s? Correct answer: 33 . 3333 C (tolerance ± 1 %). Explanation: Let : a = 2 C / s 3 , b = 3 C / s 2 , c = 7 C / s , t 1 = 2 s , and t 2 = 4 s . Since q = bracketleftbigg a 3 t 3- b 2 t 2 + c t bracketrightbigg vextendsingle vextendsingle vextendsingle vextendsingle t 2 t 1 , then q 2 = ( 2 C / s 3 ) 3 (4 s) 3- ( 3 C / s 2 ) 2 (4 s) 2 + (7 C / s)(4 s) = 46 . 6667 C , and q 1 = ( 2 C / s 3 ) 3 (2 s) 3- ( 3 C / s 2 ) 2 (2 s) 2 + (7 C / s)(2 s) = 13 . 3333 C . so that q 21 = q 2- q 1 = 33 . 3333 C . Question 3, chap 27, sect 3. part 1 of 1 10 points A current of 2 A flows in a copper wire 2 mm in diameter. The density of valence electrons in copper is roughly 9 × 10 28 m- 3 ....
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hw3 solutions - homework 05 – BOBBITT JAMES – Due 4:00...

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