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Unformatted text preview: lim (kl9356) – Homework 9 – Weathers – (22202) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The quantity of charge passing through a sur face of area 1 . 87 cm 2 varies with time as q = q 1 t 3 + q 2 t + q 3 , where q 1 = 4 . 2 C / s 3 , q 2 = 4 . 6 C / s, q 3 = 3 . 6 C, and t is in seconds. What is the instantaneous current through the surface at t = 0 . 4 s? Correct answer: 6 . 616 A. Explanation: Let : q 1 = 4 . 2 C / s 3 , q 2 = 4 . 6 C / s , q 3 = 3 . 6 C , and t = 0 . 4 s . I ≡ d q dt = 3 q 1 t 2 + q 2 = 3 ( 4 . 2 C / s 3 ) (0 . 4 s) 2 + 4 . 6 C / s = 6 . 616 A . 002 (part 2 of 2) 10.0 points What is the value of the current density at t = 0 . 4 s? Correct answer: 35379 . 7 A / m 2 . Explanation: Let : a = 1 . 87 cm 2 = 0 . 000187 m 2 and t = 0 . 4 s . J ≡ I A = 6 . 616 A . 000187 m 2 = 35379 . 7 A / m 2 . 003 10.0 points A conductor with crosssectional area 9 cm 2 carries a current of 17 A. If the concentration of free electrons in the conductor is 7 × 10 28 electrons / m 3 , what is the drift velocity of the electrons? Correct answer: 0 . 00168422 mm / s. Explanation: Let : I = 17 A , n = 7 × 10 28 electrons / m 3 , q e = 1 . 60218 × 10 − 19 C , and A = 9 cm 2 = 0 . 0009 m 2 . The current in a conductor is given by I = n q v d A , where n is the number of charge carriers per unit volume, q is the charge per carrier, v d is the drift velocity of the carriers and A is the cross section of the conduction. Solving for v d , we have v d = I n q A = 17 A 7 × 10 28 electrons / m 3 × 1 1 . 60218 × 10 − 19 C × 1 . 0009 m 2 × 1000 mm 1 m = . 00168422 mm / s . 004 10.0 points The drift velocity of free electrons in a cop per wire is 7 mm / s, resistivity is 1 . 8 × 10 − 8 Ω · m, and the free electron density is 8 . 43 × 10 28 electrons / m 3 ....
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 Spring '00
 Littler
 Charge, Magnetism, Work, lim, Correct Answer, Electrical resistance

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