Spring 11 Phys 121 - HW 06

Spring 11 Phys 121 - HW 06 - dzifanu(sd26397 – hw 6 –...

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Unformatted text preview: dzifanu (sd26397) – hw 6 – opyrchal – (121102) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The current in a wire decreases with time according to the relationship I = (3 . 59 mA) × e − a t 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 . 0269358 C. Explanation: I = dq dt q = integraldisplay t t =0 I dt = integraldisplay ∞ t =0 (0 . 00359 A) e − . 13328 s- 1 t dt = (0 . 00359 A ) e − . 13328 s- 1 t- . 13328 s − 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ∞ = . 0269358 C . 002 10.0 points The drift velocity of free electrons in a cop- per wire is 6 mm / s, resistivity is 1 . 67 × 10 − 8 Ω · m, and the free electron density is 8 . 42 × 10 28 electrons / m 3 . Calculate the electric field in the conductor. Correct answer: 1 . 35158 N / C. Explanation: Let : v d = 6 mm / s = 0 . 006 m / s , n = 8 . 42 × 10 28 electrons / m 3 , ρ = 1 . 67 × 10 − 8 Ω · m , and q e = 1 . 602 × 10 − 19 C / electron . From Ohm’s law, J = σ E = E ρ = n q V d E = n q v d ρ = (8 . 42 × 10 28 electrons / m 3 ) · (1 . 602 × 10 − 19 C / electron) · (0 . 006 m / s) (1 . 67 × 10 − 8 Ω · m) = 1 . 35158 N / C . 003 10.0 points A conductor with cross-sectional area 6 cm 2 carries a current of 8 A. If the concentration of free electrons in the conductor is 9 × 10 28 electrons / m 3 , what is the drift velocity of the electrons?...
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This note was uploaded on 09/16/2011 for the course PHYS PHYS 121 taught by Professor Prodan during the Spring '09 term at NJIT.

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Spring 11 Phys 121 - HW 06 - dzifanu(sd26397 – hw 6 –...

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