Spring 11 Phys 121 - HW 06

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

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 10.0points The current in a wire decreases with time according to the relationship I = (3 . 59 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 . 0269358 C. Explanation: I = dq dt q = integraldisplay t t =0 I dt = integraldisplay t =0 (0 . 00359 A) e 0 . 13328 s - 1 t dt = (0 . 00359 A ) e 0 . 13328 s - 1 t - 0 . 13328 s 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 0 = 0 . 0269358 C . 002 10.0points 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 10.0points 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? Correct answer: 0 . 000924668 mm / s. Explanation: Let : I = 8 A , n = 9 × 10 28 electrons / m 3 , q e = 1 . 60218 × 10 19 C , and A = 6 cm 2 = 0 . 0006 m 2 .
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern