PHY303LHW5S

# PHY303LHW5S - rosen(arr956 Homework 05 Chiu(58295 This...

This preview shows pages 1–3. Sign up to view the full content.

rosen (arr956) – Homework 05 – Chiu – (58295) 1 This print-out should have 14 questions. Multiple-choice 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 2 . 06 cm 2 varies with time as q = q 1 t 3 + q 2 t + q 3 , where q 1 = 4 . 7 C / s 3 , q 2 = 7 . 9 C / s, q 3 = 5 . 7 C, and t is in seconds. What is the instantaneous current through the surface at t = 0 . 6 s? Correct answer: 12 . 976 A. Explanation: Let : q 1 = 4 . 7 C / s 3 , q 2 = 7 . 9 C / s , q 3 = 5 . 7 C , and t = 0 . 6 s . I d q dt = 3 q 1 t 2 + q 2 = 3 ( 4 . 7 C / s 3 ) (0 . 6 s) 2 + 7 . 9 C / s = 12 . 976 A . 002 (part 2 of 2) 10.0 points What is the value of the current density at t = 0 . 6 s? Correct answer: 62990 . 3 A / m 2 . Explanation: Let : a = 2 . 06 cm 2 = 0 . 000206 m 2 and t = 0 . 6 s . J I A = 12 . 976 A 0 . 000206 m 2 = 62990 . 3 A / m 2 . 003 (part 1 of 3) 10.0 points Consider two cylindrical conductors made of the same ohmic material. Conductor 1 has a radius r 1 and length 1 while conductor 2 has a radius r 2 and length 2 . Denote: The currents of the two conductors as I 1 and I 2 , the potential differences between the two ends of the conductors as V 1 and V 2 , and the electric fields within the conductors as E 1 and E 2 . V 1 vector E 1 I 1 1 r 1 V 2 vector E 2 I 2 2 r 2 If ρ 2 = ρ 1 , r 2 = 2 r 1 , ℓ 2 = 3 1 and V 2 = V 1 , find the ratio R 2 R 1 of the resistances. 1. R 2 R 1 = 2 3 2. R 2 R 1 = 2 3. R 2 R 1 = 1 4 4. R 2 R 1 = 1 2 5. R 2 R 1 = 4 3 6. R 2 R 1 = 3 2 7. R 2 R 1 = 4 8. R 2 R 1 = 3 9. R 2 R 1 = 3 4 correct 10. R 2 R 1 = 1 3 Explanation:

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

View Full Document
rosen (arr956) – Homework 05 – Chiu – (58295) 2 The relation between resistance and resis- tivity is given by R = ρ ℓ A = ρ ℓ π r 2 . Then since r 2 = 2 r 1 and 2 = 3 1 , the ratio of the resistances is R 2 R 1 = ρ ℓ 2 π r 2 2 π r 2 1 ρ ℓ 1 = 2 r 2 1 1 r 2 2 = (3 1 ) r 2 1 1 (2 r 1 ) 2 = 3 4 . 004 (part 2 of 3) 10.0 points When the two conductors are attached to a battery of voltage V, determine the ratio E 2 E 1 of the electric fields.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern