Homework_05-solutions

Homework_05-solutions - toupal (rgt374) – Homework 05 –...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: toupal (rgt374) – 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 . 29 cm 2 varies with time as q = q 1 t 3 + q 2 t + q 3 , where q 1 = 5 . 2 C / s 3 , q 2 = 4 . 1 C / s, q 3 = 9 . 4 C, and t is in seconds. What is the instantaneous current through the surface at t = 1 . 4 s? Correct answer: 34 . 676 A. Explanation: Let : q 1 = 5 . 2 C / s 3 , q 2 = 4 . 1 C / s , q 3 = 9 . 4 C , and t = 1 . 4 s . I ≡ dq dt = 3 q 1 t 2 + q 2 = 3 ( 5 . 2 C / s 3 ) (1 . 4 s) 2 + 4 . 1 C / s = 34 . 676 A . 002 (part 2 of 2) 10.0 points What is the value of the current density at t = 1 . 4 s? Correct answer: 1 . 51424 × 10 5 A / m 2 . Explanation: Let : a = 2 . 29 cm 2 = 0 . 000229 m 2 and t = 1 . 4 s . J ≡ I A = 34 . 676 A . 000229 m 2 = 1 . 51424 × 10 5 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 ℓ 1 while conductor 2 has a radius r 2 and length ℓ 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 ℓ 1 r 1 b V 2 vector E 2 I 2 ℓ 2 r 2 b If ρ 2 = ρ 1 , r 2 = 2 r 1 , ℓ 2 = 3 ℓ 1 and V 2 = V 1 , find the ratio R 2 R 1 of the resistances. 1. R 2 R 1 = 3 4 correct 2. R 2 R 1 = 1 2 3. R 2 R 1 = 2 3 4. R 2 R 1 = 3 2 5. R 2 R 1 = 1 4 6. R 2 R 1 = 3 7. R 2 R 1 = 2 8. R 2 R 1 = 4 9. R 2 R 1 = 1 3 10. R 2 R 1 = 4 3 Explanation: toupal (rgt374) – 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 ℓ 2 = 3 ℓ 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 = ℓ 2 r 2 1 ℓ 1 r 2 2 = (3 ℓ 1 ) r 2 1 ℓ 1 (2 r 1 ) 2 = 3 4 ....
View Full Document

This note was uploaded on 04/26/2009 for the course PHY 58310 taught by Professor Chiu during the Spring '09 term at University of Texas.

Page1 / 6

Homework_05-solutions - toupal (rgt374) – Homework 05 –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online