This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: yang (ey942) ohw13 turner (56705) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 6) 10.0 points Consider two conductors made out of the same material; i.e. , they have the same re sistivity. The material is ohmic. V 1 vector E 1 I 1 1 r 1 b V 2 vector E 2 I 2 2 r 2 b If A = r 2 , 2 = 1 , r 2 = 2 r 1 , 2 = 2 1 , and V 2 = V 1 , what is the ratio E 2 E 1 , where E 1 and E 2 are the electric fields inside of conductor 1 and of conductor 2, respectively? 1. E 2 E 1 = 1 2. E 2 E 1 = 4 3. E 2 E 1 = 1 8 4. E 2 E 1 = 1 4 5. E 2 E 1 = 8 6. E 2 E 1 = 1 2 correct 7. E 2 E 1 = 2 Explanation: We know that the potential across a dis tance l with a constant electric field E is V = E l . The ratio is E 2 E 1 = V 2 2 V 1 1 = 1 2 = 1 2 . 002 (part 2 of 6) 10.0 points What is the ratio J 2 J 1 of the current densities? 1. J 2 J 1 = 2 2. J 2 J 1 = 4 3. J 2 J 1 = 1 2 correct 4. J 2 J 1 = 8 5. J 2 J 1 = 1 8 6. J 2 J 1 = 1 4 7. J 2 J 1 = 1 Explanation: vector J = vector E The material is ohmic, and 1 = 2 , so J 2 J 1 = 2 E 2 1 E 1 = 1 2 003 (part 3 of 6) 10.0 points What is the ratio v d, 2 v d, 1 of the magnitudes of the drift velocities? 1. v d, 2 v d, 1 = 8 2. v d, 2 v d, 1 = 2 3. v d, 2 v d, 1 = 1 4. v d, 2 v d, 1 = 1 8 5. v d, 2 v d, 1 = 1 4 6. v d, 2 v d, 1 = 1 2 correct 7. v d, 2 v d, 1 = 4 Explanation: vector J = n qvectorv d yang (ey942) ohw13 turner (56705) 2 Since the conductors are made out of the same material, the density of charge carriers must be the same ( n 1 = n 2 ), so v d, 2 v d, 1 = J 2 n 2 q J 1 n 1 q = J 2 J 1 = 1 2 . 004 (part 4 of 6) 10.0 points What is the ratio 2 1 of the conductivities?...
View
Full
Document
 Spring '10
 Turner

Click to edit the document details