{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution_pdf 19

# solution_pdf 19 - malsam(wgm329 – hw19 – turner...

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: malsam (wgm329) – hw19 – turner – (56705) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the set up shown in the figure where a solenoid has a steadily increasing magnetic flux which generates identical in- duced emf s for the two cases illustrated. B B B B A #1 B #2 i i Case 1: Case 1: Two identical light bulbs are in series. The corresponding electrical power consumed by bulb 1 and bulb 2 are P 1 and P 2 , respectively. B B B B A #1 B #2 D C O i 2 i 1 Case 2: Case 2: Let the points C and D be on the symmetry line of the diagram. Connect points C and D by a wire, which equally divides the magnetic flux. The corresponding electrical power consumed by bulb 1 and bulb 2 are P ′ 1 and P ′ 2 , respectively. What is the ratio of P ′ 1 P 1 ? It may be helpful to first find an expression for P 1 and then write down the loop equation for ACODA in case 2. 1. P ′ 1 P 1 = 2 2. P ′ 1 P 1 = 1 correct 3. P ′ 1 P 1 = 4 4. P ′ 1 P 1 = 8 5. P ′ 1 P 1 = 1 3 6. P ′ 1 P 1 = 1 8 7. P ′ 1 P 1 = 1 4 8. P ′ 1 P 1 = 3 9. P ′ 1 P 1 = 1 2 10. P ′ 1 P 1 = 0 Explanation: Let E and R be the induced emf and resis- tance of the light bulbs, respectively. For case 1, since the two bulbs are in series, the equivalent resistance is simply R eq = R + R = 2 R and the current through the bulbs is I = E 2 R . Hence, for case 1, the power consumed by bulb 1 is P 1 = parenleftbigg E 2 R parenrightbigg 2 R = E 2 4 R . For case 2, the loop equation for ACODA is E 2 − I 1 R = 0 . Solving for I 1 yields I 1 = E 2 R . Hence the power dissipated by bulb 2 for case 2 is P ′ 1 = parenleftbigg E 2 R parenrightbigg 2 R = E 2 4 R . malsam (wgm329) – hw19 – turner – (56705) 2 This is identical to the expression for P 1 . Therefore, P ′ 1 P 1 = 1 . keywords: 002 (part 1 of 3) 10.0 points Four long, parallel conductors carry equal cur- rents of I . An end view of the conductors is shown in the figure. Each side of the square has length of ℓ . A B D C × P x y ℓ Which diagram correctly denotes the direc- tions of the magnetic fields from each conduc- tor at the point P ? The current direction is out of the page at points A , B , and D indi- cated by the dots and into the page at point C indicated by the cross. 1. P B A B B B C B D 2. P B A B B B C B D 3. P B A B B B C B D 4. P B A B B B C B D correct 5. P B A B B B C B D Explanation: The direction of the magnetic field due to each wire is given by the right hand rule. Place the thumb of the right hand along the direction of the current; your fingers now curl in the direction of the magnetic field’s circular path: A B D C × P B A B B B D B C 003 (part 2 of 3) 10.0 points At point P , as far as the magnitudes are concerned, B A = B B = B C = B D ≡ B i , where B i is introduced to represent any of the four magnetic fields....
View Full Document

{[ snackBarMessage ]}

### Page1 / 9

solution_pdf 19 - malsam(wgm329 – hw19 – turner...

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

View Full Document
Ask a homework question - tutors are online