hw19-solutions

# hw19-solutions - Reilly(mar3978 hw19 turner(56725 This...

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

Reilly (mar3978) – hw19 – turner – (56725) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Consider the set up shown in the fgure where a solenoid has a steadily increasing magnetic ±ux 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 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 ±ux. 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 frst fnd 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 = 0 3. P 1 P 1 = 1 8 4. P 1 P 1 = 8 5. P 1 P 1 = 1 3 6. P 1 P 1 = 1 2 7. P 1 P 1 = 4 8. P 1 P 1 = 3 9. P 1 P 1 = 1 4 10. P 1 P 1 = 1 correct Explanation: Let E and R be the induced emf and resis- tance oF the light bulbs, respectively. ²or 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 = p E 2 R P 2 R = E 2 4 R . ²or case 2, the loop equation For 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 = p E 2 R P 2 R = E 2 4 R .

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

View Full Document
Reilly (mar3978) – hw19 – turner – (56725) 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 ±gure. Each side of the square has length of . A B D C × P x y Which diagram correctly denotes the direc- tions of the magnetic ±elds 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 5. P B A B B B C B D correct Explanation: The direction of the magnetic ±eld 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 ±ngers now curl in the direction of the magnetic ±eld’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 ±elds.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

hw19-solutions - Reilly(mar3978 hw19 turner(56725 This...

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

View Full Document
Ask a homework question - tutors are online