finalexamSPRING2004

finalexamSPRING2004 - Flores, Michael Final 1 Due: May 15...

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: Flores, Michael Final 1 Due: May 15 2004, 1:00 pm Inst: Sonia Paban 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Given: A current in the long, straight wire which lies in the plane of the rectangular loop, that also carries a current, as shown in the figure. 10 cm 13 cm 54cm 3 . 1A 13A x y Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Correct answer: 2 . 46005 10- 5 N. Explanation: Let : c = 0 . 1 m , a = 0 . 13 m , = 0 . 54 m , I 1 = 3 . 1 A , and I 2 = 13 A . c a I 1 I 2 x y Note: The magnetic forces on the top and the bottom segments of the rectangle cancel. We now concern ourselves with the seg- ments of the rectangle which are parallel to the long wire. Using Amperes law, the mag- netic fields from the long wire at distances c and ( c + a ) away are B c =- I 1 2 c k B ca =- I 1 2 ( c + a ) k . The forces on the left and right vertical segments of the rectangle are, respectively, F c = I 2 B c and F ca = I 2 B ca . The forces are oppositely directed since the current I 2 has a direction in the loop at distances c op- posite from the current I 2 in the loop at a distance c + a , away from the long wire. The net force is ~ F = I 1 I 2 2 + 1 c + a- 1 c = (3 . 1 A)(13 A)(0 . 54 m) 2 + 1 (0 . 1 m) + (0 . 13 m)- 1 (0 . 1 m) =- 2 . 46005 10- 5 N | F | = 2 . 46005 10- 5 N . 002 (part 2 of 2) 10 points What is the direction of the net force? Flores, Michael Final 1 Due: May 15 2004, 1:00 pm Inst: Sonia Paban 2 1. no net torque and no net force 2. down 3. right 4. up 5. left correct 6. out 7. only a torque, no net force 8. in Explanation: Currents in the same direction attract, while currents in the opposite direction re- pel. As the arm of the loop with length closest to the long wire with current I 1 is at- tractive, with a larger magnitude of force than the repulsive force of the arm at a larger dis- tance from the long wire, the total force the loop feels is directed toward the long wire. Since right is the positive x direction, the net force is in the left direction. Note: Observe the sign of the force in the previous part. 003 (part 1 of 1) 10 points The Earth is 1 . 49 10 11 meters from the sun. If the intensity of the solar radiation at the top of the Earths atmosphere is 1690 W / m 2 , what is the total power output of the Sun? Correct answer: 4 . 71486 10 26 W. Explanation: The intensity of the radiation at a distance R from the source is I = P 4 R 2 . Solving for the total power output, we have P = I 4 R 2 = (1690 W / m 2 )4 (1 . 49 10 11 meters) 2 = 4 . 71486 10 26 W ....
View Full Document

This note was uploaded on 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.

Page1 / 12

finalexamSPRING2004 - Flores, Michael Final 1 Due: May 15...

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