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finalexamSPRING2004

finalexamSPRING2004 - Flores Michael – Final 1 – Due...

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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 Ampere’s law, the mag- netic fields from the long wire at distances c and ( c + a ) away are B c =- μ I 1 2 π c ˆ k B ca =- μ I 1 2 π ( c + a ) ˆ 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 = μ I 1 I 2 ‘ 2 π • + 1 c + a- 1 c ‚ ˆ ı = μ (3 . 1 A)(13 A)(0 . 54 m) 2 π × • + 1 (0 . 1 m) + (0 . 13 m)- 1 (0 . 1 m) ‚ ˆ ı =- 2 . 46005 × 10- 5 Nˆ ı | 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 Earth’s 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 π R 2 . Solving for the total power output, we have P = I 4 π R 2 = (1690 W / m 2 )4 π (1 . 49 × 10 11 meters) 2 = 4 . 71486 × 10 26 W ....
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