TEST02-solutions - Version 005 TEST02 TSOI (92515) This...

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Version 005 – TEST02 – TSOI – (92515) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A indicates the magnetic feld B point- ing up From the surFace oF the paper and a indicates the magnetic feld B pointing down into the surFace oF the paper. Acopperbarhasaconstan tve loc ity v in the plane oF the paper and is perpendicular to amagneticfeld B as shown in the fgures. How are charges distributed on the bar? 1. v B B + + - - 2. v B B - - + + 3. v B B - - + + cor- rect 4. v B B - - + + Explanation: Positive charges will move in the direction oF the magnetic Force, while negative charges move in the opposite direction. Study the case where the magnetic feld B is pointed into the plane oF the paper and the bar is moving From right to leFt ( ). To produce the indicated charge separa- tion, the positive charges in the conductor ex- perience downward magnetic Forces while the negative charges in the conductor experience upward magnetic Forces leaving the charge separation shown in the fgure. F F v B B - - + + Using the right-hand rule " F = q"v × " B ,the top will be negative and the bottom will be positive. The other three choices are incorrect since they do not abide by the right-hand rule. 002 10.0 points In this problem assume that the batteries have zero internal resistance. The currents are ±owing in the direction indicated by the arrows. 11 . 24 Ω 3 . 2V 11 V I 1 I 2 I 3 ²ind the current through the 11 . 6Ωresistor and the 3 . 2VbatteryatthetopoFthecircuit. 1. 2.74242 2. 2.06061 3. 1.22414 4. 4.05357
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Version 005 – TEST02 – TSOI – (92515) 2 5. 2.26582 6. 3.1746 7. 1.11111 8. 1.59412 9. 1.5098 10. 3.0 Correct answer: 1 . 22414 A. Explanation: R 1 R 2 E 1 E 2 I 1 I 2 I 3 Let : R 1 =11 . , R 2 =24Ω , E 1 =3 . 2V , and E 2 =11V . At nodes, we have I 1 - I 2 - I 3 =0 . (1) Pay attention to the sign of the battery and the direction of the current in the Fgure. Us- ing the lower circuit in the Fgure, we get E 2 + I 2 R 2 (2) so I 2 = - E 2 R 2 = - 11 V 24 Ω = - 0 . 458333 A . Then, for the upper circuit E 1 - I 2 R 2 - I 1 R 1 . (3) E 1 + E 2 - I 1 R 1 I 1 = E 1 + E 2 R 1 = 3 . 2V+11V 11 . = 1 . 22414 A . Alternate Method: Using the outside loop -E 1 -E 2 + I 1 R 1 (4) I 1 = E 1 + E 2 R 1 . keywords: 003 10.0 points Two coils are suspended around a central axis as shown in the Fgure below. One coil is connected to a resistor with ends labeled a and b .Theotherco i lisconnectedtoabattery E .Theco i lsaremov ingre lat ivetoeachother as indicated by the velocity vectors v . Use Lenz’s law to answer the following ques- tion concerning the direction of induced cur- rents and magnetic Felds.
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This note was uploaded on 06/10/2011 for the course PHY 302l taught by Professor Morrison during the Summer '08 term at University of Texas at Austin.

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TEST02-solutions - Version 005 TEST02 TSOI (92515) This...

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