TEST02-solutions-1 - Version 069 TEST02 tsoi (57210) This...

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Version 069 – TEST02 – tsoi – (57210) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points AcircularcoilismadeoF N turns oF copper wire as shown in the fgure. A resistor R is in- serted in the copper wire. Initially, a uniForm magnetic feld oF magnitude B i points hor- izontally From leFt-to-right through the per- pendicular plane oF the coil. When viewed From the right the coil is wound counter-clockwise. R Magnetic ±ield B ( t ) During a time interval t the feld uniFormly changes at a constant rate, until a reversed feld is reached equal in magnitude to the initial feld. The electrons in the resistor, R, shown in the fgure 1. move in a direction that is undetermi- nated From the inFormation given. 2. do not move in a preFerred direction, since they have thermal kinetic energy. 3. move leFt-to-right. 4. move right-to-leFt. correct Explanation: As the leFt-to-right magnetic feld decreases (and eventually ²ipping sign and increasing in magnitude) it Follows From Lenz’s law (op- position to the change in magnetic feld will tend to keep the current constant and ²owing in the same direction) that the induced emF will produce an leFt-to-right magnetic feld arising From induced currents in the coil. By the right hand rule, the induced current ²ows counter-clockwise when viewed From the right. The coils are wound counter-clockwise as the wire goes From the right to leFt termi- nals. The current must enter the loop From the right terminal and exit at the leFt termi- nal. Since the current is continuous, the current must ²ow through the resistor in the leFt-to- right direction, which means that electrons are ²owing right-to-leFt in the resistor. That is, the electrons ²ow in the opposite direction From the current since they have a negative charge. 002 10.0 points Calculate the resistance in an RL circuit in which L =1 . 67 H and the current increases to 85 . 2% oF its fnal value in 4 . 57 s. 1. 2.13199 2. 1.15712 3. 1.35266 4. 1.30311 5. 0.698163 6. 2.93158 7. 7.1511 8. 3.30293 9. 6.53223 10. 6.25451 Correct answer: 0 . 698163 Ω. Explanation: Let : L . 67 H , R =85 . 2% , and t =4 . 57 s . At time t the current is I ( t )= I 0 ± 1 - e - Rt/L ² = I 0 R . So the resistance is R = - L t ln(1 -R ) = - 1 . 67 H 4 . 57 s ln(1 - 0 . 852) = 0 . 698163 Ω .
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Version 069 – TEST02 – tsoi – (57210) 2 003 10.0 points The fgure below shows a cylindrical coaxial cable oF radii a , b ,and c in which equal, uni- Formly distributed, but antiparallel currents i exist in the two conductors. O i out ± i in F E D C r 1 r 2 r 3 r 4 c b a Which expression gives the magnitude B ( r 3 )a t D oF the magnetic feld in the re- gion b<r 3 <a ? 1. B ( r 3 )= μ 0 ir 3 2 πb 2 2. B ( r 3 μ 0 i 2 πr 3 3. B ( r 3 μ 0 i ( a 2 - r 2 3 ) 2 3 ( a 2 - b 2 ) correct 4. B ( r 3 μ 0 i ( a 2 + r 2 3 - 2 b 2 ) 2 3 ( a 2 - b 2 ) 5. B ( r 3 μ 0 i ( a 2 - b 2 ) 2 3 ( r 2 3 - b 2 ) 6. B ( r 3 μ 0 3 2 πc 2 7. B ( r 3 μ 0 3 2 πa 2 8. B ( r 3 μ 0 i 3 9. B ( r 3 μ 0 i ( r 2 3 - b 2 ) 2 3 ( a 2 - b 2 ) 10. B ( r 3 )=0 Explanation: Ampere’s Law states that the line inte- gral ± - B · d - , around any closed path equals μ 0 I ,where I is the total steady current pass- ing through any surFace bounded by the closed path.
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This note was uploaded on 03/22/2012 for the course PHYS 302L taught by Professor Tsoi during the Spring '10 term at University of Texas at Austin.

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TEST02-solutions-1 - Version 069 TEST02 tsoi (57210) This...

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