X2.6a 1150PHYS-1150(6) (Kaldon-40451) Name __________S O L U T I ON_________WMU - Fall 2007Exam 2A - 100,000 points Book Title ________________________________________________Rev. 10/22/07 Mo.4.r1State Any Assumptions You Need To Make – Show All Work – Circle Any Final AnswersUse Your Time Wisely – Work on What You Can – Be Sure to Write Down EquationsShort Answers Should Be Short! – Feel Free to Ask Any QuestionsEXAM 2 [FORM - A]PHYS-1150 (KALDON-6)FALL 2007WMUkNmCCNm=´×=´×-8 988 10885 1092201222..eeCTm A=´=´ ×--1602 104101907./mpPhysics Tricks or TreatsCookies, Candies, Problems…Boo!Physics 1150 / Exam 2 Fall 2007 Page 2You Can’t Spell “Resistance” Without W (50,000 points)1.) Consider a resistor made from a cylinder of resistivity r = 8.125 W ·m, with a length of2.50 cm and a diameter of 1.00 cm. (a) What would be the resistance R of this cylinder?Dcm mrD mRLAmmm== ====×=100 0 0100 0 005008125002500 005002586122..; ....rpWWbgbgbg(b) Hook this resistor up to a 16.55 volt battery. Find the voltage drop, the current and power lost of thisresistor. If you didn’t get an answer to (a), use R = 812 W.VvoltsVIRIVRvoltsAPIV A volts WOR PVRvoltsW==== === === =16 5516 5525860 0064000 006400 16 55 0105916 5525860105922........WWbgbgbg(c) Take five identical 625 W resistors and our 16.55 volt battery and hookthem up in the circuit as shown. Find the equivalent resistance of thecircuit, R12345 , by reducing the resistor network.Resistors R2 and R3 don’t contribute due to shortcircuits which route the current around them. Parallel Series Done!111 162516252625312 5625 312 5 937 545 4 545145 1 45RRRRRRR=+= + ===+ = + =WW WWWW W...Physics 1150 / Exam 2 Fall 2007 Page 3(d) This circuit cannot be reduced by network reduction. UseKirchhoff’s Laws and write down the necessary junction and loopequations – BUT DO NOT SOLVE THEM. In order to make lifeeasier for grading, take all resistor currents as pointing either tothe right or down as needed. Draw the direction of your currentsand loops on the circuit diagram for full credit.Junctions(corrected)iiiORiiiiiiiiiiii123 12334624561500=+ --==+++==+bgLoops+- - - =-+ + =-++ =ViR iR iRiR iR iRiR iR iR11 33 6 622 44 3344 55 66000(e) An RC circuit consists of a 1150 W resistor and a 511 mF capacitor. Find thetime t in seconds it takes for the fully charged capacitor to discharge to two-thirdsof its maximum value, i.e. q = 0.667 Qmax .t == ´ ======-=-===-----RC FqQe q QQQeeetRCtRCtRCtRCtRCtRC1150 511 10 0 58770667066706670667040500 4050 0 4050 0 5877023806Wbgchdibgbgbgbg.sec;...ln ln .....sec.secPhysics 1150 / Exam 2 Fall 2007 Page 4These Aren’t Your Grandmother’s Kitchen Magnets (50,000 points)2.) (a) If a magnetic field B1 = 0.150 T is turned on, a negative charge -q will go around incircular path of radius r as shown. Sketch on the diagram the directions of the vectors forthe magnetic field and the magnetic force.The charged particle problem above has the following properties: charge q = -0.0350 C,speed v = 2220 m/s, and radius r = 0.120 m. Find (b) the magnitude of the magnetic forceand (c) the mass of the charged particle.(b) FqvBCmsTNB==-=-0 0350 2220 01501166./.() .bgbgbg(c) FFmvrmFrvNmmskgBcB=====´-22271166 012022202839 10../.bgbgbgTo get only charged particles traveling at 2220 m/s, we must pass them through avelocity selector. If the electric field E = 1250 N/C as shown, find (d) the magnitudeand direction of the magnetic force FB and (e) the magnitude and direction of theapplied magnetic field B, to make this velocity selector work.(d) FFqECNCNBE====0 0350 125043 75./.bgb gFB points UP (+y-direction)(because FE has to point down because q is negative)(e) FqvBBFqvNCmsTBB====43 7500350 222005631../.bgb gB points OUT of the page by
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