X2.5a 1150PHYS-1150(5) (Kaldon-40527) Name __________S O L U T I O N_________WMU - Fall 2006Exam 2A - 100,000 points Book Title ________________________________________________Rev. 10/23/06 Mo.4State 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-5)FALL 2006WMUkNmCCNm=´×=´×-8 988 10885 1092201222..eeCTm A=´=´ ×--1602 104101907./mpMmm… Yummy Physics ConstantsGoes Good With Cookies…I love this job!Physics 1150 / Exam 2 Fall 2006 Page 2“R” Is Just The 18th Letter Of The Alphabet… Except In Physics! (50,000 points)1.) Consider a 625 W resistor made from a cylinder with a length of 2.50 cm and a diameterof 1.00 cm. (a) What would be the resistivity r of this resistor’s material?Ar m m mRLARALmmm== = = ´===´=×--pprr22252520 00500 0 00007854 7 854 10625 7 854 10002501964.. ....bgbgchbgWW (b) Hook this resistor up to a 16.55 volt battery. Find the voltage drop, the current and power lost of thisresistor.V voltsVIRIVRvoltsAP IV A volts W==== === =16 5516 556250 026480 02648 16 55 0 4382......Wbgbg(c) Take four identical 625 W resistors and our 16.55 volt battery and hookthem up in the circuit as shown. Find the equivalent resistance of thecircuit, R1234 , by reducing the resistor network.R2 is shorted out, so doesn’t count.111 1625162526251312 5625 312 5937 534 3 41234 134 1 34RRRRRRR=+= + = ===+= +=WW W WWWW...Physics 1150 / Exam 2 Fall 2006 Page 3(d) For an infinite system of 100 W resistors, as shown below, find theequivalent resistance. Hint: This isn’t impossible. You may need to trysome shorter, finite systems first.(e) An RC circuit consists of a 1130 W resistor and a 311mF capacitor. Find the time t in seconds it takes for thefully discharged capacitor to charge to two-thirds of itsmaximum value, i.e. q = 0.667 Qmax .qQqQ eQQ eeeetRCtRCFtRCtRCtRCtRCtRC==-FHIK=-FHIK=-=- =FHIK=-==-=-==-----066710667 10667 11 0 667 0 33303330 33303330 333 1130 311386 400 0 38640000.....ln ln .ln .ln .ln .,sec.secbgbgbgbgW mmRR R+= = =+=+=+=+=+=+=+=+=+=+=+=22100 200100 100 20011001200166 67100 66 67 166 711001166 7162 50100 62 50 162 511001162 516190100 6190 161911001161916182100 6182 161811001161816180100 6180 1618.... ...... ... ... ...... ...... ... ... .WWWW WWW WWWWWW WWWWWW WWW WWW WWWWWWW WWWWbgPhysics 1150 / Exam 2 Fall 2006 Page 4It Must Be My Animal Magnetism (50,000 points) (a)2.) (a) If a magnetic field B1 is turned on, a negative charge -q will go around in circularpath of radius r as shown. If an electric field E2 is turned on as well, the charge will go ina straight line. Sketch on the diagram the directions of the vectors for the magnetic field,the magnetic force, the electric field and the electric force.The charged particle problem above has the following properties: mass m = 0.0100 kg,charge q = -0.0350 C, speed v = 222 m/s, and radius r = 0.120 m. Find the magnitudesof (b) the magnetic force and magnetic field, and (c) the electric force and electric field. (b) FFmvrkg m smNFqvBBFqvNCmsTBcBB== = ==== =2200100 22201204107410700350 222528 6./../.bgbgbgbgbg (c)(d) A coil consists of 4550 turns of wire in a length of 15.0 cm and a diameter of 8.00 cm. The B-field in thesolenoid measures B = 0.00135 T, as shown. Find the magnitude of the current I in the coil and indicatewhether the current comes out the LEFT or the RIGHT of the coil.BNIIBNTmTm AA===´×=-mmp0070 00135 0150410 45500 03542ll../.bgbgchbg(e) For the same solenoid coil as in (d), with B = 0.00135 T, a second wire is centered directly in front of thecoil so it gets hit with the coil’s B-field. If the current in this second wire is i = 0.500 A as shown, find themagnitude and direction of the magnetic force, FB , on this second wire. Use the diameter of the coil as thelength of the second wire.Magnetic force on the wire, FB , points OUT of the page.FF NFqEEFqNCNCEBEE======4107410700350117 300.,/FilB A m TNNB====´-0 500 0 0800 0 001350 00005400 5400 105..