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# Electric And Magnetic Interactions PHYS 27200

Purdue

GPA 3.95

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This 12 page Class Notes was uploaded by Mrs. Justyn Cremin on Saturday September 19, 2015. The Class Notes belongs to PHYS 27200 at Purdue University taught by Laura Pyrak-Nolte in Fall. Since its upload, it has received 124 views. For similar materials see /class/207909/phys-27200-purdue-university in Physics 2 at Purdue University.

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Date Created: 09/19/15

PHYS 272 Practice Exam Problem In the diagram below the electric eld is uniform and has components lt200 300 0gt NC Point A is at lt0 0 0gt m B is at lt0 2 0gt In What is AV along a path fromA to B 0 V 300 V 500 V 400 V 1000 V 300 V 500 V 600 V 1000 V pw wwaxwwr B Problem VB VA 2 500 V VC VB 200 V What is VC VA SIMPENN 700 V 300 V 0 V 300 V 700 V A B X X Problem A bar magnet has a length L cross sectional area A and magnetic dipole moment of p08 A m2 It is cut into two equal pieces each with length L2 and cross sectional area A Each of these bar magnets has a magnetic dipole moment ill The ratio tilp0 is 10 20 40 05 025 No enough information 0MPW t Problem An electron and a proton are near each other moving with velocities as shown in the plane of the page What is the direction of the magnetic force on the electron due to the proton I vs o gt Vp 1 Out of the page 2 Into the page 3 gt 4 lt 5 l 6 7 No force on electron due to proton at this instant Problem There are two coils separated by a large distance 2L L gtgt R R is the radius of the coil The x aXis passes through the center of each coil and the plane of each coil is perpendicular to the x axis In one coil the conventional current ows clockwise while in the other the conventional current ows counter clockwise What is the magnitude and direction of the magnetic field generated by a current of magnitude I owing in these coils at the positions conventional current x20 and at 2625 where 5 ltltL Note arrows show direction of ow in each loop Ay 5 E L A L Li a 2 2 R21 1 x 0 IBI 4 73 in the positive 6 direction 72 2 2 x 5 B uio 39u70 in the positive 6 direction 47 L63 47v L af 2 27rR21 2 x 0 IBI 4 T in the negative 6 direction 72 2 2 x 5 B Mi uio 272R I in the negative 6 direction 47 L63 47v L 63 3 x 0 5 0 2 2 x 5 B in the negative 6 direction 47v L63 47v L S3 4 x 0 E 0 2 2 x 5 B amp in the positive 6 direction 47v L63 47v L 63 5 x 0 0 2 2 x 5 B 39uo 272R 1 272R I in the negativexdirection EL63 47v L af Charge 1 s mnvmg mm wasth speedV Wm 13 m dmcunnafthz magma mm a mm m 1 x 2 7x a y b 7y 5 1 5 7 mm at m shave thlzm m magma same MW machmzs we sing canafwue m gamma slang magmm mm wnhm m call Snppase um h magnmldz at m magmm mm m m cenmxafacmlaf dnlsk Seamaan machmz ma mm MRI call canhe mamas athmcal xe valzmmamwgizlaap wmwamm hams m afthzmagnzuc eh m a imam 2R mm m cemex at m mp mm mm T a m 3 EM 5 m5 Problem A capacitor consists of two metal plates separated by distance 5 At a particular instant it is charged as shown and connected to two wires forming part of a circuit in a quasi steady state Estimate AVCD VC iVD I I 1 I E 39 390 I 39 39 ptucnhirhmwwmim I rrrrrrrrrrrrrrrrrrr quotI s d 1 Ells izld 2 EIIHEZId 371E11s zld 47130713ch Problem Many commercial resistors are made of carbon which has a Very low conductivity ScamDn 3x10 Am39ZV m m long carbon resistor with a crossisectional area of 01 mm2 has a resistance of about 17 ohms How long would a copper wire of the same diameter haVe to be to obtain the same resistance 5supper 58x107 Qmquot 1 1m 2 10mm 3 10m 4 01m 5 001m thlzm mm cumin shnwnhelaw Illafthz wquot a mad awahmme hmam 35mm has a much smallzx cmss secunnal ales a m palms Shaw h smaaysxam 212cm mm a m lacauam Manama mclmmg m hm segmem A n 1np scanmymawpmsesammnusesamhmuayamashnw mm sn ace charge aumnuoanma pmams m 212cm mm mndlzw a m m meme clza y m m enmes human Ryan at hgh sums charge dznsny and Ryan at law sn ace charge dznsny c pam39snfthz Innawithdmztaa hzhckwmmtha anhuhnwmlsil mm s m lam am am velncxtymthz Mm m mamm hnmn7smwym warm Pmblem The mmun helm15 made mm hettenes and three hghtbulbs Bum 1 m ohms Bu 3 30 ohm a m pmn39s A er the swnahxs clused and the steadystate ts es39abhshed the currents Lhmugh bulbs 1 2 andi HIE1End lespectwely wme lamp and heae equa uns theteeum he subed t demaner these thee cunems butdu hetsehe the equetehe 1213210111112 magma whatnunem ampere lumps and mdes yau using b 10 points Each battery in the circuit provides an emf of 15 Volts Treat each as an ideal battery When the switch is closed and the bulbs are glowing bulb 1 has a resistance of 10 ohms bulb 2 has a resistance of 40 ohms bulb 3 has a resistance of 30 ohms and the copper conducting wires have negligible resistance Solve your equations in part a for the currents I 12 and 13 c 5 points Give a numerical value for the power delivered by the batteries and the power dissipated by each of the three light bulbs What is the relationship of the power delivered by the batteries to the power dissipated by the light bulbs Fundamental Principles Equations and principles you must know 1 Electric field ofa point charge 2 Relationship between electric field and electric force 3 Magnetic field of a moving point charge Conservation of charge The Superposition Principle The Energy Principle The Momentum Principle Uel s V57 qAV AU AV VfiVi 7 iEodlz 72EXAXEyAyEZAZ Specific results Electric field of a uniformly charged spherical shell outside shell like point charge inside shell 0 l rodl a J distance r from the center lEmdl z if r ltlt L 4 50 72 LZ2 4755 7 A 1 qz E 39 on ax1s I quotgl 4n20Z2R232 A QA z J A QA z QA E 17 E z li z fltltR I Cllskl 280 Z2R212 I dlskl 280 R 280 1 z lEcapac ml z Q A for Q and 7Q disks IEfnngel z just outside capacitor 80 so 2R IE 39 lz 1 along dipole axis forxgtgt S39 zqs along J axis fory gtgtS dipole 4180 x3 7 dlpole 4n 0y3 7 i nAV 1 IqInAT v uE A u A u u AB OIA1 zxr for a short segment of currentcarrying conductor lBWiml 0 z 02I for rltlt L 4quot 4quot rArz Lz2 4quot 7 2 2 A u u lBloopl 0 HER 02mR on axiszgtgt R u IA Ii39ltR2 current loop EZ2R232 z 41 Z3 A 02 leipolel on ax1s Physical constants 2 7 2 7 2 L 9x109N m so 885x10 12L 1x10 7 Tm 47550 C2 Nm2 4n Cms Tm2 Cms u0 415x 10 7 c 3x108 ms 8 16x10 19 C mmmzmneutron 17gtlt10 27 kg melectmn 9gtlt10 31 kg 6gtlt1023 moleculesmole Atomic radius z 10 10 m Proton radius z 103915 m g 98 Nkg Electric field necessary to ionize air about 3x106 NC horizontal component of Earth s magnetic field about 2gtlt1075 T Key to Practice Exam Problem Answer KDOONCDU39IAWNH WWU39IO lwbbNb

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