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# General Physics PHYS 101P

GPA 3.66

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This 9 page Class Notes was uploaded by Hayley Jenkins on Thursday October 29, 2015. The Class Notes belongs to PHYS 101P at College of William and Mary taught by Staff in Fall. Since its upload, it has received 36 views. For similar materials see /class/231186/phys-101p-college-of-william-and-mary in Physics 2 at College of William and Mary.

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Date Created: 10/29/15

REVIEW AND SYNTHESIS CHAPTERS 19 21 Review Exercises 1 Strategy The maximum torque occurs when the plane of the loop is parallel to the magnetic field of the solenoid Use Eqs l9l3a and 1917 Solution Find the maximum possible magnetic torque on the loop 17 NIIIAIBS NIIIAIMOnSIS 100220 An00800 m24n gtlt107 T mA8500 m1250 A Since the magnetic field inside the solenoid is along the axis of the solenoid the magnetic torque on the loop is at its maximum value When the plane of the loop is parallel to the axis of the solenoid 2 a Strategy Use Eq 1914 and the RHR Draw a diagram Let the magnetic field due to the lefthand current b V be BL and that due to the righthand current be BR Solution According to the RHR the magnetic fields are as shown Since the triangle is equilateral 6 60 The angle is 90 60 30 Due to symmetry the horizontal components of the magnetic fields cancel and the vertical components add so the net magnetic field is directed in the positive y direction The magnitudes of both magnetic fields are the same Since sin 30 050 each vertical component is half of the magnitude of the field so the magnitude of the total magnetic field is the same as that due to either wire Compute the magnitude of the field M01 4n x107 T mA50 A 210032 m B 31x105 T 27w Thus the magnetic field at the third comer is 3 lgtlt105 T along the yaxis Strategy The positive zaxis is directed out of the page Use Eq 195 Solution Find the magnetic force on the proton E E ii L R 0 9 32 cm 32 cm 9 6 D 32 cm EB 6 X T3 so by the RHR the force on the proton is directed along the negative xaxis The magnitude of the force is FB evBsinB 1602x1019 C18gtlt107 ms3125x105 Tsin90 90x1017 N Thus EB 90gtlt1017 Nalong the xaxis 841 Review and Synthesis Chapters 79 27 College Physics 3 a Strategy Use Eq 1914 and the RHR Draw a diagram Let the magnetic field due to the lefthand current be BL and that due to the righthand current be BR Solution According to the RHR the magnetic fields are as shown y Since the triangle is equilateral 6 600 The angle is l 900 600 300 Due to symmetry the vertical components of the 5 E magnetic fields cancel and the horizontal components add so the net 6 magnetic field is directed in the positive xdirection The magnitudes of both magnetic fields are the same Since cos 300 z xg 2 the 250 Cm 250 cm magnitude of the magnetic field is about xg of the magnitude of the field due to either wire Compute the magnitude of the field 250 cm I 7 BMguo J 4nx10 T mA120 A 166X104 T 2m 27100250 m Thus the magnetic field at the third corner is I l66gtlt 10 4 T along the xaXis b Strategy Use F IixB Solution If the current of the third wire is into the page the force on the wire is directly along the negative y aXis If the current is out of the page the force on the wire is directly along the positive y aXis So the current in the wire should flow out of the page c Strategy Use Newton s second law and F IEXB V Solution Find the current in the third wire 3 2 SF F mgILBMgO SCIAg0150x10 kgm980 ms 884A y B B 1663x104 T 4 Strategy Use Lenz s law and the RHR Solution The current in the large loop ows clockwise According to the RHR the magnetic field generated by the current is into the page at loop B and out of the page at loop C As the resistance in the variable resistor is increased the magnetic field generated by the decreasing current in the large loop decreases which decreases the magnetic flux directed into the page through loop B and out of the page at loop C The direction of the current induced in loop B flows such that the magnetic field it generates adds to the manetic flux to o u ose its decrease39 that is the current generates an increasing magnetic field into the page Thus The direction of the current induced in loop C flows such that the magnetic field it generates adds to the magnetic flux to oppose the decrease Thus Ithe current in loop C flows counterclockwisel39 it generates an increasing magnetic field out of the page 842 College Physics 5 9 gt1 Review and Synthesis Chapters 19 21 Strategy Use Eq 1914 and FB qVX Let the positive ydirection be up Solution According to the RHR the magnetic field generated by the power line at pointP is in the positive ydirection Looking at the side view we see that the angle between the velocity of the muon and the magnetic field is 180 25 155 The charge of the muon is negative so according to the RHR and FB qVX T3 the direction of the force on the muon is out of the plane of the paper in the side view or to the right in the endon view Compute the magnitude of the force I 19 7 7 0 FB evB 5mg W M0 5mg 1602x10 C70x10 ms4nx10 T mA160 ASIn155 27w 2n0850 m 18x 103917 N Thus FB 18 x103917 N out of the plane of the paper in the side view or to the right in the end on View Strategy Use Eqs 21 6 and 21 7 and the fact that the capacitance of a capacitor is directly proportional to the area of is p ates Solution Find the new rms current in terms of the old C L 1 gt501f li li fo Ci 114139 U 60141 I mi 114139 1 Strategy Use Eqs 21 4 and 209 Solution 3 The power is related to the voltage and the current by Pav I msts so the current flowing through the P 5 copperwires is Irms VAW 7x103 A rms b The power dissipated in the wires due to their resistance is Pav ImszR 6667 A212 2 530 MW But 530 MW gt 800 kw so 1 all of the power is dissipated since 530 MW is greater than the total power output of the power plant 80x105 w 17 A 48000 V d The power dissipated in the wires due to their resistance is Pav ImszR 167 A212 S2 Compute the percent this power loss is of the total power output of the plant 3333 7 1000 0420 800000x o 9 Find the number of secondary turns the transformer requires to step the 48 kV voltage down to 120 V P av Vn39ns c The current flowing through the wires at a voltage of 48 kV rms is Inns N2 2 120v i7 soN 7N 710000 25 N1 2 1 1 48000V 843 Review and Synthesis Chapters 7 9 27 8 a 390 College Physics Strategy Use Faraday s law Eq 1817 and the definitions of resistance and electric current The initial magnetic flux through the loop is BA and the final magnetic flux through the loop is BA Solution Find the amount of charge that ows through the circuit ACID NBNMRC Re ii so At At At q At q At R1 R2 1 l 2 l l A 2NBA 2 50 14T 045m 85C q R1 R2 X X 1009 50 9 Strategy Use Kirchhoff s rules Let the current through the circuit be I CCW 11 RI through the loop the current through the 1009 resistor be 1 to the left and 1 the current through the 509 resistor be I 2 to the left R2 12 Solution Find the amount of charge that ows through the 509 resistor A61 A 11 A512 Accord1n to the unctionrule I I 1 soA A A g J At 1 2 At At q q1 q2 A611 A612 Aq qu A612 Accord1n to the loo rule 0IR I R R R R R so g P 1122 At1Al2 At 1At2 R 1009 A 1A 85C 57C qz 1212 q 1009509 9 Strategy Use Lenz s law and F IEXB Solution According to the RHR the magnetic field points into the page at the loop a b C IA counterclockwise current is induced because the flux through the loop is increasing as it nears the wirel This counterclockwise current generates a field directed out of the page According to the RHR the magnetic force on the part of the loop closest to the wire is away from the wire and that on the art of the loo 3 farthest from the wire is toward the wire Since the magnetic field is greater nearer the wire e net force on the 100 g No current is induced because there is no change in flux through the loopl Since there is no current in the loop Ithere is no magnetic force acting on the loopl IA clockwise current is induced because the flux through the loop is decreasing as it moves away from thel This clockwise current generates a field directed into the page According to the RHR the magnetic force on the part of the loop closest to the wire is toward the wire and that on the part of the loop farthest from the wire is away from the wire Since the magnetic field is greater nearer the wire he net force on the loo 3 is toward the wire 844 College Physics Review and Synthesis Chapters 7 9 27 10 Strategy Use Eqs 188 209 and 214 11 Solution 3 b C d e V a V b V 0 Find the total resistance in the copper wires 3 R p167gtlt108 9mm 051 9 n0050 m2 The current in the w1res 1s ms av plant les The average power dissipated by heating of the wires is 2 P 0519 22 MW av plant 25x106 w 1200 v 2 Pheat11ms R fIIlS Find the number of turns in the secondary coil yt N 150 2 soN 150N 1501000 150000 1 N1 2 1 lt gt P 6 avplant 25X10 W 14A THIS The new rms current in the wires is I ms The average power dissipated in the transmission lines using the transformer is 2 P 0519 25x106 W 1501200 V av plant 2 Pheat11ms R fIIlS Strategy Use F E X T3 Solution The current ows into the page through the rod so by the RHR the force on the rod is Strategy Use F EX l3 Newton s second law and Eq 45 Solution Find the normal force on the rod ZFy N mg0 soNmg The force of kinetic friction is fk MkN Mkmg Find the acceleration of the rod LB ZFx FB fk ILB Mkmg max so ax ng In Find the speed of the rod LB vfxz vm2 v2 0 ZaXAx 2 ngAx so In LB 200 AO500 m0750 T 2 2 Ax 2 0350 980 800 136 v m 11g J lt ms gtlt mgt Strategy and Solution As the rod slides down the rails the applied emf has to increase because of the increased resistance in the longer rail lengths in the circuit and because of the increasingly large induced emf as the rod moves faster VBL 845 Review and Synthesis Chapters 19 21 12 a A Uquot 13 a A Uquot v College Physics Strategy The maximum power output is achieved when X c X L The airfilled capacitor Ca is related to the dielectricfilled capacitor Cd by Cd KCa Use Eqs 21 7 and 21 1 0 Solution Find the value of the airfilled capacitor 1 X wL X 71 L C szL 4712220 Hz25500650 H 7 7 so Ca de wKCa Strategy Use Eq 21 1 8 Solution Find the resonant frequency of the circuit before the insertion of the dielectric wo2mi 1 Sof0 1 cha 271 LCa 2n0650H0146x10396 F Strategy At resonance Xc XL so I 0lt R1 Thus when the resistance is doubled the current is cut in half Use Eq 21 4 Solution 1fthe initial power dissipated is P IizRi then the final power dissipated is l the power is cut in half when the resistance of an Therefore 2 2 I 1 Pf If2Rf 1 2R1 112R1 2 2 RLC circuit is doubled Strategy When the circuit is not at resonance the situation is more complicated because the current is inversely proportional to the impedance Z and the impedance is not simply equal to the resistance Use Eq 2114b Solution Find the initial and final impedances in terms of the initial resistance Zr ZR ltXL Xcgt2 J R Form a proportion to find the final current in terms of the initial current I Z i7l Z501f i Ii Zf J5R 5 5 Form a proportion to find the final power dissipated in terms of the initial power dissipated 2 2 ilj 2R L wi 5011 Pi 1312i 1312 33 5 5 Thus the power is 45 of its original value 14 Strategy Use Eq 2114b and the definitions of capacitive and inductive reactance Solution Form a proportion to compare the initial and final impedances Zr 1 So Z JRZ wiL 2 le wa j 100 of 271020 Hz00150 Hym j 2 the impedance increases by a factor of 123 123 2 1 W1C 100 32 2n60 Hz00150 H W 846 College Physics Review and Synthesis Chapters 19 21 15 a Strategy Use Eq 2118 Solution Compute the resonant frequency of the RLC series circuit wo ano 1 1 1 Sofo 7 445 Hz VLC ZWLC 27140146 H877x10399 F b Strategy and Solution At resonance the capacitive and inductive reactances are equal If the frequency is less than the resonant frequency the capacitive reactance will be greater than the inductive reactance because Xc 0lt f1 WhileXL 0lt When Xc gt XL the voltage lags the current Therefore the leads the voltage c Strategy Use Eq 21 1 5 and the definitions of capacitive and inductive reactance Solution Find the phase angle of the circuit 050L 20 1 1 A A m mm 0w mKOSOF 20 so R R R R R R C 15 L 15 l 0146H tan391 7 391 7 67 C 25552 877x10399F d Strategy Use Eqs 21 1 4a and 21 1 6 Solution Find the impedance so Z cos I Find the rms current R n cos 480 Vcos 674 V IZJ I ZJ I 7301 7 01A 9n nns rms COS nns ZR 2255 g e Strategy Use Eq 21 4 R cos 7 Z Solution The average power dissipated in the circuit is P W ImszR 0512 A2255 g 1 Strategy Use Ohrn s law and Eqs 216 and 219 Solution The maximum voltages are VR IR lm sR J2051 A255 S2 VL IXL JEIrmSwL J2051A2n050445 Hz0146 H 150 v and 1 J5o51 A V lX J I 7 590v C C quotquot5 wC 2n050445 Hz877x10399 F 16 Strategy qus IrmsXL andXL wL Solution Form a proportion to find the final rms current If st Xi wL wiLi 7 7 117067solf 0671 11 Xf erf 213301i 847 Review and Synthesis Chapters 7 9 27 College Physics 17 Strategy Since the measurement is made relatively close to the electron beam the beam can be approximated as a long straight wire Use Eq 1914 and the definition of current Solution Let N be the number of electrons then the amount of charge passing the point in 130 microseconds is Aq B 18 a b c V d Ne Find the magnetic field strength Kieran measures amp4nx107TmAl40gtlt1011l602gtlt1019C Al 2n00200 ml30gtlt106 s Strategy Q C AV for a capacitor where C GOAd Use conservation of energy 27w 1 At 27w 173x10 7 T 27w Solution The kinetic energy of an ion is equal to the potential energy gained mvz Find the magnitude of the charge 2 594 lmv2 eAV so AV 2 2e mv 0Amv2 d 26 2de To accelerate a positive ion the righthand plate must have the lower potential so e west plate is positive and the east late is neative Strategy Use Eq 195 and FE 61173 QCAV Solution FE must oppose PB in the velocity selector According to FB 6 va and RHR 1 FE is downward for a positive ion so the electric force FE opposing PB is directed upward or north in the figure Since the ion is positive the electric field must be upward Set the magnitudes of the forces equal and solve for the magnitude of evB eE so E vB Thus E Strategy Form a proportion to determine the direction of the de ection Solution F ev B v B L amp lt 1 so F lt F and the ions are deflected upward in the selector B 1011 E ion FE ion evB v l l Ion Haj ectory i E B G Strategy Apply Newton s second law to the circular path Solution Compute the mass 2 B 2 B 6 BD 2Fr evBmv som r v v 2v 848 College Physics Review and Synthesis Chapters 19 21 MCAT Review H N Equot P 5quot 9 gt1 9quot Strategy and Solution The current flows counterclockwise through the apparatus By the RHR the magnetic field generated by the current is directed upward The correct answer is Strategy Refer to the data in the table Solution Two rows of data are given for a mass of 001 kg The current is increased by 150 from 100 A to 150 A Due to the increase in the current the exit speed of the projectile increases by 150 from 20 kms to 30 kms Thus the exit speed is directly proportional to the current So if the current were decreased by a factor of two the exit speed would be decreased by a factor of two as well The correct answer is Strategy and Solution Since for a given current the force is constant along the entire length of the railgun lengthening the rails would increase the exit speed because of the longer distance over which the force is present The correct answer is Strategy and Solution Since the resistance of the rails is directly proportional to their resistivity lowering the resistivity of the rails would decrease the power required to maintain the current that flows through them The correct answer is Strategy The average power is equal to the change in kinetic energy divided by the time interval Solution Find the average power supplied by the railgun P imvz 0 mv2 010 kg100 ms2 3quot Ar Ar ZAI 22o s The correct answer is Strategy and Solution F ma 0lt I2 so a 0lt Izm The speed of the projectile is directly proportional to is 25W acceleration so v 0lt I 2m Form a proportion using data from the table to find the approximate speed of the 008kg projectile v Izzm I 2 m I 2 m i272 2 71 so v2 2 ilvl For 1 we use the data giveninthe third row ofthe table V1 11 m1 11 m2 11 m2 2 2 I v2 2 v1 04 kms14 kms I1 m2 100 008 The correct answer is Strategy and Solution The power from the power plant is given by P IV so for a given amount of power increasing the voltage decreases the current required Since the power lost as heat is given by P IZR reducing the current reduces the power lost as heat The correct answer is Strategy and Solution The magnetic field lines due to the section of currentcarrying wire are circles which is the only possibility given the symmetry of the situation The direction of the field is determined by the RHR Pointing the thurnb of the right hand in the direction of the current in the wire then curling the fingers inward toward the palm the direction of the field is indicated by the direction of the curl of the fingersiin this case counterclockwise as seen from above The correct answer is 849

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