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# GENERAL PHYSICS 2 PHYS 2020

GPA 3.96

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This 58 page Class Notes was uploaded by Mrs. Peter Toy on Friday October 30, 2015. The Class Notes belongs to PHYS 2020 at University of Colorado at Boulder taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/232111/phys-2020-university-of-colorado-at-boulder in Physics 2 at University of Colorado at Boulder.

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Date Created: 10/30/15

Faraday CT39s Fl In which situation is the magnetic ux through the loop the smallest A Same ux in all three situations gt B C D gt gt 5 gt area 2A H y edge 0n area A constant area A tilted 60 uniform face 0n B field F2 A loop of wire is moving rapidly through a uniform magnetic eld as shown True A or FalseB there is a nonzero emf induced in the loop 8 A loop of wire is spinning rapidly about a stationary aXis in uniform magnetic field as shown TrueA or FalseB there is a nonzero emf induced in the loop 969 969 69696969 969 696969633 69696969 969 69696969 F3 A loop of wire is sitting in a uniform constant magnet field as shown Suddenly the loop is bent into a smaller area loop During the bending of the loop the induced current in the loop is A zero B clockwise C counterclockwise F4 A bar magnet is positioned below a horizontal loop of wire with its North pole pointing toward the loop Then the magnet is pulled down away from the loop As Viewed from above is the induced current in the loop clockwise or counterclockwise A eyeb all t N A clockwise B counter clockwise U FS A square loop is rotating in a fixed external magnetic field into the page At the instant shown the loop is out of the plane of the page with left side of the loop above the page and coming out of the page the right side in going in The induced current is axis of rotation C44 Bin C Neither F6 A loop of wire is near a long straight wire which is carrying a large current I which is decreasing The loop and the straight wire are in the same plane and are positioned as shown The current induced in the loop is A counterclockwise B clockwise C zero I to the right but decreasing loop F7 Two loop of wires labeled A and B are placed near each other as shown A large current I in loop A is suddenly turned on This causes an induced current in loop B which causes A A net replusive force the two loops repel B A net attractive force the two loops attract C whether the force is attractive or repulsive depends on whether the current in first loop is CW or CCW D No net force F8 A wire loop moving right enters a region where there is a constant uniform magnetic field pointing into the page B As the loop enters the Bfield the current induced in the loop is A CW B CCW As the loop enters the Bfield the direction of the net force on the loop is A right gt B lt C up T D Ldown E there is no net force F9 An electric motor consists of a coil free to turn on an aXis and located in an magnetic field B created by an arrangement of permanent magnets With the current and field directions as shown which way will the coil rotate A CW axis B CCW C the coil won t rotate when at this particular orientation FlO A solenoid has an increasing current causing an increasing Bfield in its interior An endon View of the solenoid is shown below An electron is directly below the solenoid While the current I in the solenoid is increasing what is the direction of the force on the electron 69 A 8 Bin D B electron C Hint What is the direction of the Efield at the position of the electron Think Lenz s Law Fll A long solenoid of radius R contains a uniform magnetic field B which is increasing at a steady rate Bt A t where A is a constant For distances r lt R from the center of the solenoid how does the magnitude of the Electric field a d 4 depend on r Hint zdl a daj A E zero B E oc lr C E oc r DEocr2 EEocr3 Refractng telescope eyepiece objective Re ecting telescope obj ective eyepiece Concave mirror focal length f posmve Convex mirror I I a V I focal length f neg atlve Converging lens convex f e M knife edge blocks rays Schlieren Imaging point source mirror 1 mirror 2 0 object M G knife edge blocks most rays camera 171 SIP Phys 2020 Fa 01 Giancoli CH 17 Electric Potentials and Capacitors We ve been talking about electric forces and the related quantity E Fq the E field or quotforce per unit chargequot In 2010 after talking about forces we moved on to work and energy Ch 6 Quick Review of work and energv F The work done by a force F moving something through a displacement quotdquot is e gt Work Fd or more carefully W 13 d F dc 03 d Eg if you an quotexternal forcequot lift a book at constant speed up a distance d Newton H says Fnet ma F eXtT ie Fext Fg 0 7 because remember if speed is constant gt a0 l or Fext mg Fig 1 mg You do work Wext Fextd mgd The sign is because 6 is 0 degrees your force is UP and so is the displacement vector The gravity field does Wfield Fgd mgd The minus sign is because 6 is 180 degrees the force of gravity points DOWN while the displacement vector is UP The NET work done by all forces is WextWfield 0 that s just the workenergy principle which says Wnet AKE 0 here You did work Where did it go NOT into KE it got quotstored upquot it turned into potential energy PE In other words Fext did work which went into increased gravitational potential energy For gravity we defined this potential energy to be PE mgy so APE mgyfinal yinitial mgd Wext The change in PE is all we ever cared about in real problems 172 SIP Phys 2020 Fa 01 Now let s drop the book and see what happens I There is no more quotexternal forcequot touching the book like quotmequot 1 7 in the previous example only gravity acts Neglect friction F g1 mg Energy conservation says 2 f f This formula gives a quick and easy way to find vf The concept of energy and energy conservation is very useful Another way of rewriting that equation is 1 PEiKEiPEfKEf ie mgd0 0 Emv PEf PEiKEf KEZ0 ie APE AKE 0 or AEmt 0 End of quick review of work and energy see Ch 6 for more details There is an electric quotanaloguequot of the above examples Consider 2 charged parallel metal plates called a quotcapacitorquot a fixed distance d apart Between the plates E is uniform v V quot 116 i l constant and points from the 39 39 39 towards the plate lI Il Ev v v v f Imagine a charge q initially located near the bottom plate The force on that charge is FEqE down do you see why Let s totally neglect gravity here Now LIFT quotqquot from the bottom to the top at constant speed You do work Wext Fextd qu The Electric field does Wfield FEd qu Do you understand those signs Think about them Just like the previous case you did work but where did it go As before it didn t turn into KE it turned into potential energy We say the charges electrical potential energy has increased APE qEyfinal yinitial qE d Wext where y is the distance above the negative plate We lifted the charge from a region of LOW PE near the quotquot plate to a region of HIGH PE near the quotquot plate Note quotupquot and quotdownquot are irrelevant here you could turn the picture on its side or even upside down It s not gravity in this story it s 100 electrical energy 173 SIP Phys 2020 Fa 01 Just like we defined EFq dividing out q gives forceunit charge let s now define something we call quotelectrical potentialquot or just quotpotentialquot PEq Calling this quantity quotpotentialquot is really a pretty bad name because this quotpotentialquot is quite DIFFERENT from quotpotential energyquot Potential has units of energycharge JoulesCoulomb JC We call 1 JC 1 Volt 1V People use the symbol quotVquot for the unit volt as well as for the quantity itself Another bad choice but we have to live with it A change in potential is called a quotpotential difference AVsz Vl APEq andfromthiswesee MEquV Example A car battery maintains 12 V between the terminals If the headlights contain a 36 W bulb how much charge is the battery moving through the bulb each second And how many electrons is that Answer 36 W 36 Watt 36 Js Each second 36 Joules of energy are dissipated in a bulb This energy all comes from the loss of potential energy as charges ow from one terminal through the bulb to the other terminal If a charge quotqquot drops 12V the energy lost is APE qAV or q12V Each second 36 J are lost ie 36 J q12 V or q 36 J12 V 36 J12 JC 3 C That s a lot of electric charge being moved by a car battery The number of electrons going through the bulb each second is 3Cl 6El 9 Celectron 2E1 9 electrons A heck of a lot I was a little sneaky about signs the charge of an electron is negative just think about it Here s a related question for you given that its negative electrons that ow out of a battery which way do they go from the quot terminal through the bulb to the quotquot or the other way The answer is from to Electrons are repelled from quotquot and attracted towards 174 SIP Phys 2020 Fa 01 For a parallel plate capacitor we just found two pages ago APEqu so AV APEq quq Ed Here s another sketch of a capacitor b V is high here d DeltaV VbVa Ed gt 0 l a 4 V is low here With gravity you can choose to call quotzeroquot potential energy wherever you want You might choose sea level or the tabletop or the ground It s the same story with electricity you can pick any spot you want and call the electrical potential energy 0 there We usually call this point quotthe groundquot Let s call point quotaquot in the diagram above quotthe groundquot or quot0 potentialquot Now put a charge quotqquot at the point quotbquot in that figure It will have a potential given by Vat point b Ed It has a potential energy at point b of PE qVat b qu It has quotquot potential energy there which makes sense It s like a pebble up in the air it can do work just let it go The upper plate repels a quotquot charge the lower plate attracts it if you let it go it will run quotdownhillquot in energy from high potential to low Notes 0 E points from high V to low V always quotquot charges want to head towards low V if you ll let them quotquot charges want to head towards HIGH V if you ll let them D 0 We can talk about the potential at a point or the potential energy at a point but the numerical value depends on where we chose to call 0 We can talk about potential differences between points and then it does NOT matter where we chose to call 0 175 SIP Phys 2020 Fa 01 What if you re near a point charge 0 rather than a capacitor H H 0Q r gt 2 What s the potential at the point quotaquot 1st where do we want to call 0 potential It s not so obvious here A standard choice is far away off at infinity Out there PEO V0 seems reasonable Now we need to think about moving a test charge quotqquot from far away where PE0 V0 to the point quotaquot Because the work you do bringing it from PE0 to the point is precisely its potential energy It s like how much work you do lifting a book from the ground ie PE 0 up to a height d it s mgd the final potential energy Now Work Fdistance and Fqur 2 Unfortunately this force changes as you move in from far away r is changing So you really need calculus to figure out the work The answer though is very simple and maybe you can even guess it just multiply Fr Wext k Q qr Notice that s an r downstairs not an rA2 So the PE at point quotaquot is exactly that PEat quotaquot k Q qr or Vat quotaquot PEat quotaquotq k Qr Note that we chose V0 to be off at infinity to get that formula If Q and q are both quotquot then PEqur gt0 This makes sense two positive charges want to quotfly apartquot they ll DO work if you ll let them The system has positive potential energy Like a rock up in the air Also just like with capacitors the potential V is big quotquot when you re near a positive charge Q the closer the bigger If Q and q are opposite signs then PElt0 This is also correct you would have to do work ON opposite charges to quotpry them apartquot the system has a negative potential energy We might say the system is bound 176 SIP Phys 2020 Fa 01 What if there are a bunch of point charges what s the potential The voltage at any point is just the sum of the voltages arising from each of the individual particles this is quotsuperpositionquot again It s really quite easy to find the voltage at a point because of this Example Three charges 1 2 and 3 with charges Q 2Q and 2Q respectively are 2 392Q arranged as shown 0 What is the potential V at the origin Answer VVfrom lVfrom 2Vfrom 3 r kQr k2Qr k2Qr 3 2Q kQr The answer is positive if you put a positive charge there it would be happy to run away off to infinity if you d let it The math is reasonably simple No vectors or components to worry about finding E at the origin would be a lot more trouble Example A test charge quotqquot is moved from point quotaquot to quotbquot in the figure There are two other charges present Q and Q fixed in position at the corners of a square as shown How much work does this take Answer Wext APE PEI PEG Remember at any point the potential energy is just PE qV PEb q Vat b wed Q Lag QM r r r PE WW aklt29gtk9q a r 1 I I Subtracting we find Wext APE PEb PEG 0 It doesn t take any net external work at all Depending on how you move you might do some work part of the way and work part of the way but in the end you do zero total work going from this particular quotaquot to quotbquot Trying to figure out the work by thinking of forcedistance along the path would be HARD because force changes all the time Using voltages makes this much easier to figure out 177 SIP Phys 2020 Fa 01 It s often useful to find all the points in a diagram that have the same voltage Eg consider a capacitor again Everywhere along the top 39I39 39I39 39I39 39l39 39I39 surface the potential is the V is high here same VEd d V is in between hereltgt Everywhere along the bottom the potential is the l V is low here same V20 And every point on that dashed line has the same potential something between 0 and Ed We call such a line an quotequipotentialquot equal potential line If you move a test charge along an equipotential line eg the horizontal dashed line in the figure the potential is the same everywhere so no work is required It s like walking along a at surface where there s no change in the gravitational potential Or eg traversing sideways on a ski slope There can t ever be a component of E parallel to an equipotential line if there was a nonzero E parallel you d do work moving along it since W F paralleld q E paralleld This means that in drawings E field lines are always perpendicular to equipotential lines The equipotential lines are 439 39I39 39I39 39I39 39I39 39l39 shown as quotdashedquot in this Est IL Vv 7 v Vhigherv figure I I middlev Theyre like contour lll les ona topo map which show quot gravitational equipotentialquot lines Constant height on a topo means constant PE grav lowerV V ult IA I Anywhere along a dashed line the potential is constant Inside any static conductor we know E0 That means no work is required to move charges around anywhere inside or along the surface So metals are equipotentials throughout their volume Real life is 3D those lines are really surfaces 178 SIP Phys 2020 Fa 01 More examples of equipotential lines Remember voltage near a charge 39 10Wer V01t lge Q is given by V kQr I equipotentlal lme The farther away you get the Q I l lower the voltage higher voltage eQUiPOtential line If quotrquot is fixed ie a circle V is constant Again in 3D these would be equipotential surfaces rather than lines In this case they d be spheres surrounding the charge Q Another example here s a dipole with a couple of equipotential lines shown Look back in Ch 16 we sketched the electric field lines before Now we re adding in the equipotential surfaces We usually choose to define that center equipotential line as V0 Remember it s up to you to pick where V0 is and that line extends out to in nity That s a pretty common choice Far away from everything the potential is considered zero To find the numerical value of the potential on one of those dashed surfaces you d do a quick calculation just like 2 pages ago find the distance rl to Q and r2 to Q and then V kQrl kQr2 Anywhere on the center line you re equidistant from both charges so rlr2 and the two terms cancel V 0 It s all consistent 179 SIP Phys 2020 Fa 01 ENERGY and units Example Release a proton charge e from point a in the figure Suppose the voltage difference between the plates is 5000 V a typical voltage in e g a normal TV set How fast is the proton going at point I Answer I d use conservation of energy here 1 2 PEiKEiPEfKEf 1e eVa0 eVb 5mvb Solving for vb we get vb 2eVa Vb m J216103919C5000VC167103927kg 106 m s Can you check that the units worked out ok otet at V V 5000 J C Le a 1s a otentia N h a b 39 V39 HIGHERp 39l However the change in potential of the proton as it moves is AV Vb Va 5000 J C Think about the signs objects spontaneously move to LOWER potential energy if they can That means objects like to go to lower potential ie lower voltage What was the final KB of the proton in this example It s easy enough to find using conservation of E KEfKEi PEi PEf 0eVa eVb e5000V 16 10 19C 5000J C 8 10 16 J Many people prefer to change units here like converting 254E2 m to 1 inch metric to nonmetric We can define a new unit of energy leV 16El9 J The quoteVquot is also called an quotelectron Voltquot but it is NOT a volt which is a unit of potential or JC recall It s just a name eevee The eV is defined so as to be the energy loss of a particle of charge quotequot like a proton dropping across 1 Volt of potential Since energy change is qDelta V this is an energy change of el V 16El9 Cl JC 16El9 Jjust like we said In the little problem at the top of the page we could ve done it without a calculator if we d used eV s instead of J Namely KEfKEi PEi PE 0eVa eVb e5000V 5000eV You can check that this agrees by doing a simple unit conversion 16 103919 J 1 eV 5000eV 5000eV 810 J which is what we got above 1710 SIP Phys 2020 Fa 01 CAPACITORS Any time you have two metal surfaces near each other separated by air or an insulator you have a capacitor My favorite example is the quotparallel plate capacitorquot We ve already looked at it quite often A capacitor can be charged or not When we say a Q capacitor has quotcharge Qquot we mean Q on one plate and T Q on the other like in the picture here Of course the d net charge is really zero but we still say it s charged i Q If we say quota capacitor has voltage Vquot we really mean it has a difference in voltage between the plates AV Ed We drop the quot A quot for convenience ie VEd It s a bad notation but standard The more charge Q you put on the plates the stronger the E field between the plates hence the bigger the voltage drop In fact if you double Q you discover that E doubles too and so must V That means Q 2 V or C is a constant and we call it the quotcapacitancequot Of course you can also turn it around and say C QV A capacitor can store charge and energy for you that s why they re useful Think of a closet with lots of bowling balls up on the shelf at high gravitational potential You can let them fall down whenever you like to do lots of work or damage later Your closet shelf has only a limited capacity to hold bowling balls Similarly metal plates of a given voltage difference have a certain limited capacity to hold charge and that s what quotCquot tells you A big capacitance large C means you hold lots of charge with a little voltage A small C means you can only hold a little bit of charge for a given voltage a bit like a small shelf in the closet Of course given C you can make Q as big as you want by making the potential bigger So our quotclosetquot metaphor breaks down The units of capacitance are CoulombsVolt CV We call 1 CN l Farad 1 F The units are getting confusing now because 1 V l JC so 1 Fl CAZJ too 1 F is a really big capacitor It would hold 1 C a lot of charge with only one volt difference between the plates Most normal capacitors are more like micro or nano Farads 1711 SIP Phys 2020 Fa 01 For large parallel plates of area A the E field between them is given by a simple formula whichl won t derive E41tk2 A The formula says the more Q you have the bigger E is That at least makes some sense doesn t it Also the bigger the area of the plates the more the charge is spread out and that should weaken the E field so the A dependence in that formula also seems reasonable The formula doesn t depend on quotdquot how far apart the plates are This is perhaps a bit surprising but it s correct E is UNIFORM it doesn t change with distance from either plate 2 Earlier we defined so L 885 10 12 C 41tk Nm Q A so you can also say 2 a 1 E 80 We know VEd for a parallel plate capacitor Combining this with the E LE so A Finally C QV that defines C So that last equation for V yields can you check the algebra yourself formula for E just above gives V 4nk 1A A C g 41tkd 0d This formula is only for parallel plate capacitors Since VEd and E is uniform the closer the plates the smaller the voltage drop Smaller V for a fixed charge means CQV is BIGGER If you can gets the plates closer together you get a better bigger C capacitor So that quot ldquot dependence in the formula makes some physical sense Similarly if the area of the plates is bigger the formula says you have more quotcapacityquot to hold charge That makes good intuitive sense to me39 you have more area to spread those charges out onto so you can easily hold more charge 1712 SIP Phys 2020 Fa 01 If you connect a wire up to a capacitor like on the left the Q charges on the capacitor plates are 0 free to travel The and attract one another the capacitor will discharge quickly possibly with some nice Q sparks There s energy in there 0 It takes work to charge up a capacitor Consider e g a battery charging up a capacitor A battery quotwantsquot the plates to be at a certain voltage difference V Eg a 12 V car battery always wants things connected to it to be 12 V apart To do this it forces charges to go from one plate onto the other plate fighting against an E field all the while How much work does the battery do charging the capacitor up from 0 to V A The first charge moves easily it takes almost no Batteryjust work because at first the plates have no charge Pushed the lst charge from one There s no E field and charges are pretty much plate to the other free to move around wherever they want On the other hand if the capacitor is fully charged with a voltage V across it you d eXpect that moving Q charges across that voltage V would cost energy QV The real answer is the average of 0 and QV on average while charging the voltage is V2 so Energy to charge up Energy stored in capacitor U QV2 The symbol quotUquot is for some reason often used for energy In this case it s electrical potential energy Don t confuse potential energy with potential Since QCV we can rewrite this as U CV 22 or UQA22C Depending on what s held constant these forms are sometimes useful 1713 SIP Phys 2020 Fa 01 Example A 12 V car battery is hooked up to a capacitor with plates of area 03 mquot2 a distance 1 mm apart How much charge builds up on the plates Answer The battery will charge the capacitor up until the voltage is 12 V Since QCV we only need to find C But we know that C so Ad So we re all set Z Z QCV eo V 88510 C 2039Zm 12v d Nm 10 m 3210 gc V 32 nC N m I used 1 Vl JC check for yourself that that nasty combination of units simplifies like I claimed to Coulombs How much energy is stored in the capacitor now UQV2 32E9 C 12V 2 02 micro Joules Not so much Aside Where exactly is the energy stored in a capacitor The answer is that it s stored in the E field Wherever you have electric fields there is stored energy The energy is stored in the quotspacequot between the plates in the form of electric field energy In diagrams we will represent capacitors with a simple symbol like this even if the capacitor in reality isn t C physically parallel plate Sometimes the C is left I off too 17 14 SIP Phys 2020 Fa 01 DIELECTRICS Any insulating material paper plastic etc can be called a dielectric Most real capacitors have dielectric materials between the plates This helps to keep the plates apart After all the 0 plates are oppositely charged and so attract each other strongly If they ever touched the capacitor would Dielectric discharge or quotshort ou quot and be useless So with a dielectric in there you can make quotdquot quite small Q and remember that helps make the capacitance bigger But the main reason for putting a dielectric in there is something different it actually decreases the E field in the capacitor Why Real materials like dielectrics are quotpolarizedquot by a strong E field That means the E field in the capacitor effectively pulls some quotquot charge towards the top of the dielectric nearer Q the quotquot plate and some quotquot towards the bottom nearer the quotquot plate In the region throughout the middle it looks 39 39 like some of the quotQquot around you has been weakened or shielded or canceled out The net effect is that E is reduced throughout the dielectric Q And since VEd the voltage between the plates is reduced Since CQV if Q is fixed and V is reduced C gets bigger In this way dielectrics make the capacitance bigger Your capacitor can hold MORE charge for a given voltage with a dielectric in there because the surface of the dielectric effectively shields out some of the E field from the middle region Recall the formula C L so 1 4 41tk d 61 That s if there s no dielectric If there IS a dielectric we just argued C is bigger It turns out for most dielectric materials C is bigger by some constant factor which depends only on the material ie sch l A Cwith dielectric Km KEO 1 K will be some constant for any given dielectric material Giancoli has a table Bigger K means you get a bigger C Paper e g has K3 roughly 1715 SIP Phys 2020 Fa 01 Capacitors are everywhere in circuits radios computers TV s It s handy to have a quotcharge storagequot device The final brief topic of this chapter is a real world application of capacitors CRT s or quotcathode ray tubesquot vertical capacitor hot metal quotquot cathode screen quotquot anode with holes horizontal capacitor A quotcathode rayquot is an oldfashioned name for electrons The quotcathodequot is a heated piece of metal set at a very low voltage The quotanodequot is set at a high potential so the quotcathodequot and quotanodequot basically form a capacitor Electrons boil off the hot cathode and then they are accelerated towards the high voltage quotquot charged anode The anode is a grid with lots of holes so many electrons can y right on by and cruise towards the screen They pass through a pair of capacitors vertically and horizontally oriented which have a voltage that quotsweepsquot As electrons pass through these capacitors they feel the force from the E field which bends the path of the electron Since the voltage is swept the electrons are also swept They y on by and hit the screen which glows where the electrons hit So you see them sweeping by and this makes the whole screen glow By turning the anode on and off you can make the electrons go through or not thus making bright or dark spots which allows you to make an image This device is used in old computer monitors and TV oscilloscopes EKG traces you still see them all over We ll play with an oscilloscope in lab where you can control the voltages by hand and mess around with manipulating electron beams Faraday CT39s F l In which situation is the magnetic ux through the loop the smallest A Same ux in all three situations gt B C D gt gt 539 gt area 2A U y edge 0n A area A 0095mm region tilted 60 uniform B field Answer B When the loop is edge0n t0 the eld the ux through the loop is zero F 2 A loop of wire is moving rapidly through a uniform magnetic eld as shown True A or FalseB there is a nonzero emf induced in the loop 6969 6969696969 Answer False The magnetic ux through the loop is not changing constant B field constant area A constant angle between B field and plane of loop so the emf is zero A loop of wire is spinning rapidly about a stationary aXis in uniform magnetic field as shown TrueA or FalseB there is a nonzero emf induced in the loop 69 Answer True There is a changing ux through the loop since the angle between the B field and the plane of the loop is changing A changing ux creates an emf F 3 The magnetic ux through a loop of wire is shown At which point is the emf induced in the loop a maximum VB p A uX Answer The rate of change of ux and hence the emf is maximum at point B F 4 A loop of wire is sitting in a uniform constant magnet field as shown Suddenly the loop is bent into a smaller area loop During the bending of the loop the induced current in the loop is A zero B clockwise C counterclockwise 0 Bin Answer The ux is decreasing as the loop area decreases To fight the decrease we want the induced B to add to the original B By the right hand rule version 11 a clockwise induced current will make an induced B into the page adding to the original B The answer is Clockwise F S A bar magnet is positioned below a horizontal loop of wire with its North pole pointing toward the loop Then the magnet is pulled down away from the loop As viewed from above is the induced current in the loop clockwise or counterclockwise A eyeb all N A clockw1se B counter clockwise U Answer The B eld from a bar magnet points out of the North pole As seen from above the eld through the loop is out toward the observer As the magnet is pulled away the ux is decreasing To ght the decrease the induced B eld should add to the original B eld and also be out toward the observer The induced current will be counterclockwise in order to make an induced B eld out F 6 A square loop is rotating in a xed external magnetic eld into the page At the instant shown the loop is out of the plane of the page with left side of the loop above the page and coming out of the page the right side is below the page going away The direction of the induced current is aXis of rotation P Bin C Neither Answer At the moment shown the ux through the loop is decreasing since the amount of B eld quotthreadingquot the loop is decreasing To ght the decrease the induced eld should add to the original eld The answer is B F 7 A loop of wire is near a long straight wire which is carrying a large current I which is decreasing The loop and the straight wire are in the same plane and are positioned as shown The current induced in the loop is A counterclockwise B clockwise C zero I to the right but decreasing loop Answer CW At the position of the loop the B eld created by the long wire is into the page The ux is decreasing so Lenz39s law says the induced B eld should into the page to add to the original ux and ght the decrease Righthandrule II says that to make B point into page we need a clockwise induced current F S Two loop of wires labeled A and B are placed near each other as shown A large current I in loop A is suddenly turned on This causes an induced current in loop B which causes A A net replusive force the two loops repel B A net attractive force the two loops attract C whether the force is attractive or repulsive depends on whether the current in rst loop is CW or CCW D No net force Answer A repulsive force the two loops repel Assume a CCW current I in Loop A When the current I is turned on there is an increasing B eld pointing out of the page in the middle of both loops A and B This increasing B eld outward induces a m current in loop B so as to create an induced B eld into the page ghting the increase in ux Now you have to remember that antiparallel currents repel parallel currents attract The two currents are antiparallel so the loops repel If the current in loop A was the other way CW then everything would be reversed and the two currents would still be antiparallel F 9 A wire loop moving right enters a region where there is a constant uniform magnetic eld pointing into the page As the loop enters the B eld the current induced in the loop is A CW B CCW As the loop enters the B eld the direction of the net force on the loop is A right gt B lt C up T D Ldown E there is no net force Answers Question 1 CCW As the loop enters the eld the ux increases To ght the increasing ux the induced B eld out of the page should be opposite the original B into the page A CCW current will create an induced B out of the page This current is an example of an quoteddy curren quot Question 2 Method I the easy way The magnetic forces on eddy currents are always in the direction which opposes the motion Otherwise you could make a perpetual motion machine Method 11 harder way Work out all the directions As the loop enters the eld the ux thru the loop is increasing By Lenz39s law the induced current in the loop will be CCW so that the induced B eld out of the page ghts the change in ux There will be a magnetic force on the right side of the loop The direction of that force is to the left F 10 The vertical wire shown is moving to the right in a uniform magnetic eld which is into the page There is a current upward meaning that there is a ow of electrons downward The xed positive ions in the wire are moving to the right along with the wire The negative conduction electrons are moving to the right and down What is the direction of the net magnetic force on this segment of the wire A Up B Down C Left D Right E None of these conventional Ben current I umform motion of wire and of xed positive ions p electron current 3971 motion of conduction electrons in wire Answer The net force on the wire is Left The force due to the rightward motion of the electrons is exactly cancelled by the force due to the rightward motion of the protons Opposite sign charges so oppositedirection forces Only the downward component of the velocity of the electrons produces a force leftward which is unbalanced by the force on the protons F 11 An electric motor consists ofa coil free to turn on an aXis located in an magnetic eld B created by an arrangement of permanent magnets With the current and eld directions as shown which way will the coil rotate A CW axis B CCW C the coil won t rotate when at this particular orientation Answer The coil will turn counterclockwise B F 12 Atransformer is connected to a battery as shown The voltage difference across the resistor R is A V NzNl B V NlNz C V D zero E not enough information to answer N2 iron COTS Answer It39s a trick question Transformers only work with timevarying voltages AC voltages The DC voltage V from the battery produces a DC current in the primary coil but produces no voltage of any kind in the secondary coil Transformers work because of Faraday s Law the changing ux produced by the AC current in the primary coil produces an emf in the secondary coil If the ux is not changing there is no emf The answer is Zero F 13 The primary coil of a transformer is connected to a battery a resistor and a switch The secondary coil is connected to an ammeter When the switch is thrown closed the ammeter shows A a zero current B a nonzero current for a brief instant C a steady nonzero current Answer There is a nonzero current brie y as the switch is closed but then there is no current after the switch has closed As the switch is closed the current in the primary changes from zero to some nonzero value While the current is changing there is a changing Bfield and a changing ux which causes an emf in the secondary and a current ow in the secondary iron core F 14 A stepdown transformer is attached to an AC voltage source and a resistor as shown How does the current in the resistor I R compare to the current in the drawn from the AC source Iinpm With AC circuits we always use rrns values of I and V A I R gt 17in B I R lt 17in C 1 R 17in 11n 1R D Depends on the value of Ii gt 3 Answer When the voltage is stepped down the current must be stepped up since powerin powerom so P IV is constant F 15 An electrical engineer at a power plant wants to reduce the energy wasted during power transmission from the plant to the city The power output P0IV of the plant is fixed at lOOMW The engineer decides to double output voltage V By what factor does the power lost in the cable P1051 I2 Rcable decrease Hint if PIV is fixed when V goes up I goes down I Rcabl e VAC I Rcit 3 adj lsted to keep PIV constant A No decrease B factor of 2 decrease C factor of 4 decrease D factor of 8 decrease Answer The power lost in the cable decreases by a factor of 4 If V doubles then I decreases by a factor of 2 since P IV is constant If I decreases by a factor of 2 then I2 decreases by a factor of 4 F 16 A long straight wire carrying an increasing current I passes along a diameter of a wire loop The straight wire and the loop are in the same plane but are not in electrical contact No electrical contact between loop The induced current in the loop is and stralght w1re A zero B counterclockwise C clockwise loop 1 increasing Answer zero The total ux in the loop due to the straight wire is zero at all times The B eld on the right half of the loop is into the page the B fleld on the left half is out of the page The de nition of ux is d3 B A cos 9 where 9 is the angle between the normal to the loop and the B fleld For half of the loop the angle 9 0 cosG 1 for the other half of the loop 9 180 cosG l The two halves have opposite sign ux and they cancel Since the ux is always zero which is a constant the time rate of change of ux is zero so no emf no induced current F 17 A rectangular loop is placed in a uniform magnetic eld with the plane of the loop perpendicular to the direction fo the eld If a current is made to ow through the loop in the sense shown by the arrows the eld exerts on the loop A a net force only B a net torque only C a net force and a net torque D neither a net torque nor a net force TH I I Answer No net torque and no net force Using the right hand rule for each side of the square loop the force on each side is away from the center of the loop The net force and net torque are both zero Violet light of wavelength 9 passes through a double slit separation is quotdquot and forms an interference pattern on a screen If the Violet light is replaced With red light of wavelength 2 the original pattern on the screen is reproduced if the slit separation is changed to PINK d2 GREEN d4 BLUE 2d YELLOW 4d PURPLE no change is necessary Red light and green light are both shining on the same double slit or grating Which pattern has the bright spots spread farther apart GREEN Green light bright spots are farther apart PINK Red light bright spots are farther apart YELLOW All bright spots are equally far apart Extra 1 Laser light wavelength 9 illuminates a mask with a single narrow slit What can you conclude about the slit width D OQQOO BLUE D is greater than 9 YELLOW D is smaller than 9 PINK Not enough information given to decide between the above Extra 2 The pattern below appears on a screen when laser light shines on two narrow slits What pattern will you see if the left slit is covered BLUE Pattern stays the same YELLOW Pattern same but gets dimmer PINK Maxima on left half vanish go black GREEN Maxima on right half vanish go black PURPLE Every other maximum vanishes B Y No more dark lines just fades slowly bright in the middle gt darker at the edges ALL COLORS Something entirely different Laser light illuminates two different samples one is quotsingle slitquot with Width D one is quottwoslitquot dif action grating With tiny slits that are quotdquot apart a Which pattern arises om the quottwoslitquot grating l l l l l l OOOOOOO PURPLE OOQOQ YELLOW b Roughly how are quotdquot and quotDquot related BLUE 1 D YELLOW d 2D PINK 1 12 D Chapter 16 Concept tests Q161 Two uniformly charge spheres are attached to frictionless pucks on an air table The charge on sphere 2 is three times the charge on sphere 1 Which force diagram correctly shows the magnitude and direction of the electrostatic forces on the two spheres d B 2Q A I0 I6 I6 l9 I6 l E I0 I6 Answer The forces are equal in magnitude and opposite in direction by Newton39s Third Law and by the form of the Coulomb force equation Q162 A charge Q is xed in space A second charge q is brought a distance r away Then q charge is removed and another charge 2q is brought a distance 2r away Then the 2q charge is removed and a charge 5q is brought a distance 2r away A B C QO O q 4 D r QO O 2q 21 QO O 5q 21 Which charge feels the largest force A q B 2q C 5q D The two ofthe charges feel the same size force Answer 5q The force on the 5q is of magnitude 54 qurz Q163 Two vectors A and B are shown Consider the vector sum C A B What is Cy the y component of C y A3 B2 C 2 D 4 E None of thesedon39t know Answer Cy Ay By 13 2 Q164 An electric dipole consists of two equal and opposite charges q and q separated by some xed distance 6 A charge Q is brought near the dipole and is positioned so that the distances to the q and the q charges are identical as shown below OQ B What is the direction of the net electrostatic force on the Q charge Answer Q165 An electric dipole Q and Q separated by a distance 6 is placed along the XaXis as shown A positive test charge q is placed at position A to the right of the dipole Q Q 0 a q gt0 The test charge feels a force that is A zero B to the right C to the left If the test charge q is removed electric eld at position A is A zero B to the right C to the left If a negative test charge is placed at A it feels a force A zero B to the right C to the left Answers The force on the q is to the left The Efield at A is to the left A negative test charge at A feels a force to the right Q 39Q q o O lt gt X 1 2 F2 F1 Q 1 3 Qt l q F1 F2 Fnet Q166 Two positive charges each of size Q are equal distances from the origin as shown What is the direction of the electric eld at the point in empty space which forms a square with the two charges and the origin y Field Q here X Q Y cF ield Q here Answer A B C D E No Field there A Blue B Green C Yellow D Pink E Purple No field there Etot E2 y I l E1 Q 1 2 Q y A E2 1 A Q E1 Q167 Two charges Q and Q are located on the XaXis as shown what is the magnitude of the electric eld at point A Q Q A C C gtx ltTN RN Rgt 0 0 o A M R R R A E Bl a B kQ 1 C Zero D 1 E None ofthese Answer A Magnitude is the size of the E eld Magnitude is positive by de nition Q gt gt O l m E A E2 L I V I r X 1 2 Q22 0 Is paper transparent or opaque to infrared light such as is produced by your transmitter B transparent C opaque HINT Physics is an experimental science Answer Paper is as opaque in the infrared as it is in the visible This can be veri ed by placing paper over your transmitter and attempting to vote Actually more than one student reported that they were able to vote with paper covering the transmitter Just as a bright light source can shine through a single sheet of paper it appears that some of the IR from the transmitters can leak through the paper if the paper is brought very close to the transmitter head Q22 1 Which has higher frequency UV or IR radiation B UV C IR D They have the same frequency Answer c hf UV has a shorter wavelength 7 then IR Smaller 7 means higher frequency UV has the higher frequency Q22 2 Under cover of night a Girlscout signals her friends on a distant hill by alternately placing slabs of green or yellow jello over her ashlight This signal is most accurately described as B Frequency modulation C Amplitude modulation Answer Frequency modulation By placing various colored jellos over the light the scouts are altering the color of the light Different colors correspond to different frequencies CT22 3 A point souce of radiation emits power P0 isotropically in all directions uniformly A detector of area ad is located a distance R away from the source What is the power p received by the detector P0 B 4ch2 d 2 a O FOR dz detector ad ad D P0 P0 E 75R2 d Answer Wad At a distance R from the source the power P0 is now spread out 1 uniformly over a sphere of radius R area 41TR2 So at distance R the energy ux or intensity which is powerarea is P04TER2 The power received by the detector is power P0 area of detector 2 ad area 411R CT22 4 Two radio dishes are receiving signals from a radio station which is sending out radio waves in all directions with power P Dish 2 is twice as far away as Dish 1 but has twice the diameter Which dish receives more power B Dish 1 C Dish 2 D Both receive the same power W x Answer both receive the same power Dish 2 has twice the diameter but 4 times the area area of disk ndz 4 The energy ux powerarea at any distance R is P04TER2 So at Disk 2 the ux is 14 the ux at Disk 1 Power received powerarea gtlt detector area see concept test above The factor 4 increase in disk area is just cancelled by the factor of l 4 decreased in ux

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