MULTIPLE CHOICE SOLUTIONS--E&M TEST I

MULTIPLE CHOICE SOLUTIONS--E&M TEST I 1.) The capacitrs in the circuit are fully charged. At t 0, the dielectric between the plates f C 1 is quickly remved and that capacitance is halved. As a cnsequence: a.) After a lng perid f time, the charge n C will V C 1 C increase. [After a lng perid f time, the vltage acrss C will again be V. As the value f C hasn't changed, its charge Q C V will nt have changed. This respnse is false.] b.) After a lng perid f time, the charge n C will decrease. [Frm abve, this is false.] c.) Just an instant after the dielectric is remved, the vltage acrss C 1 will g t V. [The initial charge n C 1 is Q 1 C 1 V C C 1 V. At t 0 when the dielectric is remved, there will nt yet have been time fr the free charge n C 1 's plates t mve, but the reverse-riented induced charge n the dielectric surface will be gne (reverse-riented in the sense that psitive charge is induced n the dielectric surface next t the capacitr's negative plate--the presence f this induced charge creates an electric field that fights the electric field generated by the free charge n the plates). That remval will effectively increase bth the magnitude f the electric field and vltage difference between the plates. At just a hair past t 0, (C 1 /) (Q 1 )/(V ), and the plate vltage will have temprarily dubled. This respnse is true.] d.) Nne f the abve. [Npe.].) The charges shwn in the cnfiguratin frm an equilateral triangle. Where will a negative charge mst likely feel a net frce in the -j directin? +Q Pint A a.) Pint A. [A negative charge at Pint A will be attracted t Pint B the +Q charge and repulsed by the tw -Q charges. Because the tw -Q (at triangle's charges are farther away and at an angle, their net effect will be less. center) As Pint A is relatively clse t the +Q charge, it will hld sway and -Q -Q the net frce at A will be in the -j directin. Other pssibilities?] b.) Pint B. [At Pint B, the electric field will be almst zer. Pint C Nevertheless, the +Q charge and the tw -Q charges will apply frce in the +j directin, s this ptin is definitely nt a pssibility.] c.) Pint C. [At Pint C, the electrical effect due t the tw -Q charges will cancel leaving a negative charge feeling attractin t the +Q charge. This frce will be in the +j directin. This respnse is false.] d.) Bth Pints A and C. [Npe.] V (vlts) e.) Nne f the labeled pints. [Npe.] 10.) An electrical ptential field alng the x-axis is defined by the graph shwn. The assciated electric field is:...6 x (meters) -10 5

a.) Initially negative, then psitive. Als, the field is zer at x.5 meters. [The relatinship between an electric field E and its assciate electrical ptential field V is summarized by the relatinship E V. In ne dimensin, the relatinship states that minus the rate at which the electrical ptential functin changes (i.e., V's slpe) is equal t the electric field functin. Fr ur electrical ptential graph, the slpe is first negative, then psitive. That means that the electric field is first psitive, then negative. The turnarund pint where bth the slpe f V and the electric field E are zer is at x. meters. This respnse is false.] b.) Initially psitive, then negative. Als, the field is zer at x.5 meters. [Npe.] c.) Initially psitive, then negative. Als, the field is zer at x. meters. [This is the ne.] d.) Nne f the abve. [Npe.].) A hllw sphere f inside radius a and utside radius a has a vlume charge kr density sht thrugh it f ke, where k r and k are cnstants. The electric flux thrugh a sphere whse radius is a will be: a.) π k kr ( e ), and that functin wuld have been different if the inside kε radius f the hllw had been (1/)a. [This is wrng fr a number f reasns. First, the charge enclsed is psitive. This means the field directin will prduce a psitive electric flux. Als, the charge enclsed calculatin shuld be evaluated frm a t a. The functin shwn is an indefinite integral that hasn't been evaluated. If yu didn't happen t ntice either f thse prblems, yu prbably did the math, which fllws: q flux φ E ds electric enclsed ε qenclsed φelectric ε a c a πk a ρdv c a ε kc ke c ε π a ( c a k c) πk ( πcdc) k e dc ε 1 kc ( e ) k [ ] ε k π k ε a c a k( a) ka ( e e ) 1 k ε k( a) ka ( e e ). a a 5

( ) b.) π k k( a) ka e e, and that functin wuld have been different if the inside kε radius f the hllw had been (1/)a. [Frm abve, this is the crrect expressin. Additinally, halving the inside radius wuld have increased the charge enclsed and, as a cnsequence, the flux thrugh a sphere f radius a. This respnse is true.] πk kr c.) ( e ), and that functin wuld nt have been different if the inside radius f kε the hllw had been (1/)a. [On the surface, at least the sign appears t be crrect here (thugh appearances can be deceiving). This unevaluated integral will nt d, thugh. ALWAYS evaluate integrals when yu can. This respnse is false.] d.) π k k( a) ka ( e e ), and that functin wuld nt have been different if the kε inside radius f the hllw had been (1/)a. [ight expressin, wrng analysis f what wuld happen if the inside radius had been smaller. This respnse is false.] e.) The integral invlved in this prblem is nt standard, hence the prblem cannt be slved withut access t a table f integrals. [This is a thrw-away fr thse wh lked at the prblem and thught what are the dds? This respnse is false.] 5.) A psitive charge mves with knwn velcity v int regin I in which exists an unknwn B-field. It accelerates as shwn in the sketch, then enters regin II in which there exists nt nly B but als an unknwn electric field E. a.) The directin f the B-field is tward the bttm f the page, and there is n need fr the presence f an electric field t keep the charge mving in the directin shwn in regin II, E 0. [There clearly is an initial dwnward frce n the charge as it enters regin I, but magnetic frces d nt rient themselves in the directin f magnetic fields. The directin f a magnetic frce will be PEPENDICULA t bth the directin f B and the directin f v. In shrt, if B v v flattened right hand in this directin regin II (bth B & E) E had been directed tward the bttm f the page, the frce qvxb wuld have pushed the charge int the page. This respnse is false.] b.) The directin f the B-field is int the page, and the directin f the E-field is t the left. [Instead f trying each f the pssibilities listed in the varius respnses, the easiest way t d this part f the prblem is t use the crss prduct assciated with qvxb backwards. That is, run yur flattened right hand in the directin f the initial v, then rient yur thumb hitch hiker style in the directin f the required frce (in this case, dwnward). The directin in which yur curled F hitch hiker thumb in this directin regin I (B-field B nly) fingers end up is the directin f the magnetic field. The directin that satisfies this criterin fr this prblem is ut f the page. That is the directin f the magnetic field. As such, this respnse is false.] c.) The directin f the B-field is int the page, and the directin f the E-field is t the right. [At the very least, the directin fr the B-field is wrng. This respnse is false.] fingers will curl ut f the page--this is the directin f the magnetic field 55

d.) The directin f the B-field is ut f the page, and the directin f the E-field is t the left. [This has the crrect directin fr the magnetic field. What abut the directin f the electric field? With a B-field directed ut f the page, the charge mving in the vertical will feel a magnetic frce riented t the left (qvxb). Because the charge is psitive, the directin f the electric frce and the directin f the electric field will be the same. T cunteract a magnetic frce t the left, therefre, the electric field directin must be t the right. This respnse is false.] e.) The directin f the B-field is ut f the page, and the directin f the E-field is t the right. [This is the ne.] f.) Nne f the abve. [Npe.] 00 Ω 6.) Given the circuit infrmatin shwn in the circuit, hw large must the resistr be? a.) 500 Ω. [Fr practice, let's use Kirchff's Laws n this ne. The currents and lps are defined in the accmpanying sketch (nte that a nde equatin was used t define the current thrugh the 1000 Ω resistr--this is nice because it means we have nly ne current unknwn alng with the unknwn ). Summing the vltage differences arund L-I yields (00 vlts) - (00 Ω)(.75 amps) - (1000 Ω)(.75 - i ) 0. Slving yields i.5 amps. Summing the vltage differences arund L-II yields (.75 - i )(1000Ω) - i 0. Substituting in fr i yields a resistance 500 Ω. This respnse is true.] b.) 1000 Ω. [Npe.] c.) 000 Ω. [Npe.] d.) Nne f the abve. [Npe.] V00 vlts V00 vlts i.75 amps 00 Ω i.75 amps L-I.75-i 1000 Ω L-II 1000 Ω i 7.) The MS vltage acrss the 10 kω resistr in the L circuit shwn is vlts. The apprximate MS vltage acrss the inductr is: a.) vlts. [The amplitude f the pwer supply is 10 vlts, s the MS value acrss the pwer supply is.707(10 vlts) 7.07 vlts. If the MS vltage acrss the resistr is vlts, the MS vltage acrss the nly ther element in the circuit must be 5.07 vlts. This respnse is false.] b.) 5 vlts. [This is the ne.] c.) 8 vlts. [If yu mistk the MS vltage acrss the pwer supply fr 10 vlts, yu gt this incrrect respnse.] d.) Nne f the abve. [Npe.] L, r V(t) 10 sin (77 t) 0 Ω 8.) The capacitrs in the circuit shwn are initially uncharged. At t 0, the switch is clsed. At t. secnds, it is bserved that the current being drawn frm the battery is apprximately amps. The capacitance f C is apprximately: a.). mf. [The initial current in the circuit, given the fact that bth capacitrs will initially act like shrts, is i V/ (100 v)/(0 Ω) 5 amps. The fact that the 100 v 1. mf C 56

current has drpped by 60% means that apprximately ne time cnstant has passed during the. secnd perid. Fr this circuit, ne time cnstant equals C eq. The equivalent capacitance is C + 1.x10 - farads, s we can write (. secnds) (0 Ω)(C + 1.x10 - ), r C 8.8x10 - farads. This respnse is false.] b.) 8.8 mf. [This is the ne.] c.) 1. mf. [Npe.] d.) Nne f the abve. [Npe.] B-field 9.) A cil f resistance faces a unifrm B-field cming ut f the page that dubles at a cnstant rate every 10 secnds. At t 0, the field strength is.5 teslas. As time prgresses: a.) The EMF generated in the cil will be cnstant,and the induced current in the cil will be clckwise. [Because the magnetic field increases unifrmly, the magnetic flux will increase unifrmly, φ m will be t cnstant, and the induced EMF will be cnstant. The first part f the respnse is true. As fr the directin f induced current flw, with an increasing magnetic flux, the current will flw in such a directin as t prduce an induced magnetic field thrugh the cil that is ppsite the directin f the external field. The induced current that prduces a B-field int the page will flw in a clckwise directin. This respnse is true.] b.) The EMF generated in the cil will increase, and the induced current in the cil will be cunterclckwise. [The EMF is cnstant. This respnse is false.] c.) The EMF generated in the cil will decrease, and the induced current in the cil will be clckwise. [The EMF is cnstant. This respnse is false.] d.) The EMF generated in the cil will increase, and the induced current in the cil will be clckwise. [The EMF is cnstant. This respnse is false.] e.) There will be n EMF r induced current in the cil. [Npe.] 10.) A slid cylinder f radius a has a vlume charge density sht kr thrugh it f ke, where k r and k are cnstants. The electric field functin fr r < a is: a.) k kr e. [Defining a Gaussian cylinder f length L krε and radius r > a, we need t determine the ttal charge inside the charged area. Using a cylindrical shell f radius c, thickness dc, circumference πc, and differential vlume dv (πc dc)l, Gauss's Law yields: qenclsed E ds ε E( π rl) E r c 0 ρdv ε kc ke πcldc c ( πrl) ε ( ) a 57

k c ( ) k e dc c 0 E rε E r 1 k e k rε [ ] kc ( ) c 0 ( ) k kr E e 1. krε The respnse in this sectin is that f an indefinite integral. Because we knw the limits f integratin, we must d the evaluatin. This respnse is false.] b.) k kr ( e 1). [This lks like the ne.] k rε ( ) k kr ka c.) e e. [Npe. The limits are messed up.] kεr d.) Nne f the abve. [Npe.] r 11.) A 1 amp fuse is placed in series with an C circuit in which the AC vltage amplitude is 1500 vlts. The net resistance is 10 Ω, and the capacitance is 0 mf. Apprximately what is the largest frequency at which the pwer supply can perate withut blwing the fuse? a.) 50 Hz. [Under nrmal circumstances, Ohm's Law shuld be used with MS values. In this case, thugh, the.707 factr in the V MS i MS expressin will simply cancel ut. As such, we will use amplitude values instead (this is actually useful here as the current f interest is nt 10 amps MS but 10 amps maximum). Using the apprpriate expressin fr impedance f an C circuit, we can write Ohm's Law as Vmax imaxz imax net + 1 πνc. Putting in the apprpriate values, we get [1500 vlts] [1 amp][[10 Ω] + 1/[πν (x10 - farads)] ] 1/. Slving yields ν.5x10 hertz. This respnse is true.] b.) 500 Hz. [Npe.] c.) 5000 Hz. [Npe.] d.) Nne f the abve. [Npe.] 1.) An electric field is set up as shwn. a.) The field will d mre wrk when a psitive charge ges frm A t B than when the same charge ges frm A t C. [As W q V, the amunt f wrk the field des is related slely t the size f the charge and the electrical ptential difference between the tw pints. The ptential difference between A and B will be the same as the ptential difference between A and C as bth B and C are n the same equiptential line (draw a line between B and C--that line will be perpendicular t the electric field lines, a characteristic f equiptential lines). In shrt, the field will d the same amunt f wrk n a charge mving frm A t B as it des n a charge mving frm A t C. This respnse is false.] A D B C E 58

b.) The field will d the same amunt f wrk when a psitive charge ges frm A t B as when the same charge ges frm A t C. [Frm abve, this is true.] c.) The field will d mre wrk when a psitive charge ges frm A t B than when the same charge ges frm A t D because the distance between A and D is greater. [The line between A and D is an equiptential line--the wrk dne by the field as the bdy mves frm A t D will be ZEO. This respnse is false.] d.) Nne f the abve. [Npe.] Ω 1.) Apprximately 1 amp flws int the bx. The circuit that exists inside the bx will be: a.) Fur Ω resistrs in series. [Frm Ohm's Law, i V/. If the ttal current i drawn frm the battery is t be apprximately equal t 1 amp, and if V 10.1 vlts, eq must be smewhere arund 10 Ω's. Knwing that the external resistr is Ω's, the equivalent resistance frm inside the bx must be apprximately 8 Ω's. Fur Ω resistrs in 10.1 vlts series will prvide 8 Ω's t the system, and this respnse is true. Are there ther true respnses?] b.) Fur Ω resistrs in parallel. [Fur Ω resistrs in parallel will have an equivalent resistance f ( Ω)/ 8 Ω's (nte that I've used a shrtcut t determine the equivalent resistance f the parallel cmbinatin--if all f the resistrs in the cmbinatin are equal, eq /n, where is the value f the cmmn resistance and n is the number f resistrs in the cmbinatin--if yu dn't believe that this will wrk, try it n a few examples). This will d the trick, als. Are there ther true respnses?] c.) An 8 Ω resistr in parallel with a 10,000 Ω resistr. [Fr parallel circuits, eq will always be smaller than the smallest resistr in the cmbinatin. If there is an enrmus difference in resistr size amngst the resistrs in the cmbinatin, it is the small resistrs that determine eq. This shuld be clear, but if it is nt, see fr yurself. Fr this situatin, eq [1/(8 Ω) + 1 /(1000 Ω)] -1 [.15 +.001] -1 [.16] -1 7.995 Ωs. This is, t a very gd apprximatin, equal t 8 Ω's. This respnse is true.] d.) Bth a and b. [Nt quite.] e.) espnses a, b, and c. [This is the ne.]? 1.) In what regin is the net magnetic field equal t teslas directed ut f the page? a 1 a a.) In regin I. [As the directin f the magnetic field prduced by the amp current in regin I is ut f the page, and as the 1 amp current prduces a field that is als ut f the page in that regin, regin I will have a place where the net magnetic field is regin I regin II regin III ut f the page and equal t teslas. This respnse is true. Are there thers?] b.) In regin II. [The temptatin is t assume that because the amp wire prduces a big magnetic field int the page in regin II, there will be n pint where ur criterin is satisfied. That is nt true. The 1 amp wire prduces a magnetic field that is ut f the page in regin II. Very clse t that wire, its B-field will predminate ver that f the amp wire. Furthermre, there will be a place where the net magnetic field will equal teslas. This statement is true.] 59

c.) In regin III. [Bth wires prduce magnetic fields int the page in regin III. This respnse is false.] d.) Bth in regins I and II. [This is the ne.] e.) All f the abve. [Npe.] 15.) A diple is placed in an electric field as shwn. Over time, the diple will: a.) Experience a cnstant acceleratin f its center f mass tward the right and will experience a cnstant trque that mtivates it t angularly accelerate in a clckwise directin. [The electric field in this situatin is cnstant. That means that the frce n the psitive charge directed t the left (i.e., in the directin f the electric field lines) will equal the frce n the negative charge directed t the right. In ther wrds, there will be n net frce n the diple and it will nt translatinally accelerate at all. That, in itself, makes this statement false.] b.) Experience n acceleratin f its center f mass but will experience a varying trque that mtivates it t angularly accelerate in a clckwise directin. [The first part f this respnse is true. As fr the secnd part: There will be a trque n the diple that will vary in magnitude, depending upn the angular psitin f the diple relative t the electric field (when the diple is aligned with the electric field, the trque will be zer; when the diple is perpendicular t the electric field, the trque will be a maximum). That trque will be in the cunterclckwise directin, and this respnse is false.] c.) Experience a varying acceleratin f its center f mass tward the left and will experience a varying trque that mtivates it t angularly accelerate in a clckwise directin. [Npe.] d.) Experience n acceleratin but will experience a varying trque that mtivates it t angularly accelerate in a cunterclckwise directin. [This is the ne.] e.) Nne f the abve. [Npe.] 16.) Charges are placed as shwn at the crners f a rectangle. a.) At the center f the rectangle, the x-cmpnent f the electric a field will be psitive (t the right) and the y-cmpnent will be negative (dwnward). [All f the charges are psitive. That means that a psitive Q test charge will be repulsed by each. As mst f the charge is n the right a side f the system, the x-cmpnent f the electric field at the center will be t the left in the negative directin. As such, this respnse is false.] b.) At the center f the rectangle, the x-cmpnent f the electric field will be negative (t the left) and the y-cmpnent will be negative (dwnward). [The first part is crrect. In the y-directin, the electric field cmpnent due t Q will verpwer the field cmpnent due t Q, and the field cmpnent due t Q will verpwer the field cmpnent f Q. As Q and Q are n the bttm, the net y-cmpnent shuld be upward in the psitive directin, and this respnse is false.] c.) At the center f the rectangle, the x-cmpnent f the electric field will be psitive (t the right) and the y-cmpnent will be psitive (upward). [Npe.] d.) At the center f the rectangle, the x-cmpnent f the electric field will be negative (t the left) and the y-cmpnent will be psitive (upward). [This is the ne.] Q Q Q 550

B-field current-carrying wire 17.) A meter lng wire carries a.5 amp current as shwn. If the wire is bathed in a 10 - tesla magnetic field, the magnitude f the frce n the wire is: a.) Zer newtns. [Using F ilxb, the magnitude f the frce will be F (.5 a)( m)(10 - T)(sin 90 ) 10 - newtns. This respnse is false.] b.).5x10 - newtns. [If yu mistakenly tk the angle between L and B t be 0, yu gt this incrrect answer. NOTE: When either L r B is in the plane f the paper and the ther is perpendicular t that plane, the angle between the tw vectrs will ALWAYS be 90.] c.) 10 - newtns. [This is the ne.] d.) Nne f the abve. [Npe.] i 0 switch A B 18.) In the system shwn, the switch has been set n cntact A fr a lng time. At t 0, the switch flips frm cntact A t cntact B. The pwer dissipated by 00 v.01 f 00 Ω the resistr during the discharge will be: a.) 00 watts. [This is a tricky questin. When a prblem refers t pwer dissipated, it is referring t the amunt f wrk per unit charge being dne at a particular pint in time (in a resistr, fr 5 Ω instance, that quantity is equal t i ). If the questin had asked fr the average pwer dissipated during the discharge, we might have had a sht, but as wrded, this questin makes n sense. Nte that if the prblem had asked fr the amunt f energy dissipated by the resistr during discharge, that value wuld have equaled the ttal energy stred in the capacitr. That quantity is.5cv.5(.01)(00 v) 800 jules. In any case, this respnse is false.] b.) 00 watts. [Npe.] c.) 800 watts. [As this is the amunt f energy stred in the capacitr, it is als the amunt f energy dissipated by the resistr during the discharge. Unfrtunately, that wasn't the questin and this respnse is false.] d.) Nne f the abve. [This is the ne.] 19.) The magnetic flux thrugh a square lp is measured and fund t be 10 - webers when the lp's nrmal is riented at an unknwn angle θ relative t the directin f the B-field (see sketch). When the nrmal is parallel t the directin f the B-field, the magnetic flux thrugh the lp B-field lp nrmal is 10 - webers. The angle θ is: lking dwn n the cil a.). [The magnetic flux in the first rientatin is equal t BA cs θ 10 - webers. When the cil is facing the magnetic field, we have BA cs 0 10 - webers. As the magnetic field B is cnstant and the area A is 0 551

cnstant, the prduct f the tw will always be equal t 10 - webers. Using that infrmatin in the first equatin yields BA cs θ (10 - teslas) cs θ 10 - webers, r θ cs -1 (.1). Accrding t my calculatr, that equals 8.. This respnse is false.] b.) 7. [Npe.] c.) 78. [Npe.] d.) Nne f the abve. [This is the ne.] 0.) An electric field is defined by E kr r, where r is a unit vectr in the radial directin. a.) The units f k must be kilgrams/(meter. secnd. culmb). [k's units must be such that when they are multiplied by meters (i.e., r 's units), we get the electric field units (newtns/culmb). That means that k's units are newtns/culmb. meter. As the units fr newtns are kilgram. meters/secnd, the units fr k must be kilgrams/(meter. secnd. culmb). This respnse is true. Are there thers?] b.) The units f k must be vlt/meter. [Electric fields are related t electrical ptential fields by the expressin E. d V. Frm this, the units fr E can be written as vlts/meter. k's units must be such that when they are multiplied by meters, we get the electric field units (vlts/meter). Thse units will be vlts/meter, and this respnse is true. Nte that the verall crrect respnse fr this prblem will prbably be all f the abve, but t be sure, we ught t examine espnse c t be sure.] c.) The units f k must be newtns/(meter. culmb). [eiterating what has been said abve, k's units must be such that when they are multiplied by meters, we get the electric field units (newtns/culmb). Cmbining, we get k's units as newtns/culmb. meter, and this respnse is true.] d.) All f the abve. [Lks like this is the ne.] e.) Nne f the abve. [Npe.] 1.) The vltage acrss the.5 Ω resistr is: a.) 1.5 vlts. [If we knew the vltage V acrss the battery, we culd subtract frm V the vltage acrss the 1 Ω resistr (V i) t get the vltage acrss the.5 Ω resistr. Unfrtunately, we dn't have that infrmatin. What we d knw is that the current thrugh the.5 Ω resistr is equal t the current int the parallel cmbinatin, and the vltage acrss the parallel resistrs is the same. That parallel vltage equals (5 amps)(1 Ω) 5 vlts. The current thrugh the Ω resistr is that.5 Ω V 5 amps 1 Ω Ω vltage divided by Ω's, r i (5 vlts)/( Ω).5 amps, and the ttal current int the parallel cmbinatin (i.e., the current thrugh the.5 Ω resistr) is.5 amps + 5 amps 7.5 amps. With that, we can write: V.5Ω res (7.5 amps)(.5 Ω).75 vlts. This respnse is false.] b.) 7.5 vlts. [This is the numerical value fr the current thrugh the.5 Ω resistr, nt the vltage acrss the.5 Ω resistr. This respnse is false.] c.) 5 vlts. [Npe.] d.) Nne f the abve. [This is the ne.] 55

.) A.5 kg mass has a 10 culmb charge n it. It is placed in an electrical ptential field at x meters where the vltage is vlts. eleased frm rest, the mass is allwed t accelerate freely. At x meters, its velcity is 1 m/s. a.) The vltage at x meters is 1.6 vlts. [An easy way t lk at this prblem is thrugh the cnservatin f energy. That therem states that KE 1 + U 1 + W ext KE + U. emembering that a charge q at a pint whse electrical ptential is V has ptential energy qv, we can write: qv x.5mv + qv x, r (10 cul)( vlts).5(.5 kg)(1 m/s) + (10 cul)v x. Slving, we get V x -1.6 vlts. As such, this respnse is false, thugh ne might be tempted t ignre the negative sign and assume it is crrect. In fact, the negative sign isn't a prblem. It is changes f electrical ptential that are imprtant (just as it is changes in ptential energy that are imprtant). The electrical ptential at a specific pint has significance nly in its relatinship t the electrical ptential f ther pints.] b.) The vltage at x meters is -1.6 vlts. [As nnsensical as a negative electrical ptential may seem, this is the answer. If yu think abut it, it isn't that utrageus. The electric field is fairly big (the bdy accelerated frm rest t 1 m/s ver a mere ne meter), which suggests that the electrical ptential difference between x and x must be fairly big. Psitive charges accelerate frm higher t lwer electrical ptential which means that the electrical ptential at x must be smaller than the electrical ptential at x. It isn't utside the realm f pssibility that that smaller electrical ptential culd range int the negative numbers... which it did. In any case, this respnse is true.] c.) The vltage at x meters is -1.7 vlts. [If yu frgt t square the velcity term, yu came ut with this. It is incrrect.] d.) Nne f the abve. [Npe.].) A triangular lp and a rectangular lp each have the same area. Each is frced t apprach a current-carrying wire with the same cnstant velcity (see sketch and ignre gravity). a.) The directin f the induced current in bth is clckwise, and the induced current in the triangle is greater than the induced current in the rectangle. [The lng wire is prducing a magnetic current-carrying wire v triangular lp i rectangular lp field that is int the page in the regin f the lps (use the right-thumb rule t determine this). The cils are appraching the lng wire, s the magnetic flux thrugh each is increasing. That means the induced current will set up s as t create an induced magnetic field thrugh the cils that is ut f the page. T d that, the current must be in the cunterclckwise directin. Frm that alne, this respnse is false.] b.) The directin f the induced current in bth is cunterclckwise, and the induced current in the triangle is greater than the induced current in the rectangle. [The induced current directin is crrect here. What abut the secnd part? Due t its shape, the magnetic flux change will be greater in the rectangle than it is in the triangle. As such, this respnse is false.] c.) The directin f the induced current in bth is clckwise, and the induced current in the rectangle is greater than the induced current in the triangle. [Npe.] v 55

d.) The directin f the induced current in bth is cunterclckwise, and the induced current in the rectangle is greater than the induced current in the triangle. [This is the ne.] e.) The induced currents are in ppsite directins. [Npe.].) Fr the circuit shwn: a.) -8i 9-18i 7-6i 60. [The lps used fr all f the equatins written fr this questin are highlighted in the sketch. Summing the vltage changes arund LOOP a yields -7i 9-18i 7-6i - 11i 9-60 0. Cmbining like terms and placing the 60 vlt quantity n the right side f the equal sign, we get -8i 9-18i 7-6i 60. This respnse is true. Are there thers?] b.) 16i 8-7i + 8i 6-0. [LOOP b yields 16i 8-7i - 0 + 8i 6 0. This respnse wuld wrk if the sign f the 0 vlt term was crrect. This respnse is false.] c.) 7i - 9i + 7i 5 5. [LOOP c yields 7i - 9i - 5 + 7i 5 0. This matches Ω 6 Ω i 1 i i 5 v 6 Ω 7 Ω 9 Ω LOOP c 7 Ω 11 Ω i i 5 16 Ω LOOP b 0 v 18 Ω i 6 i 8 8 Ω LOOP a and this respnse is true. Are there ther pssibilities?] d.) There are at least tw crrect lp equatins abve. [A nasty thing t d, making yu check each lp instead f finding the first crrect ne and stpping there. In any case, this respnse is true.] e.) Nne f the abve. [Npe.] C i 7 60 v i 9 7 6 Ω 5.) The capacitance f the capacitr in the circuit is 10 nf and the resistance is 1000 Ω. At 00 cycles per secnd, the apprximate capacitive reactance f the circuit is: a.) 1.5x10-1 Ω. [The circuit's capacitive reactance is X C πνc 1/[(π)(00 Hz)(10x10-9 f)] 7.96x10 Ω. This respnse is false.] b.) 000 Ω. [Npe.] c.) 80,000 Ω. [This is the ne.] d.) Nne f the abve. [Npe.] V(t) V sin (π t) 6.) The equivalent capacitance fr the capacitr cmbinatin shwn is: a.) 11/8 picfarads. [The tw parallel cmbinatins have equivalent capacitances f pf and 16 pf respectively. Adding the inverse f pf, 16 pf, and 8 pf 8 pf 8 pf 8 pf 8 pf 8 pf 55

8 pf, then inverting that result yields an equivalent capacitance f (8/11) pf. This respnse is false.] b.) 8/11 picfarads. [This is the ne.] c.) 1.67 picfarads. [If yu reversed the equivalent capacitance relatinships treating the parallel cmbinatins like series cmbinatins etc., yu gt this incrrect answer.] d.) Nne f the abve. [Npe.] 7.) In the circuit shwn, what will the ammeter read? a.) Zer amps. [There is bviusly a vltage difference acrss the 0 Ω resistr, s the current thrugh that resistr and that branch will nt be zer and this respnse is false.] b.). amps. [In a way, this is s easy it's tricky. Althugh the circuit lks awful, the nly thing that is required t determine the reading f the ammeter is knwing the current thrugh. That feat can be easily dne if yu knw the vltage acrss and the magnitude V100 v 15 Ω 1 15 Ω 0 Ω f. The resistr's magnitude was given. The vltage acrss is simply the vltage acrss the battery (nte that there is nthing between and either the high vltage r lw vltage terminal f the battery). As such, i (100 vlts)/(0 Ω). amps. This respnse is true.] c.) 10 amps. [Npe.] d.) Nne f the abve. [Npe.] A 15 Ω 15 Ω 5 8.) Three current-carrying wires riented perpendicular t the page are psitined at the crners f a triangle as shwn. Assume the current magnitudes are the same fr all f the wires. Wires C and D will prduce a magnetic frce n wire A. In what directin will that frce be? wire A wire C a.) b.) c.) d.) e.) Nne f these. wire D [Cmmentary: The net frce n A will be the sum f the magnetic frces prduced by the interactin between A's current and the magnetic fields generated by C and D. Wrking with C first: The directin f C's magnetic field at A is determined using the right-thumb rule. That is, rient yur right thumb in the directin f current in C (i.e., int the page). The curl f yur fingers will give yu the sense f the circulatin f C's B-field arund C's wire. Evaluated at wire A, C's magnetic field directin is straight up. The directin f the frce n A due t that field is determined using F ilxb. The right-hand rule used fr crss prducts (with yur flattened right hand pinting in the directin f the first vectr (v), rtate yur flattened hand s that yu can curl yur fingers in the directin f the secnd vectr (B)--yur thumb extended hitch hiker style will pint in the directin f the crss prduct (F)) suggests that the directin f this frce will be directly hrizntal and t the right. A similar analysis yields a magnetic field due t D at A that is t the right and up, and the frce crss prduct fr 555

that situatin yields a frce directin t the right and dwn. The vectr sum f the tw frce vectrs yields a net vectr that is dwn and t the right. A graphic summary f all f this is shwn belw. The vectr that best mimics this cnfiguratin is espnse c.] directin f magnetic field due t current in wire C (this is tangent t a circle centered n wire C and passing thrugh wire A) Nte that the frce n wire A is perpendicular t bth the current directin and the directin f the magnetic field directin f magnetic field due t current in wire D (this is tangent t a circle centered n wire D and passing thrugh wire C wire A) wire A B c F d B d F c wire D 9.) The net impedance in an L circuit is 000 Ω. The resistr-like resistance in that circuit is 1000 Ω. The inductr is remved and placed in a secnd circuit whse vltage amplitude is twice that f the riginal circuit and whse frequency is the same. The resistr-like resistance in that circuit is 500 Ω. The apprximate impedance in that circuit will be: a.) 1000 Ω. [T begin with, the amplitude f the pwer surce in the circuit has n bearing n the resistive nature f the circuit. As such, the fact that the amplitude f the pwer surce has been dubled means nthing. A measure f the resistive nature f the circuit is called the net L C impedance. The mst general expressin fr the impedance is + ( X X ). Because there is n capacitr in ur circuit, that expressin becmes net + ( πν L) (nte that the inductive reactance X L has been replaced by its equivalent πνl). Plug in numbers fr the riginal circuit int that expressin, we get (000 Ω) [(1000Ω) + X L ] 1/, r XL () 1/ x10 Ω. Because the frequency is the same in bth circuits, the inductive reactance X L will be the same in bth situatins. This infrmatin cupled with the impedance expressin as used n the secnd circuit yields Z [(500Ω) + X L ] 1/ [(500Ω) + [() 1/ x10 Ω] ] 1/ [(50000Ω) + (000000 Ω)] 1/ 180 Ω. This respnse is false.] b.) 1800Ω. [This is the ne.] c.) 000 Ω. [Npe.] d.) Nne f the abve. [Npe.] Q 1 Q 0.) Charges Q 1 and Q are placed as shwn. We dn't knw whether they are psitive r negative. It is knwn that charge q is psitive and that, as placed, it feels n electric frce. a.) Charge Q 1 must be psitive and larger in magnitude than Q, which must be negative. [Fr q t feel n frce, Q must be smaller than Q 1. Hw s? Q 1 is farther away than Q. The nly way the frce prvided by Q 1 can cunteract the frce prvided by Q is if Q 1 is ppsitely charged and larger in magnitude than Q. As such, the first part f this statement is q 556

true. As fr the actual sign, it desn't matter as lng as the signs are ppsite. That is, if Q 1 were psitive, it wuld prduce a frce n q in the +x directin. An equal and ppsite frce wuld have t be prvided by Q fr the zer net frce requirement t be satisfied. If Q 1 were negative, it wuld prduce a frce n q in the -x directin. An equal and ppsite frce wuld have t be prvided by Q fr the zer net frce requirement t be satisfied. In either case, the zer frce requirement is met and neither charge cmbinatin is necessary fr the bserved data t be true. This statement is false.] b.) Charge Q 1 must be negative and larger in magnitude than Q, which must be psitive. [This is tricky. The "must be" part f the respnse makes it false, as it desn't matter which charge is psitive and which is negative.] c.) Charge Q 1 must be negative and smaller in magnitude than Q, which must be psitive. [Again, the "must be" part f the respnse makes it false.] d.) If q's charge had been negative, it wuld have been necessary t place it n the ther side f Q 1 t find a pint where it wuld feel n frce. [Nt s. The electric fields prduced by Q 1 and Q had t add t zer fr the frce n q t equal zer. Given what we have deduced abut the charges invlved, that will nly happen at ne place--t the right f Q. In fact, it desn't matter whether q is psitive r negative. When placed at a pint where the E-field is zer, the frce n any q will be zer. This respnse is false.] e.) Nne f the abve. [This is the ne.] 1.) Tw large, parallel plates are ppsitely charged, then discnnected frm the battery surce that charged them up. The distance between the plates is then dubled. a.) The electric field will halve as will the electrical ptential difference between the plates. [If we use Gauss's Law t determine the electric field between ppsitely charged parallel plates, we find an electric field that is cnstant and equal t σ ε, where σ is the charge per unit area n ne plate (the nly assumptin made here is that the dimensins f the plates are large in cmparisn t the distance between the plates; that is, the field expressin is gd in the regin away frm the edges f the plate--if the plate dimensins are n par with the distance between the plates, edge effects take ver and the expressin is n lnger valid). Nwhere in that derivatin is the plate separatin used. As such, the electric field is nt dependent upn plate separatin. The first part f this respnse is false.] b.) The electric field will halve but the electrical ptential difference between the plates will stay the same. [Npe, the electric field stays cnstant.] c.) The electric field will stay the same and the electrical ptential difference between the plates will stay the same. [The first part is crrect. As fr the secnd part: The electrical ptential in a cnstant electric field gets prprtinally smaller as ne mves dwnstream in the field. That means that if the plate separatin dubles and the electric field desn't change (it is still σ ε ), the electrical ptential difference between the plates will duble. The secnd part f this respnse is false.] d.) The electric field will stay the same but the electrical ptential difference between the plates will duble. [This is the ne.] e.) Nne f the abve. [Npe.] 557

.) A variable pwer supply prduces an AC vltage equal t 5 sin (πt) vlts. What is the frequency f the surce? a.) hertz. [The mst general frm fr the vltage acrss a pwer surce is V( t) V sin( ωt+ φ) V sin(( π ν) t+ φ), where V is the amplitude f the vltage functin, φ is the phase shift (usually zer), and ν and ω are the surce's frequency and angular frequency, respectively. As the angular frequency and frequency are related by ω πν, the frequency, ω in this case will be ν π π π cycles/ secnd. This respnse is true. Are there thers?] b.) cycles/secnd. [This questin is bviusly checking t see if yu knw that hertz and cycles/secnd are the same thing. They are. This respnse is als true. Are there thers?] c.) 79 hertz. [If yu multiplied the angular frequency by π instead f dividing by that amunt, yu gt this incrrect respnse.] d.) Bth a and b. [This is the ne.] e.) Nne f the abve. [Npe.].) A wire f radius a is made f an dd mixture f metals that effectively allws fr a current flw that varies frm pint t pint acrss a cut-away f the wire. In fact, the current alng the central axis is zer, with the current magnitude getting larger as ne mves ut frm there. The current density functin (i.e., the current per unit area) is j kc, where c is a distance frm the wire's central axis t the pint f interest. What is the magnitude f the magnetic field a distance r units frm the central axis, where r < a? a.) k µ c /(r). [This is an Ampere's Law prblem (see auxiliary sketch). If the current thrugh a circular Amperian path f radius c was equally distributed thrughut the crss-sectin f the wire, that current wuld equal the current density j (in amps per unit area) times the area A f the face bunded by the path. Mathematically, this wuld be i ja (j)(πc ). Unfrtunately, the current density varies as ne gets farther and farther away frm the central axis. As such, we must define an imaginary lp f differential area da lcated a distance r units frm the central axis and thrugh which the current density is j kr. The differential current di thrugh that differential area will equal di jda (kr)(πr dr). Integrating that ver the entire Amperian face will yield the ttal current thrugh that face. Ding this in the cntext f Ampere's Law yields: B dl µ i thru Bdl ( ) cs0 µ B ( dl) µ ( kr)( πr dr) B( π r) πkµ r dr B k µ c. r This respnse is false.] c r 0 c r 0 di crss-sectin f the wire (current ut f page) Amperian path wire's edge differential path length dl B dr r a j kr with current density j kr, current thru da is i jda (kr)(πr dr) differential area da 558

b.) k µ c /(r). [This is the ne.] c.) k µ c /(r). [Npe. Please NOTE: If yu hadn't knwn hw t d this prblem, a tricky way t apprach it wuld have been t d a unit analysis (this is smetimes called a dimensinal analysis) n the pssible answers. Yu culdn't be sure the cnstants were crrect, but at least a match in units wuld be a big start tward eliminating prspects.] d.) Nne f the abve. [Npe.].) A charge +Q is suspended at the center f a hllw sphere that is, itself, charged unifrmly t -Q. The sphere's inside radius is a and its utside radius is a. a.) The electric field lines utside the sphere will be riented radially utward, and there will be a place between a and a at which the electric field is zer. [The ttal charge enclsed inside a Gaussian surface that is utside the sphere will be -Q. That means that the electric field will be ne f a negative charge. Electric field lines always enter negatively charged structures, s the electric field lines in this situatin shuld be inward, nt utward. This respnse is false.] a +Q a -Q unifrmly distributed thrughut the vlume b.) The electric field lines utside the sphere will be riented radially inward, and there will be a place between a and a at which the electric field is zer. [Frm abve, the first part f this statement is true. As fr the secnd part, there will be a Gaussian radius between a and a where the ttal charge enclsed inside the Gaussian surface will be zer. On that surface, the electric field must evaluate t zer and this statement is true. Are there thers?] c.) The electric field will be cntinuus acrss the bundary defined by the inside radius (i.e., at a). [We culd guess at this, but it wuld prbably be better t actually d the math and see what we get. We knw that the functin that defines an electric field generated Q by the pint charge at the center f the hllw will be E. Evaluating that at a yields π εr Q Q E π ε( a) 16πεa. Inside the regin between a and a, the electric field can be determined using Gauss's Law. Create a Gaussian sphere in the regin. The vlume charge distributin inside the slid is the ttal charge divided by the ttal vlume in which charge Q Q resides, r ρ ( / ) π( a) ( / ) π( a) ( / ) π( 19a. Frm this, we can write: ) q E ds enclsed ε Q+ ρdv a EdS ( ) cs0 ε r Q Q + π πcdc a ( / ) ( a ) ( ) 19 E ( ds) ε 6Q Q 19a E( π r ) ε r r c a ( cdc) 559

6Q c Q 19a a E πr ε Q Q ( r ( a) ) 19a E πr ε Q Q ( r 8a ) 19a E. πr ε This mderately clumsy-lking expressin des smething unexpected. At r a, the (r - 8a Q Q ) term ges t zer and the electric field at r a is fund t be E π ε( a) 16πεa. This is exactly what we gt befre, s evidently, in this case, the electric field is cntinuus acrss the bundary at r a. Please nte: This electric field cntinuus acrss a bundary situatin is nt always true. If the slid had been a cnductr, the electric field at r a due t Q Q the levitated pint charge wuld have been E π ε( a) 16πεa, but the electric field at r a evaluated using the expressin fr the electric field inside a cnductr wuld have been zer. In that case, the electric field is nt cntinuus. Still, this respnse is true.] d.) Bth b and c. [This is the ne.] e.) Nne f the abve. [Npe.] r 5.) A 10 mf capacitr is charged by a 100 vlt battery, then islated (i.e., remved frm the circuit). It is then cnnected in parallel with an uncharged capacitr C. After the charge n the 10 mf capacitr redistributes itself, the vltage acrss C is measured at 80 vlts. The capacitance C is: a.) 1 mf. [Nte: This is a gd example f a prblem that isn't very difficult but that isn't exactly straightfrward, either. Yu just have t diddle with it t get it. Diddling n: The initial charge n the 10 mf capacitr (nte that this is 10 - farads) will be Q 1 C 1 V 1 (10 - f)(100 v) 1 culmb. That is the ttal amunt f charge in the system. When the 10 mf capacitr is cnnected acrss C, that charge redistributes itself between the tw capacitrs until the vltage acrss each is the same (in this case, 80 vlts). Assume the new charge n C is Q. The new charge n the 10 mf capacitr will be (1 - Q). That means that (1 - Q) C 10mf V (10 - farads)(80 vlts).8 culmbs. In turn, that means that Q. culmbs. Knwing the charge Q n C and the vltage acrss C, we can write C Q/V (. culmbs)/(80 vlts).5x10 - farads. This respnse is false.] b.) 8 mf. [Npe.] c.).5 mf. [This is the ne.] d.) Nne f the abve. [Npe.] 560

6.) If yu place a charge -Q n a hllw cnducting sphere, the electric field lines fr the situatin will lk like: a.) b.) c.) d.) e. Nne f these. e.) [Cmmentary: In a static electricity situatin, free charge will always distribute itself s as t have n electric field inside a cnductr. This eliminates espnse b. In an attempt t get away frm its kind, and because there is n additinal charge levitating at the center f the sphere, free charge will migrate t the utside f a cnducting surface (a Gaussian surface anywhere within the cnfines f the sphere will have zer charge enclsed). That means there will be electric field lines utside the sphere, but nne inside the sphere (this eliminates espnses c and d). As the charge is negative, the electric field lines will be inward, which means that espnse a des the task fr us here.] 7.) A.5 kg mass has a 10 culmb charge n it. It is placed at Pint A in an electric field and released frm rest, freely accelerating t Pint B. The electrical ptential f B is 0 vlts. a.) If the field des 00 jules f wrk in the prcess, the velcity f the mass when at B is (00) 1/ m/s. [This is tricky. The temptatin is t try t use the cnservatin f energy and write qv A + 00.5mv + qv B. If yu d, yu will have inadvertently used the 00 jule quantity twice. Hw s? The amunt f wrk the field des as the bdy mves frm A t B can be taken int cnsideratin in ne f tw ways. Either yu can use ptential energy quantities (yu knw V B -- yu can get V A using W field -(qv B - qv A ) 00 jules) and put W extraneus. 0, r yu can ignre the ptential energy quantities and accunt fr the wrk dne by the field by setting W extraneus W field 00 jules. YOU CANNOT DO BOTH. Using the latter apprach, we get (00 jules).5mv B, r vb (800) 1/ m/s. This respnse is false.] b.) If the field des 00 jules f wrk in the prcess, the velcity f the mass when at B is (800) 1/ m/s. [This is the ne.] c.) If the field des 00 jules f wrk in the prcess, there isn't enugh infrmatin t determine the velcity at A because we dn't have enugh infrmatin t calculate the vltage at A. [This isn't the case.] d.) Nne f the abve. [Npe.} 8.) A graph f the current thrugh the primary cil f a transfrmer lks like the sketch shwn t the right (nte that at t 0, the current is negative). A graph f the EMF set up in the secndary will lk like: current (amps) i time (secnds) -i 561

a.) EMF (vlts) b.) EMF (vlts) c.) EMF (vlts) time (sec) time (sec) d.) EMF (vlts) time (sec) time (sec) e.) Nne f the abve. [Cmmentary: An induced EMF is set up whenever there is a changing magnetic flux thrugh the crss-sectin f a cil. In the case f a transfrmer, the changing flux is essentially the same fr bth the primary and secndary cils. As such, a change f current in the primary creates a changing magnetic flux in the secndary. This induces an EMF in the secndary cil. The relatinship between the current change in the primary and the EMF in the secndary is ε s -L(di p /dt). In ther wrds, what we are lking fr here is minus the slpe f the current functin. Lking at the current graph, the slpe starts ut at apprximately zer, then gets prgressively larger until it hits a sectin ver which the slpe is the same (i.e., the slpe is a cnstant). The slpe then gets less and less until, at the tp, it is zer. Either f these tw bservatins eliminates graphs a and b. Graphs c and d generally d a nice jb f graphing the slpe f the current functin, but we want minus that slpe, s graph c gets the nd.] 9.) In the circuit belw, each f the resistrs characterized by is the same size. Assuming the pwer supply's EMF is ε while its very small internal resistance is r i, which circuit will increase the temperature f water the fastest? Assume that all three heating resistrs are immersed. a.) b.) c.) r i r i r i d.) e.) f.) T a gd apprximatin, there are at least tw circuits that will prvide the maximum heating pwer t the circuit. r i r i 56

[Cmmentary: The circuit that will raise the temperature f water the fastest will be the circuit whse heating resistrs dissipate the mst pwer. As pwer dissipatin in a resistr is gverned by the current thrugh the resistr, we are lking fr the situatin in which at least mst (if nt all) f the heating resistrs have the greatest current pssible thrugh them. That will happen when the equivalent resistance f the circuit is a minimum. Circuit a is essentially a shrt circuit--n current will pass thrugh the heating resistrs--hence it is nt a viable candidate. In examining the ther circuits, remembering that the equivalent resistance f a parallel cmbinatin is always smaller than any f the individual resistrs within the cmbinatin, it lks as thugh Circuit c will d the trick. NOTE: When in parallel, all three heating resistrs will have the maximum pssible vltage acrss them, hence the greatest pwer utput fr warming water.] 0.) A cil is placed in a changing magnetic field. A graph f the B-field is shwn n each f the grids belw. Due t the changing B-field, an induced current is generated in the cil. Which graph depicts the apprpriate current functin, given the B-field functin? a.) B b.) B c.) B d.) B e.) Nne f these. t t t t i i i i [Cmmentary: The relatinship between the induced EMF and the induced current is εi, where is the resistance in the cil. As EMF and induced current are prprtinal, the relatinship between the changing magnetic field and the induced EMF will be the same as the relatinship between the changing magnetic field and the induced current. The frmer f thse db relatinships is ε ( NA cs θ). As such, we need an induced current graph that lks like dt minus the slpe f the magnetic field graph. Just after t 0, the B-field graph has a negative slpe the magnitude f which gets larger with time, then retreats tward zer. Minus that yields psitive values. In this case, the graph that des the trick is Graph c.] 56