PHYS 110B - HW #2 Spring 2004, Solutions by David Pace Any referenced equations are from Griffiths Problem statements are paraphrased

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1 PHYS 11B - HW # Spring 4, Solution by David Pace Any referenced equation are from Griffith Problem tatement are paraphraed [1.] Problem 7. from Griffith A capacitor capacitance, C i charged to potential V o. It i then connected to a reitor of reitance and at time t = it begin dicharging. eference figure 7.5 a and b throughout the relevant part of thi problem. a For the capacitor find it charge, Qt. For the reitor find it current, It. b Ue equation.55 in Griffith to write the original energy tored in the capacitor. Integrate equation 7.7 to how that the energy diipated in the reitor i equivalent to the energy lot by the capacitor. c Imagine a new cenario in which the capacitor i charging up from zero initial charge. Conider the circuit of figure 7.5 b in which a battery of contant voltage V o i connected at time t =. Find the Qt and It a in part a. d Ue V o I to calculate the total energy output by the battery. Calculate the energy diipated in the reitor. Find the final tored energy of the capacitor. Determine the fraction of battery output energy that goe into the tored energy of the capacitor. Solution a The charge on any capacitor i given by Q = CV, where C i the capacitance and V i the potential acro it. In thi problem the charge i changing in time and thi require that the potential be time-dependent a well. The capacitance i a function of the geometry of the ytem and i contant unle the actual phyical capacitor i altered, which it i not in thi cae. A figure 7.5 a how, the potential acro the reitor will be the ame a that acro the capacitor. Therefore, the charge on the capacitor i related to the current through the reitor by, V t = It Eq. 7.4 Qt = CV t 1 where the time dependencie are hown. Current i defined a the time rate of change of electric charge, I = dq/, o we can ue 1 to obtain a differential equation for the charge. Q = C dq Since the charge of the capacitor i decreaing we hould write the current a, I = dq/. After making thi adjutment it i poible to olve for the charge. Qt = C dqt dq 3 = 1 Q 4 1

2 Thi i a imple differential equation to which the anwer i known you can alway take the derivative of Q to verify that it give you the original equation, Qt = A exp t 5 where A i a contant. Solving for A require uing the initial condition on the ytem, which i that the initial charge on the capacitor i the total charge, Qt = = CV o. Qt = = CV o = A exp 6 A = CV o 7 Qt = CV o exp t and the charge decreae to zero a time goe to infinity. The current ha been defined a the negative time derivative of the charge, therefore it i given by, It = d CV o exp t 1 = CV o exp t 9 It = V o exp t 1 The current begin at it larget value and goe to zero a time progree. b The problem require that we ue, Calling the initial energy tored in the capacitor W o, W = 1 CV Eq W o = 1 CV o 1 To find the energy diipated in the reitor we are aked to integrate the following, P = I Eq Power i the energy per unit time, o the energy i found through integration a uggeted, dw = P = I dw = I 14 W = I = = V o V o exp t 15 exp exp 16 = 1 CV o 17

3 which i equal to the energy originally tored in the capacitor, W o. c Connecting a battery to charge the capacitor reult in a different decription of the potential in the circuit. The battery force it potential acro the reitor-capacitor combination, making the total potential acro thoe two element equal to V o. V cap + V re = V batt 18 Q cap C + I = V o 19 Alo in thi problem, the charge i now increaing and the current hould be written a, I = dq/. Thi preent the differential equation to olve. Q C + dq dq Integrate both ide = V o = 1 dq CV o Q = where A i again a contant to be determined. V o Q C 1 = 1 CV o Q 3 lncv o Q = t + contant 4 CV o Q = A exp t 5 Q = CV o A exp t 6 Initial condition require that Qt = =. Solving for A, Qt = = = CV o A exp 7 = CV o A 8 A = CV o 9 The charge i, Qt = CV o [1 exp t ] 3 3

4 which reache a final value of Q = CV o a t. The current i the time derivative of 3, It = V o exp t 31 Notice that the current till goe to zero a t. When the capacitor i fully charged there will be no more current in the ytem. d Ue the given equation to find the energy output of the battery, W batt = V o I = V o Vo exp t 3 = V o [exp exp] 33 = CV o 34 The energy diipated in the reitor i found from 14, but the value of and I here are the ame a in part b and therefore the energy i the ame. The final energy of the capacitor i given by, W re = 1 CV o 35 W = 1 CV = 1 Q C 36 The final value of Q i, Qt = CV o [1 exp ] = CV o 37 Inerting 37 into 36 give the final energy of the capacitor, W cap = 1 C C V o = 1 CV o 38 Half of the work done by the battery i.e. it expended energy end up tored in the capacitor. [.] Problem 7.7 in Griffith A metal bar lide on parallel conducting rail. The liding i frictionle and the bar are eparated by a ditance l. The rail are connected by a reitor of value, and there i a uniform background magnetic field B, which point into the page throughout the entire region. eference figure 7.16 in Griffith. a The bar move right at peed v. Find the current in the reitor, and it direction. b Find the magnetic force on the bar and it direction. 4

5 c The initial peed of the bar i vt = = v o. Find the peed at a later time, t. d Show that the total energy diipated in the reitor i mv o/. Solution a The current in the reitor i given by, I = V/. In thi problem the only potential i due to the emf, E the explanation for how V = E may be found in Griffith, page 93. To find the E acro a loop ue, E = dφ Eq where Φ i the magnetic flux through the urface encloed by the loop. In turn, Φ i found from, Φ where the integral i taken over the urface of interet. B d a Eq Here the magnetic field i uniform, let B = B o ẑ, o the flux i thi field multiplied by the area. Φ = B o lx 41 where x i defined a the horizontal length of the rectangular area between the liding bar and the ret of the ret of the rail circuit. Since thi flux i poitive that mean I have defined that d a point in the ẑ direction parallel to B. By the right-hand rule, defining the direction of d a in thi way aume that the current found i flowing in the clockwie direction. Placing your right thumb along the direction of d a, your finger curl in the direction of the uppoed current. If the calculation of current come out negative, then it direction will be oppoite thi uppoed direction. We do not know the direction of the current until thi value i determined. Defining z a into the page, I alo define x a pointing to the right horizontal and y pointing down vertical. Back to finding E. E = dφ ince v = dx/ by definition of v being to the right. = d B olx 4 = B o l dx 43 = B o lv 44 Alo, now that E ha been determined it i poible to give the direction of the current. The current i proportional to E I = E/, o a negative E lead to a negative current. Our E wa defined by a clockwie loop; it negative value mean the proper direction i counterclockwie. Following a counterclockwie loop in figure 7.16 mean the current flow down the reitor. The current i alway repreented by a poitive value ince we tated the direction, I = B olv 5 45

6 b ecall that the force on a current in a magnetic field i, F mag = Id l B Eq where d l point in the direction of the current. The current i in the ŷ direction in the bar, according to the coordinate ytem I am uing. In thi problem all of the quantitie are uniform, which allow, F mag = I l B 47 = Il ŷ B o ẑ 48 = IlB oˆx 49 eplace I with 45, it value in term of parameter given in the problem, thi force i acting againt the velocity of the bar. F mag = l Bov ˆx 5 c From b we olved for the magnetic force on the bar, which i acting to low it down. We have, F total = m a = m d v = l Bov ˆx 51 The direction of v i given in the problem to be, by the coordinate ued here, in the ˆx direction. It i poible now to remove all vector dependence from the equation above. dv = Bov l m Take the general olution of thi differential equation and ue the initial condition to olve for the unknown contant, A, 5 v = A exp l Bo m t 53 vt = = v o = A exp A = v o 54 vt = v o exp l Bo m t 55 d Following the method ued previouly to determine the energy delivered to a reitor, thi energy here i, 6

7 W re = I = = B ol = B ol v o B o l v v o exp m l B o 56 l B o m t 57 [exp exp] 58 = 1 mv o 59 and the total energy delivered to the reitor i the expected value. [3.] Problem 7.8 from Griffith A quare loop of wire i a ditance from an infinite wire of current I. The quare loop ha ide of length a. eference figure 7.8 in Griffith. a Determine the magnetic flux through the loop. b The loop i pulled directly away from the wire entirely along the coordinate in the cylindrical ytem at peed v. Find the emf generated in the wire and the direction of the current flow. c What happen if the loop i moving to the right along z in the cylindrical coordinate ytem at peed v intead of along a in part b? Solution a The magnetic flux i found uing 4. The magnetic field i that due to a very long i.e. may be treated a infinite wire carrying current I, B = µ oi π ˆφ Eq where thi i in cylindrical coordinate with the current directed along the z-axi. Defining the current direction a above, the + ˆφ direction inide the quare loop i coming out of the page. Thi i known becaue if you place your right thumb in the +ẑ direction, then your finger wrap around in the + ˆφ direction. Thi agree with the field of B above. The flux through the loop i, Φ = B d a = µo I π ˆφ d dz ˆφ 61 = µ oi π +a a 1 d dz 6 = µ oia ln + a ln 63 π = µ oai + a π ln 64 7

8 While it i important to know the direction of the relevant vector in determining the flux, thi quantity i alway a calar value. b Determining the emf i traightforward ince we already have the flux. Note that our flux value in 64 defined the direction of d a to be ˆφ. Thi i coming out of the page in the region of the loop. Uing the right-hand rule with thi direction give a path going counterclockwie in the loop. If we get a poitive emf in the next few tep, then the direction of the current it induce will alo be counterclockwie in the loop. A negative emf mean the current will be in the oppoite direction. The loop i moving away from the wire along, therefore = t, E = dφ = d µ o ai + a π ln = µ oai π d + a ln = µ [ oai 1 π + a = µ [ oai 1 d 1 + a ] π + a = µ oai π = µ o a Iv π + a d + + a d ] [ 1 d a ] 69 + a ince d/ = v by definition. Notice that in thi cae the emf depend not only on the velocity, but alo on the poition. Once the loop i very far away from the current-carrying wire, the magnetic field i eentially zero and there i no magnetic flux through the loop. egardle of the velocity at thi location, the magnetic flux doe not change from it zero value. The emf hould be expected to reach zero once the ditance away from the current get very large. Thi emf i poitive, therefore our direction choice for the flux calculation tell u that the current induced in the loop will be counterclockwie. c The velocity of the loop in thi part i directed along the z axi. The poition i therefore contant, t. The flux through the loop i unchanged, but now the time derivative of thi flux i zero. There i no emf in thi cae. [4.] Problem 7.11 from Griffith A quare loop of aluminum i placed with it top half in a region of uniform magnetic field, B. Thi field point into the page. The loop i releaed and fall under the influence of gravity. eference figure 7.19 in Griffith. For the following tak ue a field of 1T, ma denity of aluminum i ρ = kg/m 3, and the reitivity of aluminum i according to table 7.1 in Griffith η = Ωm. - Calculate the terminal velocity of the loop. Compute a numerical value in unit of m/. 7 8

9 - Determine the velocity of the loop, v = vt. - Calculate the time taken for the loop to reach 9% of it terminal velocity. Compute a numerical value in unit of. - State what happen if the loop i cut uch that the circuit i broken. Solution Let the length of the ide of the loop be l. Solving for the motion of the loop require olving, F tot = m a = m g + F mag 71 where g i the acceleration due to gravity and F mag i the magnetic force experienced by the loop. The coordinate ytem ued here i z into the page, o that B = B o ẑ, the +ˆx direction i to the right, and the +ŷ direction i downward. With the uniform field and the dicrete length through which current may flow, the magnetic force i given by 47. To find the current we mut firt know the emf. The emf i found imilarly to the method ued in problem 7.7. The magnetic flux through the loop i, Φ = B o ly 7 The emf i, E = d B oly = B o lv 73 where dy/ = v ince the y length i decreaing a the loop fall. The emf i poitive and the wa determined uing d a in the ˆφ direction. Thi implie a current going clockwie in the loop, which i in the +ˆx direction in the top length of the loop. The magnitude of thi current i, where i the reitance of the loop. Now we are ready to ue 47. I = E = B olv 74 F mag = I l B = B olv lˆx B oẑ 75 = B ol v ŷ 76 According to the coordinate ued here, thi magnetic force i acting againt the gravitational force. The terminal velocity of the loop occur when the acceleration i zero becaue the gravitational and magnetic force balance out. Setting a = let u change v v t in the force 9

10 equation, 71, where v t i the terminal velocity. Back to 71, uing g = gŷ, = m g + F mag 77 m g = B ol v t ŷ 78 mg = B ol v t 79 v t = gm B ol 8 To get an actual number for thi velocity we need to define the reitance and ma of the loop. The reitance of an object with uniform cro ection i, = ηl A 81 where L i the length of the object and A i the area of the cro ection. In thi problem we have, noting that there are four egment of the loop, = 4lη A where the area i left unknown. Griffith hint in the problem tatement that the dimenion will cancel out, o there i good reaon to believe thi A will not matter for the final anwer. The ma of the loop i given by it volume multiplied by the ma denity. Again, the following factor of four relate to the four length that comprie the loop, 8 m = 4lAρ 83 where thi A i the ame cro ectional area found in the reitance calculation. eturn to the expreion for terminal velocity and ue 8 and 83, v t = gm B ol = 9.8m/4lAρ 4lη A 1 T l 84 = 9.816ρη kg Ω T m 85 = m/ 86 The terminal velocity of the loop i approximately 1.1 cm/. = m/ 87 It i not obviou that the unit in 85 implify to m/, o here i a proof. It begin by writing the reitance and magnetic field unit in term of the fundamental unit i.e. meter, econd, Coulomb, kilogram. 1

11 kg Ω T m = m 88 Ω = V A 89 V = J C = N m C = kg m m C = kg m C 9 A = C 91 The magnetic field unit, the Tela, become, Ω = kg m C 9 T = Bringing all of thi together give, kg m N A m = C m = kg C 93 kg Ω T m = kg kg = m C kg m C m Solving for the velocity require returning to 71 and uing a. The mathematic follow the method ued in problem 7., part c. dv dv m g Bol ln B ol m v = g B ol m v 96 g B ol v = 97 m = t + contant 98 ln g B ol m v g B ol m v = A exp = B ol t + contant m 99 B ol m t 1 vt = gm B ol A exp 11 B ol m t 11

12 where A i a contant. Ue the initial condition to find A. The loop i releaed after being placed partially in the magnetic field. We may ay, vt = =. Then, vt = = = gm A exp 1 Bol A = gm B ol 13 eplacing A give the velocity a, vt = gm B ol [ ] 1 exp B ol m t 14 Finding the time at which 9% of the terminal velocity i reached i eay if we notice that the expreion for the velocity contain the terminal velocity a a factor. vt = gm B ol = v t [1 exp [ ] 1 exp B ol m t ] B ol m t The velocity ha reached 9% of the terminal velocity when, v v t =.9 17 Thi time occur at, v =.9 = 1 exp B ol v t m t 18 exp B ol m t =.1 19 B ol t = ln.1 11 m m ln.1 t = 111 Bol Inert 83 and 8 into 111 to get a numerical value for the time, 4lη 4Alρ A t = ln.1 11 l = =

13 where the unit are imply given and not proven. Finally, cutting a lit in the ring break the circuit and prevent current from flowing. Without current there can be no magnetic force on the loop. In thi cae the loop fall according to gravity, with no effect from the magnetic field aluminum i a non-magnetic metal. [5.] Problem 7.1 from Griffith A long olenoid i carrying an alternating current. The radiu of the olenoid i, a, and the field inide due to it current i Bt = B o coωtẑ. A loop of wire i placed coaxially inide the olenoid. The radiu of thi loop i a/ and it reitance i. Determine the current in the loop, I = It. Solution The current will be found through the uual I = E/ relation. Since the loop i coaxial with the olenoid it i natural to let the d a of the loop point in the ẑ direction o a to be parallel with the magnetic field. The emf i, E = d a/ π B o coωtẑ d dφẑ 115 = d B o coωtπ a/ 116 The current i, [6.] Problem 7.16 from Griffith = πa B o ω inωt = πa ωb o 4 inωt 118 I = πa ωb o 4 inωt 119 A current I = I o coωt flow in a wire. Thi current return along a concentric tube outide the wire. a What i the direction of the induced electric field radial, longitudinal, or circumferential? b Aume E =. Find the electric field, E, t. Solution a Note that Griffith i ure to point out the imilarity between Ampere law and Faraday law page 36. The electric field in thi problem can be found from, E d l = dφ Eq Ampere law i, B d l = µ o I enc 11 13

14 In the wire there i only magnetic field between the conductor. We can ue Ampere law a in previou aignment to get the field inide the coaxial cable a, B = µ oi π ˆφ = µ oi o coωt ˆφ 1 π Thi mean that the magnetic flux i determined through urface that have vector direction in the ˆφ, parallel to B. From Ampere law, if an encloed current wa along the ˆφ direction, then the right-hand rule would tell u that the magnetic field it induce i along the z-axi. Here the changing flux i along the ˆφ direction and therefore the electric field it induce i in the z direction. The electric field i in the ẑ direction, longitudinal. Note that the magnetic flux i not a vector; it ha no direction. There i, however, a choice in direction made when computing the flux through a urface. It i thi direction that you hould keep in mind when thinking about the direction of the induced electric field. b To get the electric field we firt determine the magnetic flux. Φ = B d a = µo I o coωt ˆφ d dz π ˆφ 13 = µ oi o coωt π a l = µ oi o coωt a l ln π d dz Now the electric field can be found uing 1. The electric field i already known to be in the ẑ direction, which account for the implification in the left ide of the following tep, E z l = d µ o I o coωt a l ln 16 π E z = µ oi o a π ln ω inωt 17 = µ oi o ω inωt π a ln 18 The geometry ued in applying Faraday law to thi problem i given in figure 1. Thi i imilar to uing Ampere law while olving for the magnetic field of a olenoid. The magnetic field due to the current in the center wire i hown. Thi field change direction in time. The lightly haded region i the area through which the magnetic flux change. A the problem tatement ay, the electric field may be aumed to be zero a. Thi mean we may neglect the contribution of E along the vertical length of the loop at the far right. The electric field i, E = µ oi o ω inωt π a ln ẑ 19 14

15 Figure 1: Geometry for uing Faraday law to find the electric field in problem [7.] Problem 7.17 from Griffith A long olenoid ha radiu a and n turn per unit length. Thi olenoid pae through a loop of wire with reitance. eference figure 7.7 in Griffith. a epreent the current in the olenoid, I, by di/ = k where k i a contant. Find the current that flow in the loop. Doe it flow to the right or left through the reitor in figure 7.7? b Now let the olenoid current remain contant. You pull the olenoid out from the loop, turn it around, and then reinert it through the loop. What total charge pae through the reitor in thi proce? Solution a To find the current induced in the loop we will have to find the emf generated by the olenoid. There i no magnetic field outide the olenoid. The field inide the olenoid i given by, B = µ o niẑ Eq where I arbitrarily aign the direction. Thi field i uniform. The magnetic flux i the flux through the urface area of the olenoid. Φ = B area = µ o niπa

16 The emf i, E = dφ = d µ oniπa 13 = µ o nπa di 133 = µ o nπa k 134 The definition of B and the area vector being in the +ẑ direction meant that the direction of the current derived from the reult above hould be in the + ˆφ direction. Thi i parallel to the direction of the olenoid current in figure 7.7 ince we ue the olenoid current to determine the magnetic field. The negative emf found above require that the current induced in the loop mut be oppoite the olenoid current. The induced current flow to the right acro the reitor. One could alo ue Lenz law to determine the current direction in the loop. The current in the olenoid i increaing over time, therefore the magnetic field in the olenoid i a well. The urface area of the olenoid i contant o the magnetic flux through the loop with the reitor mut be increaing. The current induced in the loop will try to counteract thi flux change by generating a magnetic field againt the change. The induced magnetic field trie to cancel out the increaing field by directing itelf oppoitely. To get a magnetic field in the ẑ direction, the current in the loop mut flow to the right through the reitor. The value of the current i, I = E = µ onπa k 135 b The change in magnetic flux through the loop doe not change uniformly in time. Fortunately, it doe not matter how quickly the flux change, the total change in flux reult in a pecific amount of charge flow regardle of the time over which it change. Thi i omewhat dicued in example 7.8 of Griffith. Solve thi problem by treating all of the parameter a dicrete. Thi mean that we are not concerned with how thing change through time, rather, we are concerned with the total change of the relevant quantitie. In a dicrete conideration we have a current, I = Q t Thi Q i the total amount of charge that pae through the reitor in a time t. Continuing thi dicrete analyi, 136 Q = I t 137 = E t 138 = 1 Φ t = Φ t

17 where it i now demontrated mathematically that the time over which thi proce occur doe not affect the final olution. We need to find the total change in the magnetic flux due to the removing and replacing of the olenoid. Φ = Φ final Φ initial 14 The current and hence the magnetic field in the olenoid remain contant. The initial and final fluxe will be equal in magnitude but have different ign due to the orientation of the olenoid. We have, The total change in magnetic flux i, Φ initial = µ o niπa Φ final = µ o niπa 141 The total charge paing through the reitor i, Φ = µ o niπa 14 Q = Φ = µ oniπa