MISN-0-138 AMPERE S LAW by Kiby Mogn 1. Usefullness................................................ 1 AMPERE S LAW 2. The Lw................................................... 1. The Integl Reltionship............................... 1 b. Detemining Signs (±).................................. 3 3. Simple Applictions...................................... 3. Mgnetic Field Ne Long Thin Wie.................. 3 b. Outside Long ylindicl onducto................ 4 c. Inside Long ylindicl onducto.................. 4 d. Infinite Plne of Adjcent Wies.........................5 4. Emple Devices. The Solenoid............................................ 6 b. lculting the Field of Solenoid...................... 6 c. The Tooid.............................................. 7 5. Using uent Density. Intoduction.............................................8 b. The uent Though Sufce......................... 8 c. Ampee s Lw in Tems of the uent Density.......... 9 d. Emple: Hollow onducting ylinde.................. 9 Acknowledgments.......................................... 10 Glossy..................................................... 10 A. Line Integls............................................11. Pojection of n Ae...................................12 Poject PHYSNET Physics ldg. Michign Stte Univesity Est Lnsing, MI 1
ID Sheet: MISN-0-138 Title: Ampee s Lw Autho: Kiby Mogn, Hndiomputing, 319 E. Heny, hlotte, MI 48813 Vesion: 2/1/2000 Evlution: Stge 0 Length: 1 h; 32 pges Input Skills: 1. Vocbuly: cuent density (MISN-0-118); tesl (MISN-0-122). 2. Using the Ampee-Lplce-iot-Svt eqution, clculte the mgnetic field due to cuent in long stight wie (MISN- 0-125). 3. Tke the line integl of vecto function ove specified integtion pth (Appendi). Output Skills (Knowledge): K1. Vocbuly: line integl, solenoid, tooid, Ampee s lw. K2. Stte Ampee s lw fo cuents nd define ech symbol. K3. Descibe how one cn detemine the diection of mgnetic fields poduced by cuent segment. K4. Stte Ampee s lw fo cuent densities nd define ech symbol. Output Skills (Poblem Solving): S1. Use Ampee s lw to clculte the mgnetic field due to symmeticl configutions of cuent such s long stight wies, infinite plnes of cuent, solenoids nd tooids. S2. Given cuent density in conducto of simple geometic shpe, use Ampee s lw to clculte the ssocited mgnetic field. Post-Options: 1. The Ampee-Mwell Eqution: Displcement uent (MISN- 0-145). 2. Mwell s Equtions (MISN-0-146). THIS IS A DEVELOPMENTAL-STAGE PULIATION OF PROJET PHYSNET The gol of ou poject is to ssist netwok of eductos nd scientists in tnsfeing physics fom one peson to nothe. We suppot mnuscipt pocessing nd distibution, long with communiction nd infomtion systems. We lso wok with employes to identify bsic scientific skills s well s physics topics tht e needed in science nd technology. A numbe of ou publictions e imed t ssisting uses in cquiing such skills. Ou publictions e designed: (i) to be updted quickly in esponse to field tests nd new scientific developments; (ii) to be used in both clssoom nd pofessionl settings; (iii) to show the peequisite dependencies eisting mong the vious chunks of physics knowledge nd skill, s guide both to mentl ogniztion nd to use of the mteils; nd (iv) to be dpted quickly to specific use needs nging fom single-skill instuction to complete custom tetbooks. New uthos, eviewes nd field testes e welcome. PROJET STAFF Andew Schnepp Eugene Kles Pete Signell Webmste Gphics Poject Diecto ADVISORY OMMITTEE D. Aln omley Yle Univesity E. Leond Jossem The Ohio Stte Univesity A. A. Stssenbug S. U. N. Y., Stony ook Views epessed in module e those of the module utho(s) nd e not necessily those of othe poject pticipnts. c 2001, Pete Signell fo Poject PHYSNET, Physics-Astonomy ldg., Mich. Stte Univ., E. Lnsing, MI 48824; (517) 355-3784. Fo ou libel use policies see: http://www.physnet.og/home/modules/license.html. 3 4
MISN-0-138 1 MISN-0-138 2 AMPERE S LAW by Kiby Mogn 1. Usefullness Ampee s lw is pt of Mwell s equtions: it eltes mgnetic fields to electic cuents tht poduce them. 1 Using Ampee s lw, you cn detemine the mgnetic field ssocited with given cuent o the cuent ssocited with given mgnetic field, poviding thee is no timechnging electic field pesent. Ampee s lw is pticully useful in situtions whee thee eists high degee of geometicl symmety, just s is the cse with Guss s lw. 2 Fotuntely, mny pplictions hve such symmety. 2. The Lw 2. The Integl Reltionship. Guss s lw nd Ampee s lw hve some similities, lthough Guss s lw involves sufce integl, Guss s lw: E ds = 4πk e Q S, 1 See Mwell s Equtions (MISN-0-146). 2 See Guss s Lw nd Spheiclly Distibuted hges (MISN-0-132). S ` cuent in - diection cuent in + diection Figue 2. A ight-hnd ule ssigns sign to ech cuent bounded by the loop. while Ampee s lw involves line integl : 3 Ampee s lw: Hee: k m is the mgnetic foce constnt. d l = 4πk m I. (1) denotes integtion long closed imginy line.4 The closed imginy line fo ny pticul poblem is usully clled the integtion loop o pth fo tht poblem. The line must pss though the point whee you wnt to know the mgnetic field. I denotes the net electic cuent pssing though ny (imginy) sufce whose boundy is the sme closed line used in (see Fig. 1). 5 dl is n infinitesiml element of length long the integtion line. ` d`l Figue 1. An integtion pth in mgnetic field. The ssocited cuents e not shown. The diection of integtion ound the line is bity, but once tken it fies the diection of cuent tht must be clled positive. The elevnt ule will be tken up lte. 3 See the Appendi of this module fo discussion of line integls. 4 The wod closed mens tht the line hs no end so it must be closed loop. 5 All sufces bounded by the sme line give the sme vlue fo I. 5 6
MISN-0-138 3 MISN-0-138 4 wie I out of pge ` d`l ` Figue 3. nd d l e both tngent to, the cicul integtion pth. 2b. Detemining Signs (±). The lgebic sign (±) of ny cuent enclosed by the integtion loop in Ampee s lw is detemined by ighthnd ule: A cuent is tken to be positive if it points in the diection of the thumb on the ight hnd when the finges of tht hnd encicle the loop in the diection tht the line integl is tken (see Fig. 2). If it is in the opposite diection, the cuent must be tken s negtive. 3. Simple Applictions 3. Mgnetic Field Ne Long Thin Wie. The mgnetic field t some point in spce, ssocited with cuent I in long stight wie, cn be clculted using Ampee s lw. The integtion pth we choose is cicle, centeed on the wie (see Fig. 3) nd going though the point whee we wish to know. y symmety, we epect the mgnetic field to hve the sme mgnitude t ll points on the cicle nd we epect the mgnetic field to be tngent to the cicle t ech point on tht cicle. 6 Since d l is lso tngent to the cicle, d l = dl nd the loop integl is simply dl nd dl is just the cicumfeence of the cicle. lling the dius of the cicle, which is lso the distnce fom the wie to the point whee we wish to know, Ampee s lw gives: ()(2π) = 4πk m I, (2) 6 See The Mgnetic Field of uent: The Ampee-Lplce Eqution (MISN-0-125) fo poof tht this is so. so: ` R ` sufce of wie Figue 4. icul pth of integtion fo cuent unifomly distibuted thoughout the coss section. ( ) I = 2k m. (long stight wie). (3) If the wie is not infinitely long, nd no wie is, this vlue of is ccute to the etent tht is much less thn the distnce fom the field-point to eithe end of the wie. 7 3b. Outside Long ylindicl onducto. Ampee s lw cn be used to show tht the mgnetic field t points outside long cicul cylinde cying cuent unifomly distibuted ove its coss section is the sme s if ll the cuent wee concentted in line long the is. Fo points outside the cylinde, cicul pth of integtion will enclose ll of the cuent nd, gin by symmety, nd d l e pllel. y the sme nlysis tht ws used fo the long wie, we find: ( ) I = 2k m ; > R, (4) whee is the distnce fom the cente of the wie nd R is the dius of the cylinde. 3c. Inside Long ylindicl onducto. The mgnetic field t point inside cylindicl conducto cying cuent depends on how the cuent is distibuted. If it is unifomly distibuted ove its coss section nd cicul pth of integtion is gin chosen (see Fig. 4), the fction of the cuent enclosed by the pth will be π 2 /πr 2, so tht Ampee s lw gives: ( ) π 2 ()(2π) = 4πk m I πr 2, (5) 7 See Guss s Lw Applied to ylindicl nd Pln hge Distibutions (MISN- 0-133) fo the electosttic equivlent, line of fied chge, whee the much less thn condition is lso discussed. 7 8
MISN-0-138 5 MISN-0-138 6 ` l d d l o ` Figue 5. Rectngul integtion pth fo the infinite plne of wies. Figue 6. The mgnetic field outside solenoid of finite length. ( ) I = 2k m R 2. (6) Note tht the mgnetic field is linely popotionl to, the distnce of the field point fom the is. Note: Fo the cse whee the cuent esides only on the sufce of the cylinde, no cuent would be enclosed by the integtion pth nd the mgnetic field would be zeo t ll points inside such sufce conducto. 3d. Infinite Plne of Adjcent Wies. Ampee s lw cn be used to find the mgnetic field due to conducto consisting of n infinite plne of djcent wies. The wies e infinitely long (o e long enough to be egded s such) nd ech cies cuent I. y symmety, you would epect to be pllel to the plne: then ectngul integtion pth of length l which etends distnce d on ech side of the plne would be good choice (see Fig. 5). Along the sides of the pth, noml to the plne, is pependicul to d l so d l is zeo thee. Then Ampee s lw yields: d l = 2l = 4πk m nli, (7) whee n is the numbe of wies pe unit length nd nl is the totl numbe enclosed. Solving fo gives: = 2πk m ni. (8) Figue 7. The integtion pth fo vey long solenoid hving = 0 outside. This eqution indictes tht the field ssocited with n infinite plne of cuent is independent of the distnce fom the plne. 4. Emple Devices 4. The Solenoid. A solenoid is tightly wound cylindicl heli of cuent-cying wie, used to mke n electicl signl cuse onediectionl mechnicl foce (fo emple, opeting plunge). Solenoids e fequently encounteed in science nd technology; thee e t lest sevel in evey c. The mgnetic field inside solenoid cn be esily found using Ampee s lw. The etenl mgnetic field due to solenoid of finite length is quite simil to tht of b mgnet (see Fig. 6). Howeve, if the solenoid is vey long, (i.e., if its length is much gete thn its dius), the field outside is essentilly zeo, nd inside the solenoid it is unifom nd pllel to the solenoid s is (see Fig. 7). 8 4b. lculting the Field of Solenoid. The mgnitude of inside solenoid cn be found by pplying Ampee s lw to the ectngul integtion pth shown in Fig. 7. Outside the solenoid is zeo. Inside, is t ight ngles to the ends of the ectngle so the only non-zeo contibution to the integl is long the length l tht is inside the solenoid. Theefoe: d l = l. (9) 8 Look t the solenoid in Fig. 6 nd notice tht the mgnetic field lines e much moe dense inside the solenoid thn outside it. Imgine mking the solenoid longe nd longe, duing which the density inside emins constnt but the density outside becomes moe nd moe spce. 9 10
MISN-0-138 7 MISN-0-138 8 I I cicle of integtion Figue 8. A tooid. ` Figue 9. The cicul pth of integtion inside tooid. The net cuent though the ectngle is nli, whee n is the numbe of tuns pe unit length ove the entie length l. Ampee s lw then gives fo the mgnetic field: indicting unifom field. = 4πk m ni, (solenoid) (10) 4c. The Tooid. A tooid is solenoid tht hs been bent into cicle, ssuming the spce-sving shpe of doughnut (see Fig. 8). The mgnetic field inside tooid cying cuent I cn be found using Ampee s lw. y symmety, the mgnetic field is tngent to the cicul integtion pth shown in Fig. 9. Theefoe: d l = ()(2π), (11) nd the enclosed cuent is NI, whee N is the totl numbe of tuns on the solendoid. Then: ( ) NI = 2k m. (12) Notice tht, unlike the solenoid, the mgnetic field inside the tooid is not constnt ove the coss section of the coil but vies invesely s the distnce. Fo points outside tooid, it cn be shown tht the field is essentilly zeo if the tuns of wie e vey close togethe. Help: [S-1] j` ds n^ Figue 10. The cuent density j is not lwys long the noml ˆn to n bity sufce element ds. 5. Using uent Density 5. Intoduction. Just s it is often useful to use the concept of chge density in electosttics, in mgnetics we often use the concept of cuent density. hge density is scl nd hs thee vieties: line, sufce nd volume. uent density is vecto nd hs one viety. uent density hs non-zeo vlue only t those spce-points whee thee e chges flowing so thee is n electic cuent: it is the net mount of chge going though the spce-point pe unit time, pe unit e pependicul to the diection of the cuent. The diection of the cuent density vecto is the diection of the electic cuent t the spcepoint in question. The univesl symbol fo the cuent density is j( ), whee the gument indictes tht the cuent density my chnge s one moves fom one spce-point to nothe. 9 5b. The uent Though Sufce. We now ssume we know the cuent density j t vious spce-points of inteest nd we wnt to find the I used in Ampee s lw, Eq. (1). We stt with j t the point of sufce element ds hving noml unit vecto ˆn. We wnt to know how much cuent di is pssing though this element of sufce. Since j is the cuent pe unit e noml to the cuent, we must multiply by n element of e da noml to the cuent (see Fig. 10). If we know ds, ˆn S, nd ĵ, we cn get da by (see Appendi ): Substituting di = jda we get: da = ĵ ˆn ds. di = j ˆn ds. (13) Fo the specil cse of unifom cuent flowing pependicul to plne sufce of e A, the eqution simplifies to I = ja; simple sttement hee. 9 Of couse j my lso be function of time but we e not deling with tht cse 11 12
MISN-0-138 9 MISN-0-138 10 fo the mgnetic field within the conducting mteil. Help: [S-3] b tht the cuent is the cuent density times e. Figue 11. A hollow conducting cylinde with non-unifom cuent density j = k/. Finlly, we integte both sides of Eq. (13) to get: I = j ˆn ds. (14) S 5c. Ampee s Lw in Tems of the uent Density. Ampee s lw my be ewitten in tems of the cuent density, using Eqs. (1) nd (14), giving: d l = 4πk m j ˆn ds. (15) Hee is the closed pth ound the peimete of the sufce S. 5d. Emple: Hollow onducting ylinde. Wht is the mgnetic field t points inside hollow conducting cylinde which is mde such tht its cuent density vies invesely s the distnce fom the cente of the cylinde? The conducto is shown in Fig. 11 nd the cuent density in this poblem is: S j = k ; < < b, (16) whee k is constnt. If cicul integtion pth is chosen, the cuent enclosed by it, the ight side of Eqs. (1) nd (15), is: I = k 2π d = 2πk( ). < < b Help: [S-2] (17) Ampee s lw then gives: ( ) = 4πk m k, < < b (18) Acknowledgments This module is bsed on n elie vesion by J. Kovcs nd O. McHis. The Model Em is tken fom tht vesion. Ry G. Vn Ausdl povided editoil ssistnce. Peption of this module ws suppoted in pt by the Ntionl Science Foundtion, Division of Science Eduction Development nd Resech, though Gnt #SED 74-20088 to Michign Stte Univesity. Ampee s lw: Glossy the integl fom of one of Mwell s equtions: d l = 4πk m I. It eltes the integl of the mgnetic field ound closed loop to the net cuent flowing though ny sufce bounded by the integtion loop. Ampee s lw is univeslly tue, but is useful only when thee is high degee of symmety. cuent density: vecto whose mgnitude t spce point is the cuent pe unit e noml to the diection of the cuent t tht point nd whose diection is the diection of the cuent t tht point. line integl: the integl of function long specified pth in spce. In Ampee s lw one evlutes the line integl of the tngentil component of the mgnetic field ound closed pth tht: (i) goes though the point t which one wishes to know the mgnetic field; nd (ii) is such tht it hs constnt vlue fo the integnd so the integl cn be pefomed tivilly. solenoid: tooid: tightly wound cylindicl heli of cuent-cying wie. solenoid bent into the shpe of doughnut. 13 14
MISN-0-138 11 A. Line Integls MISN-0-138 12 whee l b is the length of the pth fom to b. ^ i ` l i b (ii) is lwys pependicul to the pth: b d l = 0.. Pojection of n Ae 1 2 3... If pln (flt) e S is pojected onto nothe plne, the e A on the pojected-onto plne is given by: Figue 12. A = ĵ ˆn S. (19) The line integl b d l, fo the pth shown bove, cn be ppoimted by dividing the pth into mny smll segments l i nd fo ech segment the poduct i cos θ i l i cn be found. Hee i cos θ i is the component of tngent to the cuve. The integl cn be clculted ppoimtely by summing these segments tems, fo emple, on compute. Howeve, the ect vlue of the line integl is given by the limit: b d l = lim n i=1 i cos θ i l i. If is joined to b, the pth becomes closed nd the esultnt integl d l is ound the closed pth. Often, the clcultion of this integl is highly simplified by utilizing pth tht tkes dvntge of symmeties in the poblem. Two emples of such simplifictions e: (i) is constnt nd lwys tngent to the pth: b d l = b dl = b dl = l b, Hee ˆn is unit vecto noml to the plne of the oiginl e nd ĵ is unit vecto noml to the pojected-onto plne (see the sketch). This is entiely equivlent to the sttement tht the es e elted by the cosine of the ngle between the plnes (gin see the sketch): A j^ A = S cos θ. (20) y pojection we men tht fom evey point on the peiphey of the oiginl e S we dop pependicul to the pojected-onto plne. The locus of those points on the pojected-onto plne define the peiphey of the pojected e A. Equtions (19)-(20) e esily poved by consideing infinitesimllywide stight line elements of the e A tht e noml to the line of intesection of the two plnes. Fo ech such element thee is pojection of it onto the pojected-onto plne, nd the es of the two elements e obviously elted by the cosine of the ngle between the plnes. Since the es themselves e simply the integls of the infinitesiml es, nd since the ngle between the plnes is independent of whee one is in one of the es, the cosine cn be pulled outside the integl nd Eqs. (19)-(20) e poved. If the e S is cuved (non-pln) then Eqs. (19)-(20) pply only to infinitesiml es (which cn be consideed to be pln fo these n^ S 15 16
MISN-0-138 13 MISN-0-138 PS-1 puposes): da = ĵ ˆn ds. PROLEM SUPPLEMENT Note: Poblems 8, 9, nd 10 lso occu in this module s Model Em. 1. Thee infinitely long pllel wies ech cy cuent I in the diection shown below. Wht is dl fo ech of the thee pths 1, 2, nd 3? 3 1 2 2. The mgnetic field in cetin egion of spce is given by = A 0 ˆ whee A 0 = 3 T/m, is the -coodinte of the point, nd ˆ is unit vecto in the -diection. In this egion, conside ectngul pth in the -y plne whose sides e pllel to the nd y es espectively s shown below. y=3m D y=1m A =1m =5m. Evlute the line integl of fom A to. b. Do the sme long the line fom to. 17 18
MISN-0-138 PS-2 MISN-0-138 PS-3 c. Fo to D. d. Fo D to A. e. Evlute the dl ound this closed pth. f. Detemine the net cuent tht must be cossing the -y plne though the ectngle AD. 3. A long cylindicl conducto of dius R hs unifom cuent density j sped ove its coss section. Detemine the mgnetic field poduced t points < R nd > R nd sketch the mgnitude of s function of. 4. A vey long non-conducting cylinde hs N conducting wies plced tightly togethe ound its cicumfeence nd unning pllel to its is s shown below:. Show tht the mgnetic field inside the conducto ( < < b) is: = 2k mi ( 2 2) (b 2 2 ) b. Epess in tems of the cuent density j. c. Show tht when 0 you get the sme nswe s in poblem 3. 6. A long coil cble consists of two concentic conductos. The outside conducto cies cuent I equl to tht in the inside conducto, but in the opposite diection. b R c Find the mgnetic field t these points: 5. If ech wie cies cuent I, find the mgnetic field t points inside nd outside the cylinde. b 7.. inside the inne conducto ( < ), b. between the conductos ( < < b), c. inside the oute conducto (b < < c), nd d. outside the cble ( > c). Unifom cuent density j` diected out of the pge ½t to A hollow cylindicl conducto of dii nd b hs cuent I unifomly sped ove its coss section. An infinite, plne, conducting slb of thickness t cies unifom cuent density of j mpees pe sque mete diected out of the pge in the bove digm. 19 20
MISN-0-138 PS-4 MISN-0-138 PS-5 8.. Apply Ampee s lw to detemine the mgnetic field t height h bove the cente line of the slb fo h > t/2. Eplin cefully how you mke use of symmety in setting up you integtion pth. b. Suppose you integtion pth hd been ectngul loop with two sides pllel to the slb sufce (s you must hve used), but with one pllel pth distnce h bove the cente line nd the othe distnce h below the cente line (both h nd h e gete thn t/2). Eplin in this cse, nd without pio knowledge of you finl nswe, why Ampee s lw cnnot tell you t points h bove the slb. Then show how the use of symmety guments solves the poblem. c. Use the nswe to pt () nd Ampee s lw to detemine the mgnetic field t points distnce y below the sufce of the slb, inside the mteil. Wht is the field t the cente line? Sketch the diection of the field t vious points inside the slb. y(m) 3.0 1.0 0 D A 1.0 (m) 5.0 d. Do the sme fo the line fom D to A. [P] e. Evlute the loop integl dl fo this closed pth. Use the esults of pts ()-(d) to find you nswe. [J] f. Fom you nswe to pt (e), detemine the net cuent tht must be cossing the -y plne though this ectngle AD. [A] 9. Repet Poblem 8, pts () though (f) fo the cse whee the mgnetic field in this egion is now given by 10. (, y) = (A 0 + A 1 y) ˆ whee A 0 = 2.0 T, A 1 = 0.50 T/m nd y is the y-coodinte of the point.. [] b. [K] c. [O] d. [M] e. [H] f. [L] A A Pth 1 Pth 2 = 1.0 10 1 ˆ tesls eveywhee. In cetin egion of spce, the mgnetic field intensity is unifom nd hs the vlue of 10 tesls diected in the positive -diection t evey point in the egion. In this egion conside ectngul pth in the -y plne fom point A to point pllel to the -is, to pllel to the y-is nd D bck to A pllel to the y-is (see the sketch bove).. Evlute the line integl of fom A to. [N] b. Do the sme fo the line fom to. [] c. Do the sme fo the line fom to D. [I] R R = dius of the cylindicl conducting wie j = the cuent pe unit e (distibuted unifomly) diected into the pge A = the distnce fom the cente to point A outside the conducto 21 22
MISN-0-138 PS-6 MISN-0-138 PS-7 = the distnce fom the cente to point inside the conducto. Pth 1 (solid line) is cicul pth suounding the cylinde concentic with the cylindicl conducto pssing though point A. Pth 2 (dshed line) is n bity pth suounding the conducto, lso pssing though point A.. Wht is dl fo ech of the pths 1 nd 2? [F] b. Eplin how symmety enbles you to evlute t point A only if you use pth 1. [G] ief Answes: 4. < R: = 0 NI > R: = 2k m ( 2 2 ) 5. = 2πk m j 1. icul pth: Net cuent = I d l = 4πkm I Rectngul Pth: Net uent = I + ( I) = 0 1 d l = 0 Iegul Pth: Net uent = I I I = I d l = 4πkm I 1 6.. < : = 2k m I 2 I b. < < b: = 2k m ( I c 2 2 ) c. b < < c: = 2k m c 2 b 2 7.. d. > c: = 0 D h 2.. 36 m T b. zeo c. 36 m T d. zeo e. zeo f. d l = 0 so I = 0 though ectngle AD. 3. < R: = 2πk m j jr 2 > R: = 2πk m 1 t A Symmety tells you tht the field, t ll points on the line D, hs the sme vlue diected to the left nd this is lso the sme s the field t ll points on line A (but thee, diected to the ight). = 2πk m jt, independent of h if h > t/2 (the slb is infinitely long). h 23 24
MISN-0-138 PS-8 MISN-0-138 PS-9 b. If the distnce below the cente line hd been h in the sketch [see pt ()], then Ampee s lw would give you + = 4πk m jt, whee is the field vlue t points h below the cente line. Only if h = h cn you gue tht = [s in pt ()] nd then detemine. c. P y I. 4.0 10 1 T m Help: [S-4] J. d l = 0 ound the closed pth. K. Zeo L. 3.2 10 6 A, diected into the pge. M. Zeo N. 4.0 10 1 T m Q z O. 14 T m P. Zeo ( ) t (t P ) = 4πk m j 2 y diected to the left. (t the cente line) = 0. ( ) t (t Q) = 4πk m j 2 z diected to the ight. oth y nd z e less thn t/2. A. A zeo net cuent.. Zeo. +1.0 10 1 T m F. ecuse both pths completely encicle the cuent, (4πk m jπr 2 ) fo both pth 1 nd pth 2. dl is G. Fo pth 1, symmety tells you tht is the sme (nd tngent to the pth) t evey point on the pth, so d l = dl = 2π = 4πk m jπr 2, so t point A: H. 4.0 T m = 2πk m jr 2 25 26
MISN-0-138 AS-1 MISN-0-138 AS-2 SPEIAL ASSISTANE SUPPLEMENT S-3 (fom TX-4e) d l = 4πkm I S-1 (fom TX-4c) ()(2π) = 8π 2 k m k( ) ( ) = 4πk m k S-4 (fom PS-poblem 8) An integl is just the limit of sum: d l = lim l 0 l. Note tht l is negtive long the pt of the pth lbeled D. Theefoe the sum is negtive fo tht pt of the pth nd hence so is the pth integl fo tht segment of the pth. To do it fomlly, note tht long tht pt of the pth we hve: Fo ny shped pth enclosing the entie tooid, the net cuent is zeo. y Ampee s lw, d l = 0, which implies = 0 since the pth is bity. S-2 (fom TX-4e) d l = ˆdl nd = ˆ so: D ˆ ( ˆdl) = D dl = + D d = 1.0 m 5.0 m d = 4.0 m. I = j ˆn ds with j = k ˆ, giving: S I = ( ) 2π k 0 ˆ (ˆ) d dθ = 2πk ( ) 1 d = 2πk d = 2πk = 2πk( ). 27 28
MISN-0-138 ME-1 MISN-0-138 ME-2 1. MODEL EXAM 3. A A Pth 1 Pth 2 3.0 D R y(m) 1.0 A 0 1.0 (m) = 10 ˆ tesls eveywhee. In cetin egion of spce, the mgnetic field intensity is unifom nd hs the vlue of 10 tesls diected in the positive -diection t evey point in the egion. In this egion conside ectngul pth in the -y plne fom point A to point pllel to the -is, to pllel to the y-is nd D bck to A pllel to the y-is (see the sketch bove).. Evlute the line integl of fom A to. b. Do the sme fo the line fom to. c. Do the sme fo the line fom to D. d. Do the sme fo the line fom D to A. e. Evlute dl ound this closed pth. Use the esults of pts ()-(d) to find you nswe. f. Fom you nswe to pt (e), detemine the net cuent tht must be cossing the -y plne though this ectngle AD. 2. Repet Poblem 1, pts () though (f) fo the cse whee the mgnetic field in this egion is given by = (A 0 + A 1 y) ˆ 5.0 R = dius of the cylindicl conducting wie j = the cuent pe unit e (distibuted unifomly) diected into the pge = the distnce fom the cente to point A outside the conducto = the distnce fom the cente to point inside inside the conducto. Pth 1 (solid line) is cicul pth suounding the cylinde concentic with the cylindicl conducto pssing though point A. Pth 2 (dshed line) is n bity pth suounding the conducto, lso pssing though point A.. Wht is dl fo ech of the pths 1 nd 2? b. Eplin how symmety enbles you to evlute t point A only if you use pth 1. ief Answes: 1. See Poblem 8 in this module s Poblem Supplement 2. See Poblem 9 in this module s Poblem Supplement 3. See Poblem 10 in this module s Poblem Supplement whee A 0 = 2 T, A 1 = 0.5 T/m nd y is the y-coodinte of the point. 29 30
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