CHAPTER 6 MAGNETIC EFFECT OF AN ELECTRIC CURRENT

CHAPTER 6 MAGNETIC EFFECT OF AN ELECTRIC CURRENT 6. Introduction Most of us re fmilir with the more obvious properties of mgnets nd compss needles. A mgnet, often in the form of short iron br, will ttrct smll pieces of iron such s nils nd pper clips. Two mgnets will either ttrct ech other or repel ech other, depending upon their orienttion. If br mgnet is plced on sheet of pper nd iron filings re scttered round the mgnet, the iron filings rrnge themselves in mnner tht reminds us of the electric field lines surrounding n electric dipole. All in ll, br mgnet hs some properties tht re quite similr to those of n electric dipole. The region of spce round mgnet within which it exerts its mgic influence is clled mgnetic field, nd its geometry is rther similr to tht of the electric field round n electric dipole lthough its nture seems little different, in tht it intercts with iron filings nd smll bits of iron rther thn with scrps of pper or pith-blls. The resemblnce of the mgnetic field of br mgnet to the electric field of n electric dipole ws sometimes demonstrted in Victorin times by mens of Robison Bll-ended Mgnet, which ws mgnet shped something like this: FIGURE VI. The geometry of the mgnetic field (demonstrted, for exmple, with iron filings) then gretly resembled the geometry of n electric dipole field. Indeed it looked s though mgnet hd two poles (nlogous to, but not the sme s, electric chrges), nd tht one of them cts s source for mgnetic field lines (i.e. field lines diverge from it), nd the other cts s sink (i.e. field lines converge to it). Rther thn clling the poles positive nd negtive, we somewht rbitrrily cll them north nd south poles, the north pole being the source nd the south pole the sink. By experimenting with two or more mgnets, we find tht like poles repel nd unlike poles ttrct. We lso observe tht freely-suspended mgnet (i.e. compss needle) will orient itself so tht one end points pproximtely north, nd the other points pproximtely south, nd it is these poles tht re clled the north nd south poles of the mgnet. Since unlike poles ttrct, we deduce (or rther Willim Gilbert, in his 6 book De Mgnete, Mgneticisque Corporibus, et de Mgno Mgnete Tellure deduced) tht Erth itself cts s gint mgnet, with south mgnetic pole somewhere in the Arctic nd north mgnetic pole in the Antrctic. The Arctic mgnetic pole is t present in Bthurst Islnd in northern Cnd nd is usully mrked in tlses s the North Mgnetic Pole,

though mgneticlly it is sink, rther thn source. The Antrctic mgnetic pole is t present just offshore from Wilkes Lnd in the Antrctic continent. The Antrctic mgnetic pole is source, lthough it is usully mrked in tlses s the South Mgnetic Pole. Some people hve dvocted clling the end of compss needle tht points north the north-seeking pole, nd the other end the south-seeking pole. This hs much to commend it, but usully, insted, we just cll them the north nd south poles. Unfortuntely this mens tht the Erth s mgnetic pole in the Arctic is relly south mgnetic pole, nd the pole in the Antrctic is north mgnetic pole. The resemblnce of the mgnetic field of br mgnet to dipole field, nd the very close resemblnce of Robison Bll-ended Mgnet to dipole, with point source (the north pole) t one end nd point sink (the south pole) t the other, is, however, deceptive. In truth mgnetic field hs no sources nd no sinks. This is even expressed s one of Mxwell s equtions, div B =, s being one of the defining chrcteristics of mgnetic field. The mgnetic lines of force lwys form closed loops. Inside br mgnet (even inside the connecting rod of Robison mgnet) the mgnetic field lines re directed from the south pole to the north pole. If mgnet, even Robison mgnet, is cut in two, we do not isolte two seprte poles. Insted ech hlf of the mgnet becomes dipolr mgnet itself. All of this is very curious, nd mtters stood like this until Oersted mde n outstnding discovery in 8 (it is sid while giving university lecture in Copenhgen), which dded wht my hve seemed like n dditionl compliction, but which turned out to be in mny wys gret simplifiction. He observed tht, if n electric current is mde to flow in wire ner to freely suspended compss needle, the compss needle is deflected. Similrly, if current flows in wire tht is free to move nd is ner to fixed br mgnet, the wire experiences force t right ngles to the wire. From this point on we understnd tht mgnetic field is something tht is primrily ssocited with n electric current. All the phenomen ssocited with mgnetized iron, nickel or coblt, nd lodestones nd compss needles re somehow secondry to the fundmentl phenomenon tht n electric current is lwys surrounded by mgnetic field. Indeed, Ampère speculted tht the mgnetic field of br mgnet my be cused by mny circulting current loops within the iron. He ws right! the little current loops re tody identified with electron spin. If the direction of the mgnetic field is tken to be the direction of the force on the north pole of compss needle, Oersted s observtion showed tht the mgnetic field round current is in the form of concentric circles surrounding the current. Thus in figure VI., the current is ssumed to be going wy from you t right ngles to the plne of your computer screen (or of the pper, if you hve printed this pge out), nd the mgnetic field lines re concentric circles round the current,

3 FIGURE VI. In the reminder of this chpter, we shll no longer be concerned with mgnets, compss needles nd lodestones. These my come in lter chpter. In the reminder of this chpter we shll be concerned with the mgnetic field tht surrounds n electric current. 6. Definition of the Amp We hve seen tht n electric current is surrounded by mgnetic field; nd lso tht, if wire crrying current is situted in n externl mgnetic field, it experiences force t right ngles to the current. It is therefore not surprising tht two current-crrying wires exert forces upon ech other. More precisely, if there re two prllel wires ech crrying current in the sme direction, the two wires will ttrct ech other with force tht depends on the strength of the current in ech, nd the distnce between the wires. Definition. One mp (lso clled n mpère) is tht stedy current which, flowing in ech of two prllel wires of negligible cross-section one metre prt in vcuo, gives rise to force between them of 7 newtons per metre of their length. At lst! We now know wht n mp is, nd consequently we know wht coulomb, volt nd n ohm re. We hve been left in stte of uncertinty until now. No longer! But you my sk: Why the fctor 7? Why not define n mp in such mnner tht the force is N m? This is good question, nd its nswer is tied to the long nd tortuous history of units in electromgnetism. I shll probbly discuss this history, nd the vrious CGS units, in lter chpter. In brief, it took long time to understnd tht electrosttics, mgnetism nd current electricity were ll spects of the sme bsic

4 phenomen, nd different systems of units developed within ech topic. In prticulr so-clled prcticl unit, the mp (defined in terms of the rte of deposition of silver from n electrolytic solution) becme so entrenched tht it ws felt imprcticl to bndon it. Consequently when ll the vrious systems of electromgnetic units becme unified in the twentieth century (strting with proposls by Giorgi bsed on the metre, kilogrm nd second (MKS) s long go s 895) in the Système Interntionl (SI), it ws determined tht the fundmentl unit of current should be identicl with wht hd lwys been known s the mpère. (The fctor, by the wy, is not relted to their being two wires in the definition.) The mp is the only SI unit in which ny number other thn one is incorported into its definition, nd the exception ws forced by the desire to mintin the mp. [A proposl to be considered (nd probbly pssed) by the Conférence Générle des Poids et Mesures in 8 would re-define the coulomb in such mnner tht the mgnitude of the chrge on single electron is exctly.67 % 9 C.] One lst point before leving this section. In the opening prgrph I wrote tht It is therefore not surprising tht two current-crrying wires exert forces upon ech other. Yet when I first lerned, s student, of the mutul ttrction of two prllel electric currents, I ws very stonished indeed. The reson why this is stonishing is discussed in Chpter 5 (Specil Reltivity) of the Clssicl Mechnics section of these notes. 6.3 Definition of the Mgnetic Field We re going to define the mgnitude nd direction of the mgnetic field entirely by reference to its effect upon n electric current, without reference to mgnets or lodestones. We hve lredy noted tht, if n electric current flows in wire in n externlly-imposed mgnetic field, it experiences force t right ngles to the wire. I wnt you to imgine tht there is mgnetic field in this room, originting, perhps, from some source outside the room. This need not entil gret del of imgintion, for there lredy is such mgnetic field nmely, Erth s mgnetic field. I ll tell you tht the field within the room is uniform, but I shn t tell you nything bout either its mgnitude or its direction. You hve stright wire nd you cn pss current through it. You will note tht there is force on the wire. Perhps we cn define the direction of the field s being the direction of this force. But this won t do t ll, becuse the force is lwys t right ngles to the wire no mtter wht its orienttion! We do notice, however, tht the mgnitude of the force depends on the orienttion of the wire; nd there is one unique orienttion of the wire in which it experiences no force t ll. Since this orienttion is unique, we choose to define the direction of the mgnetic field s being prllel to the wire when the orienttion of the wire is such tht it experiences no force.

5 This leves two-fold mbiguity since, even with the wire in its unique orienttion, we cn cuse the current to flow in one direction or in the opposite direction. We still hve to resolve this mbiguity. Hve ptience for few more lines. As we move our wire round in the mgnetic field, from one orienttion to nother, we notice tht, while the direction of the force on it is lwys t right ngles to the wire, the mgnitude of the force depends on the orienttion of the wire, being zero (by definition) when it is prllel to the field nd gretest when it is perpendiculr to it. Definition. The intensity B (lso clled the flux density, or field strength, or merely field ) of mgnetic field is equl to the mximum force exerted per unit length on unit current (this mximum force occurring when the current nd field re t right ngles to ech other). The dimensions of B re MLT = MT Q. LQT Definition. If the mximum force per unit length on current of mp (this mximum force occurring, of course, when current nd field re perpendiculr) is N m, the intensity of the field is tesl (T). By definition, then, when the wire is prllel to the field, the force on it is zero; nd, when it is perpendiculr to the field, the force per unit length is IB newtons per metre. It will be found tht, when the ngle between the current nd the field is θ, the force per unit length, F ', is F ' = IB sin θ. 6.3. In vector nottion, we cn write this s F' = I B, 6.3. where, in choosing to write I B rther thn F' = B I, we hve removed the twofold mbiguity in our definition of the direction of B. Eqution 6.3. expresses the right-hnd rule for determining the reltion between the directions of the current, field nd force. 6.4 The Biot-Svrt Lw Since we now know tht wire crrying n electric current is surrounded by mgnetic field, nd we hve lso decided upon how we re going to define the intensity of mgnetic field, we wnt to sk if we cn clculte the intensity of the mgnetic field in the vicinity of vrious geometries of electricl conductor, such s stright wire, or plne coil, or solenoid. When we were clculting the electric field in the vicinity of

6 vrious geometries of chrged bodies, we strted from Coulomb s Lw, which told us wht the field ws t given distnce from point chrge. Is there something similr in electromgnetism which tells us how the mgnetic field vries with distnce from n electric current? Indeed there is, nd it is clled the Biot-Svrt Lw. θ I δs r FIGURE VI.3 P δb Figure VI.3 shows portion of n electricl circuit crrying current I. The Biot-Svrt Lw tells us wht the contribution δb is t point P from n elementl portion of the electricl circuit of length δs t distnce r from P, the ngle between the current t δs nd the rdius vector from P to δs being θ. The Biot-Svrt Lw tells us tht I δs sin θ δ B. 6.4. r This lw will enble us, by integrting it round vrious electricl circuits, to clculte the totl mgnetic field t ny point in the vicinity of the circuit. But cn I prove the Biot-Svrt Lw, or is it just bld sttement from nowhere? The nswer is neither. I cnnot prove it, but nor is it merely bld sttement from nowhere. First of ll, it is not unresonble guess to suppose tht the field is proportionl to I nd to δs, nd lso inversely proportionl to r, since δs, in the limit, pproches point source. But you re still free to regrd it, if you wish, s specultion, even if resonble specultion. Physics is n experimentl science, nd to tht extent you cnnot prove nything in mthemticl sense; you cn experiment nd mesure. The Biot-Svrt lw enbles us to clculte wht the mgnetic field ought to be ner stright wire, ner plne circulr current, inside solenoid, nd indeed ner ny geometry you cn imgine. So fr, fter hving used it to clculte the field ner millions of conductors of myrid shpes nd sizes, the predicted field hs lwys greed with experimentl mesurement. Thus the Biot-Svrt lw is likely to be true but you re perfectly correct in sserting tht, no mtter how mny mgnetic fields it hs correctly predicted, there is lwys the chnce tht, some dy, it will predict field for some unusully-shped circuit tht disgrees with wht is mesured. All tht is needed is one such exmple, nd the lw is disproved. You my, if you wish, try nd discover, for Ph.D. project, such circuit; but I would not recommend tht you spend your time on it!

7 There remins the question of wht to write for the constnt of proportionlity. We re µ free to use ny symbol we like, but, in modern nottion, we symbol we use is. Why 4π the fctor 4π? The inclusion of 4π gives us wht is clled rtionlized definition, nd it is introduced for the sme resons tht we introduced similr fctor in the constnt of proportionlity for Coulomb s lw, nmely tht it results in the ppernce of 4π in sphericlly-symmetric geometries, π in cylindriclly-symmetric geometries, nd no π where the mgnetic field is uniform. Not everyone uses this definition, nd this will be discussed in lter chpter, but it is certinly the recommended one. In ny cse, the Biot-Svrt Lw tkes the form µ I δs sin θ δ B =. 6.4. 4π r The constnt µ is clled the permebility of free spce, free spce mening vcuum. The subscript llows for the possibility tht if we do n experiment in medium other thn vcuum, the permebility my be different, nd we cn then use different subscript, or none t ll. In prctice the permebility of ir is very little different from tht of vcuum, nd hence I shll normlly use the symbol µ for experiments performed in ir, unless we re discussing mesurement of very high precision. From eqution 6.4., we cn see tht the SI units of permebility re T m A (tesl metres per mp). In lter chpter we shll come cross nother unit the henry for quntity (inductnce) tht we hve not yet described, nd we shll see then tht more convenient unit for permebility is H m (henrys per metre) but we re getting hed of ourselves. Wht is the numericl vlue of µ? I shll revel tht in the next chpter. Exercise. Show tht the dimensions of permebility re MLQ. This mens tht you my, if you wish, express permebility in units of kg m C lthough you my get some queer looks if you do. Thought for the Dy. I, δs I, δs The sketch shows two current elements, ech of length δs, the current being the sme in ech but in different directions. Is the force on one element from the other equl but opposite to the force on the other from the one? If not, is there something wrong with Newton s third lw of motion? Discuss this over lunch.

8 6.5 Mgnetic Field Ner Long, Stright, Current-crrying Conductor O x dx I P θ FIGURE VI.4 Consider point P t distnce from conductor crrying current I (figure VI.4). The contribution to the mgnetic field t P from the elementl length dx is r db µ. I dx cos θ =. 6.5. 4π r (Look t the wy I hve drwn θ if you re worried bout the cosine.) Here I hve omitted the subscript zero on the permebility to llow for the possibility tht the wire is immersed in medium in which the permebility is not the sme s tht of vcuum. (The permebility of liquid oxygen, for exmple, is slightly greter thn tht of free spce.) The direction of the field t P is into the plne of the pper (or of your computer screen). We need to express this in terms of one vrible, nd we ll choose θ. We cn see tht r = secθ nd x = tn θ, so tht dx = sec θdθ. Thus eqution 6.5. becomes µ I db = sin θdθ. 6.5. 4π Upon integrting this from π/ to + π/ (or from to π/ nd then double it), we find tht the field t P is B µ I =. 6.5.3 π Note the π in this problem with cylindricl symmetry.

9 6.6 Field on the Axis nd in the Plne of Plne Circulr Current-crrying Coil I strongly recommend tht you compre nd contrst this derivtion nd the result with the tretment of the electric field on the xis of chrged ring in Section.6.4 of Chpter. Indeed I m copying the drwing from there nd then modifying it s need be. δs I x θ P FIGURE VI.5 The contribution to the mgnetic field t P from n element δs of the current is µ I δs in the direction shown by the coloured rrow. By symmetry, the totl 4π( + x ) component of this from the entire coil perpendiculr to the xis is zero, nd the only µ I δs component of interest is the component long the xis, which is times sin θ. 4π( + x ) The integrl of δs round the whole coil is just the circumference of the coil, π, nd if we write sin θ =, we find tht the field t P from the entire coil is / ( + x ) B µ I =, 6.6. 3/ ( + x ) or N times this if there re N turns in the coil. At the centre of the coil the field is B µ I =. 6.6. The field is gretest t the centre of the coil nd it decreses monotoniclly to zero t infinity. The field is directed to the left in figure IV.5. We cn clculte the field in the plne of the ring s follows.

Q r O φ x θ P Consider n element of the wire t Q of length d φ. The ngle between the current t Q nd the line PQ is 9º (θ φ). The contribution to the B-field t P from the current I this element is The field from the entire ring is therefore µ. I cos( θ φ) dφ. 4π r

µ I 4π π cos( θ φ) dφ r, where r = + x x cos φ, + r x nd cos( θ φ) =. r This requires numericl integrtion. The results re shown in the following grph, in which the bsciss, x, is the distnce from the centre of the circle in units of its rdius, nd the ordinte, B, is the mgnetic field in units of its vlue µ I /() t the centre. Further out thn x =.8, the field increses rpidly. Mgnetic field in the plne of ring current..5.5 B.5...3.4.5.6.7.8.9 x

6.7 Helmholtz Coils Let us clculte the field t point hlfwy between two identicl prllel plne coils. If the seprtion between the coils is equl to the rdius of one of the coils, the rrngement is known s Helmholtz coils, nd we shll see why they re of prticulr interest. To y FIGURE VI.6 x P x c begin with, however, we ll strt with two coils, ech of rdius, seprted by distnce c. There re N turns in ech coil, nd ech crries current I. The field t P is µ NI B =. + 3/ 3/ [ ( ) ] [ ( ) ] 6.7. + c x + c + x At the origin (x = ), the field is B µ NI =. 6.7. 3/ ( + c ) (Wht does this become if c =? Is this wht you d expect?) If we express B in units of µni/() nd c nd x in units of, eqution 6.7. becomes B = +. 3/ 3/ [ + ( c x) ] [ + ( c + x) ] 6.7.4

3 Figure VI.7 shows the field s function of x for three vlues of c. The coil seprtion is c, nd distnces re in units of the coil rdius. Notice tht when c =.5, which mens tht the coil seprtion is equl to the coil rdius, the field is uniform over lrge rnge, nd this is the usefulness of the Helmholtz rrngement for providing uniform field. If you re energetic, you could try differentiting eqution 6.7.4 twice with respect to x nd show tht the second derivtive is zero when c =.5. 8 5.755 For the Helmholtz rrngement the field t the origin is. µ NI µ NI =. 5 FIGURE VI.7.8 c =..6.4 c =.5. B.8.6 c =..4. -.5 - -.5.5.5 x 6.8 Field on the Axis of Long Solenoid?????????????????????????????????????????????? θ O x δx FIGURE VI.8

4 The solenoid, of rdius, is wound with n turns per unit length of wire crrying current in the direction indicted by the symbols nd?. At point O on the xis of the solenoid the contribution to the mgnetic field rising from n elementl ring of width δx (hence hving n δx turns) t distnce x from O is This field is directed towrds the right. Let us express this in terms of the ngle θ. 3 µ nδx I µ ni. δx δ B = =. 6.8. 3/ 3/ ( + x ) ( + x ) We hve x = tn θ, δx = 3 sec θδθ, nd = cos θ. Eqution 6.8. 3/ ( + x ) becomes δb = µ ni cos θ. 6.8. If the solenoid is of infinite length, to find the field from the entire infinite solenoid, we integrte from θ = π/ to nd double it. Thus 3 π / Thus the field on the xis of the solenoid is B = µ ni cos θdθ. 6.8.3 B = µni. 6.8.4 This is the field on the xis of the solenoid. Wht hppens if we move wy from the xis? Is the field little greter s we move wy from the xis, or is it little less? Is the field mximum on the xis, or minimum? Or does the field go through mximum, or minimum, somewhere between the xis nd the circumference? We shll nswer these questions in section 6.. 6.9 The Mgnetic Field H If you look t the vrious formuls for the mgnetic field B ner vrious geometries of conductor, such s equtions 6.5.3, 6.6., 6.7., 6.8.4, you will see tht there is lwys µ on the right hnd side. It is often convenient to define quntity H = B/µ. Then these equtions become just

5 H I =, 6.9. π I H =, 6.9. NI H =, + 3/ 3/ [ ( ) ] [ ( ) ] 6.9.3 + c x + c + x H = ni. 6.9. It is esily seen from ny of these equtions tht the SI units of H re A m, or mps per metre, nd the dimensions re QT M. Of course the mgnetic field, whether represented by the quntity B or by H, is vector quntity, nd the reltion between the two representtions cn be written B = µh. 6.9. In n isotropic medium B nd H re prllel, but in n nisotropic medium they re not prllel (except in the directions of the eigenvectors of the permebility tensor), nd permebility is tensor. This ws discussed in section.7. with respect to the eqution D = εe. 6. Flux Recll from Section.8 tht we defined two extensive sclr quntities Φ E = E nd Φ = D d A, which I clled the E-flux nd the D-flux. In n entirely similr D mnner I cn define the B-flux nd H-flux of mgnetic field by B da ΦB = da 6.. nd Φ = H d A. 6.. H The SI unit of Φ B is the tesl metre-squred, or T m, lso clled the weber Wb. A summry of the SI units nd dimensions of the four fields nd fluxes might not come miss here. E V m MLT Q D C m L Q

6 B T MT Q H A m L T Q Φ E V m ML 3 T Q Φ D C Q Φ B Wb ML T Q Φ H A m LT Q 6. Ampère s Theorem In Section.9 we introduced Guss s theorem, which is tht the totl norml component of the D-flux through closed surfce is equl to the chrge enclosed within tht surfce. Guss s theorem is consequence of Coulomb s lw, in which the electric field from point source flls off inversely s the squre of the distnce. We found tht Guss s theorem ws surprisingly useful in tht it enbled us lmost immeditely to write down expressions for the electric field in the vicinity of vrious shpes of chrged bodies without going through whole lot of clculus. Is there perhps similr theorem concerned with the mgnetic field round currentcrrying conductor tht will enble us to clculte the mgnetic field in its vicinity without going through lot of clculus? There is indeed, nd it is clled Ampère s Theorem. H δs? I FIGURE VI.9

7 In figure VI.9 there is supposed to be current I coming towrds you in the middle of the circle. I hve drwn one of the mgnetic field lines dshed line of rdius r. The strength of the field there is H = I/(πr). I hve lso drwn smll elementl length ds on the circumference of the circle. The line integrl of the field round the circle is just H times the circumference of the circle. Tht is, the line integrl of the field round the circle is just I. Note tht this is independent of the rdius of the circle. At greter distnces from the current, the field flls off s /r, but the circumference of the circle increses s r, so the product of the two (the line integrl) is independent of r.? FIGURE VI. Consequently, if I clculte the line integrl round circuit such s the one shown in figure VI., it will still come to just I. Indeed it doesn t mtter wht the shpe of the pth is. The line integrl is H ds. The field H t some point is perpendiculr to the line joining the current to the point, nd the vector ds is directed long the pth of integrtion, nd H ds is equl to H times the component of ds long the direction of H, so tht, regrdless of the length nd shpe of the pth of integrtion: The line integrl of the field H round ny closed pth is equl to the current enclosed by tht pth. This is Ampère s Theorem.

8 So now let s do the infinite solenoid gin. Let us clculte the line integrl round the rectngulr mperin pth shown in figure VI.. There is no contribution to the line integrl long the verticl sides of the rectngle becuse these sides re perpendiculr to the field, nd there is no contribution from the top side of the rectngle, since the field there is zero (if the solenoid is infinite). The only contribution to the line integrl is long the bottom side of the rectngle, nd the line integrl there is just Hl, where l is the length of the rectngle. If the number turns of of wire per unit length long the solenoid is n, there will be nl turns enclosed by the rectngle, nd hence the current enclosed by the rectngle is nli, where I is the current in the wire. Therefore by Ampère s theorem, Hl = nli, nd so H = ni, which is wht we deduced before rther more lboriously. Here H is the strength of the field t the position of the lower side of the rectngle; but we cn plce the rectngle t ny height, so we see tht the field is ni nywhere inside the solenoid. Tht is, the field inside n infinite solenoid is uniform. l?????????????????????????????????????????????? H FIGURE VI. It is perhps worth noting tht Guss s theorem is consequence of the inverse squre diminution of the electric field with distnce from point chrge, nd Ampère s theorem is consequence of the inverse first power diminution of the mgnetic field with distnce from line current. Exmple. Here is n exmple of the clcultion of line integrl (figure VI.) (, ) (, ) FIGURE VI. I

9 An electric current I flows into the plne of the pper t the origin of coordintes. Clculte the line integrl of the mgnetic field long the stright line joining the points (, ) nd (, ). In figure VI.3 I drw (circulr) line of force of the mgnetic field H, nd vector dx where the line of force crosses the stright line of interest. FIGURE VI.3 (, ) x H dx (, ) θ I The line integrl long the elementl length dx is H. dx = H dx cos θ. Here I H = / π( + x ) nd cos θ =, nd so the line integrl long dx is / ( + x ) I dx. Integrte this from x = to x = nd you will find tht the nswer is I/8. π( + x ) Figure VI.4 shows nother method. The line integrl round the squre is, by Ampère s theorem, I, nd so the line integrl n eighth of the wy round is I/8. You will probbly immeditely feel tht this second method is much the better nd very clever. I do not deny this, but it is still worthwhile to study crefully the process of line integrtion in the first method.

(, ) FIGURE VI.4 I Another Exmple / 3 I An electric current I flows into the plne of the pper. Clculte the line integrl of the mgnetic field long stright line of length whose mid-point is t distnce / 3 from the current.

If you re not used to line integrls, I strongly urge you to do it by integrtion, s we did in the previous exmple. Some reders, however, will spot tht the line is one side of n equilterl tringle, nd so the line integrl long the line is just I. 3 We cn ply this gme with other polygons, of course, but it turns out to be even esier thn tht. For exmple: I θ Show, by integrtion, tht the line integrl of the mgnetic H-field long the thick line is θ just times I. π After tht it won t tke long to convince yourself tht the line integrl long the thick line θ in the drwing below is lso times I. π I θ

Another Exmple A stright cylindricl metl rod (or wire for tht mtter) of rdius crries current I. At distnce r from the xis, the mgnetic field is clerly I/(πr) if r >. But wht is the mgnetic field inside the rod t distnce r from the xis, r <? FIGURE VI.5 Figure VII.5 shows the cross-section of the rod, nd I hve drwn n mperin circle of rdius r. If the field t the circumference of the circle is H, the line integrl round the circle is πrh. The current enclosed within the circle is Ir /. These two re equl, nd therefore H = Ir/(π ). More nd More Exmples In the bove exmple, the current density ws uniform. But now we cn think of lots nd lots of exmples in which the current density is not uniform. For exmple, let us imgine tht we hve long stright hollow cylindricl tube of rdius, perhps liner prticle ccelertor, nd the current density (mps per squre metre) vries from the middle (xis) of the cylinder to its edge ccording to ( r) = ( r / ). The totl 3 current is, of course, I = π ( r) rdr = π, nd the men current density is =. 3 The question, however, is: wht is the mgnetic field H t distnce r from the xis? Further, show tht the mgnetic field t the edge (circumference) of the cylinder is, nd tht the field reches mximum vlue of 6 3 t r = 3. Well, the current enclosed within distnce r from the xis is 4 6 I r r = π ( x) xdx = π r ( 3 ),

3 nd this is equl to the line integrl of the mgnetic field round circle of rdius r, which is π rh. Thus r H = r ). ( 3 At the circumference of the cylinder, this comes to. Clculus shows tht H reches mximum vlue of function of x/. 6 6 3 t r = 3. The grph below shows H /( ) s 4..8.6.4. H/..8.6.4....3.4.5.6.7.8.9 r/ Hving whetted our ppetites, we cn now try the sme problem but with some other distributions of current density, such s 3 4 kr kr kr kr ( r) =, ( r) =, ( r) =, ( r) = The men current density is = r ( r) dr, nd the totl current is π times this. r The mgnetic field is H ( r) = x ( x) dx. r Here re the results:.

4 kr k r kr 3 3. =, =, H ( r) =. /.9.8.7.6.5.4 k =. k =. k =.4 k =.6 k =.8 k =..3.....3.4.5.6.7.8.9 r/.5 H/.45.4.35.3.5..5. k =. k =. k =.4 k =.6 k =.8 k =..5...3.4.5.6.7.8.9 r/ H reches mximum vlue of 3 cylinder only if k >. 4 3 3 t r = 6k 4 k, but this mximum occurs inside the

5 3 kr r kr 4. =, = ( k), H ( r) =. k =. /.9.8.7.6.5.4.3 k =. k =.4 k =.6 k =.8 k =......3.4.5.6.7.8.9 r/ H/.5.45.4.35.3.5. k =. k =. k =.4 k =.6 k =.8 k =..5..5...3.4.5.6.7.8.9 r/ H reches mximum vlue of the cylinder only if k >. 3 7k t r =, but this mximum occurs inside 3k

6 3. = H ( r) = kr 5k r, 5 = 5k kr 3/ 3/ 5 / [ 8 ( k) + ( k) ] + 3 kr 5/.,.9.8 k =. k =. k =.4.7 k =.6.6 k =.8 /.5.4.3 k =......3.4.5.6.7.8.9 r/ H/.5.45.4.35.3.5. k =. k =. k =.4 k =.6 k =.8 k =..5..5...3.4.5.6.7.8.9 r/ I hve not clculted explicit formuls for the positions nd vlues of the mxim. A mximum occurs inside the cylinder if k >.989.

7 4. = H ( r) = kr 3kr, = kr 3k 3/. 3/ [ ( k) ],.9 k =. k =. k =.4.8 k =.6.7 k =.8.6 k =. /.5.4.3.....3.4.5.6.7.8.9 r/.5.45.4.35 k =. k =. k =.4 k =.6 k =.8 k =..3 /.5..5..5...3.4.5.6.7.8.9 r/ A mximum occurs inside the cylinder if k >.8665. In ll of these cses, the condition tht there shll be no mximum H inside the cylinder ( ) tht is, between r = nd r = is tht >. I believe this to be true for ny xilly symmetric current density distribution, though I hve not proved it. I expect tht firly simple proof could be found by someone interested.

8 Additionl current density distributions tht reders might like to investigte re = + x / = + x / = + x / = + x / = x / e = e x / 6. Boundry Conditions We recll from Chpter 5, Section 5.4, tht, t boundry between two medi of different permittivities, the norml component of D nd the tngentil component of E re continuous, while the tngentil component of D is proportionl to ε nd the norml component of E is inversely proportionl to ε. The lines of electric force re refrcted t tn θ ε boundry in such mnner tht =. tn θ ε The sitution is similr with mgnetic fields. Tht is, t boundry between two medi of different permebilities, the norml component of B nd the tngentil component of H re continuous, while the tngentil component of B is proportionl to µ nd the norml component of H is inversely proportionl to µ. The lines of mgnetic force re tn θ µ refrcted t boundry in such mnner tht =. tn θ µ B x H x µ θ θ B y µ B x /µ H y H x µ θ θ µ H y /µ B y FIGURE VI.6

9 The configurtion of the mgnetic field inside n infinitely long solenoid with mterils of different permebilities needs some cre. We shll be guided by the Biot-Svrt lw, µ Idssin θ nmely B =, nd Ampère s lw, nmely tht the line integrl of H round 4π r closed circuit is equl to the enclosed current. We lso recll tht the mgnetic field inside n infinite solenoid contining single homogeneous isotropic mteril is uniform, is prllel to the xis of the solenoid, nd is given by H = ni or B = µ ni. The esiest two-mteril cse to consider is tht in which the two mterils re rrnged in prllel s in figure VI.7.?????????????????????????????????????????????? µ µ FIGURE VI.7 One cn see by pplying Ampère s lw to ech of the two circuits indicted by dshed lines tht the H-field is the sme in ech mteril nd is equl to ni, nd is uniform throughout the solenoid. It is directed prllel to the xis of the solenoid. Tht is, the tngentil component of H is continuous. The B-fields in the two mterils, however, re different, being µ ni in the upper mteril nd µ ni in the lower. We now look t the sitution in which the two mterils re in series, s in figure VI.8. We ll use horizontl coordinte x, which is zero t the boundry, negtive to the left of it, nd positive to the right of it.?????????????????????????????????????????????? µ µ FIGURE VI.9

3 We might t first be tempted to suppose tht B = µ ni to the left of the boundry nd B = µ ni to the right of the boundry, while, by n ppliction of Ampère s lw round ny of the dshed circuits indicted, H = ni on both sides. Tempting though this is, it is not correct, nd we shll see why shortly. The B-field is indeed µ ni long wy to the left of the boundry, nd µ ni long wy to the right. However, ner to the boundry it is between these limiting vlues. We cn clculte the B-field on the xis t the boundry by the sme method tht we used in Section 6.8. See especilly eqution 6.8, which, with the present geometry, becomes π / π / B = µ ni cos θdθ + µ ni cos θdθ. 6.. It should come s no surprise tht this comes to B = ( µ + µ ) ni. 6.. It is the sme just to the left of the boundry nd just to the right. The H-field, however, drops suddenly t the boundry from µ + ni immeditely to µ the left of the boundry to µ + ni immeditely to the right of the boundry. µ In ny cse, the very importnt results from these considertions is At boundry between two medi of different permebilities, the prllel component of H is continuous, nd the perpendiculr component of B is continuous. Compre nd contrst this with the electricl cse: At boundry between two medi of different permittivities, the prllel component of E is continuous, nd the perpendiculr component of D is continuous.