Chapter 18 MAGNETIC FIELDS

Transcription

1 Ch. 18--Magnetc Felds Chapter 18 MAGNETIC FIELDS A.) A Small Matter of Specal Relatvty: 1.) Assume we have a partcle of charge q movng wth an ntal velocty v q parallel to a currentcarryng wre as shown n from the laboratory frame of reference Fgure a.) Consder the stuaton from the perspectve of the laboratory frame of reference (.e., the frame n whch you and I st and n whch the wre s motonless):.) The postve charges (the protons) FIGURE 18.1 are fxed n the wre whle the negatve charges (the electrons) have some non-zero average velocty v e. q v wre n the lab frame: the wre s statonary; the current--postve charge--flows to the left; a test-charge q moves rght wth velocty v..) There are as many electrons as protons n the wre before the current begns (.e., the wre s electrcally neutral)..) As many electrons leave the wre as come onto the wre whle current flows. As such, the wre s perceved to be electrcally neutral even when current s flowng. b.) Consder now the stuaton from q's frame of reference: Note: From ths frame of reference, the charge q wll be statonary whle everythng else s movng around t. 187

2 .) In q's frame of reference (see Fgure 18.2), the wre and all postve charges (protons) wll move to the left wth velocty v q. Meanwhle, negatve charges (electrons) wll move to the left wth velocty v q - v e (we are assumng v q > v e ). from q's frame of reference The acton s summarzed n Fgure wre and protons movng wth velocty v q toward left electrons movng wth velocty v q - v e toward left q statonary blow-up secton of wre FIGURE 18.2.) Notce that the protons move faster than electrons from ths perspectve. 2.) Ensten's Theory of Relatvty suggests that when one object passes a second object, the second object wll appear to the frst to have contracted n length. Called "length contracton," the phenomenon s mmedately evdent only at very hgh speeds but does occur mcroscopcally at low speeds. 3.) Because all of the charge n the wre moves relatve to q's frame of reference, the dstances between the charges should appear to be closer (relatvstc length contracton) than would otherwse have been the case f vewed from the lab frame. What's more, the protons wll appear to be more tghtly packed because they are movng faster than the electrons (see Fgure 18.3). from q's frame of reference protons well packed due to hgher velocty v q p + p + p + p + p + p + p + p + p + p + p + p + e- e- e- e- e- e- e- e- e- electrons not as well packed due to lower velocty v q - v e q statonary From q's frame, an electrc feld exsts due to the predomnance of postve charges n the wre. As q s postve, ths electrc feld wll push q away from the wre. FIGURE

3 Ch. 18--Magnetc Felds a.) In other words, the wre wll appear to have more protons than electrons on t. That means charge q wll perceve an electrc feld due to the predomnance of postve charge, and that electrc feld wll motvate q to accelerate away from the wre. b.) If we set up an experment n whch a postve charge s made to move parallel to a current-carryng wre and opposte to the current's drecton, we wll observe a force on q pushng t away from the wre. The force s due to the relatvstc effect we have been dscussng, but observers n the prevous century dd not know that (Ensten's Theory of Relatvty wasn't publshed untl 1905). Workng strctly from emprcal observaton, they assumed there must exst a new knd of force--a magnetc force--actng on the movng charge. The theory developed on behalf of that belef s today called "the classcal theory of magnetsm." It s the subject we are about to consder. B.) Some Early Observatons: There are a number of observed phenomena that led early scentsts to formulate the classcal theory of magnetsm. In no partcular order: 1.) When suspended, certan metallc ores are found to have the pecular ablty to orent themselves north/south. They evdently algn themselves wth some sort of feld, a feld that n the early days of "modern scence" was eventually called a magnetc feld. a.) In expermentng wth a pece of such ore, t has been observed that ths north/south orentaton s always the same. That s, the same face always algns tself to the north whle the opposte face always algns to the south. To dstngush between the two, one s called "the North Seekng Magnetc Pole N" and the other s called "the South Seekng Magnetc Pole S." These observatons were, n early tmes, the bass for what s today called a compass. 2.) When a compass s put n the vcnty of a "magnetzed" pece of metallc ore, the compass s found to pont n dfferent drectons at dfferent places. compass N lne defnng compass drecton at varous ponts magnetc bar S FIGURE 18.4a 189

4 a.) Followng the needle drecton for the varous sample ponts shown n Fgure 18.4a on the prevous page, a lne can be drawn. Dong ths for a number of dfferent postons around the bar allows us to sketch what are called "magnetc feld lnes" (see Fgure 18.4b). N magnetc feld lnes leave North Pole and enter South Pole bar magnet S FIGURE 18.4b b.) Magnetc feld lnes are smlar to electrc feld lnes n the sense that where the feld lnes are close together, the feld s sad to be large, but: c.) Magnetc feld lnes are DIFFERENT from electrc feld lnes n one very mportant way. The drecton of an electrc feld lne s defned as the drecton a postve test charge wll accelerate f released n the electrc feld. In other words, electrc felds are really nothng more than slghtly modfed force-feld lnes (E = F/q). The drecton of a magnetc feld lne s defned as the drecton a compass wll pont f a magnetc feld s present. As wll be shown shortly, magnetc felds are NOT modfed force felds (though they are dstantly related to force). d.) A constant magnetc feld s denoted by feld lnes that are equdstant and parallel as shown n Fgure 18.4c. B 3.) The strength of a magnetc feld coupled wth the drecton of the magnetc feld s combned together to defne the magnetc feld vector B. More wll be sad shortly about B, ts relatonshp to the force on a charge movng n a magnetc feld, and ts unts. depcton of constant magnetc feld lnes FIGURE 18.4c 190

5 Ch. 18--Magnetc Felds 4.) Whle expermentng wth electrcal crcuts n 1820, a man named Oersted observed that when a compass was placed near a current-carryng wre, the compass responds. Expermentng further: wre a.) Oersted found that magnetc feld lnes CIRCLE around a current-carryng wre (see Fgure 18.5a). Notce that the drecton of a current-produced magnetc feld can B-feld of currentcarryng wre crculates thumb of rght hand upward n drecton of current FIGURE 18.5a be determned by usng the followng "werd" rght-hand rule (from here on, ths rule wll be termed the rght-thumb rule): Poston the thumb of the rght hand so that t follows the drecton of current flow--the drecton the fngers curl s the drecton of the magnetc feld's crculaton around the wre (Fgure 18.5b). fngers curl around n drecton of B-feld b.) Oersted concluded that magnetc felds are somehow related to CHARGE IN MOTION. wre drecton of B-feld FIGURE 18.5b 5.) From expermentaton, t has been observed that f a postve charge q s placed n a magnetc feld B: B a.) The charge wll feel NO FORCE due to the presence of the magnetc feld f the charge s statonary (see Fgure 18.6a); b.) The charge wll feel NO FORCE due to the presence of the magnetc feld f the charge q at rest charge q feels no force when statonary n B-feld FIGURE 18.6a 191

6 s movng wth velocty v along the magnetc feld lnes (Fgure 18.6b); c.) The charge WILL FEEL A FORCE due to the presence of the magnetc feld f the charge's velocty vector s orented at any angle other than zero or 180 o relatve to the magnetc feld vector B (see Fgure 18.6c). Furthermore, the drecton of the force wll be perpendcular to the plane defned by the magnetc feld vector and the velocty vector. In the case shown n Fgure 18.6c, that drecton s perpendcular to the plane of the page. d.) The charge wll feel a maxmum force f the velocty vector v s perpendcular to the magnetc feld vector B. e.) Puttng all of the above nformaton together, the expermentally determned relatonshp that exsts between the magntude of the force F B on a charge q q v q v B charge q wll feel no force when movng parallel to B-feld q FIGURE 18.6b v 0 B charge q movng at an angle 0 wth B wll feel a force perpendcular to the page FIGURE 18.6c movng wth velocty vector v at an angle θ wth the magnetc feld B s: F B = q v B sn θ. Ths s the magntude of a cross product, whch mples: F B = q v x B. Note: The drecton of a cross product s always perpendcular to the plane defned by the two vectors beng crossed. That s exactly the drecton we needed for our magnetc force vector. f.) IMPORTANT CONCLUSION: Magnetc felds are centrpetal n nature--they change the drecton of charged bodes but do not make them speed up or slow down. Addtonally, the relatonshp between magnetc felds as defned (.e., havng a drecton determned by the way a compass orents tself) and magnetc forces as observed expermentally s not an obvous one. More about ths later. 192

7 Ch. 18--Magnetc Felds C.) Contnung Wth Observatons--The Bar Magnet: 1.) If a magnetc feld s created by charges n moton, what knd of moton creates the magnetc feld n an apparently motonless bar magnet? Possbltes: Electrons confned to the atom are constantly n moton-- they both orbt about the nucleus and spn about ts axs. Let's consder both: a.) Orbtal Moton: Whle the orbtal moton of electrons around the nucleus surely produces a magnetc feld, the drecton of an electron's moton wll be "ths way" as much as "that way" (electrons travel around the atom at speeds upward of 150,000 mles per second). Consequently, the net magnetc feld produced by electron orbtal moton s, on average, zero. b.) Spnnng On Axs: Electron spn also produces a magnetc feld. Due to quantum mechancal effects, electrons spn n only one of two drectons. These drectons are usually referred to as "spn up" and "spn down." In most elements, there are as many electrons spnnng up as spnnng down whch means the net magnetc feld generated by all the spnnng electrons s zero..) There are some elements whose number of electrons spnnng n one drecton s notceably dfferent from the number spnnng n the opposte drecton. Iron, for nstance, has sx more electrons spnnng one way than the other. As a consequence, the net magnetc feld due to electron spn n an ron atom s not zero. Put another way, every ron atom s a mn-magnet unto tself..) Elements that exhbt ths magnetc characterstc are called ferromagnetc materals. The most common are ron, nckel, and cobalt. 2.) Ferromagnetc materals do not always exhbt magnetc effects. Iron nals, for nstance, do not usually attract or repulse one another as would be expected f they were magnetzed. The queston s, "Why?" a.) Take a structure made of ron (a steel bolt, for example). Wthn t, there exst mcroscopc sectons called domans. A doman s a volume n whch each atom has algned ts magnetc feld n the same drecton as all the other atoms n the secton. b.) Fgure 18.7a (next page) shows a sde-vew blow-up of the domans that resde on the face of a pece of ron. Notce that each doman has ts magnetc feld n some arbtrary drecton. Because none 193

8 of the doman-felds are algned, the net (read ths average) magnetc feld on the face s essentally zero. Ths s an example of a ferromagnetc materal that does not appear to be magnetzed. c.) If the bolt s placed n a relatvely strong magnetc feld, the domans wll algn themselves wth the external feld and, n dong so, wll algn themselves wth one another (see Fgure 18.7b). In that case, each face of the bolt wll ether be a North Pole or South Pole. That s, we end up wth a magnetzed pece of ron. ndvdual, nonalgned domans enlarged secton ndvdual domans algned makng face a South Pole sde-vew of non-magnetzed ron bar FIGURE 18.7a sde-vew of magnetzed ron bar enlarged secton FIGURE 18.7b D.) Observatons--The Earth's Magnetc Feld: 1.) Through expermentaton, t was found that North Seekng Magnetc Poles always attract South Seekng Magnetc Poles. Lke poles (.e., N-N or S-S poles) repulse. One of the consequences of ths s the pecular stuaton we have wth respect to the earth's magnetc feld. 2.) By defnton, the North Seekng Magnetc Pole of a compass ponts toward the northern geographc regon of the earth. But f North Seekng Magnetc Poles are attracted to South Seekng Magnetc Poles, there must exst a South Magnetc Pole n the northern geographc hemsphere. In fact, that s exactly the case. The earth's magnetc feld lnes leave Antarctca and enter the Arctc (they actually enter n the Hudson Bay regon--see Fgure 18.8a on the next page for the theoretcal dstrbuton of magnetc feld lnes around the earth). 194

9 Ch. 18--Magnetc Felds 3.) Solar wnds are streams of hgh energy subatomc partcles that are constantly beng emtted by the sun. Due to these solar wnds, the earth's magnetc feld lnes are actually compressed n toward the earth on the earth's sun-sde whle beng extruded out away from the earth on the earth's dark sde. See Fgure 18.8b. Earth's magnetc feld lnes n theory earth's geographc north pole (earth's axs of rotaton) magnetc feld lnes leavng earth's North Magnetc Pole n geographc southern hemsphere magnetc feld lnes enterng earth's South Magnetc Pole n northern geographc hemsphere equator FIGURE 18.8a 4.) The earth's magnetc feld s beleved to be caused by moton of molten ron at the earth's core. By lookng at core samples of the earth's geologcal hstory over long perods of tme, t has been found that the earth's magnetc feld changes drecton perodcally (sometme between 200,000 to 400,000 years per cycle). Although scentsts are not completely sure why, the current theory s that long-perod oscllatory sun earth's magnetc feld lnes dstorted by solar wnds FIGURE 18.8b varatons n the moton of the earth's ron-rch molten nteror create ths effect. 195

10 E.) Approach for Determnng a Magnetc Feld--Ampere's Law: 1.) We have made our observatons; now t s tme to examne the math that has grown up around those observatons. Theoretcans have developed a number of ways for determnng magnetc feld functons for current confguratons. In the 1820's, one of these approaches was created by Andre Ampere. 2.) Ampere's Law states the followng: a.) Defne a closed path n a regon n whch a B-feld exsts. b.) Defne a dfferental length-vector dl over a dfferental secton of the path. c.) Determne the dot product between the magnetc feld B evaluated at the dfferental secton and the vector dl. d.) Sum all such dot products around the closed path. e.) That sum wll always be proportonal to the amount of current that passes through the face of the path (the face of the path s the area enclosed wthn the boundares of the path). f.) Puttng ths all n mathematcal terms, we get: B dl =µ o thru, where µ o s the proportonalty constant called the permeablty of a vacuum and s equal to 4πx10-7 teslas. meter/amp (.e., 1.26x10-6 T. m/a). path. Note 1: The ntegral symbol denotes an ntegraton around a closed Note 2: The term thru s used to denote the current passng through the face of the defned Amperan path. 3.) As a pont of order, and because t puts thngs n perspectve, t should be noted that Ampere's Law does for magnetc felds what Gauss's Law dd for electrc felds. That s: 196

11 Ch. 18--Magnetc Felds a.) In Gauss's Law, we defned an arbtrary closed surface that had the rght geometry and symmetry; n Ampere's Law we defne an arbtrary closed path that has the rght geometry and symmetry. b.) In Gauss's Law, we defned a dfferental surface area vector ds at an arbtrary poston on the surface. In Ampere's Law, we defne a dfferental length vector dl at an arbtrary poston on the path. c.) In Gauss's Law, we dotted the unknown electrc feld functon E (assumed to be evaluated at ds) nto the dfferental surface area vector ds. In Ampere's Law, we dot the unknown magnetc feld functon B (assumed to be evaluated at dl) nto the dfferental length vector dl. d.) In Gauss's Law, we summed all the E. ds quanttes over the entre surface (we used a surface ntegral to do ths) to determne the total electrc flux through the surface. In Ampere's Law, we sum all the B. dl quanttes over the entre path (we use a lne ntegral to do ths) to determne the total magnetc crculaton around the path. e.) In Gauss's Law, the electrc flux through the Gaussan surface s proportonal to the charge enclosed wthn the surface. In Ampere's Law, the magnetc crculaton around the Amperan path s proportonal to the current passng through the path's face. S F.) Ampere's Law and a Straght, Infnte, Current-Carryng Wre: 1.) Consder an nfntely long current-carryng wre: a.) As mentoned above, Oersted found that a current-carryng wre produces a magnetc feld that crculates around the wre accordng to the rght-thumb rule (thumb of rght hand n drecton of current; fngers curl n drecton of the B-feld). b.) To determne the magntude of the magnetc feld, we wll use Ampere's Law. 2.) The geometry n whch Ampere's Law s most easly used s one n whch the magntude of the magnetc feld s the same at every pont along the arbtrarly defned path. Ths happens to be just such a case. 197

12 a.) Due to symmetry, the Amperan path of choce here s a crcle of arbtrary radus r centered on the wre. Fgure 18.9a shows just such a path whle Fgure 18.9b vews the stuaton lookng along the lne of the wre. B-feld generated by a current-carryng wre o r b.) Just as was the case wth Gauss's Law, Ampere's Law has two parts to deal wth: the rght-hand sde of the equaton and the left-hand sde of the equaton. We wll deal wth both separately, then put t all together. c.) The left-hand sde of Ampere's Law s: wre B crclng FIGURE 18.9a B dl, or the ntegral sum of the dot product of the magnetc feld vector (evaluated at a pont on the Amperan path) and the dfferental dsplacement along the path at that pont. wre wth current comng out of page (as vewed along the lne of the wre) r o closed Amperan path.) As the vector dl s orented n the same dl drecton as the vector B (the drecton of dl was defned that way), and as the path FIGURE 18.9b was chosen so that the magntude of B would be constant at every pont along the path, the dot product can be treated as: B dl = B( dl) cos0 = B dl = B( 2πr). d.) The rght-hand sde of Ampere's Law requres that we determne the amount of current that breaks through the face of the area defned by the path. o B 198

13 Ch. 18--Magnetc Felds.) In ths case, the total current breakng through the face s smply the current through the wre, or o. 3.) Puttng everythng together and presentng t the way you wll be expected to present t on a test, we wrte: B dl = µ o thru o Bdl ( ) cos0 = µ B dl = µ o o B( 2π r) = µ µ oo B =. 2 π r o o o thru G.) A Note About the Vector Nature of Magnetc Felds: 1.) Consder the two wres shown n Fgure 18.10a. One carres a.5 amp current out of the page (the crcle wth dot depcts an arrowhead comng out of the page) whle the other carres a.25 amp current nto the page (the crcle wth cross depcts an arrowhead gong nto the page). Assumng the dstances are as shown n the sketch, what s the net magnetc feld generated at Pont P? =.5 amps 1 =.25 amps 2.75 meters.2 m wres perpendcular to page Pont P FIGURE 18.10a a.) As for the B-feld drectons (see Fgure 18.10b for the bottom lne):.) Current 1 produces a B- feld whose drecton at Pont P s toward the top of the page (+j drecton) whle current 2 produces a B-feld whose drecton at Pont P s toward the bottom of the page (-j drecton). Note that B the magnetc felds at Pont P B 2 FIGURE 18.10b both are tangent to crcles centered on ther respectve current carryng wres, just as suggested by the rght thumb rule and the Fgure 18.9b. 199

14 b.) The magntude of the magnetc felds are:.) For current 1 : B 1 = µ ο 1 /2πr 1 = [(1.26x10-6 kg. m/coul 2 )(.5 A)]/[2π(.95 m)] = 1.06x10-7 teslas..) For current 2 : B 2 = µ ο 2 /2πr 2 = [(1.26x10-6 kg. m/coul 2 )(.25 A)]/[2π(.2 m)] = 2.51x10-7 teslas. c.) The net B-feld wll be the vector sum of the two felds: B net = B 1 + B 2 = (1.06x10-7 teslas)(+j) + (2.51 x10-7 teslas)(-j) = (1.06x10-7 teslas)(j) - (2.51 x10-7 teslas)(j) = (-1.45x10-7 teslas) j. Note: In the thrd lne, the negatve sgn belongs to the second term's unt vector ( 2 's B-feld drecton was toward the bottom of the page). It has been brought out n front of the magntude value to make t easer to track. 2.) For the wre confguraton shown n Fgure 18.11a, what s the net magnetc feld generated at Pont P? a.) As for the B-feld drectons (see Fgure 18.11b on the next page for the bottom lne): a.) Current 1 produces a B-feld whose drecton at Pont P s toward the top of the page (+j drecton). a 2 3.) Current 2 produces a B-feld whose drecton at Pont P s toward the bottom-left of the page at a 45 o angle to the left (.e., n the - + (-j) drecton). 1 Pont P FIGURE 18.11a 200

15 Ch. 18--Magnetc Felds.) Current 3 produces a B-feld whose drecton at Pont P s toward the rght of the page (+ drecton). b.) The magntudes of the magnetc felds are: 2 B 1 3 crcle centered on 2.) For current 1 : B 1 = µ ο 1 /(2πr 1 ) = µ ο 1 /(2πa)..) For current 2 : B 2 = µ ο 2 /(2πr 2 ) = µ ο 2 /[2π( 2a)]. 1 B 2 B 3 magnetc felds at Pont P FIGURE 18.11b.) For current 3 : B 3 = µ ο 3 /2πr 3 = µ ο 3 /(2πa). c.) Notng that B 2 must be broken nto ts components, the net B- feld wll be the vector sum of the three ndvdual felds, or: B net = B 1 + B 2 + B 3 µ = o 1 (+ j) 2πa + µ o 2 (cos 45 o µ )( ) + o 2 (sn 45 o )( j) 2π( 2a) 2π( 2a) + µ o 3 2πa (+) B net = µ o 2πa ( 3 ) 2 cos 45 o 2 + ( ) 2 1 sn 45 o 2 j. H.) Ampere's Law and the B-Feld Produced n an Infntely Long Col: 1.) Consder a very long (read ths nfntely long) wre col havng n wnds-per-unt-length (see Fgure 18.12a). If we spread the cols out n order to see how the current-produced magnetc feld acts at varous places wthn the geometry, we wll note some nterestng thngs. Specfcally: very long current carryng col FIGURE 18.12a 201

16 a.) The net B-feld perpendcular to the col's axs adds to zero. That s, the B-feld generated by the charge flow n the upper-left-sde secton of wre (see Fgure 18.12b) generates a magnetc feld perpendcular to the col's axs that s equal and opposte to the magnetc feld generated by charge flow n the upper-rght-sde secton of wre. The two felds smply add to zero. sde of wre closest to observer current-carryng col B-felds perpendcular to axs (due to charge flow n wres to rght and left of space) add to zero col's axs FIGURE 18.12b b.) For an nfntely long col, the B-feld parallel to the axs and outsde the col wll be zero, as the magnetc feld lnes along the central axs wll never leave that axs (agan, ths s for an nfnte col). Note: To a very good approxmaton, ths s true even for fnte-length cols (see B Fgure 18.12c). The reasonng s as follows: Assumng we are a far dstance from the col, the B-feld generated by charge flow on the top sde of a gven col s almost equal and opposte to the B B-feld generated by charge flow on the bottom sde of the col (see Fgure 18.12d). A smlar stuaton exsts for the net feld below the wre and, n fact, for all ponts outsde the permeter of the col. c.) For an nfntely long col, the net B-feld s all down the axs of the col. magnetc feld lnes for a current-carryng col (notce how ntense the feld s down the axs and how lttle t s outsde the col) FIGURE 18.12c B-felds outsde col (due to charge flow n bottom and top sectons of wres) approxmately add to zero B bot B top 2.) To determne the Amperan path: top bot col's axs FIGURE 18.12d 202

17 Ch. 18--Magnetc Felds a.) In GAUSS'S LAW, the trck was to fnd a closed surface through whch the electrc feld was ether a constant (.e., on the cylndrcal part of a Gaussan cylnder), zero (.e., nsde a conductor), or such that E. ds was zero (.e., on the flat end of a cylndrcal Gaussan surface). b.) Ampere's Law s smlar. We want a path upon whch the magnetc feld s ether a constant, zero, or such that over the path, B. dl s zero. Amperan path Path 3 dl 3 c.) The path that works for ths case s a rectangle, as shown n Fgure Path 4 dl 4 dl 2 dl 1 Path 2 3.) Wth our Amperan path defned, we are ready to use Ampere's Law. col Path 1 a.) The frst thng to notce s that the path has four sdes and, hence, wll requre four dot products (each path secton s shown on the sketch for convenence). FIGURE ) As there s no magnetc feld perpendcular to the col, the magnetc feld along Paths 2 and 4 s zero. As such, the ntegrals assocated wth those paths are zero..) As there s no magnetc feld outsde the nfntely long col, the magnetc feld along Path 3 s also zero. Note: Even f we had been workng wth a fnte col, the magnetc feld outsde the col s very small, especally f we allow the path to extend out a consderable way from the col's axs. Bottom lne: B. dl along Path 3 wll be zero (f not exactly, then to a good approxmaton) whether the col s nfntely long or not. b.) The total current passng through the area bounded by the path s equal to the current n one wre tmes the number of wres nsde the Amperan path (see Fgure on the next page)..) The number of wres nsde the Amperan path wll equal the number of wres per unt length--call ths n--multpled by the length of the path defned n the sketch (next page) as L 1, or nl

18 cross-secton of col.) That means that the net current passng through the boundary defned by the path wll be: thru = (n wnds/meter)(l 1 meters)( o amps/wnd) = nl 1 o. dl 3 L 1 dl 4 dl 2 dl 1 current comng out of page 4.) Puttng t all together, we can wrte: n wnds per unt length current gong nto page B dl = µ o thru B dl + B dl + B dl + B dl = µ ( nl ) B dl + ( 0) + ( 0) + ( 0) = µ ( nl ) BL = µ B = o 1 o L1 L2 L3 L4 L 1 1 o 1 o ( nl ) 1 o 1 o µ o n o. FIGURE I.) The Law of Bot Savart: 1.) Ampere's Law s useful when there s symmetry (.e., when the magntude of B s a constant over an Amperan path), or when the magntude of B s a constant over part of an Amperan path whle B. dl s zero over the rest of the path. When a more general stuaton s at hand, a more general approach s needed. In such cases, we turn to the Law of Bot Savart. 2.) The Law of Bot Savart states that a dfferental secton of wre wll produce a dfferental magnetc feld db at some pont near the wre such that: In ths expresson: db = µ o dlxˆr. 4π r 2 a.) db s the dfferental magnetc feld vector due to the current n dl as evaluated at the pont of nterest (see Fgure 18.15); b.) s the current n the wre; 204

19 Ch. 18--Magnetc Felds c.) dl s a vector whose magntude s equal to the length of the dfferental secton of wre and whose drecton s n the drecton of the current; d.) r s the magntude of a poston vector drawn from the dfferental secton dl to the pont of nterest (call ths Pont P); and dfferental secton of length dl Bot Savart db = (u /4π)(dl/r 2)sn 0 o lne of dl 0 r pont of nterest currentcarryng wre FIGURE e.) ˆr s a UNIT VECTOR n the drecton alluded to n Part d drectly above. Note: A number of texts wrte Bot Savart as: db = µ o dlxr, 4π r 3 where r s the entre poston vector, both drecton and magntude, and an r 3 s placed n the denomnator to compensate for the fact that r ncludes the vector's magntude. Ether presentaton s correct, although n practce you wll fnd the frst one easer to use. Bg Note: The symbols dl and r are defned dfferently n Bot Savart and n Ampere's Law. KNOW AND UNDERSTAND THE DIFFERENCES. 3.) Although the cross product operaton defnes the drecton of the magnetc feld at Pont P, there are nstances when the magntude of the magnetc feld s of sole nterest. In such cases, the magntude of the cross product yelds a dfferental magnetc feld equal to: db = µ o 4π dl sn θ, 2 r where θ s the angle between dl and r (agan, see Fgure 18.15). 205

20 J.) Example of an Easy Bot Savart Problem: 1.) Consder the current-carryng wre shown n Fgure 18.16a (the power supply has been added, but we wll gnore ts presence n the magnetc feld calculaton). What s the net magnetc feld (as a vector) at the common center of the semcrcles? a.) Fgure 18.16b shows dl vectors for each secton of the system. semcrcular crcut R 1 R 2 b.) Usng Bot Savart on the top semcrcle, we can determne the dfferental magnetc feld due to the charge flow n the segment dl 1. That expresson s: db 1 = µ o dl 1 2 4π r sn(90o ), 1 V FIGURE 18.16a poston vectors and dfferental segments defned for use n Bot Savart where r 1 = R 1. c.) To determne the total magnetc feld due to the entre upper semcrcle, we must ntegrate db 1. Substtutng r 1 = R 1 nto our expresson and dong the ntegraton, we get: dl 3 r 3 V r 1 dl 1 r 2 dl 2 B 1 = db µ o = 2 dl 4 π R 1 µ o 2πR1 = 2 4 π R 1 2 µ o =. 4R 1 around sem crcle FIGURE 18.16b d.) Usng the rght-thumb rule on the wre, we fnd the magnetc feld due to the upper semcrcle s out of the page. 206

21 Ch. 18--Magnetc Felds e.) A smlar exercse generates a magnetc feld expresson due to current flowng n the lower semcrcle. It equals: B 2 = µ o 4R 2. f.) Examne Fgure 18.16b agan. The angle between dl 3 and r 3 s 180 o. As the sne of 180 o s zero, the magnetc feld due to the left-hand, straght-wre secton wll be zero at the pont of nterest (t wll not be zero at other places, but at the center of the semcrcles t s zero). The same s true of the straght-lne secton on the rght. g.) Ths means that as a vector the net feld equals: B = B + B 1 2 µ o µ o = + ( + k ). 4R1 4R2 K.) A Second Example of Bot Savart: 1.) Reteraton: So far, we have dealt wth stuatons n whch the magnetc felds at a pont of nterest have all been n the same drecton (.e., n the prevous problem, the feld was out of the page for all dl segments). In such cases, the approach used was: a.) Defne dl (an arbtrary segment on the current-carryng wre). b.) Defne r (a vector from dl to the pont of nterest). c.) Defne the angle θ between the lne of dl and the lne of r. d.) Use the rght-hand rule on dlxr to determne the drecton of the dfferental magnetc feld at the pont of nterest. e.) Use Bot Savart to determne the magntude of the dfferental magnetc feld at the pont of nterest due to current n dl. f.) Integrate the dfferental magnetc feld expresson to determne the net magnetc feld due to all of the dl's for whch your derved db expresson s vald (n the prevous problem, there were 207

22 four such sectons--the upper and lower semcrcles and the rght and left-hand straght-lne sectons). g.) Add up all the derved B-feld expressons to get the net magnetc feld at the pont of nterest, complete wth a statement of drecton. 2.) There s another twst that hasn't yet been addressed. What happens f the major segments produce magnetc felds whose drectons are dfferent? To see how such a possblty mght occur, consder the followng stuaton: 3.) A crcular wre of radus R rests n the x- z plane (.e., n the horzontal) wth ts center at the orgn of the coordnate system beng used (see Fgure 18.17a). Current flows n the wre as shown n the sketch. Derve an expresson for the net magnetc feld a dstance y unts up the y-axs. B-feld up axs of currentcarryng wre y (0, y) arbtrary pont a.) Proceedng wth our approach: The vector dl s defned n the drecton of current flow. It s supposed to be an arbtrarly defned segment of wre, but for smplcty and ease of vewng on the accompanyng sketches, let's defne t to be the segment that cuts through the x-y plane movng nto the page (see Fgure 18.17b). z R x FIGURE 18.17a b.) The vector r s defned as a vector from the segment to the pont of nterest. In ths case, the pont of nterest s at an arbtrary pont y unts up the y-axs. Note that as defned, r s n the x-y plane. y db dl r c.) Usng the rght-hand rule to determne the drecton of the cross product between dl and r, we fnd that the drecton of the dfferental magnetc feld db produced by that dfferental segment of current s n the x-y plane as shown n Fgure 18.17b. FIGURE 18.17b d.) The angle between dl and r s 90 o (dl s nto the page whle r s n the plane of the page). Addtonally observng that r = (y 2 + R 2 ) 1/2, we can use Bot Savart to wrte: z 2 2 1/2 y r = (y + R ) R dl x 208

23 Ch. 18--Magnetc Felds µ odx l rˆ db = π 2 4 r µ o dl = 4 π y + R / [( ) ] µ o dl = 4 π y + R ( ). sn 90 o e.) The magntude of db wll be the same for any gven dl, but the drecton of db wll be dfferent from segment to segment. That means we could break db nto ts components and ntegrate each component separately, or we could be clever. mrror mage dl's db y 0 db f.) Beng clever, examne r the drecton of the magnetc dl 0 dl feld produced by a segment R x that s 180 o from our defned dl (see Fgure 18.17c). The horzontal component of that z vector s equal and opposte to FIGURE 18.17c the horzontal component of db produced by dl. As all such components wll add to zero, we can gnore the horzontal component and deal solely wth the vertcal components. Note: If you don't beleve that we can gnore the horzontal component, determne db sn φ (.e., db's horzontal component) and do the ntegral. You wll fnd that t evaluates to zero. g.) Wth the horzontal component gnored, B net = db y = db cos φ. Usng the geometry of a crcle and Fgure 18.17c, we can see that cos φ = R/r. As such, we can wrte: 209

24 B net = db y µ o dl = 4 π y + R 2 2 ( ) µ dl o R = 4 π ( y + R ) y + R µ o R = 4 π y + R ( ) µ o R = 4 π y + R µ o = 2 ( ) cos φ ( ) ( ) / / / 2 2 R. 2 2 ( y + R ) 3/ 2 dl ( 2πR) L.) The Force on a Charge Movng n a Magnetc Feld: Note: We have been examnng the mathematcs around the theoretcal determnaton of magnetc feld functons. For the next few sectons, we wll assume the avalablty of magnetc felds wthout consderng ther orgn. 1.) As has already been stated, a charge movng n a magnetc feld wll feel a force under certan crcumstances. The relatonshp between ths magnetc force F B, the magnetc feld strength B, the velocty of the charge v, and the sze of the charge q has been expermentally determned to be: F B = q v x B. a.) Ths cross product yelds both the magntude and drecton of the force on a POSITIVE CHARGE movng n the magnetc feld. b.) IMPORTANT: If the charge s NEGATIVE, the magntude of the force wll be the same BUT THE DIRECTION OF THE FORCE WILL BE OPPOSITE that determned usng the rght-hand rule. Note: From the MKS unts for force, charge, and velocty, the unts for the magnetc feld vector B must be nt/[c. (m/s)], or kg/(c. s). Ths set of MKS unts s gven the specal name "teslas." 210

25 Ch. 18--Magnetc Felds 2.) Examples: Determne the magntude and drecton of the force on a 4 coulomb charge movng wth velocty 12 meters/second n a magnetc feld whose strength s 5 teslas f the velocty and magnetc feld vectors are as shown n Fgures 18.18a through 18.18d. Note: Vectors pontng perpendcularly nto the page are depcted ether by a group of crcles wth crosses n them or smply by crosses. Vectors pontng perpendcularly out of the page are depcted by a group of crcles wth ponts at ther centers or smply by ponts. a.) For Fgure 18.18a: the magntude of the force s F B = q v B sn θ, where θ s the angle between the lne of v and the lne of B (note that the sketch s a bt trcky --θ should be the angle between the lne of v and the lne of B--that s not the angle gven n the fgure. Puttng n the numbers, we get: B-feld v 30 o q FIGURE 18.18a F B = (4 C)(12 m/s)(5 T)(sn 150 o ) = 120 newtons. The drecton s found usng the rght-hand rule for a cross product. The rght hand moves n the drecton of the lne of the frst vector (v); the fngers of the rght hand curl n the drecton of the lne of the second vector (B). Dong so yelds a force drecton for ths stuaton nto the page. You wll not normally be asked to do so, but for the sake of completeness for ths frst try, ths force can be wrtten as a vector n unt vector notaton as: F B = (120 newtons)(-k). b.) The magntude of the force n Fgure 18.18b s: B-feld v F B = q v B sn θ = (4 C)(12 m/s)(5 T)(sn 90 o ) = 240 newtons. q 30 o FIGURE 18.18b 211

26 The drecton s found usng the rght-hand rule for a cross product. In ths case, the drecton wll be toward the bottom of the page, perpendcular to v, and to the rght. If asked a queston lke ths on a test, you wll not be asked to put the fnal force vector n a unt vector notaton. You wll be asked to draw n the force drecton on the sketch n addton to determnng the force magntude. c.) The magntude of the force n Fgure 18.18c s: F B = (4 C)(12 m/s)(5 T)(sn 180 o ) = 0 newtons. q v B-feld There s no drecton assocated wth zero force. Note: Ths should gve you a bt of a hnt as to FIGURE 18.18c the order of operatons on a test problem. Determne magntudes frst before tryng to determne drecton-- tryng to get the rght-hand rule to work on a cross product whose magntude s zero can be enormously frustratng. d.) For amusement, assume q s negatve. The magntude of the force n Fgure 18.18d s: F B = (4 C)(12 m/s)(5 T)(sn 90 o ) = 240 newtons. q B-feld v nto page As the charge s negatve, the drecton s toward the bottom of the page. FIGURE 18.18d 3.) When electrc and magnetc felds are present, the net possble electrcal force actng on a charge s qvxb + qe. Ths s called Lorentz's equaton. M.) The Force on a Current-Carryng Wre n a Magnetc Feld: 1.) Consder a current-carryng wre of length L stuated n a magnetc feld as shown n Fgure on the next page. Fnd the force on the 212

27 Ch. 18--Magnetc Felds current-carryng wre as the movng charge nteracts wth the magnetc feld. B-feld a.) Assume that all the free charges n the wre move at the same average velocty. b.) If tme t s the average amount of tme requred for one charge q to move the entre length L of the wre, the average velocty of that charge wre wre of length of length L L FIGURE (and all the others) wll be L/t, where the L s a vector whose magntude s defned as the length of the wre and whose drecton s defned as the drecton of the charge's moton. c.) We know that the force on a sngle charge q movng n a magnetc feld s F q,b = qv x B. Substtutng n v = L/t, we get: F q,b = q (L/t) x B. d.) The collectve magnetc force F B on all the charges movng n the wre (we'll call the total charge Q) yelds a net force of: F B = Q (L/t) x B = (Q/t) L x B. e.) The current s defned as the rato of the total charge Q passng a partcular pont over a perod of tme t (.e., = Q/t). Wth that n mnd, we can wrte: F B = L x B. f.) Ths relatonshp defnes the net magnetc force felt by a current-carryng wre n a magnetc feld. Note: The drecton of the cross product wll yeld the drecton of the magnetc force on the wre; the magntude of the cross product wll yeld the magntude of that magnetc force. 2.) Examples: Determne the magntude and drecton of force on a.5 meter long current-carryng wre f the magnetc feld ntensty s 4 teslas, 213

28 the current n the wre s 2 mllamps, and the vector drectons are as shown n Fgures 18.20a through 18.20d. a.) The magntude of the force n Fgure 18.20a s the evaluaton of a cross product, or: F B = L B sn θ, where θ s the angle between the lne of L and the lne of B. Puttng n the numbers, we get: 30 o B-feld FIGURE 18.20a F B = (2x10-3 amps) (.5 m) (4 T) (sn 90 o ) = 4x10-3 newtons. The drecton s found usng the rghthand rule for a cross product. The rght hand moves n the drecton of the lne of the frst vector (L); the fngers of the rght hand curl n the drecton of the lne of the second vector (B). Dong so n ths problem yelds a force drecton that s perpendcular to the wre and toward the bottom of the page. wre B-feld b.) The magntude of the force n Fgure 18.20b s: FIGURE 18.20b F B = (2x10-3 amps)(.5 m)(4 T)(sn 90 o ) = 4x10-3 newtons. The drecton wll be perpendcular to B and toward the top of the page. c.) The magntude of the force n Fgure 18.20c s: F B = (2x10-3 amps)(.5 m)(4 T)(sn 120 o ) 30 o = 3.46x10-3 newtons. B-feld FIGURE 18.20c 214

29 Ch. 18--Magnetc Felds The drecton s perpendcularly nto the page. d.) The magntude of the force n Fgure 18.20d s: F B = (2x10-3 amps) (.5 m) (4 T) (sn 180 o ) = 0 newtons. wre B-feld There s no drecton for a zero-magntude force. FIGURE 18.20d 3.) Example--Newton's Second Law: A metallc bar of mass m s placed on two supports a dstance L = a unts apart that form an nclne whose angle s φ wth the horzontal (see Fgure 18.21). A battery s attached across the supports and a constant, downward magnetc feld B permeates the setup (see Fgure 18.22). If the net resstance n the crcut s R, what voltage V o s requred to ensure that the bar does not accelerate down the nclne? support a sldng mass m 0 0 metal nclne Vo R a.) We know V o = R. If we can determne the current needed to suspend the bar, we can determne V o. b.) There are three forces actng on the bar. The frst two are gravty and a normal force. The thrd s a magnetc force due to the fact that the current passng through the bar s n a B-feld. Knowng ths, we should be able to explot Newton's Second Law to derve an equaton that wll be helpful. c.) The magntude of the magnetc force s F = LxB where L's magntude s a and the angle between L and B s 90 o (L s nto the page whle B s n the plane of the page). The free body sldng mass m support 0 sde vew FIGURE B feld metal nclne FIGURE

30 dagram for the stuaton s found n Fgure to the rght wth components hghlghted. Usng Newton's Second Law: f.b.d. (wth components) of sldng mass y Σ F x : ab cos 0 N mg(sn φ) - (ab sn 90 o )(cos φ) = ma x = 0 = mg(tan φ)/ab. ab sn 0 ab mg mg sn 0 mg cos 0 x d.) As V = R, we get: FIGURE V o = [mg(tan φ)/ab] R. e.) Assumng φ = 30 o, m=.15 kg, B = 3 T, a =.2 m, and R = 5 Ω, the voltage s found to be: V o = [mg (tan φ)/ab]r =[[(.15 kg)(9.8 m/s 2 ) tan 30 o ]/[(.2 m)(3 T)]] (5 Ω) = 7.07 volts. N.) The Use of Magnetc Felds n Buldng Meters: 1.) Prelmnary observatons: Consder a pnned, sngle-looped rectangular col whose sde-lengths are equal to a and whose top and bottom lengths are equal to b. If a current passes through the wre whle n a magnetc feld (see Fgure 18.24a), the movng charges wll feel a force accordng to F B = L x B. sngle-loop, current-carryng rectangular col n a magnetc feld N pn a b pn S Note 1: The magnetc feld lnes generated by bar magnets are not constant (see Fgure 18.24b). Nevertheless, we wll assume a constant B-feld for smplcty. FIGURE 18.24a 216

31 Ch. 18--Magnetc Felds Note 2: The drecton of the force wll depend upon the current's drecton relatve to the magnetc feld vector. Lookng at the wre and charge flow from above (see Fgure 18.24b), we can make the followng observatons: a.) The secton of wre wth current movng out of the page (sde A n force on currentcarryng wre (Sde A) N sde A--current out of page along ths sde pn B-feld lnes (assumed unform, though they wouldn't really be n ths setup) sde B--current nto page along ths sde S force on currentcarryng wre (Sde B) FIGURE 18.24b Fgure 18.24b) wll feel a force whose drecton s toward the top of the page and whose magntude s ab. b.) The secton of wre wth current movng nto the page (sde B n Fgure 18.24b) wll feel a force whose drecton s toward the bottom of the page and whose magntude s ab. c.) Each of these forces wll produce a torque on the col about the pn. Wth r = b/2, we can wrte: r x F = (b/2) (ab) sn φ, where φ s the angle between r and F. d.) As there are two such wres, the total torque on the col about the pn wll be: Γ net = 2 [(b/2) (ab) sn φ] = AB sn φ (A s defned as the area--length a tmes wdth b--of the square loop's face). e.) If there are N wnds n the col (our orgnal col had only one loop), the net torque becomes: 217

32 Γ net = NAB sn φ. f.) Defnng a vector µ m called the magnetc moment whose drecton s perpendcular to the face of the col (orented so that the thumb of the rght hand s n the drecton of ths vector when the fngers of the rght hand curl n the drecton of current flow--see Fgure 18.25) and whose magntude s equal to NA, we get the relatonshp: magnetc moment m N S B feld lnes FIGURE Γ net = µ m x B, where the angle between µ m and B s φ. Note: The above expresson s analogous to the torque expresson for an electrc dpole n an electrc feld. Contnung wth the analogy, the amount of potental energy wrapped up n a current-carryng col s: U = -µ m. B. g.) Notng that currentcarryng cols n magnetc felds can have torques appled to them, consder the magnetcally engulfed, current-carryng col shown n Fgure A sprng attached to the bottom of the col produces a countertorque f the col rotates. When rotaton occurs, a needle attached to the col also rotates. If that needle s placed over a vsble, calbrated scale, we end up wth the prototypcal currentsensng meter. current-measurng meter (a galvanometer) ponter N S calbrated scale ponter and col rotate when current present FIGURE

33 Ch. 18--Magnetc Felds 2.) The most basc verson of a current-sensng meter s called a galvanometer (the sketch shown n Fgure s actually that of a galvanometer). A col n a known magnetc feld has attached to t a sprng that s just taut enough to allow the needle to rotate full-deflecton (.e., to the end of the scale) when 5x10-4 amps flow through t. In that way, f an unknown current flows through the galvanometer and the needle fxes at half deflecton, the user knows that the current s half of 5x10-4 amps, or 2.5x10-4 amps. ALL GALVANOMETERS ARE MADE TO SWING FULL DEFLEC- TION WHEN 5x10-4 AMPS FLOW THROUGH THEM. Ths unformty s the reason galvanometer scales are labeled 1 through 5 wthout any other hnt as to the meanng of the numbers. It s assumed that f you know enough to be usng a galvanometer, you know that ts unts are "x10-4 amps" (qute a concet f you thnk about t). As all galvanometers are made to the same specfcatons throughout the ndustry, they are the cornerstone n the producton of all other meters, voltmeters and large-current ammeters alke. Note: Although t s not evdent n Fgure 18.26, a galvanometer's needle always ponts toward the center of the scale when no current s passng through the meter. In that way, the needle can deflect ether to the rght or the left, dependng upon whch meter-termnal the hgh voltage s connected to. Galvanometers are the only meters that have ths "center-zero" setup. All other meters have ther zero to the left, swngng to the rght when current passes through them. That means they depend upon you, the user, to hook the hgh voltage ammeter desgn leads to the correct termnal. 3.) The Ammeter: The sketch n Fgure shows the crcut for a 12 amp ammeter (the sketch s general to all ammeters; I have arbtrarly chosen 12 amps for the sake of a number example). Notce the desgn requres a galvanometer (desgnated by the resstance R g ) and a second resstor R s. Assume the resstance of the galvanometer s 5 ohms. The ratonale behnd the desgn s as follows: =12 amps n, max Rg -4 g,max = 5x10 s = n - g R s FIGURE a.) We want the galvanometer's ponter to swng full-deflecton when twelve amps of current passes through the ammeter. In other 219

34 words, when twelve amps flow nto the meter, we want 5x10-4 amps to flow through the galvanometer. b.) The parallel desgn allows current passng through the meter to splt up. If we pck just the rght sze resstor R s (ths s called a shunt resstor because t shunts off current from passng through the galvanometer), all but 5x10-4 amps wll flow through that resstor whenever twelve amps flow nto the devce. The trck s n fndng the proper value for the shunt resstor. To do so:.) Notcng that the voltage across R g s the same as the voltage across the R s (the two resstors are n parallel), we can wrte: g,max R g = s,max R s..) We know that g,max wll be 5x10-4 amps when 12 amps flow nto the crcut, so the amount of current passng through R s must be whatever s left over, or: s,max = (12 amps) - (.0005 amps) = amps. ) Puttng t all together, we get: g,max R g = s,max R s (5x10-4 amps) (5 Ω) = ( amps) R s R s = 2.08x10-4 Ω. c.) A short pece of wre wll have resstance n ths range. In other words, a typcal ammeter s nothng more than a galvanometer wth a measured pece of wre voltmeter desgn hooked n parallel across ts termnals. 4.) The Voltmeter: The sketch n Fgure shows the crcut for a 12 volt voltmeter (the sketch s general to all voltmeters; I have arbtrarly chosen 12 volts R g -4 g,max = 5x10 V =12 volts max R FIGURE 18.28

35 Ch. 18--Magnetc Felds for the sake of a number example). Notce the desgn requres a galvanometer (desgnated by the resstance R g ) and a second resstor R 1. Assumng the galvanometer's resstance s 5 ohms, the ratonale behnd the desgn s as follows: a.) We want the galvanometer's ponter to swng full-deflecton when twelve volts are placed across the voltmeter (.e., when the voltmeter s hooked across an electrcal potental dfference of twelve volts). In other words, when 12 volts are placed across the meter, we want 5x10-4 amps of current to flow through the galvanometer. b.) The seres desgn requres that voltage across the voltmeter be splt up between the two seres resstors (.e., the voltage drop across the galvanometer plus the voltage drop across the second resstor must sum to 12 volts). If we pck just the rght sze resstor R 1, a current of 5x10-4 amps wll flow through both resstors whenever twelve volts are placed across the meter. The trck s n fndng the proper value for the second resstor. To do so:.) When the total voltage across the meter s 12 volts, the galvanometer's voltage must be g,max R g whle the second resstor's voltage must be g,max R 1 (the two resstors are n seres, hence the current s common to both). As such we can wrte: V o = ( g,max R g ) + ( g,max R 1 ) 12 volts = (5x10-4 amps) (5 Ω) + (5x10-4 amps) (R 1 ) R 1 = x10 4 Ω. c.) In short, a typcal voltmeter s nothng more than a galvanometer hooked n seres to a large resstor. As would be expected, they draw very lttle current when hooked across an element n a crcut. 5.) Bottom lne: All analog meters (.e., meters that are not dgtal) are based on the galvanometer, and all galvanometers are based on the proposton that current movng through an approprately pnned col n a magnetc feld wll feel a torque-producng force whch s proportonal to the amount of current passng through the col. 221

36 QUESTIONS 18.1) Copper wre s electrcally neutral whether there s current flowng through t or not. Does ths make sense n lght of Ensten's theory regardng length contracton? That s, f there are electrons n moton n a wre (.e., f there s a current), shouldn't the electrons bunch up due to length contracton, creatng an electrc feld that would affect even statonary charge next to the wre? What do you thnk Ensten's response would be? Note: Don't thnk about ths too hard--i'd say ffteen seconds should do ncely. You won't be tested on any of the Relatvty materal. Ths queston s more of a teaser for the chapter on Relatvty at the end of the book than anythng else (a quck and drty answer s suppled n the Solutons). 18.2) A seres of parallel, currentcarryng wres are shown n Fgure I. What s the drecton of the net magnetc feld at: a.) Pont A on the sketch; b.) Pont E on the sketch; c.) Pont C on the sketch; d.) Pont D on the sketch. e.) Ignorng gravty, f an electron s placed at Pont E, what force wll t feel? Pont C (close to wres) Pont A Pont D (very far from wres) FIGURE I wres (current out of page) Pont E wres (current nto page) 18.3) Sx partcles wth the same mass move through a magnetc feld drected nto the page (Fgure II). a.) Identfy the postvely charged, A negatvely charged, and electrcally neutral masses. (Hnt: How would F you expect a postvely charged partcle G E to move when travelng through a B- feld drected nto the page?) C b.) Assumng all the partcles have the same charge-magntude, FIGURE II whch one s movng the fastest? (Hnt: For a fxed charge, how s charge velocty and radus of moton related? Thnk!) c.) Assumng all the partcles have the same velocty, whch one has the greatest charge? (Same hnt as above, but reversed.) B-feld (nto page) D 222