Electricity and Magnetism Gauss s Law

Transcription

1 Electrcty and Magnetsm Gauss s Law Ampère s Law Lana Sherdan De Anza College Mar 1, 2018

2 Last tme magnetc feld of a movng charge magnetc feld of a current the Bot-Savart law magnetc feld around a straght wre

3 Overvew Gauss s Law for magnetsm Ampère s Law B-feld outsde and nsde a wre Solenods

4 B-Feld around a wre revsted Usng the Bot-Savart law we found that the feld around an nfntely long straght wre, carryng a current I was: at a dstance a from the wre. B = µ 0I 2πa Is the another way we could have solved ths problem? When we had an nfnte lne of charge, there was a law we could use to fnd the E-feld...

5 B-Feld around a wre revsted Usng the Bot-Savart law we found that the feld around an nfntely long straght wre, carryng a current I was: at a dstance a from the wre. B = µ 0I 2πa Is the another way we could have solved ths problem? When we had an nfnte lne of charge, there was a law we could use to fnd the E-feld... Gauss s law. Could we use somethng smlar here?

6 Gauss s Law for Magnetc Felds Gauss s Law for Electrc felds: Φ E = E da = q enc ɛ 0 The electrc flux through a closed surface s equal to the charge enclosed by the surface, dvded by ɛ 0. There s a smlar expresson for magnetc flux! Frst we must defne magnetc flux, Φ B.

7 (30.18) Magnetc Flux Defnton of magnetc flux feld B S that makes an case s (30.19), then u and the he plane as n Fgure axmum value). weber Magnetc (Wb); 1 Wb flux 5 plane s a agnetc to the plane. B S u S d A Fgure The magnetc flux through an area element da s S B? d S A 5 B da cos u, where d S A s a vector perpendcular to the surface. The magnetc flux of a magnetc feld through a surface A s Φ B = B da da S Unts: Tm 2 If the surface s a flat plane and B s unform, that just reduces to: Φ B = B A

8 Gauss s Law for Magnetc Felds Gauss s Law for magnetc felds.: B da = 0 Where the ntegral s taken over a closed surface A. We can nterpret t as an asserton that magnetc monopoles do not exst. In dfferental form: B = 0 The magnetc feld has no sources or snks. (It s dvergence-free ; we can wrte B = a, where a s the vector potental.)

9 Gauss s Law for Magnetc Felds 32-3 INDUCED MAGNETIC FIELDS B da = PART 3 re complcated than does not enclose the Fg encloses no x through t s zero. nly the north pole of l S. However, a south ce because magnetc lke one pece of the encloses a magnetc Surface II N B Surface I S tom faces and curved B of the unform and and B are arbtrary of the magnetc flux

10 B-Feld around a wre revsted Gauss s law wll not help us fnd the strength of the B-feld around the wre: the flux through any closed surface wll be zero. Another law can.

11 Ampère s Law dstrbuton of currents due to a For current-length constant currents (magnetostatcs): the elements.agan we However, f the dstrbe s law to fnd the mag- B ds = µ 0 I enc n be derved from the André-Mare Ampère wever, the law actually The lne ntegral of the magnetc feld around a closed loop s proportonal to the current that flows through the loop. 1 (29-14) product B : ds : s to be. The current enc s the ts ntegral, let us frst The fgure shows cross 2,and 3 ether drectly p lyng n the plane of The counterclockwse en drecton of ntegra- Amperan loop Only the currents encrcled by the loop are used n Ampere's law. Drecton of ntegraton Fg Ampere s law appled to an 1 That s, the current arbtrary thatamperan flows through loop that anyencrcles surface two bounded by the loop. ds θ B

12 772 Ampère s Law CHAPTER 29 MAGNETIC FIELDS DUE T Ths s how to assgn a sgn to a current used n Ampere's law Drecton of ntegraton drecton of Fg , th are perpend current s n be n that pl In Fg The sca Thus, Ampe We can now Fg A rght-hand rule for of the Ampe A current through Ampere s thelaw, loop to determne the general the drecton sgns forof your we can nte outstretched currents thumbencrcled s assgned by aan plus Amperan sgn, andloop. a current generally around the e n the opposte The stuaton drecton sthat assgned of Fg. a mnus sgn. When w

13 Queston CHECKPOINT 2 The fgure here shows three equal currents shows (two three parallel equaland currents one antparal- (two parallel and one The (29-19) fgure here antparallel) and lel) four and Amperan loops. Rank the loops accordng to the magntude of four Amperan loops. Rank the loops s law gves us B accordng ds alongto each, the magntude greatest frst. of B : ds : along each, greatest frst. a (29-20) b ortonal to r, at Eqs c d A a, b, c, d B d, b, c, a C (a and b), d, c D d, (a and c), b 1 Hallday, Resnck, Walker, page 773.

14 Queston CHECKPOINT 2 The fgure here shows three equal currents shows (two three parallel equaland currents one antparal- (two parallel and one The (29-19) fgure here antparallel) and lel) four and Amperan loops. Rank the loops accordng to the magntude of four Amperan loops. Rank the loops s law gves us B accordng ds alongto each, the magntude greatest frst. of B : ds : along each, greatest frst. a (29-20) b ortonal to r, at Eqs c d A a, b, c, d B d, b, c, a C (a and b), d, c D d, (a and c), b 1 Hallday, Resnck, Walker, page 773.

15 Ampère s Law and the Magnetc Feld from a Current Outsde a wre All of the current s You mght Suppose we want encrcled to know and the magntude thus all of the magnetc feldntude at a B o dstance r outsde s used a wre. n Usng Ampere's Ampère s law. Law? s that the ntegraton of an encr Amperan We can Wre r loop surface the stuato the ntegra m B 0 ( 1 2 ), We sh ( θ = 0) allow us to ds 3 Fg Usng Ampere s law to fnd the magnetc feld that a current produces outsde a long straght wre of crcular cross Magnetc Fgure 29-

16 Ampère s Law and the Magnetc Feld from a Current All of the Outsde current s a wre encrcled and thus all s used n Ampere's law. Wre surface r Amperan loop ( θ = 0) Fg Usng Ampere s law to fnd the magnetc feld that a current produces outsde a long straght wre of crcular cross secton.the Amperan loop s a concentrc crcle that les outsde the wre. ds B (Current 3 s not encrcled by the loop.) We can then rewrte B cos ds 0( 1 2 ). You mght wonder why, snce current 3 contrbutes to the ntude B on the left sde of Eq , t s not needed on the s that the contrbutons of current 3 to the magnetc feld ca ntegraton n Eq s made around the full loop. In contr of an encrcled current to the magnetc feld do not cancel ou We cannot solve Eq for the magntude B of the mag Ampère s Law: the stuaton of Fg we do not have enough nformaton the ntegral. However, we do know the outcome of the ntegrat m 0 ( 1 2 ), the value of whch s set by the net current passng th We shall now apply B Ampere s ds = µ 0 law I enc to two stuatons n w allow us to smplfy and solve the ntegral, hence to fnd the m Magnetc Feld Outsde a Long Straght Wre wth Cu To fnd the B-feld at a dstancefgure r from the shows wre s a long straght centerwre choose that carres a curren page. Equaton 29-4 tells us that the magnetc feld B : produc crcular path of radus r. the same magntude at all ponts that are the same dsta By cylndrcal symmetry, everywhere along the crcle B ds s constant.

17 Ampère s Law and the Magnetc Feld from a Current All of the Outsde current s a wre encrcled and thus all s used n Ampere's law. Wre surface r Amperan loop ( θ = 0) Fg Usng Ampere s law to fnd the magnetc feld that a current produces outsde a long straght wre of crcular cross secton.the Amperan loop s a concentrc crcle that les outsde the wre. ds B (Current 3 s not encrcled by the loop.) We can then rewrte B cos ds 0( 1 2 ). You mght wonder why, snce current 3 contrbutes to the ntude B on the left sde of Eq , t s not needed on the s that the contrbutons of current 3 to the magnetc feld ca ntegraton n Eq s made around the full loop. In contr of an encrcled current to the magnetc feld do not cancel ou We cannot solve Eq for the magntude B of the mag Ampère s Law: the stuaton of Fg we do not have enough nformaton the ntegral. However, we do know the outcome of the ntegrat m 0 ( 1 2 ), the value of whch s set by the net current passng th We shall now apply B Ampere s ds = µ 0 law I enc to two stuatons n w allow us to smplfy and solve the ntegral, hence to fnd the m Magnetc Feld Outsde a Long Straght Wre wth Cu To fnd the B-feld at a dstancefgure r from the shows wre s a long straght centerwre choose that carres a curren page. Equaton 29-4 tells us that the magnetc feld B : produc crcular path of radus r. the same magntude at all ponts that are the same dsta By cylndrcal symmetry, everywhere along the crcle B ds s constant. The magnetc feld lnes must form a closed loop B ds = B ds.

18 Ampère s Law and All of the the current Magnetc s Feld from a Current Outsde encrcled a wreand thus all s used n Ampere's law. Wre surface r ds Amperan loop B ( θ = 0) You mght wonder w ntude B on the left s s that the contrbuto ntegraton n Eq. 29- of an encrcled curren We cannot solve E the stuaton of Fg. 29 the ntegral. However, m 0 ( 1 2 ), the value of We shall now app allow us to smplfy a And agan we get Fg Usng Ampere s law to fnd the magnetc feld that a current produces outsde a long B straght ds = wre µ 0 Iof enc crcular cross secton. The Amperan loop s a concentrc crcle that les outsde the wre. B(2πr) = µ 0 I B = µ 0I 2πr Magnetc Feld Ou Fgure shows a page. Equaton 29-4 t the same magntude

19 Ampère s Law and the Magnetc Feld from a Current Insde a wre PART 3 We can also use Ampère s Law n another context, where usng the 773 Bot-Savart Law s harder AMPERE S LAW he wre. We can take advanmpere s law (Eqs and lar Amperan loop of radus same magntude B at every se, so that ds : has the drec- Eq by notng that B : s as s ds :. Thus, B : and ds : are oop, and we shall arbtrarly between ds : and B : s 0, so omes B(2 r). ment lengths ds around the e 2pr of the loop. urrent of Fg The rght ve Only the current encrcled by the loop s used n Ampere's law. Fg Usng Ampere s law to fnd the magnetc feld that a current produces nsde a long straght wre of crcular cross secton.the current s unformly dstrbuted over the cross secton of the wre and emerges from the page.an Amperan loop s drawn nsde the wre. B r R ds Wre surface Amperan loop Now we place the Amperan loop nsde the wre. We stll have B ds = 2πrB, but now the current that flows through the loop s reduced.

20 Ampère s Law and the Magnetc Feld from a Current Insde a wre n- d us ry c- s re ly so e We stll have B ds = 2πrB, but now the current that flow through the loop s reduced. PART AMPERE S LAW 773 Assumng the wre has unform resstvty, I enc : Only the current encrcled by the loop s used n Ampere's law. Fg Usng Ampere s law to fnd the magnetc feld that a current produces nsde a long straght wre of crcular cross B r R ds Wre surface Amperan loop

21 Ampère s Law and the Magnetc Feld from a Current Insde a wre n- d us ry c- s re ly so e We stll have B ds = 2πrB, but now the current that flow through the loop s reduced. PART AMPERE S LAW 773 Assumng the wre has unform resstvty, I enc : Only the current encrcled by the loop s used n Ampere's law. Fg Usng Ampere s law to fnd the magnetc feld that a current produces nsde a long straght wre of crcular cross B r R ds Wre surface Amperan loop I enc = πr 2 πr 2 I = r 2 R 2 I Ampére s Law B ds = 2πrB = µ 0 r 2 R 2 I So, B = µ 0I r 2πR 2 = µ ( 0I r ) 2πR R

22 Ampère s Law For constant currents (magnetostatcs): B ds = µ 0 I enc The lne ntegral of the magnetc feld around a closed loop s proportonal to the current that flows through the loop. Later we wll extend ths law to deal wth the stuaton where the felds / currents are changng.

23 Solenods Addtonal examples, vdeo solenod A helcal col of tghtly wound wre that can carry a current. turn Fg A solenod carryng current Soleno Magnetc Feld o We now turn our useful. It concerns wound helcal col that the length of t Fgure sh The solenod s ma A sngle complete loop of wre n a solenod. Ths solenod has 10 turns, means t has 10 complete loops.

24 unform, we start wth a thn rng and then sum (va ntegraton) the rcled area. Magnetc Feld nsde and around a solenod P e central axs of a e turns are shown, as are the solenod. Each turn pror the solenod s axs, the s drected along the axs. ng magnetc feld. Outsde feld there s very weak. Each turn of wre locally has a crcular magnetc feld around t. The felds from all the wres add together to create very dense feld lnes nsde the solenod.

25 Magnetc Feld of a solenod IELDS DUE TO CURRENTS P 2 P 1 Fg Magnetc feld lnes for a real solenod of fnte length. The feld s strong and unform at nteror ponts such as P 1 but relatvely weak at external ponts such as P 2. The wres on opposte sdes (top and bottom n the pcture) have currents n opposte drectons. The felds add up between them, vdual turns (wndngs) that make up the solenod. For ponts very close to a turn, but cancel out outsde of them. the wre behaves magnetcally almost lke a long straght wre, and the lnes of B : there are almost concentrc crcles. Fgure suggests that the feld tends to cancel between adjacent turns. It also suggests that, at ponts nsde the solenod :

26 apply Ampere s law, Magnetc Feld of an deal solenod In an deal solenod B : ds : (wth 0 enc, nfnte length) the feld(29-21) outsde s small (and perpendcular to the Amp. loop) and nsde s unform. enod (Smlar of Fg , to a where capactor!) B : s unform wthn the solenod and usng the rectangular Amperan loop abcda. We wrte B : ds : as Can use an Amperan loop to fnd the B-feld nsde: d h c B ds = µ 0 I enc B a b plcaton of Ampere s law to a secton of a long deal solenod carryng mperan loop s the rectangle abcda.

27 apply Ampere s law, Magnetc Feld of an deal solenod In an deal solenod B : ds : (wth 0 enc, nfnte length) the feld(29-21) outsde s small (and perpendcular to the Amp. loop) and nsde s unform. enod (Smlar of Fg , to a where capactor!) B : s unform wthn the solenod and usng the rectangular Amperan loop abcda. We wrte B : ds : as Can use an Amperan loop to fnd the B-feld nsde: d h c B ds = µ 0 I enc B a b plcaton Here, of Ampere s supposelaw there to a secton are n of turns a long per deal unt solenod lengthcarryng the solenod, mperan then loop Is enc the = rectangle Inh abcda. Bh = µ 0 Inh Insde an deal solenod: B = µ 0 In

28 he torod, drected as shown n Fg b.Let u Torods A torod s a solenod wrapped nto a torus (donut) shape. (a) The two ends of the solenod are wrapped around an attached to each other. g (a) A torod carryng a current.(b) A orzontal cross secton of the torod. The nteror agnetc feld (nsde the bracelet-shaped tube) can be und by applyng Ampere s law wth the Amperan

29 w and the symmetry of the bracelet. ee Magnetc that the lnes Feld of B : form n aconcentrc Torod crcles nsde n Fg b.Let us choose a concentrc crcle of Cross secton through a torod: Amperan loop ng a current.(b) A rod. The nteror t-shaped tube) can be wth the Amperan We can use the Amperan loop shown to fnd the feld nsde the torod s loop. (b) r B

30 y descrbe as a (hollow) solenod that rmng a sort of hollow bracelet. What nsde the hollow of the bracelet)? We metry of the bracelet. es of B : Suppose form concentrc the torod crcles has nsde N turns..let us choose a concentrc crcle of Magnetc Feld n a Torod Amperan loop B ds = µ 0 I enc b) A r can be ran (b) r B B(2πr) = µ 0 NI Insde a torod: s B = µ 0IN 2πr Ths s not ndependent of the radus! The feld s stronger closer to the nsde: t s not unform.

31 Summary Forces on parallel wres Gauss s Law for magnetsm Ampère s Law Solenods 3rd Test Frday, Mar 9. Homework Collected homework 3, posted onlne, due on Monday, Mar 5. Serway & Jewett: PREVIOUS: Ch 30, Problems: 3, 5, 9, 13, 19 NEW: Ch 30, Problems: 21, 25, 31, 33, 34, 47