Chapter VII Electrodynamics

Transcription

1 Chapter VII Electrdynamics Recmmended prblems: 7.1, 7., 7.4, 7.5, 7.7, 7.8, 7.10, 7.11, 7.1, 7.13, 7.15, 7.17, 7.18, 7.0, 7.1, 7., 7.5, 7.6, 7.7, 7.9, 7.31, 7.38, 7.40, 7.45,

2 Ohm s Law T make a current flw yu have t frce charges t mve. Hw fast they mve in respnse t an external agent depends n the nature f the medium. Fr mst substances, it is fund that the current density J is prprtinal t the electric field E as J E (1) Ohm' s Law The prprtinality factr is called the cnductivity f the substance. The reciprcal f is called the resistivity, i.e., 1 () The resistivity f sme material are listed in table 7.1.

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4 Example 7.1 A cylindrical resistr f crsssectinal are A and length L is made f material f cnductivity. If a ptential difference V is maintained between its ends, what current flws? Slutin: Assuming that the electric field is unifrm within the wire, we have frm Eq.(1) I JA EA But the electric field is related t the ptential difference as V E dl V EL I A V L The current flwing between tw electrdes is prprtinal t the ptential difference between them.

5 Example 7. Tw lng cylinders f radii a and b are separated by material f cnductivity. If they are maintained at ptential difference V is maintained between its ends, what current flws frm ne cylinder t the ther, in a length L? Slutin: Frm Eq.(1) we get I JA EA E E I sˆ Ls But the electric field is related t the ptential difference as V b a E dl I sl b I ds I ln L s L L lna b V a Again, the current flwing between tw electrdes is prprtinal t the ptential difference between them. a b

6 The results f the previus tw examples enable us t write V IR (3) Ohm' s Law The prprtinality cnstant R is called the resistance f the substance. It depends n the gemetry f the arrangements and the cnductivity f the medium between the electrdes. Frm the differential frm f Gauss s law we have E Fr steady current and unifrm cnductivity we get 1 J 0 J The charge density is zer and any unbalanced charge resides n the surface.

7 Electrmagnetic Inductin & Faraday s Law: In 183 Michael Faradays perfrmed a series f experiments: I- He pulled a lp f wire t the right thrugh a magnetic field. A current flwed in the lp. II- He mved the magnet t the left hlding the lp still. Again a current flwed in the lp. III- With bth the magnet and the lp at rest, he changed the strength f the m.field. Once again a current flwed in the lp. Changing m. field

8 As a result f these experiments, an emf is assciated with a change in magnetic flux thrugh a circuit, i.e., d (4) Faraday' s Law The result is independent f the way in which the flux is changed. It shuld be realized that Eq.(4) represents an independent experimental law and can t be derived frm ther experimental law. Nw it is knwn that the emf and the m. flux are defined as c E dl m B Eq.(4) can nw be written as c E dl d S B ds S If the circuit is a rigid statinary circuit, the time can be taken inside the integral here it becmes a partial derivative. S Eq.(5) reads ds (5)

9 c E dl S B ds t (6) Applying Stkes s therem t the left hand side f Eq.(6) we get S E ds ds (7) S B t If this is true fr any clsed surface B E t It is a generalizatin f the identity (8) E 0 Differential frm f Faraday s Law which held fr static m. field. The negative sign in Faraday s law indicates that the directin f the induced emf is such that t ppse the change that prduces it. In anther wrd, (i) If the magnetic flux thrugh the lp is increasing, the directin f the induced current is such that it prduces a magnetic field ppsite t the surce magnetic field. (ii) If the magnetic flux thrugh the lp is decreasing, the directin f the induced current is such that it prduces a magnetic filed in the same directin as that f the surce magnetic field.

10 Example 7. A lng cylindrical magnet f radius a and length L caries a unifrm magnetizatin M parallel t its axis. It passes at cnstant velcity v thrugh a circular wire ring f slightly larger diameter. Graph the emf induced in the ring as a functin f time. Slutin: Since the magnetizatin is unifrm then J b M 0 K b M nˆ M ˆ The magnetic field is the same as that f a lng slenid. Except near the ends, the field utside the slenid is zer and inside is unifrm and equal t B ni zˆ M

11 The flux is zer when the magnet is far away; it builds up t a maximum value f Ma as the leading end passes thrugh; and it drps back t zer as the trailing end emerges. The emf is the (minus) derivative f flux with respect t time s we have tw ppsite spikes.

12 Mtinal EMF In rder t get sme insight int Faraday s law, it is useful cnsider the fllwing cnfiguratin. Cnsider a straight metal wire f length l mving with speed v in a directin perpendicular t its length in a regin f unifrm magnetic field B directed perpendicular t the in which the mves. The free electrns inside the rd will experience a magnetic frce accrding t B l v E F m evi ˆ Bkˆ evb ˆj This frce drives +ve and ve charges t the ppsite ends f the wire. As a result f this charge separatin, a net electric field E will be set up inside the wire. Therefre, free charges nw will be affected by an upward electric frce F e =ee in additin t the magnetic field.

13 Charges cntinue t build up at the ends f the wire until the tw frces balanced. At this pint, mtin f charges ceases leaving the wire with tw ppsite plarities at its end, that is an emf is prduced acrss the wire. T calculate this emf we have, frm the equilibrium cnditin F m F e evb ee E vb Since the electric field in unifrm inside the rd, the ptential difference acrss the rd is related t this electric field accrding t V El vlb (9) This ptential difference is maintained acrss the ends f the rd as lng as it is mving in the field and is called the mtinal emf.

14 Let us nw prve the last equatin using Faraday's law. Cnsider nw the wire is sliding alng cnducting rails in the magnetic field such that it frms a clsed lp. As the area f the lp at sme instant is A=lx, the magnetic flux thrugh the lp is m Blx l B x l (b) Where x, the wih f the lp, is changing with time as the wire mves. Using Faraday's law, we find that the induced emf in the lp is d d dx m Blx Bl Blv Which is the same result except f the minus sign. Eq.(9) can be generalized in vectr ntatin t be written as I v V B l v (10)

15 It shuld be nted that if the wire is nt mving while the magnetic field is changing with time we will have the same charge separatin and the same ptential difference between the ends f the wire as befre. In this case there is n magnetic frce since the charges are nt mving and the frce that makes the separatin in the induced electric frce due the change f the m. field as stated by Eq.(8). In the steady state the net electric frce must be zer, i.e., F qe 0 with E E 1 E Where E 1 is due the charge separatin and E is the field due t the changing m. field. Since E 1 is static we have, frm Eq.(8) B B E E E. ds. ds t t Using Stke s therem we get B E. dl. ds t C S d Blv d S B. ds (11) S S

16 Example 7.4 A metal disk f radius a rtates with angular velcity abut a vertical axis, thrugh a unifrm m. field B, pinting up. A circuit is made by cnnecting ne end f a resistr t the axle and the ther end t the uter edge f the disk. Find the current in the resistr. Slutin: Nw since v is nt cnstant alng the disk but depends n the distance frm its center and frm Eq.(9) we have Blv And fr the current we have Bvdl a B sds 0 1 Ba I R 1 Ba R

17 Induced Electric Field Recalling Eq.(11) we have C E dl d S d B. ds E dl m (1) C That is, a changing in the magnetic flux thrugh a lp prduces an induced electric field alng that lp. Since the directin f current is the directin f the mtin f +ve charges and +ve charges mves alng the directin f the electric field, the directin f the induced electric field is determined by Lenz s law. It shuld be nted that the induced e. field given by Eq.(1) is nt cnservative. Otherwise, the result f the integratin in Eq.(1) wuld be zer.

18 Example 7.8 A unifrm magnetic field B(t), pinting straight up, fills the shaded circular regin f radius a, as shwn. If B is changing with time, what is the induced electric field every where? Slutin: Fr s<a, we draw an Amperian lp f radius s, and nting E pints alng the circumference f such a lp we get s s B(t) a C d E dl m d E s B s s db E ˆ Nw, fr s>a, we draw anther Amperian lp f radius s>a. Again E pints alng the circumference f the lp, s we have E d s B a E a s db ˆ

19 Example 7.8 A line charge is glued nt the rim f a wheel f radius b, which is then suspended hrizntally, as shwn, s that it is free t rtate. In the central regin, ut f radius a, there is a unifrm magnetic field B, pinting up. Nw smene turns the field ff. What happens? Slutin: The changing m. field (when it is turned ff) will induce an e. field curling arund the axis f the wheel as shwn. dl This e. field exerts a frce n the charges at the rim, and the wheel starts t turn. Accrding t Lenz s law, it will rtate is such a directin that is its m. field must be in the same directin as the riginal m. field, i.e., the mtin is cunterclckwise. Quantitatively we have, frm Faraday s law Nw the trque acting n a small element dl is C b E dl d B m a a db E dn qe be( dl bf b )

20 The ttal trque acting n the wheel is N b C E dl b a db Nw frm Newtn s law the trque is related t the angular mmentum as N dl L N ba db ba 0 B B

21 Inductance When a current passes in a circuit, a m. field will be created thrugh the circuit. S the m. flux depends n the current in the circuit. Thus fr a rigid statinary circuit the nly changes in m. flux result frm the change in the current, i.e., d d di di Nw, frm Faraday s law Eq.(4) we have with di L L d di (13) (14) (15) d d di di Is called the self inductance f the circuit and its SI unit is Henry (H)

22 Example 7.11 Find the self inductance f a tridal cil with rectangular crss sectin f inner radius a, uter radius b and height h, which carries a ttal f N turns. Slutin: The m. field inside the trid is B NI s The m. flux thrugh a single turn is h s a B da b NI ds NIh b h ln s s s a The ttal flux is N times the flux thrugh a single turn tt a Ih b N N ln s a Frm Eq.(15) the self inductance is then N h b L ln I s a

23 Suppse nw that yu have tw lps at rest. If yu run a steady current I 1 in lp 1, it prduces a m. field B 1. A m. flux is expected t set up thrugh lp given by B1 da (16) Since B 1 is prprtinal t I 1 s will by M1I1 (17) Where the prprtinality cnstant M 1 is called the mutual inductance f the tw lps. Nw it knwn that B A A da (18) 1 Using Stke s therem we get A1 dl (19) Nw frm Eq.(41) f chapter (5) we have

24 dl A I 4 Cmparing Eqs. (17) & (0) we get M 4 1 dl 1 dl Frm Eq.(1) we cnclude that: (I) I 1 4 dl1 (1) dl Neumann frmula (0) M 1 is purely gemetrical quantity, depends n the sizes, shapes, and relative psitins f the tw lps. (II) The integral in Eq.(1) is unchanged by switching the rles f the lps, i.e., M 1 M1 M () If yu vary the current in lp 1, the flux thrugh lp will vary accrdingly, and Faradays s law says an induced emf will set up in lp d M di 1 (3) Every time yu change the current in lp 1, an induced current flws in lp even thugh there are n wires cnnecting them.

25 Example 7.10 A shrt slenid (radius a and length l with n 1 turns per unit length) lies n the axis f a very lng slenid (radius b with n turns per unit length) as shwn. Current I flws in the shrt slenid. What is the flux thrugh the lng slenid? and what is the mutual inductance f the tw slenids? Slutin: Since the inner slenid is shrt, it puts a different amunt f flux thrugh each turn f the uter slenid. It is a miserable task t cmpute the ttal flux this way. Using the equality f the mutual inductance between the tw slenids we can assume that the current flws in the lng slenid and we want t cmpute the flux thrugh the inner ne. The m. field inside the lng slenid is cnstant B ni The flux thrugh a single turn in the inner slenid is Ba n Ia

26 The ttal flux is N times the flux thrugh a single turn tt n l n n li 1 1 a This is als the flux a current I in the shrt slenid wuld put thrugh the lng ne. The mutual inductance in bth cases is, frm Eq.(17) M a nn1l I Example 7.1 Cnsider the RL circuit shwn. Explain the behavir f the current in the circuit as yu cnnect r discnnect the battery. 1 S R L Slutin: Thrwing the switch t psitin 1 and applying Kirchhff s rule t the circuit we get di IR L 0 The slutin f the differential equatin given in abve is I I max1 t e

27 with I max R and L R Is the time cnstant f the RL circuit If the battery is suddenly remved, by thrwing the switch t pint in the circuit and applying Kirchhff's rule again we get di IR L 0 I e R t The tw functins are pltted as shwn. It is clear that the current takes sme time t reach its final value. This is because f the back emf induced in the circuit due t the self inductance f the circuit. Had there been n inductance in the circuit, the current have went immediately t its final value. In practice every circuit has sme self inductance and the current appraches its final value asympttically. I cnnecting the battery discnnecting the battery t

28 Energy in Magnetic Fields The wrk dne n a small charge dq, against the back emf, in ne trip arund the circuit is -dq. The ttal wrk dne per unit time is dw But frm Eq.(14) we have dq I L di dw dq di IL Starting with zer current and building it up t a final value I, the ttal wrk dne is W L I 0 IdI It is knwn that the flux thrugh the circuit is given by B da 1 LI A da A. dl (4) Remember that the flux thrugh the circuit is LI then the last equatin gives LI A. dl

29 Substituting back in Eq.(4) we get But I J. da W 1 I A. dl (5) Eq.(5) can be generalized t be written as W 1 A J d (6) Using the differential frm f Ampere s law, we get W 1 (7) A B d A B B A A B A B B B A B Using the rule Eq.(7) becmes W 1 B d A B d Using the divergence therem fr the last integral we get

30 1 W B d If we integrate ver all the space, B and A g t zer leaving A B da (8) W 1 B all space d (9) Example 7.13 A lng caxial cable carries current I that flws dwn the surface f the inner cylinder, radius a, and back alng the uter cylinder, radius b. Find the magnetic energy stred in a sectin f length l. Slutin: Frm Ampere s law the m. field between the cylinders is B I ˆ s W 1 I s sdsddz

31 But the m. filed is zer except fr the regin between the cylinders b a l b a s ds l I sdsdz s I W 4 0 a b l I W ln 4 Cmparing this result with Eq.(4) we can find the self inductance f the cable as a b l L ln

32 Maxwell s Equatins Let us recall all the laws that specify the divergence and the curl f the electric and the magnetic fields:. ( i) E ( Gauss' s law) ( ii) B 0 ( n name) B ( iii) E ( Faraday' s law) t ( iv) B J r ( Ampere ' s law ) It is knwn that the divergence f a curl is zer. Let us check this rule t frmula (iii) E B t B t 0 ( frm frmula( ii)) The rule is satisfied (the divergence f the curl is abslutely zer)

33 Let us nw check this rule t frmula (iv) B Jr Frm the cntinuity equatin it clear that the R.H.S f the last equatin is nt always zer. It is zer nly fr steady current (magnetstatics). This means that Ampere s law in its frm given by Eq.( iv) is nt always zer. As anther example t shwn that Ampere s law fails fr nn-steady current, cnsider the prcess f charging a capacitr. Applying the integral frm f Ampere s law t the clsed path P we have B dl I enc But I enc is the current piercing the surface bunded by the Amperian lp. Here the lp bunds tw surface S 1 and S. The current punctures S 1 is I, while there is n current passing thrugh S, i.e., I enc =I fr S 1 and I enc =0 fr S. Thus we have a cntradictry situatin.

34 T slve this cntradictin Maxwell starts frm the cntinuity equatin Using Gauss s law: J r J r t E E E t t (30) S Maxwell pstulated an additinal term t the right side f Ampere s law, i.e., E B J t E B dl Ienc ds (3) t r (31) With this mdificatin it is clear nw if ne takes the divergence f Eq.(31) and taking int cnsideratin Eq.(3) he will nt encunter any cntradictin. Apart frm amending the defect in Ampere s law, this mdificatin reveal that: A changing electric field induces a magnetic field. This additinal term is called the displacement current and given by

35 J d E t r (33) Returning t the capacitr example we knw that inside the capacitr we have Q E A E Q I 1 t A t A Nw fr the flat surface, E=0 and I enc =I. Fr the curved surface I enc =0, but E t ds I S we get the same answer fr either surface, thugh in the first case it cmes frm the genuine current while in the secnd case frm the displacement current. Maxwell Equatin nw can be written as ( i) E ( Gauss' s law) ( ii) B 0 ( n name) B ( iii) E ( Faraday' s law) t E ( iv) B J r ( Ampere' s law t )

36 Maxwell s Equatins in Matter When dealing with plarized and magnetized materials there will be accumulatins f bund charge and current ver which yu can t cntrl. Maxwell s equatins shuld be rewritten such that it refers nly t the free charges and currents, which we cntrl directly. We have learned that the plarizatin prduces a bund charge density given by b P (34) Similarly, the magnetizatin prduces a bund charge density given by J b M (35) In nn-static case a change in P invlves a flw f bund charges (call it plarizatin current, Jp) which must be added t the ttal current. Cnsider a tiny blck f plarized material. This plarizatin intrduces a charge density b =P at the ends f the blck. If P changes the charge changes accrdingly giving a net current given by

37 da t P da t di b And fr the current density we have (36) t P J P T check that Eq.(36) is cnsistent with the cntinuity equatin we have t t P t P J b P T cnclude we say that the ttal charge density is the sum f tw parts: (37) P f b f And the current density is the sum f three parts: (38) t P M J J J J J f P b f

38 Gauss s law is written nw as P E f 1 (39) D f (40) P E D with f P E Ampere law (with Maxwell s term) is written nw as t E t P M J B f t P E J M B f (41) t D J H f (4) M B H with

39 In terms f free charges and currents, Maxwell s equatins read ( i) D f ( iii) E B t ( ii) B 0 ( iv) H J f D t (43) Fr linear media we have P ee and M mh B s D E and H Eq.(33) can be generalized t written as (45) (44) J d D t (46)

40 Bundary Cnditins. Maxwell s equatins can be written in integral frm as ( i) ( ii) Applying frmula (i) t the pillbx shwn we btain D S S D ds B ds D 0 1n n f ( iii) ( iv) Let us study the B.C n the interface between tw media that carries charge density r current density K. (48) Applying frmula (ii) t the same pillbx we btain B n B 1 n 0 f C (49) d E dl C d H dl J f S S B ds D ds n 1 n (47) Regin I Regin II

41 Nw applying frmula (iii) t the clsed lp shwn, we get ) (51 ˆ 1 n K H H f t S ds B d l E l E 1 But as the wih f the lp is negligible s is the m. flux thrugh it (50) 0 1 t E t E Applying frmula (iv) t the same clsed lp shwn and nting again that the e. flux is zer, we get Fr linear media eqs. (48-51) can be written as (5) ˆ ) ( 0 ) ( 0 ) ( ) ( n K B B iv B B ii E E iii E E i f t n n t t f n n