1.2 Faraday s law A changing magnetic field induces an electric field. Their relation is given by:

Transcription

1 Elctromagntic Induction. Lorntz forc on moving charg Point charg moving at vlocity v, F qv B () For a sction of lctric currnt I in a thin wir dl is Idl, th forc is df Idl B () Elctromotiv forc f s any forc on a chargd particl othr than that du to othr chargs. b Elctromotanc f s dl. Convntionally calld mf, but it is not rally a forc. Th a Chins translation 電動勢, is mor appropriat. b Th mf in a sction of wir du to B-fild is ( v B) dl (3) Exampl- a A rod of lngth L is sliding down a slop in a uniform B- fild at spd v. Find mf in th rod. Solution: Th amplitud of ( v B) is vbcosθ and its dirction is pointing straight out of th papr plan, paralll to th lngth of th rod. So mf = LvBcosθ Exampl-A B Th sam rod is spinning at angular spd ω in th B-fild around on of its nd. Th angular momntum is paralll to th B-fild. Find mf btwn th two nds of th rod. Solution: B B Tak a small lngth of th rod dr at distanc r from th fixd nd. Th spd of it is ωr so its L B mf is Bωrdr. Total mf along th whol lngth is rdr BL ans.. Faraday s law A changing magntic fild inducs an lctric fild. Thir rlation is givn by: B d d E dl ds B ds t dt (4), dt l whr S is th surfac nclosd by th closd lin l. S S

2 B In diffrntial form: E (5) t Th total lctric fild now consists of two kind s of lctric filds. On is du to chargs, and th othr is du to chang B-fild. Th dirction of th inducd E-fild is such that it would gnrat an lctric currnt to countr th chang of th B-fild. It can b shown that whn a wir loop is moving rlativ to th B-fild, Eqs. (3) and (4) ar quivalnt. Howvr, Eq. (4) is mor gnral, and works vn whn thr is no rlativ motion, or th wir loop can b at th location whr thr is no B-fild. Exampl- Th currnt in a long solnoid (N turns pr unit lngth) is dcrasing linarly with tim t, I = I - kt. Find th inducd E-fild. Solution By symmtry argumnt, w can s that th B-fild is along th axis of th solnoid. Tak an Ampr s loop- outsid th solnoid, w find that th B-fild is constant. But far away th B-fild must b zro, so th B-fild outsid is zro. Now tak loop-, BL = NIL, so B = N(I kt) Not that outsid th solnoid thr is no B-fild, but th changing B-fild inducs an E-fild so that if th circular wir carris charg, it will spin. Not also that Eq. (4) is of th sam in mathmatical B form as Ampr s law, whn is viwd as lctric currnt t and th inducd E-fild as th magntic fild. Applying Eq. (4), th lft hand sid = πre, whr r is th radius of th loop, and th right hand sid = NkA, whr A is th cross sction ara of th kna solnoid. So E. Ans. r B B.3 Slf and mutual inductanc (of coils) Considr a coil (a loop of wir) carrying currnt I. If I changs with tim, its magntic fild, and thrfor th magntic flux Φ through th coil (S Eq. (4)) will chang, inducing an mf. As th fild (hnc flux) is proportional to I, w can dfin a quantity L which dpnds only on th gomtrics of th coil, calld slf-inductanc, such that Φ = LI (6), d di bcaus L (6A). dt dt Th mutual inductanc btwn two coils, M and M, ar similarly dfind. Φ = M I, and Φ = M I. (Φ is th magntic flux through coil- du to th currnt I in coil-).

3 3 It can b shown that M = M, and thy dpnd only on th coils gomtrics. Exampl-3 A vry small coil of radius a is placd at distanc z abov a larg coil of radius b along its axis, as shown in th figur. If thr is currnt I flowing in th small loop, find th magntic flux through th big loop. Solution: Th small loop can b tratd as a magntic dipol and its magntic fild can b xprssd xactly. Th flux through th big loop can thn b intgratd out. But thr is a simplr approach by using M = M. So lt instad th sam currnt I flow in th larg loop, and find th magntic fild at th cntr of th small loop, and trat such B-fild as uniform ovr th ntir loop. a z b Tak a diagonal pair of small sctions of wir on th big loop, as shown, it is asy to s that th combind B-fild is pointing along th vrtical dirction, and th amplitud is I dl b db. 4 ( b z ) b z Evrything in th xprssion is constant so th intgral ovr dl is l = b. I b So B, Ba 3 / ( b z ), and a b M. Ans. 3 / ( b z ) Exampl-4 Th magntic flux through a coil of rsistanc R changs from Φ to Φ, Find th total charg passing through any cross sction of th wir. Solution: From Eq. (6) and (6A) d dq RI R, so Q = (Φ - Φ )/R. ans. dt dt Maxwll s Equations. Maxwll s displacmnt currnt What will th B-fild b if J is changing with tim? So far w assum that th currnt I in wirs chang with tim but I is still th sam along th wir so thr is no charg piling up anywhr. In mor prcis trms, w assumd that J still holds. Lt us look at th Ampr s law again, B dl J ds S I nc, or B J (7) Thr ar infinit numbr of surfacs bound by th loop. Any surfac bound by th loop will do as long as J, bcaus th currnt through any on will b th sam as th othrs. Th gnral charg consrvation implis

4 4 J t (8). J implis that thr is charg accumulation somwhr,. t Not that ( B) J, so th Ampr s law must b modifid whn J. Maxwll introducd a scond trm to Eq. (7), E B J (9). t It is asy to show that now th divrgnc ( ) of both sids of Eq. (9) is always zro. This nw trm in Eq. (9) is calld th Maxwll s displacmnt currnt. It srvs as anothr sourc to gnrat B-fild.. Th Maxwll s Equations for E&M filds (i) E (ii) B (iii) B E t (iv) B J E t In a mdium B D E, H (). (i) D (ii) B B (iii) E t (iv) H J f D t (M ).3 Enrgy of th EM filds Poynting s vctor is th nrgy flow dnsity (watts/m ) In vacuum S E B () In mdium S E H (A) Enrgy dnsity W ( E D B H) () Enrgy is stord in th fild.

5 5 Exampl-4 Currnt flows in a sction of conductor wir. Solution: Th wir is nutral so th E-fild is uniform insid th wir (V/L) and zro outsid, whr V is th voltag diffrnc btwn th two nd surfacs, and L is th distanc btwn thm. Th B- fild can b found using Ampr s law.

6 .4 Boundary conditions 6

7 .4 Boundary conditions (continu..) 7

8 8.5 Plan wav solutions of M s Equations (EM wavs) i( kr t ) For plan wavs, th filds ar of th form E or B A, whr k is th wavvctor of th wav, and ω is th frquncy. Thy ar calld plan wavs bcaus th qual-phas surfac, k r k x x k y y k z z constant forms a plan. W now look for solutions of Eq. M abov. Not that for plan wavs, E ik E, E ik E, and E ie. t In short, ik, and i. t In vacuum, thr is no currnt or charg. Th M s quations bcom k E (3a), k B (3b) k E B (3c), k B E (3d) Th quations abov imply that k E B. Choos k kz, E Ex, B By, whr E and B ar constants, and apply k to both sids of Eq. (3c) and put th rsult in Eq. (3d), w gt k (4), c whr c is th spd of EM-wav propagation in vacuum. (c = m/s) Equation (4) mans that for a givn frquncy ω, which dscribs th tim variation of th EM filds, th spatial variation paramtr k is fixd, as givn by Eq. (4). Th dpndnc of k on ω is calld th disprsion rlation, and is dtrmind by th mdium in which EM wavs propagat. In non-conducting mdium with no fr charg and currnt, th M s quations bcom (aftr rplacing D and H by E and B using Eq. ()) k E (5a), k B (5b) k E B (5c), k B E (5d). So th rsults in vacuum still apply xcpt that n k (6). c Th spd of wav propagation is c/n, whr n is th rfractiv indx in optics. Notic that th k in Eq. (6) is diffrnt from that in Eq. (4), for th sam frquncy ω. This again shows that dpndnc of k on ω is dtrmind by th mdium in which EM wavs propagat.

9 9 Th nrgy flow, Poynting s vctor S, of th EM wav, using Eq. (), is (HW-3). S is what w usually rfr to as th light intnsity. In conducting mdium E, whil vrything ls rmains th sam as in a non-conducting mdium (Th fr J f charg dnsity du to th fr currnt can b shown to dcay xponntially with tim, so can b takn as zro if w wait long nough). Equation (5d) bcoms ik B E ie i( i ) E (7). Th solution is thn k i k k k k is ral, so k is complx Solving Eq. (8) k R k I k k R I k, I c i kzt ki z i krzt E E E R i k kr ik I (8) c R (9) Th E-wav is () k z Attnuation factor I, skin dpth dp (pntration dpth). Good conductors: k c dp i ( i). Poor conductors: thn ikzt For E E k I dp I R () I H i ik R I E E k i kz t so E & H ar / 4 out of phas to ach othr. (good conductors, k k ) R I Enrgy flow s R{ E H so s attnuats } k R E ki z 3 (),

10 krki E k z I s E E-M wav is loosing nrgy to mdia. ki z ki z E (3) Enrgy loss = hat Hat = E J R( E J k z E E E I ) R( ) s (4) Plasma (nutral ionizd gas) Eithr on typ of charg is mobil, fr lctrons in idal conductors, or both can mov, lik in th uppr atmosphr. Usually, if lctrons ar prsnt, thy will dominat bcaus of thir much lightr mass (/84 of a Hydrogn nuclus) than any othr ions. Considr th cas with on typ of mobil charg particls with charg q and mass m, and numbr dnsity N pr unit volum. Th forc du to EM filds on a particl is f q( E v B) (5). But B = E/c in vacuum so if v << c th B-fild forc can b nglctd. dv Nwton s law: qe m imv (6) dt iq N Currnt dnsity of plasma J p qnv E (7) m On can s in Eq. (7) that th particls with lightr mass contribut mor to th currnt. Applying Ampr s law, iq N q N ik B i E E i( ) E (8) m m Hr is du to th polarization of th particls, such as th immobil ions. Equation (8) is similar to th on in insulating (dilctric) mdium xcpt that now th ffctiv dilctric constant is givn by q N p ( ) ( ) ( ) (9), m is th plasma charactristic frquncy. q N whr p (3) m n ( ) ( ) (3) Th dilctric constant, and rfractiv indx, is now a function of frquncy. So w should rfr it as dilctric function, instad of dilctric constant. Whn th frquncy of th EM-wav is blow, th dilctric function () is ngativ, so th rfractiv indx n () is imaginary, and k = nω/c is also imaginary. Th EM-wav cannot propagat. Whn ω > p th plasma bhavs lik normal dilctric mdium, xcpt that n () may b lss than. p