dt d Chapter 30: 1-Faraday s Law of induction (induced EMF) Chapter 30: 1-Faraday s Law of induction (induced Electromotive Force)

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1 Chaptr 3: 1-Faraday s aw of induction (inducd ctromotiv Forc) Variab (incrasing) Constant Variab (dcrasing) whn a magnt is movd nar a wir oop of ara A, currnt fows through that wir without any battris! ; moving a magnt mans changing th -fid through th oop ara th magntic fux changing th currnt dirction cratd so that to oppos (يعاآس) th chang in th fux Chaptr 3: 1-Faraday s aw of induction (inducd MF) Th nd dfcts momntariy whn th switch is cosd; initiay (th instant of cosing switch), th magntic fid incrass during a crtain vry sma priod of tim (th nd dfcts) unti -fid bcoms stady (th nd gos back to zro) Chaptr 3: 1-Faraday s aw of induction (inducd MF) Th ctromotiv forc (mf),ε, inducd in a circuit is dircty proportiona to th tim rat of chang of th magntic fux through th circuit. Th inducd mf For N oops, = = N whr, -v sign is du to th inducd fux (from inducd currnt) that opposing th chang in th fux r r =. da Chaptr 3: 1-Faraday s aw of induction (inducd MF) = d = ( Acosθ ) To induc an mf w can chang ovr tim if th foowing: th magnitud of th ara ncosd by th oop th ang btwn and th norma to th ara any combination of th abov

2 Chaptr 3: 1-Faraday s aw of induction (MF in a coi) x: you hav a -fid to th pan. f th -fid changs inary from to.5 T in.8 s. What is th magnitud of th inducd mf in th coi whi th fid is changing? Th inducd mf is 18 cm -fid squar coi N= turns A=(.18 m)² =.34 m At t = s, = (no magntic fid) At t =.8 s, =.A = Acos = A= (.5 T)(.34 m²) =.16 T.m² Chaptr 3: 1-Faraday s aw of induction (xponntiay Dcaying Magntic Fid) A wir oop of ara A in a at - fid ( t) = max (into pag) and A = Acos = A Find th inducd mf in th oop as a function of tim. = ( t) = A at max = = Amax = aamax at d at Chaptr 3: -Faraday s aw of induction (Motiona MF) * Whn changs and th conductor is stationary inducd mf (xamps bfor) *Whn a conductor movs through a constant magntic fid motiona mf, As th wir movs, r r r = qv F ctrons movs through th conductor bcaus of F chargs insid th conductor movs in th dirction of F (on v chargs) and avs positiv chargs bhind. Chaptr 3: -Faraday s aw of induction (Motiona MF ΔV) * Whn changs and th conductor is stationary inducd mf (xamps bfor) *Whn a conductor movs through a constant magntic fid motiona mf, As th wir movs, r r r qv F = chargs insid th conductor movs in th dirction of F (on v chargs) and avs positiv chargs bhind. As thy accumuat on th bottom, an inducd ctric fid is st up insid buid in potntia V. at quiibrium F = F qv = q = v Votag drop V across th conductor du to accumuatd +v and v chargs on conductor sids is V = = v a potntia diffrnc is maintaind btwn th nds of th conductor as ong as th conductor continus to mov through th uniform magntic fid.

3 Chaptr 3: -Faraday s aw of induction (Motiona MF in a circuit) F du to currnt in th wir f th moving conductor is part of a cosd conducting path (cosd circuit of rsistanc ). Th ara ncosd by th circuit is A = x = A = x d d dx = = ( x) = v = v = = f th bar is movd with constant vocity, F = F app = v P = F app v = ( ) v = = Chaptr 3: -Faraday s aw of induction (Magntic Forc on a Siding ar) conducting bar movs on two frictionss para rais in a uniform magntic fid dirctd into th pag. Using Nwton s aws, w can find th vocity of th bar as a function of tim for a bar having initia vocity v i (witch is givn by F aap bfor it rmovd). dv v Fx = F = ma = m =, but = dv dv m v v m = = Th bar has a mass, m, and an initia vocity v i F = - Sign indicat th forc is to ft Vocity wi dcras unti stop according to v v t τ = i whr m τ = Chaptr 3: 3-nz s aw Th poarity of th inducd mf is such that it tnds to produc a currnt that crats a magntic fux to oppos (يعاآس) th chang in magntic fux through th ara ncosd by th currnt oop. As th bar is sid to th right, th fux through th oop incrass. This inducs an mf that wi rsut in an opposing fux. Sinc th xtrna fid is into th scrn, th inducd fid has to b out of th scrn. Which mans a countrcockwis currnt Chaptr 3: 3-nz s aw (concpt qustion) Figur shows a circuar oop of wir faing toward a wir carrying a currnt to th ft. What is th dirction of th inducd currnt in th oop of wir? (a) cockwis (b) countrcockwis (c) zro (d) impossib to dtrmin Sinc th oop of wir is approaching th straight wir - fid (into pag) incras fux incrass nducd mf in, wir oop, that produc currnt in a dirction to crat -fid to rduc magntic fux (rduc -fid) it must produc -fid out from pag hnc currnt must b countrcockwis (answr )

4 Chaptr 3: 3-nz s aw (Moving Magnt and Stationary Coi) ight moving magnt incrass fux through th oop. t inducs a currnt that crats it own magntic fid to oppos th fux incras. ft moving magnt dcrass fux through th oop. t inducs a currnt that crats it own magntic fid to oppos th fux dcras. Chaptr 3: 3-nz s aw (appications of ns s aw) Whn th switch is opnd again, this tim fux dcrass, so a currnt in th opposit dirction wi b inducd to countr act this dcras. Whn th switch is cosd, th fux gos from zro to a finit vau in th dirction shown. To countract this fux, th inducd currnt in th ring has to crat a fid in th opposit dirction. Aftr a fw sconds, sinc thr is no chang in th fux, no currnt fows. Chaptr 3: 3-nz s aw (oop moving through a -fid) A rctanguar mtaic oop of dimnsions and w and rsistanc movs with constant spd v to th right, as shown. Th oop passs through a uniform - fid dirctd into th pag. Pot as functions of x (A) th magntic fux through th ara ncosd by th oop, () th inducd motiona mf, and (C) th xtrna appid forc ncssary to countr th magntic forc and kp v constant. Chaptr 3: 4-nducd MF and ctric Fids Changing Magntic Fux through a conducting oop nducd mf Hnc, f th magntic fux changs with tim = A conducting oop of radius r with changs with tim nducd -Fid is cratd in th conductor in th currnt dirction Th work for moving a charg q by -fid pacd at r ovr on cyc ( r) nducd -fid W = q = F π q = q( πr ) ( πr) = = d d 1 1 = = ( πr ) πr πr r d = For a circuar oop

5 Chaptr 3: 4-nducd MF and ctric Fids n gnra, f w hav a changing fux w wi hav an inducd mf = A conducting oop of radius r with changs with tim inducd -fid. Th -fid at any point can b cacuatd by constructing a cosd oop and using th in intgra r r =.ds = For any cosd path Gnra Form of Faraday s aw Whr is th fux for th ara ncosd by th oop Th inducd -fid arising from changing magntic fux is not consrvativ, bcaus if it is consrvativ w r wi hav.ds r = x: ctric Fid nducd by a Changing Magntic Fid in a Sonoid da. ds = = d d ( t) = max cosωt. ds = ( π ) = π dµ n( t) ( πr ) = π ( πr) = π µ n ω sinωt A ong sonoid of radius has n turns of wir pr unit ngth and tim varying currnt = µ n ( t) = µ n max cosωt Find outsid th sonoid at a distanc r > and insid th sonoid at r < max µ nmaxω = sinωt r> r = πr d max µ n maxω = r sinωt r< ( πr) = πr = πr µ n ω sinωt Summary Probms *Faraday s aw of induction inducd mf *conducting bar of ngth movs at a vocity v through a magntic fid motiona mf inducd in th bar *Th appid forc to kp constant v is F Whr is rsistanc connctd to moving bar app = F v = = *nz s aw stats that th inducd currnt and inducd mf in a conductor ar in such a dirction as to st up a magntic fid that opposs th chang in th magntic fux *A gnra form of Faraday s aw of induction is Which impis inducd -fid is not consrvativ A = N. 5

6 3 First w nd to find th chang in th of th sonoid A Which is th sam fux that chang in th coi θ ut, = 4 5 a) F = q ind From faraday's aw r r da ε =.ds = = da d ( πr ) = = A b) F= whn = whn d/ =

7 6 nductanc Sf-nductanc Circuits nrgy in a Magntic Fid r r ε =.ds = = da 3: 5-Sf-nductanc Currnt in a Coi Whn th switch is cosd, th battry (sourc mf) starts pushing ctrons around th circuit. Th currnt starts to ris, crating an incrasing magntic fux through th circuit. This incrasing fux crats an inducd mf in th circuit. Th inducd mf wi crat a fux to oppos th incrasing fux. Th dirction of this inducd mf wi b opposit th sourc mf. This rsuts in a gradua incras in th currnt rathr than an instantanous on (taks ongr tim). Th inducd mf is cad th sf-inducd mf or back mf. Whn changs, an mf is inducd in th coi. f is incrasing (and thrfor incrasing th fux through th coi), thn th inducd mf wi st up a magntic fid to oppos th incras in th magntic fux in th dirction shown. f is dcrasing, thn th inducd mf wi st up a magntic fid to oppos th dcras in th magntic fux.

8 nductanc x: nductanc of a Sonoid From Faraday s aw, th sf-inducd mf is = N Find th inductanc of a uniformy wound ong sonoid having N turns and ngth A ut, and So, a sf-inducd mf is aways proportiona to th tim rat of chang of th currnt. N = = = N = whr is a proportionaity constant cad th inductanc of th coi From abov quation inductanc of a cosy spacd coi of N turns Aso inductanc can b writtn as (S.. unit is Hnry = H = V.s/A) N = µ n = µ NA = A = µ N = = µ N A = µ n V N = n and V = A N x: 3.. Cacuating nductanc and mf air-cor sonoid. N = 3 turns, = 5. cm and its cross-sctiona ara is A = 4. cm. (A) Cacuat th inductanc () Cacuat th sf-inducd mf if is dcrasing at th rat of 5. A/s. Soution:- (A) () 3.. Circuits (& in sris) x = dx = At t =, = x = x + dx = dx = x t x dx = x x n x x sf-inducd mf = = t Appy K = x t = x = = t t / ( 1 ) = t /τ ( 1 ) τ = Tim constant of circuit

9 3.. Circuits (& in sris) Th tim rat of chang of th currnt is th drivativ of Hnc at th instant of cosing swich, t /τ = 1 Wi qua to zro Hnc, th votag across th rsistor wi qua to zro and votag across th inductor wi qua to th mf of th battry aacording to K = () = = = To summariz. At t = is ik opn switch = At t = is ik cosd switch = ε/ t =, th currnt in th circuit is ( ) Th chang in th currnt 3.3. nrgy in a Magntic Fid = = + attry Powr sistor Powr w wi hav th powr nductor Powr (rat of nrgy divrd to th inductor For nrgy dnsity = µ n A (nrgy pr unit voum), considr a sonoid = µ n voum du nductor Powr= = Th nrgy stord in th inductor U = du = = 1 U = 1 1 U = = µ n A A = µ n µ U u = = A µ Probms = 4 ma =.4 A, N = 45, r = 7.5 mm =.75 m, and = 1 cm =.1 m. N/ = (πr²)

10 (a) At tim t, whr Aftr ong tim, currnt fina vau = (b) Currnt maximum vau, whn th coi ffct vanish (aftr ong tim) w can considr a circuit with ony (ohms aw) max = = ε/, O (a) Whn rach its haf maximum vau = ε/, th nrgy stord