hrb==.- trn't Pltysics 48 (Fail 201I) Chupter 30: Induction and fnductunce Reading: pages O utline:

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1 Pltysics 48 (Fil 201I) Chupter 30: Induction nd fnductunce hrb==.- trn't -7-- "For every minute you re ngry, you lose 60 seconds ofhppiness. " - Rlph Wldo Emerson "Consider how much more often you suffer f'om your nger nd griej thn.from lhose very things.for which you re ngry nd grieved. " - Mrcus Aurelius "Holding on to nger is like grsping hol col with the intent of throwing it t someone else; you re the one who gets burned. " - Buddh Reding: pges O utline: \-/ = introduction Frdy's Lw of induction Lenz's Lw = emf induced in moving conductor = induced electric frelds Frdy's Lw = inductors nd inductnce inductnce of soienoid self-induction = RL circuits + energy stored in mgnetic field energy density of mgnetic field + mutul induction (red on your own) Problem Solving Techniques You should know how to compute the flux through given re. In some cses, the mgnetic field is uniform nd you simpiy multiply the perpendiculr component by the re. In other cses, the fteld is not uniform nd you must cny out n integrtion ofthe perpendiculr component over the re. In some problems, the mgnetic field is given while in others the current tht produces the field is given nd you must use wht you lerned in the lst chpter to write n expression for it s function ofposition. For exmple, the field my be produced by current in long stright wire. '- Once n expression for the mgnetic flux is found, you differentite it with respect to time to find the emf round the boundry of tlre region. Look crefully t the expression for the flux nd decide wht is chnging with time. It might be the mgnitude of the mgnetic field, the mgnitude of the re of the region, or the orienttion of the re with respect to the fie1d. In some problems, the boundry ofthe region is conductor nd you re sked to find the current in it. First, note l1 the emfs round the boundry. One of them is the emf ssocited with the chnging

2 mgnetic flux but there my be others, produced by btteries or genertors, for exmple. Add them with their correct signs nd divide the totl by tjre resistnce in the loop. Obviously, you must be very creful bout the signs ofthe vrious emfs here. Some problems sk for the electric field ssocited with chilnging mgnetic field. Use Frdy's lw in the forrn P E ' i = -Aolt. ll of the problems hve cylindricl symmetry, with the electric field lines forming circles round the cylinder xis. For them, you integrte the tngentil component of the electric lield round one of rhe field lines. with result I f '6i :2rrE.where r is the rdius ofthe circle. Find the rte ofchnge ofthe mgnetic flux through the circle nd solve the Frdy's 1w eqution for E. Some problems sk you to use the definition of self-inductnce (L : N@/i) to compute one of the quntities tht pper in it. To cny out this tsk, you my need to compute the mgnetic flux (D of the mgnetic field, perhps by crying out n integrtion.-other problems del with the emf generted by n inductor when the current chnges. Use B: -L di/dt. Some problems del with RZ series circuits. If source of emf, such s bttery, is in the circuit nd the cunent is 0 t time I : 0 (becuse, for exmple, switch is closed then), the cunent is given by t() = (E/x) (t - n-'f "),where the inductive time constnt is 11 : L/R.If the current is lo u : 0. -llt, t\e '' nd no source of emf is in the circuit, clculte You my be sked to compute the potentil difference cross the resistor (ir) or cross the inductor (L dildt). Be sure you cn tell which end is t the higher potentil. You my lso be sked for the rte with which the source of emf is supplying energy (l' ), the rte t which the resistor is dissipting energy (;2R), or the rte t which the inductor is storing energy (Li dildt).lf you re sked for the I energy densities use ur : i so* nd u : El2po. Some problems del with the clcultion of mutul inductnce. Here you ssume curent ln one circuit nd compute the mgnetic flux it produces in nother circuit. Questions nd Exhple Problems from Chpter 30 Question I In the figure below, long stright wire with current i psses (without touching) three rectngulr wire loops with edge lengths L, 1.5L, nd 2L. The loops re widely spced (so s to not ffect one nother). Loops 1 nd 3 re symmetric bout the long wire. Rnk the loops ccording to the size of the current induced in them if current i is () constnt nd (b) incresing, gretest first. (qj W (b)

3 Question 2 If the circult conductor in the figure below undergoes therml expnsion while it is in uniform mgnetic field, current wiil be induced clockwise round it. Is the mgnetic field directed into the pge or out of it? A;""* t o,^ta'aa ) Question 3 If the vrible resistnce R in the left-hnd circuit ofthe figure below is incresed t stedy rte, is the cunent induced in the right-hnd loop clockwise or counterclockwise? -) j-a d.{.d.4e,,s",'-g.-}o $'rr^ "t c^r-t"'r"g I dj^r^".d &^^r^s 7.tt-o-i ",*r.'$r'r*,c--o-c-&.'l*.,r" Problem 1 ln the figure below, the mgnetic flux through the loop increses ccording to the reition (De:6.0f , is in milliwebers nd t is in seconds. () Wht is the mgnitude of the emf lnduced in the loop when t = 2.0 s? (b) Is the direction of the current through R to the right or left? do = 6.o L\+?.ot ^lr"r*1",jt ^ 6r= (6'0 ry) f- ( r'o *$e) t l/r yj- = -JAt (6.or*+?.0 x) = - lz-?.d B.-* lf-] = I l),t,"1.o --z l (*'-ou) l}'(e.o)r?.o =31 t- t- t) t llt= 3l.V=3t"to-3V I 42,,*.*, Bo*,^^r*.6Y* f^?" ""A ;*-.""'V Qrr**r"- $^^,*;" y; fro'nrjt-t+ ""t't /r,.,'$ fv*'*" " 3 {.,.,}J ;h itr gr'4r. )

4 Problem 2 A squre wire loop with 2.00 m sides is perpendiculr to uniform mgnetic field, with hlf the re of the loop in the field s shown in the figure below. The loop contins n idel bttery with emf e. = 20.0 V. If the mgnitude of the field vries with time ccording to B : , with B in tesls nd t in seconds, wht re () the net emf in the circuit nd (b) the direction of the current throush the btterv? st B = tl. o'llot-(o.t?otlr) r lu/lf = - 6,x1or/s t =) -Ar;".r B''u.,,t-x,nr.*+l^.t ; d,^" of-ro..r^-nrg "*L Et;,r..., t=-rt6"/t I rr'.d- t = t.rv I,.,-,-^} -{A r-r} 14 n^qx N=1 =- %t(vns = -(dbar) P k = t/'t^ -- - (- o"r?or/,) /1 ( ].o o. ; 1. - t rrr, t - l. /1,t! g n^)l ^n,r,^c'.octv* o.!-tffi"1 cl*tl-uftt Problem 3 In the figure below, wire forms closed circulr loop, with rdius R:2.0 m nd resistnce 4.0 O. The circle is centered on long stright wire; t time t:0, the current in the long^stright wire is 5.0 A rightwrd. Therefter, the current chnges ccording to i : 5.0 A - (2.0 Ns")f. (The stright wire is insulted, so there is no electricl contct between it nd the wire of the loop.) Wht re the msnitude nd direction of the current induced in the loop t times t > 0? p-"fo ^t,6 e Vn ) -;^r* fr\, delr-t/en1 l.,t',tt,e.*r*a /o\ )&\p,p-*"5 tfi,*"g'*.,"n"rq**}ie -r,tr,!-, ttu fu^k Q"=o ))4.o8/-u!e \,D-.t> ^/o lt9\,,,u-f.{.,*r Ml o*"j d i9* q- r,rori ffo!*^1 M$" SAt s6ul!-8* t*' r" n'"*.*& 4'$9*j )Vu eu,^e"'j -iin JJ14 fu^r:f** t. e"t "l s rcx^'''^it".)^zl"-rrv;g tt'"*\

5 - t = (Lt,so A/r') t'- (ro'oa/s)t Probfem n Zn For the sitution shown in tl6 figure below, : 12.0 cm nd b : 16.0 cm. The current in the long stright wire is given by i = 4.50t' , where i is in mperes nd t is in seconds. () Find the emf in the squre loop t t: 3.00 s. (b) Wht is the direction ofthe induced current in the loop?,{-f \, 't / -\!?= B..JA = ( Fl l-,u". b 9^. / 2\rT t_" =- d6/dt = -> #* S'^ g'"ib^ A )*-P L-p,v')^i4', J,,rr,* 92 +t*, +^,w"1 P'oth;\ Ptr -I-.F j- ^ dir- b-o. C = - Jd",, (- / d't o\-q BdAc-o" = ("(n"')(ul.) -- 4'j) ( d' q /.rfc/- ' L-o -) d= -,6 '^t 2,-r xtt r- b-q Fe t_ tl- J ni / 5 In the figure below, long rectngulr condu&ifrlt66fr-,61viiltfildsistnce R, nd mss m, is hung in horizontl, uniform mgnetic field tht is directed into the pge nd tht exists only bove line. The loop is then dropped; during its fll, it ccelertes until it reches certin terminl speed v1. Ignoring ir drg, find tht terminl speed. BLv :,---- = A nnq Lt3,t t= 3""t d%tro u B v,""t-^.g 6 )^^-'^;^g F"-^Z= o ---) t LI3 -^^- 9o" = Ifrg 4-m3, foq L=':'l-- Lt3," = t' /n = k l' * lttu^t I : + (r%t) =*(t ---> fno, A --J4-- Btl' dv/w) = VY /2,\.--

6 Problem 6 The inductncq of close-pcked coil of 400 tums is 8.0 mh. Clculte the mgnetic flux though the coil when the current is 5.0 ma. N--foo L= g "o"to-fr l'= 5"o' td?a 0"-? /=N6/t--- Problem 7 A bttery is connected to series RL oircuit t time t: 0. At wht multiple of t1 will the cuuent be 0.100% less thn its equilibrium vlue? ^ / ), - t.1 ''/,'-?,/- l -,- 72" ) Ar.o4rt e --;. L = 7A L - -/A\t "-q,t^i^-0r,.-t* v o.gcq?/,1 " e/n (l_"r".) - - t/2, -- -'o'1(o"oor) ---J = f( E.o, t1-tli )'( S.o"ro-:n) o.ggq = 1- f"/'' t = e"9t.z n o/= o"ool Problem 8 Consider the RL circuit of the figure below. In terms of the bttery emf e, () wht is the selfinduced emf er when the switch hs just been closed on, nd (b) wht is es when t:2.0ey? (c) ln terms of s, when will es be just one-hlf the bttery emf e? ;= 7o(l-.t/'') 'T'=Lh "= -tdat = -t %rlt/n(l-e-"'/r)7,= - g qt/z, --) ) t,) = t e*t/" * t= 6, t: Jt r= -o l t r' ce. =) tr,l'-r (n&'", w&("';*'?"n&,")t e") "&.,1 (b) I e,) = e--^'o'"/"' = 2 c^, = u" tj5 cc) lt,) = t;t/'- --n 9/= [*/'" -, 17^ '= 6t/z' gr, ('lr) - -t/-r,- f = -(n"')1. -- $L"t)2.! t =;"b'lt::

7 Problem 9 In the figure below, the bttery is idel, e : 10 V, Rr = 5.0 O, Rz: 10 C2, nd L = 5.0 H. Switch S is closed t time t = 0. Just fterwrds, wht re ();1,@) lz, (c) the current is through the switch, (d) the potentil difference Vz cross resistor R2, (e) the potentil difference Vs cross the inductor L, nd (f) the rte of chnge di2/dfl A long time lter, wht re (g) 11, (h) t, (D r., (i) Vz, ft) V1, nd (I) dir/dt? kr^- t"dr-*4 (o.-r--.1ru""6 ) : c- EA Q- d'/'") I = - Z d'/2. (,4tv t- v ) t= pt.-74 " - F'/ () t.= - t;t/z' ---1 t.= -L J f=oe, \..\-@ (r) lz=,-ol ( i,.\4"4 -''-,ry') ( h) i^= /A, (t-;ttz's,o* t-u > t^= 1A^ = :l@ # ( i) t, = t,, i^ =.Fo{f (i) V^-- "8, = 1r,o*)(to.or) : IT-r;4 ()<) tt = - 26t/2" --E;;l,,+ t= "' ( l) d,/r* = /0, Lc/q (t-;'/'')] t-i. -- t: -g/f?1".).'/.- -, * o'?'lt= o[.^*e="- = Td*o'

8 Problem 10 A coil with n inductnce of 2.0 H nd resistnce of 10 O is suddenly connected to n idel bttery with e: i00 V. 41 t = 0.10 s fter the connection is mde, wht is the rte t which () energy is, being stored in the mgnetic field, (b) therml.energy is ppering in the resistnce, nd (c) energy is being delivered by the bttery? L=\.oH 4 = tolt e = lodv () P=;,:tLd4".+h, ( "t?= Av"/t= d7417trt;') = L; t,zt ) \,.'. -th" p = L leto tt-;'h,)f (F-/,d"') = e74 0-;+t., (d'ft) -----> o) {: b.los ( b) +ol q'\).lfq, P=itR = LtlnTtit/ttl'R P = E7n ( l-.oz*'r -.: * t'o'ios L.) &* 2,tu)A t P = f t = L /u (\'d''4r7 Z P = E7A (l:*/2") -- * t= o,ro5j P = :.q," to"lv /@'" f)f) Yum2 = Y"y^x^*?,;-r^", Problem 11 Prove tht, fter switch S in the figure to the right hs been thrown from to b, ll the energy stored in the inductor will ultimtely pper s therml energy in the resistor. *tl)"rat',ht- = ; (d"/t)?=;gl JE/t -- L; J AL ----) 2- = r/^ti\ ) ^u,u.,"r- nllmr"rt )"1\*'Y\ 6xl"* t i"'hh!fu L. w!^t\ 4r-^riilA,'.4 oj (E= t/t;,') I t= o)\]'fr,'ffi;. /:= (Flt =(t^ajt-( z-))j tr -- * L: A l'='/^ i^!''l f o r. E t/-;^27 /^uo,/) t'l -- l/^i"^a i=lo;z/z' ^[ r,;t'-l^a J, -= [%1 = I/1L"'"^ i.: A I ; =Z(t-;'e) )\ o6t= t"z,= 7" 9 6w' l\^ b^.l^jyl.m+.1,f4) ^h\,.= / ( o & - \x/zl o tl.l- ^J{.r^rJWL - -gr\t\q4 en@ ---:,. /fi."a wr nr'ol"u;tl7r