A, Electromagnetic Fields Final Exam December 14, 2001 Solution
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1 , Electrognetic Fiels Finl Ex Deceer 14, 2001 Solution 1. e9.8. In chpter9.proles.extr.two loops, e of thin wire crry equl n opposite currents s shown in the figure elow. The rius of ech loop is n the istnce etween the loops is 2. The loops re in free spce n plce prllel to ech other, so tht they re oth perpeniculr to the xis -C s shown (tht is, like two wheels on n xle). Clculte:. The gnetic flux ensity t point (iwy etween the two loops).. The gnetic flux ensity t point (t the center of loop No. 2). c. The gnetic flux ensity t point C if >>. Note: In ll three cses give oth irection n gnitue of the gnetic flux ensity. I (1) (2) I 2 C Solution: Clculte the gnetic flux ensity on the xis, ue to loop. This ws clculte in exple 8.3. The solution t ny point on the xis is then the superposition of the fiels of the two loops. The ltter is given s: H = I h 2 3/2 = µ 0 I h 2 3/2 The solution is written here s gnitues ecuse the gnetic flux ensity epens on the irection of the current. We will use the right hn rule to ientify these irections.. =0. The loops re ienticl n t equl istnce fro. The gnetic flux ensities they prouce t cncel ech other.. Loop (1), prouces gnetic flux ensity pointing to the right (this is tken s positive). With h=2: = µ 0 I h 2 3/2 = µ 0 I /2 T µ = 0 I /2 T Loop (2) prouces gnetic flux ensity pointing to the left (negtive). With h=0: = µ 0I /2 T
2 The totl fiel t is now: t = = µ 0 I µ 0I 2 2 3/2 2 = µ 0I 2 3/ = 0.455µ 0I t = µ 0 I µ 0I 2 2 3/2 2 = 0.455µ 0I 2 3/2 T The gnetic flux ensity points to the left. c. t lrge istnces the gnetic fiel intensities of the two loops is pproxitely the se n in opposite irections. Thus, C = e7.10. In chpter7.proles.extr. resistor is e in the for of conucting circulr wsher with inner rius equl to n outer rius equl to. The thickness of the wsher is equl to. If the conuctivity of the conuctor is σ, clculte the totl resistnce of the wsher:. If the source is connecte etween the inner n outer surfces (s shown in ()).. If the source is connecte etween the two flt surfces s in (). σ σ. Solution:. R = I I = R J = σe E = J σ J = R2πr E = Rσ2πr = E.l = Rσ2π 1 r r Thus the resistnce is
3 1 R = σ2π 1 r r 1 = σ2π ln. S = π( 2 2 ) I = R E = σrπ( 2 2 ) n is unifor. Thus, = E = σrπ( 2 2 ) n the resistnce is R = 1 σ2π ln Ω J = Rπ( 2 2 ) R = σπ( 2 2 ) Ω 3. e8.4. in chpter8.proles.extr). cylinricl conuctor of rius =10 is e of copper with conuctivity σ= S/. The conuctor is use for one conuctor of power line tht is =1000 k long. ecuse of the current in the conuctor, potentil rop of =50 exists cross the conuctor (see Figure). Clculte the gnetic fiel intensity everywhere in spce, incluing in the interior of the conuctor. Sketch the gnetic fiel intensity. σ = S/ =10 =1000k =50 Solution: To fin the gnetic fiel intensity we ust first clculte the current in the conuctor n to o so we ust hve the resistnce. The resistnce of cylinricl conuctor of length n rius is: The current in the conuctor is now: R = σπ 2 Ω
4 I = R = σπ2 Now we istinguish two regions. One is r>, the secon is 0<r. For 0<r, we use pere's lw y rwing contour of rius r<. This contour encloses n re πr 2 while the current is unifor in n re π 2. Thus, the totl current enclose y the contour is: I r = I πr2 π = 2 Ir2 = σπr2 2 The length of the contour is 2πr n we cn write: or: 2πrH = I r = σπr2 H = σr 2 H(r ) = r = 1250r For r>, ll current is enclose y the contour n we cn write: or: 2πrH = I = σπ2 H(r>) = (0.01) r H = σ2 2r = r t r=0, the gnetic fiel intensity is zero. t r= it is equl to /. etween zero n the gnetic fiel intensity grows linerly. fter tht it iinishes s 1/r. The plot of the fiel with istnce fro the center of the conuctor is shown in the figure elow / H =0.1 r e10.4. In chpter10.proles.extr. very long sheet of etl of with is plce in unifor, perpeniculr gnetic fiel s shown. volteter is connecte to the two opposite sies with stiff wires s shown. Suppose the contcts of the volteter cn slie on the sheet. Wht is the ef re y the volteter if:
5 . The volteter together with the connecting wires is ove t velocity v x in the x irection.. If the sheet is ove in the y irection t velocity v y n the volteter is sliing in the x irection t velocity (while still keeping the contcts, tht is, it is lso sliing in the y irection with the plte) v x. c. If the sheet is ove t velocity v x in the x irection n the volteter is sttionry with respect to the sheet. y x. The ef is: ef = v x The r oves in gnetic fiel.. Motion in the y irection prouces no ef on the r, ut the otion in the x irection oes: ef = v x c. ef=0. The r n conuctor for loop. loop oving in constnt gnetic fiel prouces zero ef. nother wy to look t it is tht the loop fore y the r is prllel to the flux ensity. Thus, zero flux psses through the loop. Note: there is potentil ifference etween the two eges of the stiff wire ut the volteter is now connecte in loop n the totl voltge it esures is zero. e (in chpter10.exs.extr). trnsforer is e s shown. The thickness of the trnsforer is n the cross-sectionl re of the trnsforer is the se everywhere n equl to S. The pereility of the core is lrge ut not infinite. Two coils ech hving N turns re woun; coil (1) is woun on the left "leg" of the trnsforer; coil (2) is woun on the centrl leg of the trnsforer. current I 1 =sinωt psses through coil (1), clculte the rtio etween the inuce voltge (ef) in coil (2) n coil (1).
6 (1) N c c N (2) c c µ c Figure Note: the thickness (into the pge) of the trnsforer core is given s. Solution: First, we rw the equivlent gnetic circuit s follows: R 1 Φ 1 Φ 2 Φ 3 NI1 R 2 R 3 Figure The efs in the two coils re, y efinition: ef 1 = N Φ 1 t, ef 2 = N Φ 2 t Thus, we nee to clculte the fluxes Φ 1 n Φ 2. Fro the equivlent circuit we write: Φ 1 = NI 1 R 1 + R 2 R 3, Φ 2 = NI 1 R 1 + R 2 R 3. R 2 R 3 R 2 where R 2 R 3 ens tht R 2 n R 3 re prllel to ech other. Now, we cn clculte the efs: The rtio is: or: ef 1 = N Φ 1 t = N N(I 1/t), ef 2 = N Φ 2 = N N(I 1/t). R 2 R 3 R 1 + R 2 R 3 t R 1 + R 2 R 3 R 2 N N(I 1/t). R 2 R 3 ef 2 R = 1 + R 2 R 3 R 2 ef 1 N N(I 1/t) R 1 + R 2 R 3 = R 2 R 3 R 2 ef 2 ef 1 = R 2 R 3 R 2 = R 2 R 3 R 2 R 2 +R 3 = R 3 R 2 +R 3
7 Now we clculte the reluctnces (verge pths re shown in Figure. Thus: R 1 = R 3 = + 2c µc ef 2 ef 1 = R 3 R 2 +R 3 = + 2c µ c + 2c µ c, R 2 = c µc + c µc = + 2c + 2 3c ef 2 ef 1 = + 2c + 2 3c You coul hve rgue s follows: The flux Φ 1 splits into 2 equl prts, only one of these prts (hlf the flux) linking with coil (2). Thus the utul inuctnce etween the two coils is hlf of wht it woul e if ll flux linke. Thus, the ef in coil (2) ust e hlf of wht it woul e if ll flux links. Since the nuer of turns is the se in oth coils, the ef in coil (2) ust e hlf the ef in coil (1). e9.4. (in chpter9.exs.extr). Two very thin conucting sheets re w=100 wie n re seprte istnce =1 prt. current I=20 psses through the sheets, uniforly istriute on the with of the sheets. ssue w>> n clculte:. The inuctnce per unit length of the syste of two sheets.. The force per unit length cting on the sheets. Inicte the irection of the force on ech of the sheets. I =0 w I. 1 C.. Clculte : since w>>, the whole ssely is like prllel pltes. Fro the loop ove (Figure ): w = µ 0 I = µ 0I w T
8 To clculte inuctnce we nee the flux per unit length of the ssely (see Figure C): Φ = 1 = µ 0I w w The flux linkge equls the flux (one turn only) thus, L 11 = Φ I = µ 0 w = 4π = H L 11 = µ 0 w = H. Force. The energy in the gp etween the two conuctors is: w = 2 2µ 0 J 3 If the gp is chnge y istnce l, the chnge in energy in the gp n force on the conuctors re: F = W l W = w v = w w 1 l = w w = 2 w = µ 0I 2 2µ 0 2w = 4π = N F = µ 0I 2 2w = N The two pltes ten to seprte (you cn see tht fro the right hn rule since f=j n this ust e perpeniculr to oth current n flux ensity) e7.28 In chpter7.proles.extr. cpcitor is e of two prllel pltes of re seprte y istnce. ttery is connecte holing the lower plte t groun potentil n the upper plte t potentil. Filling the spce etween the pltes re two lossy ielectric sls ech of thickness /2. The perittivities n conuctivities of the ielectrics re given y ε 1, ε 2, σ 1, n σ 2. y ε 2,σ 2 ε 1,σ 1 /2 Neglecting ny ege effects, the potentil istriution etween the pltes y e foun fro Lplce s eqution with the pproprite ounry conitions s
9 = 2σ 2 y, 0<y</2 σ 1 + σ 2 = 2σ 1 σ 1 + σ 2 y + σ 2 σ 1 σ 1 + σ 2, /2<y<. Fin the electric fiel intensity in ech ielectric region.. Fin the current I n the power P supplie y the ttery. Solution: Clculte the electric fiel intensity s the negtive grient of the potentil:. E 1 = 1 = y 2σ 2 (σ 1 + σ 2 ), E 2 = 2 = y 2σ 1 (σ 1 + σ 2 ). J 1 = σ 1 E 1 = y 2σ 1σ 2 (σ 1 + σ 2 ), J 2 = σ 2 E 2 = y 2σ 1σ 2 (σ 1 + σ 2 ) = J 1 I = J = 2σ 1σ 2 (σ 1 + σ 2 ) P = I = 2σ 1σ 2 2 (σ 1 + σ 2 ) W e8.79 In chpter8.proles.extr 8. Two concentric, very thin sphericl-shell conuctors hve rii n, <. The spce etween the shells is fille with ielectric eiu with perrnittivity ε. ttery is connecte so tht the outer shell is hel t potentil ()=, n the inner one t ()=0.. Fin the potentil n the electric fiel intensity t ll points etween the shells.. Fin the totl chrge on ech shell. Solution:. Strt with Lplce's Eq. in Cylinricl coorintes (fiel cn only vry with r so we cn write: Integrting once: Integrting gin, The potentil is zero t R=: 1 R 2 R 2 R = C 1 R2 R R = 0 = C 1 R + C 2 R = C 1 R 2
10 The potentil is t R=: Sustituting: Thus: n the generl solution is: The electric fiel intensity is: Thus: 0 = C 1 + C 2 = C 1 + C 2 = C 1 + C 1 = C 1 C 1 =, (R) = R R C 2 = E(R) = (R) = R 1 R 2. Fro Guss' lw: E(R) = R ( )R 2 Q = 4π 2 εe() = 4π2 ε = 4πε ( ) 2 ( ) C Q = 4π 2 εe() = 4π2 ε = 4πε ( ) 2 ( ) = Q C ut, ecuse the chrge is negtive on the inner shell (since the electric fiel intensity points in the negtive R irection), we hve: Q = 4πε ( ) C Q = 4πε ( ) C
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