# Chapter 30

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12th edition physic

### Text of Chapter 30

30.1:a)V, 0.270 /s) A 830 ( H) 10 25 . 3 ( ) / (41 2 dt di M and is constant. b)If the second coil has the same changing current, then the induced voltage is the same andV. 270 . 01 30.2:For a toroidal solenoid,. 2 / and , /1 1 0 B 1 22 2r A i N i N MB So,. 2 /2 1 0r N AN M 30.3:a)H. 1.96 A) (6.52 / Wb) 0320 . 0 ( ) 400 ( /1 22 i N MB b)WhenWb. 10 7.11 (700) / H) (1.96 A) 54 . 2 ( / A, 54 . 231 2 B 21 N M i i 30.4:a)H. 10 6.82 s) / A 0.242 ( / V 10 65 . 1 ) / ( /3 32 dt di M b)A, 20 . 1 , 251 2 i N Wb. 10 3.2725 / H) 10 (6.82 A) 20 . 1 ( /432 12 N M iB c)mV. 2.45 s) / A (0.360 H) 10 82 . 6 ( / and s / A 360 . 0 /32 1 2 dt Mdi dt di 30.5:s. 1 A)s / (V 1 AC)s / (J 1 A / J 1 A / Nm 1 A / Tm 1 A / Wb 1 H 12 2 2 30.6:For a toroidal solenoid,). / ( / / dt di i N LB So solving forNwe have: turns. 238s) / A (0.0260 Wb) (0.00285A) (1.40 V) 10 6 . 12 () / ( /3 dt di i NB30.7:a)V. 10 4.68 s) / A (0.0180 H) 260 . 0 ( ) / (31 dt di L b)Terminala is at a higher potential since the coil pushes current through fromb to a and if replaced by a battery it would have the+terminal at. a 30.8:a)H. 130 . 0m) 120 . 0 ( 2) m 10 80 . 4 ( ) 1800 ( ) 500 (2 /2 5 20 20 mm r A N KKLb)Without the material,H. 10 2.60 H) 130 . 0 (5001 14mm KLKL 30.9:For a long, straight solenoid: . / / and /20 0l A N L l NiA i N LB B 30.10:a)Note that pointsa andbare reversed from that of figure 30.6. Thus, according to Equation 30.8,s. / A 00 . 4H 0.260V 04 . 1 LV Vdtdi a b Thus, the current is decreasing. b)From above we have that. ) s / A 00 . 4 ( dt di After integrating both sides of this expression with respect to, twe obtain A. 4.00 s) (2.00 A/s) (4.00 A) 0 . 12 ( s) / A 00 . 4 ( i t i 30.11:a)H. 0.250 A/s) (0.0640 / V) 0160 . 0 ( ) / ( / dt di L b)Wb. 10 4.50 (400) / H) (0.250 A) 720 . 0 ( /4 N iLB 30.12:a)J. 540 . 0 2 / A) (0.300 H) 0 . 12 (212 2 LI Ub)W. 2 . 16 ) 180 ( A) 300 . 0 (2 2 R I Pc)No. Magnetic energy and thermal energy are independent. As long as the current is constant,constant. U 30.13: rAl N LI U4 212 20 2 turns. 2850A) 0 . 12 ( ) m 10 00 . 5 (J) (0.390 m) 150 . 0 ( 4 42 2 4020 AI rUN 30.14:a)J. 10 1.73 h) / s 3600 h/day (24 W) 200 (7 Pt Ub)H. 5406A) (80.0J) 10 73 . 1 ( 2 2212722 IUL LI U 30.15:Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeabilityis used in place of.0 30.16:a)free space:J. 3619 ) m 0290 . 0 (2T) (0.560V2V30202 Bu Ub)material withJ. 04 . 8 ) m 0290 . 0 () 450 ( 2T) (0.560V2V 4503020 m2m KBu U K 30.17:a) 32602002m 1 . 25T) (0.600J) 10 60 . 3 ( 2 2Volume2 B U BVolUu . b)T. 9 . 11 T 4 . 141m) (0.400J) 10 60 . 3 ( 2 22360 0 2 BVolUB 30.18:a). mT 35 . 4) m 0690 . 0 ( 2) A 50 . 2 ( ) 600 (20 0 rNIBb)From Eq. (30.10),. m / J 53 . 72) T 10 35 . 4 (2302 302 Buc)Volume. m 10 52 . 1 ) m 10 50 . 3 ( ) m 0690 . 0 ( 2 2 V3 6 2 6 rAd). J 10 14 . 1 ) m 10 52 . 1 ( ) m / J 53 . 7 (5 3 6 3 uV Ue). H 10 65 . 3) m 0690 . 0 ( 2) m 10 50 . 3 ( ) 600 (262 6 2020 r A NL J 10 14 . 1 ) A 50 . 2 ( ) H 10 65 . 3 (21215 2 6 2 LI U same as (d). 30.19:a). s / A 40 . 2H 50 . 2V 00 . 60 When . dtdiiLiRdtdi b)When. A/s 800 . 0H 50 . 2) 00 . 8 ( ) A 500 . 0 ( V 00 . 6A 00 . 1 dtdii c)At A. 413 . 0 ) 1 (8.00V 00 . 6) 1 ( s 200 . 0s) (0.250 H) 50 . 2 / 00 . 8 ( ) / ( e eRi tt L R d)AsA. 750 . 08.00V 00 . 6 Ri t 30.20:(a)mA, 30 A 030 . 01000V 30max ilong after closing the switch. ) bV 4.0 V 26 V 30 V 26 A) (0.0259 ) (1000A 0.0259e 1 A 0.030 ) e (1R Battery LR) (max1020

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.| V VRi Vi issR / L / t (or, could uses) 20 at t L VdtdiL c) 30.21:a)R L e R it/ ), 1 ( // 2121max maxand , ) (1 when 2 / so / / t / te e i i R isLt / t 3 . 1750.0H) 10 (1.25 2) ln (Rln2and ) ( ln321 b)max221max221; Li U Li U 2 / whenmax max 21i i U U 2929 . 0 2 / 1 1 e so 2 1 1/ / t tes L t 30.7 R / ln(0.2929) 30.22:a)A 13 . 2H 0.115J) 260 . 0 ( 2 2212 LUI LI U V. 256 ) (120 A) 13 . 2 ( IR b),`

.| 20) / ( 2 2 2 ) / (2121212121and LI U e Li Li U Ie it L R t L R s. 10 32 . 321ln) 2(120H 115 . 021ln2214) / ( 2 ,`

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.| RLtet L R 30.23:a)A. 250 . 0240V 600 RI b)A. 137 . 0 A) 250 . 0 () s 10 (4.00 H) 0.160 / (240 ) / (04 e e I it L R c), V 9 . 32 ) 240 ( ) A 137 . 0 ( iR V Vab cb and c is at the higher potential. d). s 10 62 . 421ln) 240 () H 160 . 0 (21ln2142 / 1) / (02 / 1 ,`

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.| RLt eIit L R 30.24:a). V 60 and 0 0 At bc abv v tb). 0 and V 60 As bc abv v tc). V 0 . 24 V 0 . 36 V 0 . 60 and V 0 . 36 A 150 . 0 When bc abv iR v i 30.25:a)) 1 (00 . 8) V 00 . 6 () 1 ( ) 1 () H 50 . 2 / 00 . 8 (2) / (2) / (0t t L R t L Re eRe I i P ). 1 ( ) W 50 . 4 () s 20 . 3 (1te P b) 2 ) H 50 . 2 / 00 . 8 (22 ) / (22) 1 (00 . 8) V 00 . 6 () 1 (t t L RRe eRR i P . ) 1 ( ) W 50 . 4 (2 ) s 20 . 3 (1tRe P c)) ( ) 1 () / ( 2 ) / (2) / ( ) / ( t L R t L R t L R t L RLe eReLL eRdtdiiL P ,`

.| ). ( ) W 50 . 4 () s 40 . 6 ( ) s 20 . 3 (1 1t tLe e P d)Note that if we expand the exponential in part (b),then parts (b)and (c)add to give part (a), and the total power delivered is dissipated in the resistor and inductor. 30.26:When switch 1 is closed and switch 2 is open: . ) / ( ln0) / (0 000L R tiIte I i tLRI it dLRii dLRidtdiiRdtdiL + 30.27:Units of s / ) s ( / H / R L units of time. 30.28:a)fLC 21 . H 10 37 . 2) F 10 18 . 4 ( ) 10 6 . 1 ( 4141312 2 6 2 2 2 C fLb)F. 10 67 . 3) H 10 37 . 2 ( ) 10 40 . 5 ( 4141113 2 5 2min2 2max L fC 30.29:a)) F 10 00 . 6 ( ) H 50 . 1 ( 2 225 LC T s. rad 105 , s 0596 . 0 b). C 10 20 . 7 ) V 0 . 12 )( F 10 00 . 6 (4 5 CV Qc). J 10 32 . 4 ) V 0 . 12 )( F 10 00 . 6 (21213 2 5 20 CV Ud). 0 ) cos( , 0 At + t Q Q q t

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.| ) F 10 00 . 6 )( H 50 . 1 (s 0230 . 0cos ) C 10 20 . 7 ( ) cos( s, 0230 . 054t Q q t C. 10 43 . 54 Signs on plates are opposite to those at. 0 te)) sin( , s 0230 . 0 t Qdtdqi t A. 0.0499H) 10 (6.00 H) (1.50s 0.0230sinH) 10 H)(6.00 (1.50C 10 7.205 54

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.| i Positive charge flowing away from plate which had positive charge at. 0 tf)Capacitor:. J 10 2.46F) 10 2(6.00C) 10 (5.432352 4 2 CqUC Inductor:. J 10 1.87 A) (0.0499 H) 50 . 1 (21213 2 2 Li UL 30.30:(a)Energy conservation says(max) = (max)C LU U

A 0.871H 10 12F 10 18V) 22.5 (CV212136max2 2max L C V iLi The charge on the capacitor is zero because all the energy is in the inductor. (b)LC T 22 at4 1period:F) 10 (18 H) 10 (122) 2 (41416 3 LC T s 10 30 . 74 at4 3period:s 10 19 . 2 ) s 10 30 . 7 ( 3433 4 T(c)C CV q 405 ) V (22.5 F) 18 (0 30.31:F 0 . 30V 10 29 . 4C 10 15039 VQCFor an L-C circuit,LC 1 andLC T 2 2 mH 601 . 0) 2 (2 CTL 30.32:rad/s 1917) F 10 20 . 3 ( ) H 0850 . 0 (16a)C 10 43 . 4s rad 1917A 10 50 . 874maxmax max max iQ Q ib)From Eq. 31.26 2142 7 2 2s 1917A 10 00 . 5) C 10 43 . 4 (

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.| LCi Q q. C 10 58 . 37 30.33:a)) s A 80 . 2 ( ) F 10 60 . 3 ( ) H 640 . 0 ( 01622 +dtdiLC q qLC dtq d C. 10 45 . 66 b). V 36 . 2F 10 60 . 3C 10 50 . 866 Cq 30.34:a).maxmaxmax max maxLC iiQ Q i J 450 . 0) F 10 50 . 2 ( 2) C 10 50 . 1 (2. C 10 50 . 1 ) F 10 50 . 2 ( ) H 400 . 0 ( ) A 50 . 1 (102 5max2max5 10max CQUQ b) 1 410s 10 18 . 3) F 10 50 . 2 ( ) H 400 . 0 (1 1222 LCf(must double the frequency since it takes the required value twice per period). 30.35:[ ] . s s AA1ssCVVCs VCH F H ] [2 2 2s LC LC 30.36:Equation (30.20) is. 0122 + qLC dtq d We will solve the equation using: .1 10 ) cos( ) cos(1). cos( ) sin( ) ( cos2 222222LCLCtLCQt Q qLC dtq dt Q dtq dt Qdtdqt Q q + + + + + + + 30.37:a).) ( cos21212 2 2Ct QCqUC + .1since ,) ( sin21) ( sin212122 22 2 2 2LCCt Qt Q L Li UL+ + b)) ( sin21) ( cos212 2 2 22Total + + + + t Q L tCQU U UL C ) ( sin121) ( cos212 2 22 +

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.|+ + t QLCL tCQ )) ( sin ) ( (cos212 22 + + + t tCQ

Total221UCQ is a constant. 30.38:a)) cos() 2 / ( + t Ae qt L R ) sin(22 ) cos(2). sin( ) cos(2) 2 / ( ) 2 / (222) 2 / ( ) 2 / ( + + +

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.| + + t eLRA t eLRAdtq dt Ae t eLRAdtdqt L R t L Rt L R t L R ). cos() 2 / ( 2 + t Ae t L R 22222222241012 2LRLCLC LRLRqLCqdtdqLRdtq d

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.|+

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.| + + b): 0 , , 0 dtdqi Q q t At .4 / / 1 22tan2andcos0 sin cos2and cos2 2L R LC LR LRQ LQR QAA ALRdtdqQ A q 30.39:Subbing,1, , ,Ck R b L m q x we find: a)Eq. (13.41):. 0 : ) 27 . 30 .( Eq 02222 + + + +LCqdtdqLRdtq dmkxdtdxmbdtx d b)Eq. (13.43):.41: ) 28 . 30 .( Eq42222LRLCmbmk c)Eq. (13.42):). cos( : ) 28 . 30 .( Eq ) cos() 2 / ( ) 2 / ( + + t Ae q t Ae xt L R t m b 30.40:.AVV CsFH2 ]]]

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CLCL 30.41: LC LCL RLC LCL RLC LRLC 61 1261 1461412 2222

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.| . 4 . 45F) 10 (4.60 H) 6(0.2851F) 10 (4.60 H) (0.2851) H 285 . 0 ( 24 4 R 30.42:a)Whens. rad 298F) 10 (2.50 H) (0.4501 1, 050 LC Rb)We want 22 2 20) 95 . 0 (411) 4 1 (95 . 0 LC RLCL R LC . 8 . 83F) 10 (2.50(0.0975) H) 4(0.450) ) 95 . 0 ( 1 (452 CLR 30.43:a) b)Since the voltage is determined by the derivative of the current, the V versus t graph is indeed proportional to the derivative of the current graph. 30.44:a)] s) 240 cos[( ) A 124 . 0 (( t dtdLdtdiL ). ) s 240 (( sin ) V 4 . 23 ( ) ) s 240 sin(( ) 240 ( ) A 124 . 0 ( ) H 250 . 0 ( t t + + b), 0 ; V 4 . 23max i since the emf and current are 90 out of phase. c), 0 ; A 124 . 0max i since the emf and current are 90out of phase. 30.45:a)). ( ln2 2) (2) (0 0 0a bNih rdr Nih hdrr Ni hdr BbababaB

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.| b)). ( ln220a b h N iNLBc).2L2) () / ) ( 1 ( ln ) / ( ln2022

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.| ++ a a b h N aa ba a ba a b a b 30.46:a).122 2 1 012

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