30.1: a) V,0.270/s)A830(H)1025.3()/( 4 12   dtdiM  and is constant. b) If the second coil has the same changing current, then the induced voltage is the same and V.270.0 1   30.2: For a toroidal solenoid, .2/and,/ 110B12 22 rAiNiNM B       So, .2/ 210 rNANM   30.3: a) H.1.96A)(6.52/Wb)0320.0()400(/ 12 2     iNM B b) When Wb.107.11(700)/H)(1.96A)54.2(/A,54.2 3 12B2 1   NMii 30.4: a) H.106.82s)/A0.242(/V1065.1)/(/ 33 2   dtdiM  b) A,20.1,25 12  iN Wb.103.27 25/H)10(6.82A)20.1(/ 4 3 21 2     NMi B c) mV.2.45s)/A(0.360H)1082.6(/ands/A360.0/ 3 212   dtMdidtdi  30.5: Ωs.1A)s/(V1AC)s/(J1A/J1A/Nm1A/Tm1A/Wb1H1 222  30.6: For a toroidal solenoid, )./(// dtdiiNL B     So solving for N we have: turns.238 s)/A(0.0260Wb)(0.00285 A)(1.40V)106.12( )/(/ 3     dtdiiN B  30.7: a) V.104.68s)/A(0.0180H)260.0()/( 3 1   dtdiL  b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a battery it would have the  terminal at . a 30.8: a) H.130.0 m)120.0(2 )m1080.4()1800()500( 2/ 252 0 2 0m m     π μ πr ANμK K L b) Without the material, H.102.60H)130.0( 500 11 4 m m   K L K L 30.9: For a long, straight solenoid: .//and/ 2 00 lANμLlNiAμiNL BB  30.10: a) Note that points a and b are reversed from that of figure 30.6. Thus, according to Equation 30.8, s./A00.4 H0.260 V04.1    L VV dt di ab Thus, the current is decreasing. b) From above we have that .)s/A00.4( dtdi   After integrating both sides of this expression with respect to , t we obtain A.4.00s)(2.00A/s)(4.00A)0.12(s)/A00.4(         iti 30.11: a) H.0.250A/s)(0.0640/V)0160.0()/(/    dtdiL  b) Wb.104.50(400)/H)(0.250A)720.0(/ 4  NiL B 30.12: a) J.540.02/A)(0.300H)0.12( 2 1 22  LIU b) W.2.16)180(A)300.0( 22  RIP c) No. Magnetic energy and thermal energy are independent. As long as the current is constant, constant.  U 30.13: πr AlNμ LIU 4 2 1 22 0 2  turns.2850 A)0.12()m1000.5( J)(0.390m)150.0(44 224 0 2 0     μ π AIμ πrU N 30.14: a) J.101.73h)/s3600h/day(24W)200( 7  PtU b) H.5406 A)(80.0 J)1073.1(2 2 2 1 2 7 2 2    I U LLIU 30.15: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability  is used in place of . 0  30.16: a) free space: J.3619)m0290.0( 2 T)(0.560 V 2 V 3 0 2 0 2   B uU b) material with J.04.8)m0290.0( )450(2 T)(0.560 V 2 V450 3 0 2 0m 2 m   K B uUK 30.17: a) 3 2 6 0 2 0 0 2 m1.25 T)(0.600 J)1060.3(22 Volume 2      B U B Vol U u . b) T.9.11T4.141 m)(0.400 J)1060.3(22 2 3 6 00 2    B Vol U B  30.18: a) .mT35.4 )m0690.0(2 )A50.2()600( 2 00      r NI B b) From Eq. (30.10), .m/J53.7 2 )T1035.4( 2 3 0 23 0 2      B u c) Volume .m1052.1)m1050.3()m0690.0(22V 3626    rA d) .J1014.1)m1052.1()m/J53.7( 5363   uVU e) .H1065.3 )m0690.0(2 )m1050.3()600( 2 6 262 0 2 0          r AN L J1014.1)A50.2()H1065.3( 2 1 2 1 5262   LIU same as (d). 30.19: a) .s/A40.2 H50.2 V00.6 0When.    dt di i L iR dt di  b) When .A/s800.0 H50.2 )00.8()A500.0(V00.6 A00.1     dt di i c) At A.413.0)1( 8.00 V00.6 )1(s200.0 s)(0.250H)50.2/00.8()/(     ee R ε it tLR d) As A.750.0 8.00 V00.6    R it  30.20: (a) mA,30A030.0 1000 V30 max   i long after closing the switch. )b V4.0V26V30ε V26A)(0.0259Ω)(1000 A0.0259 e1A0.030)e(1 RBatteryL R )( max 10 20             VV RiV ii μs μs R/L/t (or, could use s)20at   tLV dt di L c) 30.21: a) RLeRi t /),1(/ /     2 1 2 1 maxmax and,)(1when2/so/   τ/tτ/t eeiiRεi μs L t τ/t 3.17 50.0 H)10(1.252)ln( R ln2 and)(ln 3 2 1      b) max 2 2 1 max 2 2 1 ; LiULiU  2/when maxmax 2 1 iiUU  2929.02/11eso211 //   τtτt e μsLt 30.7R/ln(0.2929)    30.22: a) A13.2 H0.115 J)260.0(2 2 2 1 2  L U ILIU V.256)(120A)13.2(      IR  b)         2 0 )/(222)/( 2 1 2 1 2 1 2 1 2 1 and LIUeLiLiUIei tLRtLR s.1032.3 2 1 ln )2(120 H115.0 2 1 ln 2 2 1 4 )/(2                    R L t e tLR 30.23: a) A.250.0 240 V60 0    R I  b) A.137.0A)250.0( )s10(4.00H)0.160/(240)/( 0 4    eeIi tLR c) ,V9.32)240()A137.0(  iRVV abcb and c is at the higher potential. d) .s1062.4 2 1 ln )240( )H160.0( 2 1 ln 2 1 4 2/1 )/( 0 2/1                   R L te I i tLR 30.24: a) .V60and00At     bcab vvt b) .0andV60As  bcab vvt c) .V0.24V0.36V0.60andV0.36A150.0When  bcab viRvi 30.25: a) )1( 00.8 )V00.6( )1()1( )H50.2/00.8( 2 )/( 2 )/( 0 ttLRtLR ee R eIiP       ).1()W50.4( )s20.3( 1 t eP    b) 2)H50.2/00.8( 2 2)/( 2 2 )1( 00.8 )V00.6( )1( ttLR R ee R ε RiP     .)1()W50.4( 2)s20.3( 1 t R eP    c) )()1( )/(2)/( 2 )/()/( tLRtLRtLRtLR L ee R ε e L ε Le R ε dt di iLP          ).()W50.4( )s40.6()s20.3( 11 tt L eeP    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: .)/(ln 0 )/( 00 0 0 LRt i I t eIit L R Ii td L R i id L R i dt di iR dt di L          30.27: Units of        s/)s(/H/ RL units of time. 30.28: a) πf LC ω 2 1  .H1037.2 )F1018.4()106.1(4 1 4 1 3 1226222       Cf L b) F.1067.3 )H1037.2()1040.5(4 1 4 1 11 3252 min 22 max       Lf C 30.29: a) )F1000.6()H50.1(22 2 5  πLCπ ω π T s.rad105,s0596.0  ω b) .C1020.7)V0.12)(F1000.6( 45   CVQ c) .J1032.4)V0.12)(F1000.6( 2 1 2 1 3252 0   CVU d) .0)cos(,0At         ωtQQqt             )F1000.6)(H50.1( s0230.0 cos)C1020.7()cos(s,0230.0 5 4 ωtQqt C.1043.5 4  Signs on plates are opposite to those at .0  t e) )sin(,s0230.0 ωtωQ dt dq it  A.0.0499 H)10(6.00H)(1.50 s0.0230 sin H)10H)(6.00(1.50 C107.20 55 4               i Positive charge flowing away from plate which had positive charge at .0  t f) Capacitor: .J102.46 F)102(6.00 C)10(5.43 2 3 5 242        C q U C Inductor: .J101.87A)(0.0499H)50.1( 2 1 2 1 322   LiU L 30.30: (a) Energy conservation says (max)=(max) CL UU A0.871 H1012 F1018 V)22.5( CV 2 1 2 1 3 6 max 22 max        LCVi Li The charge on the capacitor is zero because all the energy is in the inductor. (b) LCT    2 2  at 41 period: F)10(18H)10(12 2 )2( 4 1 4 1 63     LCT s1030.7 4  at 43 period: s1019.2)s1030.7(3 4 3 34  T (c) CCVq  405)V(22.5F)18( 0  30.31: F0.30 V 10 29 . 4 C10150 3 9        V Q C For an L-C circuit, LCω 1 and LCωπT  22  mH601.0 )2( 2  C T L  30.32: rad/s1917 )F1020.3()H0850.0( 1 6     ω a) C1043.4 srad1917 A1050.8 7 4 max maxmaxmax      ω i Q ωQi b) From Eq. 31.26 2 1 4 2722 s1917 A1000.5 )C1043.4(              LCiQq .C1058.3 7  30.33: a) )sA80.2()F1060.3()H640.0(0 1 6 2 2   dt di LCqq LC dt qd C.1045.6 6  b) .V36.2 F 10 60 . 3 C1050.8 6 6       C q  30.34: a) . max max maxmaxmax LCi ω i Q ωQi  J450.0 )F1050.2(2 )C1050.1( 2 .C1050.1)F1050.2()H400.0()A50.1( 10 25 max 2 max 510 max         C Q U Q b) 14 10 s1018.3 )F1050.2()H400.0( 11 2 2 2          LC f (must double the frequency since it takes the required value twice per period). 30.35:   .ssA A 1 s s C V Ω V C s Ω V C HFH][ 222 sLCLC  30.36: Equation (30.20) is .0 1 2 2  q LC dt qd We will solve the equation using: . 11 0)cos()cos( 1 ).cos()sin()(cos 22 2 2 2 2 2 LC ω LC ωt LC Q ωtQωq LCdt qd ωtQω dt qd ωtωQ dt dq ωtQq     30.37: a) . )(cos 2 1 2 1 222 C ωtQ C q U C    . 1 since, )(sin 2 1 )(sin 2 1 2 1 2 22 2222 LC ω C ωtQ ωtQLωLiU L      b) )(sin 2 1 )(cos 2 1 2222 2 Total   ωtQLωωt C Q UUU LC )(sin 1 2 1 )(cos 2 1 222 2          ωtQ LC L ωt C Q ))(sin)((cos 2 1 22 2   ωtωt C Q Total 2 2 1 U C Q  is a constant. 30.38: a) )cos( )2/(      tωAeq tLR )sin( 2 2)cos( 2 ).sin()cos( 2 )2/()2/( 2 2 2 )2/()2/(                     tωe L R A ωtωe L R A dt qd t ωAeωtωe L R A dt dq tLRtLR tLRtLR ).cos( )2/(2      tωAeω tLR 2 2 2 2 2 2 2 2 2 4 1 0 1 22 L R LC ω LCL R L R q LC q dt dq L R dt qd                        b) :0,,0  dt dq iQqtAt . 4//12 2 tan 2 and cos 0sincos 2 andcos 22 LRLCL R ωL R Q ω L QRQ A A ωA L R dt dq QAq             [...]... kx d 2 q R dq q    0  Eq. (30. 27) : 2    0 a) Eq (13.41): 2 dt m dt m dt L dt LC 30. 39: Subbing x  q, m  L, b  R, k  k 1 b2 R2   Eq. (30. 28) : ω   2 m 4m 2 LC 4 L ( b / 2 m )t c) Eq (13.42): x  Ae cos(ωt   )  Eq. (30. 28) : q  Ae  ( R / 2 L )t cos(ωt   ) b) Eq (13.43): ω   L  L  H s  V   2   30. 40:        A C  F C V  C 30. 41:   2  1 1 R2 1  1 1 ... a 2π  a  30. 45: a) 2 30. 46: a) M  b) ε 2  N 2 μ NN A μ N N πr N2 N A N A μ N IA  B2  2 2  B1  2 2 0 1 1  0 1 2 2  0 1 2 2 I I A1 IA1 l1 l1 l1 d B2 dt 2  N2 μ0 N1 A2 di1 μ0 N1 N 2 π r2 di1  l1 l1 dt dt 2 μ N N π r di2 di di c) ε1  M 12 2  M 2  0 1 2 2 dt l1 dt dt 30. 47: a) ε   L di  L  ε /(di / dt )  (30. 0 V) /( 4.00 A / s )  7.5 H dt d   f   i  εt   f  (30. 0 V)(12.0... the inductance for any height L  L0 1   m  D  Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury dD 4 0. 6302 4 H 0. 6300 0 H dD 2 0. 6304 8 H 0.62999 H d  3D 4 0. 6307 2 H 0.62999 H dD 0. 6309 6 H 0.62998 H Where are used the values  m (O 2 )  1.52  10 3 and  m (Hg)   2.9  10 5 d) The volume gauge is much better... μ 1 2 2U Li  L  2  l 0 ln (b / a), which is the same as in Problem 30. 50 2 i 2π 30. 52: a) L1  L2  N1 B1 i1 N 2  B2 i2 2 N1 A  μ0 N1i1  μ0 N1 A    , 2π r i1  2π r     2 N 2 A  μ 0 N 2 i2  μ 0 N 2 A   i2  2π r  2π r   2 2 2  μ N N A μ N A μ0 N 2 A b) M   0 1 2   0 1  L1 L2  2π r  2π r 2π r   2 30. 53: u B  u E  ε0 E 2 B2   B  ε 0 μ0 E 2  2 2 μ0  ε0 μ0 (650... i f (1  e ( R / L )t )    ln (1  i / i f )  L  L ln (1  i / i f ) 30. 54: a) R  L  (1860 )(7.25  10 4 s)  0.963 H ln (1  (4.86 / 6.45)) 30. 55: a) After one time constant has passed: 6.00 V  (1  e 1 )  0.474 A i  (1  e 1 )  8.00  R 1 2 1  U  Li  (2.50 H)(0.474 A) 2  0.281 J 2 2 Or, using Problem (30. 25(c)): 3/ 7 U   PL dt  (4.50 W )  (e  (3.20)t  e ( 6.40)t )dt 0... the same as part (a) 30. 57: Multiplying Eq (30. 27) by i, yields: i 2 R  Li 2 di q di q dq d 1  d 1 q  i  i 2 R  Li   i 2 R   Li 2     dt C dt C dt dt  2  dt  2 C  PR  PL  PC  0     That is, the rate of energy dissipation throughout the circuit must balance over all of the circuit elements 3T  2 3T   3  q  Q cos( t )  Q cos   Q cos 8 T 8   4 30. 58: a) If t  1... steady state is reached, so the inductor has no potential drop across it Simplified circuit becomes i  ε /R  50 V  0.385 A 130  V1  (100 )(0.385 A)  38.5 V ; V2  0 V3  V4  50 V  38.5 V  11.5 V i1  0.385 A, i2  i3  11.5 V  0.153 A 75  11.5 V  0. 230 A 50  30. 65: a) Just after the switch is closed the voltage V5 across the capacitor is zero and there is also no current through the inductor,... reads zero c) Use the results of problem 30. 49 to combine the inductor network into its equivalent: R  75.0  is the equivalent resistance Eq. (30. 14) says i  (ε R )(1  e t τ ), with τ  L / R  (10.8 mH) (75.0 )  0.144 ms ε  20.0 V , R  75.0 , t  0.115 ms, so i  0.147 A VR  iR  (0.147 A)(75.0 )  11.0 V 20.0 V  VR  VL  0 so VL  20.0 V  VR  9.0 V 30. 74: (a) Steady state: i   75.0 V... (0.299 V)sin ((125.6 s 1 )(0.0180 s))   (t )  0. 230 V 30. 48: a)   L di1 di di  L2 2  Leq , dt dt dt di di di but i1  i 2  i for series components so 1  2  , thus L1  L2  Leq dt dt dt di1 di2 di  L2  Leq , where i  i1  i 2 b) Parallel: Now L1 dt dt dt Leq di Leq di di di di di1 di2   So But 1  and 2  dt dt dt dt L2 dt dt L2 dt 30. 49: a) Series: L1 1 1 di Leq di Leq di 1   ... J  U C  U max 2 4 4  3   Q  4  2   q  2C 2C 30. 61: The energy density in the sunspot is u B  B 2 / 2 μ0  6.366  10 4 J /m 3 The total energy stored in the sunspot is U B  u BV The mass of the material in the sunspot is m  ρV K  U B so 1 2 mv  U B ; 2 1 ρVv 2  u BV 2 The volume divides out, and v  2u B / ρ  2  10 4 m /s 30. 62: (a) The voltage behaves the same as the current . A.750.0 8.00 V00.6    R it  30. 20: (a) mA,30A 030. 0 1000 V30 max   i long after closing the switch. )b V4.0V26V30ε V26A)(0.0259Ω)(1000 A0.0259 e1A0. 030) e(1 RBatteryL R )( max 10 20             VV RiV ii μs μs R/L/t (or,.     LCT s1 030. 7 4  at 43 period: s1019.2)s1 030. 7(3 4 3 34  T (c) CCVq  405)V(22.5F)18( 0  30. 31: F0 .30 V 10 29 . 4 C10150 3 9        V Q C For