32.1: a) .s28.1 sm1000.3 m1084.3 8 8     c d t b) Light travel time is: s1072.2 )hour1( )s3600( )day1( )hours24( )year1( )days365( )years61.8(years61.8 8  km.108.16m108.16s)10(2.72s)m100.3( 131688  ctd 32.2: .m180)s100.6()sm100.3( 78   tcd 32.3: jjB ˆ 2cos ˆ cos() maxmax                t c z fB ωt)kzB(z,t  jB ˆ s)m1000.3( Hz)1010.6(2cos)T1080.5()( 8 144                     t z z,t  . ˆ ))srad1083.3()m1028.1((cos)T1080.5()( 15174 jB tz,tz   ) ˆ () ˆ ()( kjE c(z,t)Bz,t y  . ˆ s)rad1083.3()m1028.1((cosm)V1074.1()( 15175 iE t)zz,t   32.4: a) .Hz1090.6 m1035.4 sm1000.3 λ 14 7 8      c f b) .T1000.9 sm1000.3 mV1070.2 12 8 3 max max       c E B c) The electric field is in the x -direction, and the wave is propagating in the z- direction. So the magnetic field is in the y-direction, since .BES  Thus: . ˆ ))srad1034.4()m1045.1cos(()mV1070.2()( ˆ sm1000.3 )Hz1090.6(2cosV/m)1070.2(),( ˆ 2cos ˆ (cos),( 15173 8 143 maxmax iE iE iiE tzz,t z t πtz t c z πfEωt)kzEtz                                     And . ˆ ))srad1034.4()m1045.1cos(()T1000.9( ˆ ),( ),( 151712 jjB tz c tzE tz     32.5: a) y  direction. b) m.1011.7 )srad1065.2( )sm1000.3(22 λ λ 2 2 4 12 8       ω πcπc πfω c) Since the electric field is in the z  -direction, and the wave is propagating in the y  -direction, then the magnetic field is in the x  -direction ).( BES  So: iiiB ˆ )sin( ˆ )sin( ˆ ),( ),( 00 ty c c E tky c E c tyE ty           iB ˆ )srad1065.2( )sm1000.3( )srad1065.2( sin sm1000.3 mV1010.3 ),( 12 8 12 8 5                       tyty . ˆ ))srad1065.2()m1083.8sin(()T1003.1(), 1233 i(B tyty   32.6: a) x  direction. b) .Hz1059.6 2 )sm100.3()mrad1038.1( 2 22 11 84    π π kc f c fπ λ π k c) Since the magnetic field is in the y  -direction, and the wave is propagating in the x  -direction, then the electric field is in the z  -direction ).( BES  So: . ˆ ))srad1014.4()mrad1038.1sin(()mV48.2(),( ˆ ))srad1014.4()mrad1038.1sin(())T1025.3((),( ˆ )2sin( ˆ ),(),( 124 1249 0 kE kE kkE txty txctx fkxcBtxcBtx      32.7: a) m.361 Hz1030.8 sm1000.3 λ 5 8     f c b) 1 m0174.0 m361 2 λ 2   ππ k c) s.rad1021.5)Hz1030.8(22 65  πfω  .mV0145.0)T1082.4()sm1000.3( 118 maxmax   cBE 32.8: .T1028.1 sm1000.3 mV1085.3 11 8 3 max max       c E B So ,1056.2 T105 T1028.1 7 5 11 earth max        B B and thus max B is much weaker than . earth B 32.9: BEBE KK cB KK BB vBE  00  .mV779.0 )23.1()74.1( )T1080.3()sm1000.3( 98     E 32.10: a) .sm1091.6 )18.5()64.3( s)m1000.3( 7 8    BE KK c v b) .m1006.1 Hz0.65 sm1091.6 λ 6 7    f v c) .T1004.1 sm1091.6 mV1020.7 10 7 3       v E B d) .mW1075.5 )18.5(2 )T1004.1)(mV1020.7( 2 28 0 103 0       B K EB I 32.11: a) .m1081.3 Hz1070.5 sm1017.2 λ 7 14 8      f v b) .m1026.5 Hz1070.5 sm1000.3 λ 7 14 8 0      f c c) .38.1 sm1017.2 sm1000.3 8 8     v c n d) .90.1)38.1( 22 2 2  n v c K K c v E E 32.12: a) s.m1034.2)m15.6)(Hz1080.3(λ 87  fv b) .64.1 )sm1034.2( )sm1000.3( 28 28 2 2     v c K E 32.13: a) 25 max 2 max0 mW101.1som,V090.0; 2 1   IEcEI  b) T100.3so 10 maxmaxmaxmax   cEBcBE c) W840)m105.2()4()mW10075.1()4( 23252 av    rIP d) Calculation in part (c) assumes that the transmitter emits uniformly in all directions. 32.14: The intensity of the electromagnetic wave is given by Eqn. 32.29: . 2 rms0 2 max0 2 1 cEcEI   Thus the total energy passing through a window of area A during a time t is J 9.15)s0.30)(m500.0()mV0200.0()sm1000.3()mF1085.8( 228122 rms0 μAtcE    32.15: J105.2)m100.2()4()mW100.5()4( 25210232 av   rIP 32.16: a) The average power from the beam is W104.2)m100.3()mW800.0( 4242   IAP b) We have, using Eq. 32.29, . 2 rms0 2 max0 2 1 cEcEI   Thus, mV4.17 s)m1000.3)(mF1085.8( mW800.0 812 2 0 rms     c I E  32.17: cIp  rad so 23 rad mW1070.2  cpI Then W105.8)m0.5()4()mW1070.2()4( 52232   rIP av 32.18: a) Hz.1047.8 m354.0 sm1000.3 λ 8 8    c f b) .T1080.1 sm1000.3 mV0540.0 10 8 max max     c E B c) .mW1087.3 2 )T1080.1()mV0540.0( 2 26 0 10 0       EB SI av 32.19: 2 0 max 2 0 2 max 2 )4( 2 r Pc Er c E ASP av      m.V0.12 )m00.5(2 )sm1000.3()W0.60( 2 0 8 max      E .T1000.4 sm1000.3 mV0.12 8 8 max max     c E B 32.20: a) The electric field is in the y  -direction, and the magnetic filed is in the z  - direction, so . ˆ ˆ ) ˆ ( ˆˆ ˆ ikjBES  That is, the Poynting vector is in the x  - direction. b) )(cos ),(),( ),( 0 maxmax 0 tkx BE txBtxE txS    ))).(2cos(1( 2 0 maxmax kxt BE    But over one period, the cosine function averages to zero, so we have: | . 2 | 0 maxmax  BE S av  32.21: a) The momentum density s.mkg107.8 s)m100.3( mW780 215 28 2 2     c S dV dp av b) The momentum flow rate Pa.106.2 sm100.3 mW780 1 6 8 2     c S dt dp A av 32.22: a) Absorbed light: .Pa1033.8 sm100.3 mW2500 1 6 8 2 rad     c S dt dp A p av .atm1023.8 atmPa10013.1 Pa1033.8 11 5 6 rad       p b) Reflecting light: .Pa1067.1 sm100.3 )mW2500(2 2 1 5 8 2 rad     c S dt dp A p av 10 5 5 rad 1065.1 atmPa101.013 Pa1067.1       p atm. The factor of 2 arises because the momentum vector totally reverses direction upon reflection. Thus the change in momentum is twice the original momentum. c) The momentum density s.mkg1078.2 s)m100.3( mW2500 214 28 2 2     c S dV dp av 32.23:  EBEBc c E EcEES 0 0 00 0 0 0 0 2 0 0 2 00 0 1            . 2 0 0 2 0 cE c EEB    32.24: Recall that :so,   BES a) . ˆ ) ˆ ( ˆ ˆ kjiS  b) . ˆˆˆˆ kijS  c) .j)i()k(S ˆ ˆ ˆ ˆ  d) . ˆ ˆ ˆ ˆ j)k(iS  32.25: T1033.1 8 maxmax   cEB   BE is in the direction of propagation. For E  in the + x -direction, B E    is in the + z - direction when  B is in the + y -direction. 32.26: a) .m00.2 )Hz100.75(2 sm1000.3 22 λ 6 8     f c x b) The distance between the electric and magnetic nodal planes is one-quarter of a wavelength = m.00.1 2 m00.2 2 4 λ    x 32.27: a) The node-antinode distance m.1038.4 Hz)1020.1(4 sm1010.2 44 λ 3 10 8      f v b) The distance between the electric and magnetic antinodes is one-quarter of a wavelength m.1038.4 Hz)10(1.204 sm102.10 44 λ 3 10 8      f v c) The distance between the electric and magnetic nodes is also one-quarter of a wavelength  m.1038.4 Hz)104(1.20 m/s102.10 44 3 10 8      f vλ 32.28: .cm0.20m200.0 )Hz1050.7(2 sm1000.3 22 λ 8 8 nodes     f c x There must be nodes at the planes, which are 80.0 cm apart, and there are two nodes between the planes, each 20.0 cm from a plane. It is at 20 cm, 40 cm, and 60 cm that a point charge will remain at rest, since the electric fields there are zero. 32.29: a) mm.7.10mm)55.3(22λ 2 λ  xx b) .mm55.3     BE xx c) s.m1056.1)m1010.7()Hz1020.2(λ 8310   fv 32.30: a) )sincos2()sinsin2( ),( maxmax 2 2 2 2 ωtkxkE x ωtkxE x x txE y          . ),( sinsin2sinsin2 ),( 2 2 00max 2 2 max 2 2 2 t txE tkxE c tkxEk x txE yy          Similarly: )cossin2()coscos2( ),( maxmax 2 2 2 2 tkxkB x tkxB x x txB z           . ),( coscos2coscos2 ),( 2 2 00max 2 2 max 2 2 2 t txB tkxB c tkxBk x txB zz          b) tkxkEtkxE x x txE y  sincos2)sinsin2( ),( maxmax       . ),( )coscos2( ),( .sincos2sincos2sincos2 ),( max max max max t txB tkxB t x txE ωtkxBtkx c E tkxE cx txE z y y                 Similarly: tkxkBtkxB x x txB z  cossin2)coscos2( ),( maxmax        tkxcB c tkxB c x txB z     cossin2cossin2 ),( max 2 max     ,( )sinsin2(cossin2 ),( 00max00max00 t t xE tkxE t tkxE x txB y z           32.31: a) Gamma rays: nm.104.62m1062.4 Hz106.50 sm103.00 λ 514 21 8      f c b) .nm522m105.22 Hz1075.5 sm1000.3 λ:lightGreen 7 14 8      f c 32.32: a) Hz.106.0 m5000 sm103.0 λ 4 8    c f b) Hz.106.0 m5.0 sm103.0 λ 7 8    c f c) Hz.106.0 m105.0 sm103.0 λ 13 6 8      c f d) Hz.106.0 m 10 5.0 sm103.0 λ 16 9 8      c f 32.33: Using a Gaussian surface such that the front surface is ahead of the wave front (no electric or magnetic fields) and the back face is behind the wave front (as shown at right), we have:   .00 0 encl xx E ε Q Ed AAE     .00 xx BABd AB   So the wave must be transverse, since there are no components of the electric or magnetic field in the direction of propagation. 32.34: Assume .with),sin( ˆ )sin( ˆ maxmax    tkxBandtkxE kBjE Then Eq. (32.12) implies: .0)cos()cos( maxmax        tkxBtkxkEx t B x E z y .λ λ / 2 2 maxmaxmaxmaxmaxmaxmax cBBfB f B k EBkE      Similarly for Eq.(32.14) .0)cos()cos( max00max00         tkxEtkxkB t E x B y z . 1 / 2 2 maxmax 2 max 2 max 00 maxmax00max E c E c fλ E λ c f E k BEkB        32.35: From Eq. (32.12): 2 2 ),(),( ),( t txB t txB tx txE t zz y                            But also from Eq. (32.14):                          t txE xx txB x y z ),( ),( 00  2 2 00 ),( t txB z     . ),(),( 2 2 00 2 2 t txB x txB zz        32.36: )cos( 2 1 2 1 )cos(),( 2 max0 2 0max tkxEEutkxEtxE Ey   B z E u B tkxBtkx c Ec u         0 2 2 max 0 2 max 2 0 2 )cos( 2 1 )cos( 2      32.37: a) The energy incident on the mirror is AtcEIAtPt 2 0 2 1   J.105.20s)(1.00)m1000.5()mV028.0()sm1000.3( 2 1 102428 0    E b) The radiation pressure .Pa1094.6)mV0280.0( 2 152 0 2 0rad    E c I p c) Power 2 rad 2 24 RcpRIP   W.101.34m)(3.20Pa)10(6.94s)m1000.3(2 42158    P 32.38: a) .Hz1081.7 m0384.0 sm1000.3 λ 9 8    c f b) T.1050.4 sm1000.3 mV1.35 9 8 max max     c E B c) .mW102.42m)V(1.35)sm1000.3( 2 1 2 1 2328 0 2 max0    cEI d) .N1093.1 s)m1000.3(2 )m(0.240T)10(4.50m)V35.1( 2 12 8 0 29 0        c EBA c IA pAF 32.39: a) The laser intensity .mW652 m)10(2.50 W)104(3.204 2 23 3 2       πD P A P I  But m.V701 )sm10(3.00 )mW2(652 2 2 1 8 0 2 0 2 0       c I EcEI And T.102.34 sm103.00 mV701 6 8     c E B b) .mJ101.09m)V(701 4 1 4 1 362 0 2 max0    Euu avav EB Note the extra factor of 2 1 since we are averaging. c) In one meter of the laser beam, the total energy is: 42)(2Vol 2 tottot LDuALuuE EE   J.101.07m)/4(1.00m)1050.2()mJ1009.1(2 112336 tot    E [...]...  1 so prad  I C (ii) totally reflecting W  0 so p rad  2 C These are just equations 32. 32 and 32. 33 c) W  0.9, I  1.40  10 2 W/m 2 prad  W  0.1, I  1.40  103 W m 2  prad  (2  0.9) (1.40  103 W m 2 )  5.13  10 6 Pa 3.00  108 m s (2  0.1)(1.40  10 2 W m 2 )  8.87  10 6 Pa 8 m 3.00  10 s 32. 41: a) At the sun’s surface: 3.9  10 26 W P P P  IA  I     6.4  10 7 W m 2 2 8... initial energy differs from the proton’s by the ratio of their masses: m (9.11  10 31 kg) E e  E p e  (6.00  10 6 eV)  327 3 eV mp (1.67  10  27 kg) So the fraction of its energy that it radiates every second is: (dE dt )(1 s) 8 .32  10 5 eV   2.54  10 8 327 3 eV E 32. 54: For the electron in the classical hydrogen atom, its acceleration is: 2 2(13.6 eV)(1.60  10 19 J eV) v 2 1 mv  21 ... current carrying solenoid as having entered through its cylindrical walls while the current was attaining its steady-state value But also U  32. 44: a) The energy density, as a function of x, for the equations for the electrical and magnetic fields of Eqs (32. 34) and (32. 35) is given by: 2 u   0 E 2  4 0 E max sin 2 kx sin t 1 1    , cos t  cos  and sin ωt  sin  b) At t  4 4 4 2 2  ˆ ˆ.. .32. 40: a) The change in the momentum vector determines p rad If W is the fraction    absorbed,  P  P out  P in  (1  W ) p  ( p )  (2  W ) p Here, (1  W ) is the fraction reflected The positive direction was chosen in the direction of reflection p is the magnitude of the incoming momentum With Eq 32. 31, and taking the average, we I get prad... in the z-direction EB ρI 2 1 ρI  0 I Its magnitude is S     0  0 a 2 2a 2 2 a 3 32. 45: a) E  J  d) Over a length l, the rate of energy flowing in is SA  I 2 lI 2 2al  2 2 a 3 a 2 l lI 2 The thermal power loss is I R  I  , which exactly equals the flow of A a 2 electromagnetic energy 2 32. 46: B  2    0i q q , and  S E  dA  EA  E , so the magnitude of the 2r 0 ... the capacitor Poynting vector is S  2 cBmax 32. 47: The power from the antenna is P  IA  4r 2 So 2 0  Bmax  2 0 P  4r 2 c 2 0 (5.50  10 4 W )  2.42  10 9 T 2 8 4 (2500 m) (3.00  10 m s) dB  Bmax  2fBmax  2 (9.50  10 7 Hz) (2.42  10 9 T)  1.44 T s dt d dB D 2 dB  (0.180 m) 2 (1.44 T s)    A    0.0366 V dt dt 4 dt 4  32. 48: I  P 1 2I   0 cE 2  E   A 2  0c... ship with the same magnitude of momentum as you gave the flashlight 32. 50: P  IA  I  E P 1   0 cE 2  E  A 2 2P  A 0 c 2Vi A 0 c 2Vi 2(5.00  105 V) (1000 A)   6.14  10 4 V m (100 m 2 )ε0 (3.00  108 m s) A 0 c And B E 6.14  10 4 V m   2.05  10 4 T 8 c 3.00  10 m s GM S m GM S 4R 3 ρ 4GM S R 3 ρ FG   2  32. 51: a) r2 r 3 3r 2 b) Assuming that the sun’s radiation is intercepted... system 32. 52: a) The momentum transfer is always greatest when reflecting surfaces are used (consider a ball colliding with a wallthe wall exerts a greater force if the ball rebounds rather than sticks) So in solar sailing one would want to use a reflecting sail b) The equation for repulsion comes from balancing the gravitational force and the force from the radiation pressure As seen in Problem 32. 51,... energy because of its acceleration is: (1.6  10 19 C) 2 (1.53  1015 m s 2 ) 2 dE q2a2    1.33  10  23 J s 3 8 3 dt 6 0 c 6 0 (3.0  10 m s)  8 .32  10 5 eV s So the fraction of its energy that it radiates every second is: (dE dt )(1 s) 8 .32  10 5 eV   1.39  10 11 E 6.00  10 6 eV c) Carrying out the same calculations as in part (b), but now for an electron at the same speed and radius... positive, which makes sense since the direction of wave propagation by definition is the direction of energy flow 32. 42: a) S ( x, t )  b) dB di d B dB di A   0 nA  0 n   dt dt dt dt dt  d B di di So,  E  dl    E 2r    0 nA    0 nr 2 dt dt dt  nr di E 0 2 dt 32. 43: a) B   0 ni  b) The direction of the Poynting vector is radially inward, since the magnetic field is along . is J 9.15)s0.30)(m500.0()mV0200.0()sm1000.3()mF1085.8( 228122 rms0 μAtcE    32. 15: J105.2)m100.2()4()mW100.5()4( 25210 232 av   rIP 32. 16: a) The average power from the beam is W104.2)m100.3()mW800.0( 4242.        32. 35: From Eq. (32. 12): 2 2 ),(),( ),( t txB t txB tx txE t zz y                            But also from Eq. (32. 14):                          t txE xx txB x y z ),( ),( 00 