Physics exercises_solution: Chapter 32

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Physics exercises_solution: Chapter 32

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 32

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  1. d 3.84  10 8 m 32.1: a) t    1.28 s. c 3.00  10 8 m s b) Light travel time is: (365 days) (24 hours) (3600 s) 8.61 years  (8.61 years)  2.72  10 8 s (1 year) (1 day) (1 hour ) d  ct  (3.0  10 8 m s) (2.72  10 8 s)  8.16  1016 m  8.16  1013 km. 32.2: d  ct  (3.0  10 8 m s) (6.0  10 7 s)  180 m.  z  32.3: B (z,t )  Bmax cos(kz  ωt)ˆ  Bmax cos 2f   t   ˆ j   j   c    z   B ( z,t )  (5.80  10 4 T) cos  2 (6.10  1014 Hz)   (3.00  10 m s) 8  t  ˆ  j     B ( z,t )  (5.80  10 4 T ) cos ((1.28  10 7 m 1 ) z  (3.83  1015 rad s)t ) ˆ . j E ( z,t )  ( B y(z,t)ˆ)  (ck ) j ˆ ˆ  E ( z,t )  (1.74  10 5 V m) cos ((1.28  10 7 m 1 ) z  (3.83  1015 rad s)t)i . c 3.00  108 m s 32.4: a) f    6.90  1014 Hz. λ 4.35  10 7 m Emax 2.70  10 3 V m b) Bmax    9.00  10 12 T. c 3.00  108 m s 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 S  E  B. Thus: ˆ  z  ˆ E ( z , t )  Emax cos(kz  ωt)i  Emax cos  2πf   t  i    c    z ˆ  E ( z , t )  (2.70  10 3 V/m) cos  2π (6.90  1014 Hz) t   3.00  108 m  i    s    E ( z,t )  (2.70  10 3 ˆ V m) cos((1.45  10 7 m 1 ) z  (4.34  1015 rad s)t )i .  E ( z, t ) ˆ And B ( z , t )  j  (9.00  10 12 T) cos((1.45  10 7 m 1 ) z  (4.34  1015 rad s)t ) ˆ. j c
  2. 32.5: a)  y direction. 2πc 2πc 2 (3.00  108 m s) b) ω  2πf  λ   7.11  10 4 m. λ ω (2.65  10 rad s) 12 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 ( S  E  B ). So: B( y, t )   E ( y , t ) ˆ  E0 i ˆ  E0 sin(  y  t )i sin( ky  t )i  ˆ c c c c  3.10  10 5 V m   (2.65  1012 rad s) ˆ  B( y, t )    3.00  108 m s   sin   (3.00  108 m s) y  (2.65  10 rad s)t i 12      ˆ  B( y, t )  (1.03  10 3 T ) sin((8.83  10 3 m) y  (2.65  1012 rad s)t )i . 32.6: a)  x direction. 2π 2π f kc (1.38  10 4 rad m) (3.0  108 m s) b) k   f    6.59  1011 Hz. λ c 2π 2π 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 ( S  E  B ). So: ˆ ˆ E ( x, t )   cB ( x, t )k   cB0 sin(kx  2 f )k ˆ  E ( x, t )   (c (3.25  10 9 T)) sin((1.38  10 4 rad m) x  (4.14  1012 rad s)t )k  E ( y, t )   (2.48 V m) sin((1.38  10 4 rad m) x  (4.14  1012 rad s)t )k . ˆ c 3.00  108 m s 32.7: a) λ    361 m. f 8.30  105 Hz 2π 2π b) k    0.0174 m 1 λ 361 m c) ω  2 f  2π (8.30  105 Hz)  5.21  10 6 rad s. E max  cBmax  (3.00  10 8 m s) (4.82  10 11 T)  0.0145 V m. Emax 3.85  10 3 V m 32.8: Bmax    1.28  10 11 T. c 3.00  10 m s 8 Bmax 1.28  10 11 T So  5  2.56  10 7 , and thus Bmax is much weaker than Bearth . Bearth 5  10 T
  3. B B cB 32.9: E  vB     K E 0 K B 0 KE KB (3.00  10 m s) (3.80  10 9 T) 8 E  0.779 V m. (1.74) (1.23) c (3.00  10 8 m s) 32.10: a) v    6.91  10 7 m s. KEKB (3.64) (5.18) v 6.91  10 7 m s b) λ    1.06  10 6 m. f 65.0 Hz E 7.20  10 3 V m c) B    1.04  10 10 T. v 6.91  10 m s 7 EB (7.20  10 3 V m)(1.04  10 10 T) d) I    5.75  10 8 W m 2 . 2K B  0 2(5.18)  0 v 2.17  108 m s 32.11: a) λ   3.81  10 7 m. f 5.70  10 Hz 14 c 3.00  108 m s b) λ 0    5.26  10 7 m. f 5.70  10 Hz 14 c 3.00  10 8 m s c) n    1.38. v 2.17  10 8 m s c c2 d) v   K E  2  n 2  (1.38) 2  1.90. KE v 32.12: a) v  f λ  (3.80  10 7 Hz)(6.15 m)  2.34  108 m s. c 2 (3.00  108 m s) 2 b) K E    1.64. v 2 (2.34  108 m s) 2 1 32.13: a) I   0 cEmax ; Emax  0.090 V m, so I  1.1  10 5 W m 2 2 2 b) Emax  cBmax so Bmax  E max c  3 .0  10 10 T c) Pav  I (4r 2 )  (1.075  10 5 W m 2 ) (4 ) (2.5  103 m) 2  840 W d) Calculation in part (c) assumes that the transmitter emits uniformly in all directions.
  4. 32.14: The intensity of the electromagnetic wave is given by Eqn. 32.29: I  1  0 cE max   0 cE rms . Thus the total energy passing through a window of area A during 2 2 2 a time t is  0 cErms At  (8.85  10 12 F m) (3.00  108 m s) (0.0200 V m) 2 (0.500 m 2 )(30.0 s)  15.9 μJ 2 32.15: Pav  I (4r 2 )  (5.0  10 3 W m 2 ) (4 ) (2.0  1010 m) 2  2.5  10 25 J 32.16: a) The average power from the beam is P  IA  (0.800 W m 2 ) (3.0  10 4 m 2 )  2.4  10 4 W b) We have, using Eq. 32.29, I  1  0 cE max   0 cE rms . Thus, 2 2 2 I 0.800 W m 2 E rms   17.4 V m  0c (8.85  10 12 F m)(3.00  10 8 m s) 32.17: p rad  I c so I  cp rad  2.70  10 3 W m 2 Then Pav  I (4r 2 )  (2.70  10 3 W m 2 ) (4 ) (5.0 m) 2  8.5  10 5 W c 3.00  108 m s 32.18: a) f    8.47  108 Hz. λ 0.354 m E 0.0540 V m b) Bmax  max   1.80  10 10 T. c 3.00  10 8 m s EB (0.0540 V m) (1.80  10 10 T) c) I  S av    3.87  10 6 W m 2 . 2 0 2 0 2 E max Pc 0 32.19: P  S av A   (4r 2 )  E max  2c 0 2r 2 (60.0 W ) (3.00  10 8 m s)  0  E max   12.0 V m. 2 (5.00 m) 2 E max 12.0 V m  Bmax    4.00  10 8 T. c 3.00  10 m s 8
  5. 32.20: a) The electric field is in the  y -direction, and the magnetic filed is in the  z - direction, so S  E  B  ( ˆ)  k  i . That is, the Poynting vector is in the  x - ˆ ˆ ˆ j ˆ ˆ direction. E ( x, t ) B ( x, t ) E B b) S ( x, t )    max max cos (kx  t ) 0 0 E max Bmax  (1  cos(2(t  kx))). 2 0 But over one period, the cosine function averages to zero, so we have: E B | S av |  max max . 2 0 dp S av 780 W m 2 32.21: a) The momentum density  2   8.7  10 15 kg m 2  s. dV c (3.0  108 m s) 2 1 dp S av 780 W m 2 b) The momentum flow rate    2.6  10 6 Pa. A dt c 3.0  10 8 m s 1 dp S av 2500 W m 2 32.22: a) Absorbed light: p rad     8.33  10 6 Pa. A dt c 3.0  10 m s 8 8.33  10 6 Pa  p rad   8.23  10 11 atm. 1.013  10 Pa atm 5 1 dp 2S av 2(2500 W m 2 ) b) Reflecting light: p rad     1.67  10 5 Pa. A dt c 3.0  10 m s 8 1.67  10 5 Pa  p rad   1.65  10 10 atm. The factor of 2 arises because the 1.013  10 Pa atm 5 momentum vector totally reverses direction upon reflection. Thus the change in momentum is twice the original momentum. dp S av 2500 W m 2 c) The momentum density  2   2.78  10 14 kg m 2  s. dV c (3.0  10 m s) 8 2 0 0 2 0 E  1 0 32.23: S  E2  E  Ec  c 0 EB  EB   0 0 0 0 c 0  0 0 0 EB E 2    0 cE 2 . 0 0c
  6.    32.24: Recall that S  E  B , so : a) S  i  ( ˆ)   k . ˆ ˆ j ˆ b) S  ˆ  i   k. ˆ j ˆ ˆ c) S  (  k)  (  i )  ˆ. ˆ ˆ ˆ j d) S  i  (  k)  ˆ. ˆ ˆ ˆ j 32.25: Bmax  E max c  1.33  10 8 T      E  B is in the direction of propagation. For E in the + x -direction, E  B is in the + z -  direction when B is in the + y -direction. λ c 3.00  108 m s 32.26: a) x     2.00 m. 2 2 f 2 (75.0  10 6 Hz) b) The distance between the electric and magnetic nodal planes is one-quarter of a λ x 2.00 m wavelength =    1.00 m. 4 2 2 λ v 2.10  108 m s 32.27: a) The node-antinode distance     4.38  10 3 m. 4 4 f 4(1.20  10 Hz) 10 b) The distance between the electric and magnetic antinodes is one-quarter of a λ v 2.10  108 m s wavelength     4.38  10 3 m. 4 4 f 4 (1.20  10 Hz)10 c) The distance between the electric and magnetic nodes is also one-quarter of a wavelength  λ v 2.10  108 m/s    4.38  10 3 m. 4 4 f 4(1.20  10 Hz) 10 λ c 3.00  108 m s 32.28: xnodes     0.200 m  20.0 cm. There must be nodes 2 2 f 2(7.50  108 Hz) 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. λ 32.29: a) x   λ  2x  2(3.55 mm)  7.10 mm. 2 b) x E  x B  3.55 mm. c) v  f λ  (2.20  1010 Hz)(7.10  10 3 m)  1.56  108 m s.
  7.  2 E y ( x, t ) 2  32.30: a)  (2 Emax sin kx sin ωt )  (2kEmax cos kx sin ωt ) x 2 x 2 x  E y ( x, t ) 2 2  2 E y ( x, t )   2k E max sin kx sin t  2 2 E max sin kx sin t   0  0 2 . x 2 c t 2  2 Bz ( x, t )  2  Similarly:  2 (2 Bmax cos kx cos t )  (2kBmax sin kx cos t ) x 2 x x  B z ( x, t ) 2  2  2 B z ( x, t )   2k 2 Bmax cos kx cos t  2 2 Bmax cos kx cos t   0  0 . x 2 c t 2 E y ( x, t )  b)  (2 Emax sin kx sin t )  2kEmax cos kx sin t x x E y ( x, t )  E    2 Emax cos kx sin t   2 max cos kx sin t   2 Bmax cos kx sin ωt. x c c E ( x, t )  B ( x, t )  y   (2 Bmax cos kx cos t )   z . x t t B ( x, t )  Similarly:  z  (2 Bmax cos kx cos t )  2kBmax sin kx cos t x x B ( x, t )    z   2 Bmax sin kx cos t   2 2cBmax sin kx cos t x c c B ( x, t )  E y ( x, t  z   0  0 2 E max sin kx cos t   0  0 (2 E max sin kx sin t )   0  0 x t t c 3.00  108 m s 32.31: a) Gamma rays: λ    4.62  10 14 m  4.62  10 5 nm. f 6.50  10 Hz 21 c 3.00  108 m s b) Green light : λ    5.22  10 7 m  522 nm. f 5.75  10 Hz 14 c 3.0  108 m s 32.32: a) f    6.0  10 4 Hz. λ 5000 m c 3.0  108 m s b) f    6.0  10 7 Hz. λ 5.0 m c 3.0  108 m s c) f    6.0  1013 Hz. λ 5.0  10 6 m c 3.0  108 m s d) f    6.0  1016 Hz. λ 5.0  10 9 m
  8. 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:   Q  E  d A  E x A  encl  0  E x  0. ε0    B  d A  Bx A  0  Bx  0. 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 E  Emax ˆ sin( kx  t ) and B  Bmax k sin( kx  t   ), with       . j ˆ Then Eq. (32.12) implies: E y B   z x   kEmax cos(kx  t )  Bmax cos(kx  t   )    0. x t  2f  kEmax  Bmax  Emax  Bmax  Bmax  f λBmax  cBmax . k 2 / λ Similarly for Eq.(32.14) B E y  z   0 0  kBmax cos(kx  t   )   0  0Emax cos(kx  t )    0. x t   2f fλ 1  kBmax   0  0Emax  Bmax  0 0 Emax  2 Emax  2 Emax  Emax . k c 2 / λ c c
  9.   E y ( x, t )    B z ( x, t )   2 B z ( x, t ) 32.35: From Eq. (32.12):      t  x  t    t  t 2   B ( x, t )    E y ( x, t )  But also from Eq. (32.14):   z     00  x  x  x   t    B z ( x, t ) 2   0 0 t 2  2 B z ( x, t )  2 B z ( x, t )    0 0 . x 2 t 2 1 1 32.36: E y ( x, t )  Emax cos(kx  t )  u E   0 E 2   0 Emax cos(kx  t ) 2 2 2 2  0c 2  Emax  1 B2  uE    cos(kx  t )  Bmax cos(kx  t )  z  u B 2 2  c  2 0 2 0 1 32.37: a) The energy incident on the mirror is Pt  IAt   0 cE 2 At 2 1  E   0 (3.00  108 m s)(0.028 V m) 2 (5.00  10 4 m 2 )(1.00 s)  5.20  10 10 J. 2 2I b) The radiation pressure prad    0 E 2   0 (0.0280 V m) 2  6.94  10 15 Pa. c c) Power P  I  4R 2  cprad 2R 2  P  2 (3.00  10 8 m s)(6.94  10 15 Pa)(3.20 m) 2  1.34  10 4 W.
  10. c 3.00  108 m s 32.38: a) f    7.81  109 Hz. λ 0.0384 m E max 1.35 V m b) Bmax    4.50  10 9 T. c 3.00  10 m s 8 1 1 c) I   0 cE max   0 (3.00  10 8 m s)(1.35 V m) 2  2.42  10 3 W m 2 . 2 2 2 IA EBA (1.35 V m)(4.50  10 9 T)(0.240 m 2 ) d) F  pA     1.93  10 12 N. c 2 0 c 2 0 (3.00  10 m s) 8 P 4P 4(3.20  10 3 W) 32.39: a) The laser intensity I     652 W m 2 . A D 2 3 π (2.50  10 m) 2 1 2I 2(652 W m 2 ) But I   0 cE 2  E    701 V m. 2  0c  0 (3.00  10 8 m s) E 701 V m And B    2.34  10 6 T. c 3.00  10 8 m s 1 1 b) u Bav  u Eav   0 E max   0 (701 V m) 2  1.09  10 6 J m 3 . Note the extra factor 2 4 4 1 of since we are averaging. 2 c) In one meter of the laser beam, the total energy is: E tot  u tot Vol  2u E ( AL)  2u E D 2 L 4  E tot  2(1.09  10 6 J m 3 ) (2.50  10 3 m) 2 (1.00 m)/4  1.07  10 11 J.
  11. 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 get prad  (2  W ) C . Be careful not to confuse p, the momentum of the incoming wave, I with prad , the radiation pressure. I b) (i) totally absorbing W  1 so prad  C 2 (ii) totally reflecting W  0 so p rad  C These are just equations 32.32 and 32.33. (2  0.9) (1.40  103 W m 2 ) c) W  0.9, I  1.40  10 2 W/m 2 prad   5.13  10 6 Pa 3.00  108 ms (2  0.1)(1.40  10 2 W m 2 ) W  0.1, I  1.40  103 W m 2  prad   8.87  10 6 Pa 3.00  10 s 8 m 32.41: a) At the sun’s surface: P P 3.9  10 26 W P  IA  I     6.4  10 7 W m 2 A 4R 2 4 (6.96  10 m) 8 2 I 6.4  10 W m 27  p rad    0.21 Pa. c 3.00  10 8 m s Halfway out from the sun’s center, the intensity is 4 times more intense, and so is the radiation pressure: p rad ( Rsun / 2)  0.85 Pa. At the top of the earth’s atmosphere, the measured sunlight intensity is 1400 W m 2  5  10 6 Pa , which is about 100,000 times less than the values above. b) The gas pressure at the sun’s surface is 50,000 times greater than the radiation pressure, and halfway out of the sun the gas pressure is believed to be about 6  1013 times greater than the radiation, pressure. Therefore it is reasonable to ignore radiation pressure when modeling the sun’s interior structure.
  12. Emax Bmax ˆ 32.42: a) S ( x, t )  (1  cos 2 (kx  t ))i  S ( x, t )  0  cos 2(kx  t )  1, 2 0 which never happens. So the Poynting vector is always positive, which makes sense since the direction of wave propagation by definition is the direction of energy flow. b) dB di d B dB di 32.43: a) B   0 ni   0 n   A   0 nA . 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 b) The direction of the Poynting vector is radially inward, since the magnetic field is along the solenoid’s axis and the electric filed is circumferential. It’s magnitude EB  0 n 2 ri di S  . 0 2 dt B2 (  ni ) 2  0 n 2 i 2  n 2 i 2 la 2 c) u   0   U  u (lA)  ula 2  0 . 2 0 2 0 2 2 Li 2 2U  0n 2 i 2 la 2 But also U   Li     0n 2 ila 2 , and so the rate of 2 i i di di energy increase due to the increasing current is given by P  Li   0n 2 ila 2 . dt dt d) The in-flow of electromagnetic energy through a cylindrical surface located at the    n 2 ai di di solenoid coils is   S  dA  S 2πal  0  2al   0n 2ila 2 . 2 dt dt e) The values from parts (c) and (d) are identical for the flow of energy, and hence we can consider the energy stored in a current carrying solenoid as having entered through its cylindrical walls while the current was attaining its steady-state value.
  13. 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: u   0 E 2  4 0 E max sin 2 kx sin t 2   1  1 b) At t  , cos t  cos  and sin ωt  sin  . 4 4 2 4 2  For 0  x  ˆ ˆ ˆ , sin kx  0, cos kx  0  S  E  B   ˆ  k   i . j ˆ ˆ 2k   And for ˆ ˆ ˆ  x  , sin kx  0, cos kx  0  S  E  B   ˆ  k  i . j ˆ ˆ 2k k 3 3 1 3 1 At t  , cos t  cos  and sin t  sin  . 4 4 2 4 2  For 0  x  ˆ ˆ ˆ j ˆ ˆ , sin kx  0, cos kx  0  S  E  B  ˆ  k  i . 2k   And for ˆ ˆ ˆ j  x  , sin kx  0, cos kx  0  S  E  B  ˆ   k   i . ˆ ˆ 2k k c) the plots from part (a) can be interpreted as two waves passing through each other in opposite directions, adding constructively at certain times, and destructively at others.
  14. I I 32.45: a) E  J   2 , in the direction of the current. A a  I b)  B  d l   0 I  B  0 , counterclockwise when looking into the current. 2a ˆ ˆ ˆ ˆ ˆ c) The direction of the Poynting vector S  E  B  k     ρ, where we have used ˆ cylindrical coordinates, with the current in the z-direction. EB 1 ρI  0 I ρI 2 Its magnitude is S    .  0  0 a 2 2a 2 2 a 3 I 2 lI 2 d) Over a length l, the rate of energy flowing in is SA  2al  . 2 2 a 3 a 2 l lI 2 The thermal power loss is I R  I 2  2 , which exactly equals the flow of A a 2 electromagnetic energy.  0i   q q 32.46: B  , and  S E  dA  EA  E , so the magnitude of the 2r 0  0 r 2 EB qi q dq Poynting vector is S    .  0 2 0 r 2 3 2 0 r dt 2 3 Now, the rate of energy flow into the region between the plates is:   lq dq 1 l d (q 2 ) d  1 l 2  d  q 2  dU   S  dA  S (2rl )   0r 2 dt  2  0r 2 dt  dt  2  0 A q   dt  2C   dt .         This is just rate of increase in electrostatic energy U stored in the capacitor. 2 cBmax 32.47: The power from the antenna is P  IA  4r 2 . So 2 0 2 0 P 2 0 (5.50  10 4 W )  Bmax    2.42  10 9 T 4r 2 c 4 (2500 m) (3.00  10 m s) 2 8 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 P 1 2I 2(2.80  103 W 36 m 2 ) 32.48: I    0 cE 2  E    242 V m. A 2  0c  0 (3.00  108 m s)
  15. 32.49: a) Find the force on you due to the momentum carried off by the light: prad  I c and F  prad A gives F  I A c  Pav / c a x  F m  Pav (mc)  (200 W ) [(150 kg)(3.00  10 8 m s)]  4.44  10 9 m s 2 2 Then x  x0  v0 x t  1 a x t 2 gives t  2( x  x0 ) a x  2(16.0 m) (4.44  10 9 m s )  2 8.49  10 4 s  23.6 h The radiation force is very small. In the calculation we have ignored any other forces on you. b) You could throw the flashlight in the direction away from the ship. By conservation of linear momentum you would move toward the ship with the same magnitude of momentum as you gave the flashlight. P 1 2P 2Vi 32.50: P  IA  I    0 cE 2  E   A 2 A 0 c A 0 c 2Vi 2(5.00  105 V) (1000 A) E   6.14  10 4 V m . A 0 c (100 m 2 )ε0 (3.00  108 m s) And E 6.14  10 4 V m B   2.05  10 4 T. c 3.00  10 m s 8 GM S m GM S 4R 3 ρ 4GM S R 3 ρ 32.51: a) FG   2 .  . r2 r 3 3r 2 b) Assuming that the sun’s radiation is intercepted by the particle’s cross-section, we can write the force on the particle as: IA L R 2 LR 2 F  .  . c 4r 2 c 4cr 2 c) So if the force of gravity and the force from the radiation pressure on a particle from the sun are equal, we can solve for the particle’s radius: 4GM S R 3 ρ LR 2 3L FG  F   R . 3r 2 4cr 2 16GM S c 3(3.9  10 26 W ) R 16π (6.7  10 11 N  m 2 kg 2 ) (2.0  1030 kg) (3000 kg m 3 ) (3.0  108 m s)  R  1.9  10 7 m. d) If the particle has a radius smaller than that found in part (c), then the radiation pressure overcomes the gravitational force and results in an acceleration away from the sun, thus removing all such particles from the solar system.
  16. 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, the latter is: 2 LA GM S m 2 LA 4GM S mc Frad  . Thus : FG  Frad    A 4r c 2 r 2 4r c2 2L 4 (6.7  10 N  m kg ) (2.0  10 kg) (10000 kg) (3.0  108 m s) 11 2 2 30  A (2)3.9  10 26 W 6.48 km 2  A  6.48  10 6 m 2  6.48 km 2  2  2.53 mi 2 (1.6 km mile) c) This answer is independent of the distance from the sun since both the gravitational force and the radiation pressure go down like one over the distance squared, and thus the distance cancels out of the problem.  q2a2  2 C 2 (m s ) 2 Nm J  dE  32.53: a)  3  2   W .  6 0 c  (C N  m ) (m s) 2 3 s s  dt  b) For a proton moving in a circle, the acceleration can be rewritten: 2(6.00  10 6 eV) (1.6  10 19 J eV) 2 v 2 1 mv 2 a  21   27  1.53  1015 m s . R 2 mR (1.67  10 kg) (0.75 m) The rate at which it emits energy because of its acceleration is: dE q2a2 (1.6  10 19 C) 2 (1.53  1015 m s 2 ) 2    1.33  10  23 J s dt 6 0 c 3 6 0 (3.0  10 m s) 8 3  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. That means the electron’s acceleration is the same as the proton, and thus so is the rate at which it emits energy, since they also have the same charge. However, the electron’s 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)  3273 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. E 3273 eV
  17. 32.54: For the electron in the classical hydrogen atom, its acceleration is: 2(13.6 eV)(1.60  10 19 J eV) 2 v 2 1 mv a  21   31 11  9.03  10 22 m / s 2 . R 2 mR (9.11  10 kg)(5.29  10 m) Then using the formula for the rate of energy emission given in Pr. (33-49): dE q 2a 2 (1.60  10 19 C) 2 (9.03  10 22 m s 2 ) 2   dt 6 0 c 3 6π 0 (3.00  108 m s) 3 dE   4.64  10 8 J s  2.89  1011 eV s, which means that the electron would almost dt immediately lose all its energy! 32.55: a) E y ( x, t )  Emax e  kc x sin (k c x  t ). E y  Emax ( k c )e kc x sin(k c x  t )  Emax ( k c )e  kc x cos(k c x  t ). x  Ey2  Emax ( k c2 )e kc x sin(k c x  t )  Emax ( k c2 )e kc x cos(k c x  t ) x 2  Emax ( k c2 )e  kc x cos(k c x  t )  Emax ( k c2 )e  kc x sin(k c x  t )  2 Emax k c2 e kc x cos(k c x  t ). E y  Emax e kc x  cos(k c x  t ). t  Ey 2 E y Set   2 Emax k c2 e k x cos(k c x  t )  Emax e k x ω cos(k c x  t ). This will c c x 2 t 2k c2   only be true if  or kc  .   2 b) The hint basically answers the question. E 1 2 2(1.72  10 8 m) c) E y  y 0  k c x  1, x     6.60  10 5 m. e kc  2 (1.0  10 Hz) 0 6
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