 # Physics exercises_solution: Chapter 32

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4 ## 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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## Nội dung Text: Physics exercises_solution: Chapter 32

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. 