Tiêu đề 35. Electromagnetic waves MedAns.pdf ✅

1. (b) 2. (c) 3. (c) 4. (c) 5. (d) 6. (d) 7. (b) 8. (a) 9. (a) 10. (c) 11. (a) 12. (c) 13. (c) 14. (c) 15. (b) 16. (a) 17. (c) 18. (c) 19. (a) 20. (d) 21. (c) 22. (a) 23. (a) 24. (d) 25. (c) 26. (c) 27. (c) 28. (c) 29. (c) 30. (d) 31. (d) 32. (c)Velocity of light waves in material is v = n ...(i) Refractive index of material is  = c v ...(ii) where c is speed of light in vacuum or air. or  = c n ...(iii) Given , n = 2 × 1014 Hz  = 5000 Å = 5000 × 10–10 m c = 3 × 108 m/s Hence, from Eq. (iii), we get  = 8 14 10 3 10 2 10 5000 10−     = 3.00 33. (d)The amplitudes of the electric and magnetic fields in free space are related by 0 0 E c B = In figure, electric field vector (E) and magnetic field vector (B) are vibrating along Y and Z directions and propagation of electromagnetic wave is shown in X- direction. Hence, electric and magnetic fields are in phase and perpendicular to each other. 34. (c)Velocity of electromagnetic radiation is the velocity of light (c), ie, 0 0 1   where 0 is the permeability and 0 is the permittivity of free space. 35. (a) 36. (c) 37. (c) 38. (a) 39. (c) 40. (c) 41. (b) 42. (c) 43. (d) 44. (d) 45. (b) 46. (a) 47. (d) 48. (d) 49. (c) 50. (d) 51. (c) 52. (c) 53. (c) 54. (d) 55. (a) 56. (b) 57. (d) 58. (a) 59. (a)
60. (b) 61. (a) 62. (a) 63. (c) 64. (d) 65. (d) 66. (b) 67. (b) 68. (b) 69. (a) 70. (a) 71. (a) 72. (a) 73. (a) 74. (b) 75. (c) 76. (d) 77. (b) 78. (c) 79. (a) 80. (b)Electromagnetic wave require no medium for their propagation. 81. (c) 82. (a) 83. (b) 84. (d) 85. (d)The wavelength order of waves are given below : Waves Wavelength (in Å) (a) X- rays 1 Å to 100 Å (b) Ultraviolet rays 100 Å to 4000 Å (c) - rays 0.001 Å to 1 Å (d) Cosmic rays upto 4 × 10–4 Å Thus, Cosmic rays have the minimum wavelength 86. (c) Velocity = c  and  decreases as wavelength increases i.e., r < v  r < v 87. (c)Here : Velocity of electromagnetic waves in free space and wavelength v = 3 108 ms and  = 150 m Using the relation for the frequency of radio waves is given by v= 8 v 3 10 150   = 2 × 106 Hz - 2 MHz 88. (a)As velocity of light is perpendicular to the wavefront and light is travelling is vacuum along the y-axis, therefrore, the wavefront is represented by y = constant. 89. (d)Energy E  hc L E hv = =    we know that infraredvojDr > visiblen`';  E infrared vojDr < Evisible n`'; 90. (d) 91. (a) E 6.3j ˆ = (v/m) and f = 20 MHz direction of propagation = ˆ k || (E B ) then ( ) ( ) 8 E 6.3 B i i T ˆ ˆ c 3 10 = − = −  = 2.1 × 10–8 T (– ˆ i ) 92. (b) 93. (a,b) 94. (b) 95. (d) 96. (b) 97. (d)Maxwell's equation describe laws of electricity and magnetism. 98. (c) 99. (b) 100.(a) 101.(c) 102.(c)Frequency range for soft X-rays and hard X-rays is different. 103.(c) 104.(d) 105.(a)  = 6 × 108z k v  = = 8 8 6 10 3 10   = 2m–1 106.(b)U = 2 0 1 E 2  = 2 0 1B 2  00 = 2 2 B E 0 0 B 1 E c =   = 107.(d)Frequency of microwaves = Resonant frequency of H2O molecules. So there is resonant absorption of microwave 108.(a) 109.(b)z = R X (100) (100) 100 2 2 2 2 c 2 + = + = imax = 100 2 220 2 z vmax = = 2.2
110.(b)E0 = 2 Erms = 2 × 6 V/m B0 = 0 2 E 2 6 T C 3 10  =  = 2 ×10–8 T = 2×1.414 × 10–8 T = 2.828× 10–8 T 111.(d) c x ˆ → direction E y ˆ → direction B z ˆ →+ direction E x B 112.(b)Wavelength is maximum for red 113.(c)Q = CV dt dQ = i = C dt dv = 20 μF × s 3V = 60 μA For circuit to be completed displacement current should be equal to conduction current. 114.(b)Direction of polarization is the direction of electric field and wave propagation along E B which is direction of propagation. 115.(b) | E | c E | B | c | B | =  = = 20 × 10–9 × 3 × 108 = 6 V/m. 116.(c)Both the energy densities are equal. 117.(d) 118.(b)Intensity  = 2 P 4 r  I = 2 0 0 1 E c 2   So 2 2 0 0 P 1 E c 4 r 2 =    2 0 2 0 2P E 4 r c =   = 9 8 2 0.1 9 10 1 3 10      E0 = 6 = 2.45 V/m 119.(d)Y X U V I M R   increasing VIBGVORX Rays X Rays MW RW  c<+r s Øe esa VIBGVORX Rays X Rays MW RW Hence energy of radio wave will be minimum and maximum for X ray. 120.(a)C = Speed in air V = Speed in medium V 1 C 2 = 2  = r 1 (Non-magnetic) 1 2 r r V 1 C 2  = =   1 2 r r 1 4  =  121.(d)Energy is equally distributed in electric field & magnetic field 122.(d) E B = C  E = B.C. = 10–4 × 3 × 108 v / m = 3 ×104 v/m 123.(c)In air = 0 0 E C B In the medium of refractive index = n = E C B n It is possible if E = E0 n and B B n = 0  B E 0 0 1 , n B E n = = 124.(d) 0 0 E B = C (speed of light in vacuum ) E0 = B0C = 3 × 10–8 × 3 × 108
= 9 N/C So blfy, E = ( ) 3 10 9sin 1.6 10 x 48 10 t  +  125.(a) 45o 45o O i i 3 2 4 1 i ( ) ( ) ( ) ( ) B0 B0 1 B0 2 B0 3 B0 4      = + + +   2R i sin90o sin45o 4 R i 0 0 −  +   + 4 R i 0   (sin45o + sin90o)       +   +   +      −  −  = 1 2 1 4 R i 2R i 2 1 1 4 R i 0 0 0       − + +  + +   = 1 2 1 2 2 1 1 4 R i 0  +    = 2 2 4 R i 0       +    = 2 1 2 R i 0 126.(b) c B E = E = B × c = 15 N/c 127.(c) East North West South B Proton E = 1 Mev K.E. = 1.6 × 10–13 = 1 27 1.6 10 2 −   v 2 v = 7 2 10  m/s  Bqv = ma B = 27 12 19 7 1.6 10 10 1.6 10 2 10 − −      = 0.71× 10–3 T So, 0.71 mT 128.(a)Magnetic field vectors associated with this electromagnetic wave are given by k ˆ c E B 0 1 =  cos (kx – t) & i ˆ c E B 0 2 =  cos (ky – t) F qE q(V B)     = +  q(E E ) q(V (B B )) 1 2 1 2      = + +  + by putting the value of E1  , E2  , B1  & B2  The net Lorentz force on the charged particle is  kˆ j 0.2cos(ky t) ˆ i cos(kx t) ˆ F qE 0.8cos(kx t) = 0 − + − + −  at t = 0 and at x = y = 0 t = 0 x = y = 0 ij k] ˆ j 0.2 ˆ i ˆ F = qE0 [0.8 + +  129.(c)EM wave is in direction fo Electric field is in direction fo| E B    → direction of propagation of EM wave fo| 130. (b) 1 E 2 1 2 2 h mE hc E   = 1 1 2 2 E    = 131. (a) 1 2 2 2 1 2      = + Let us assume that one of the particles is moving along the positive x direction and the other is moving along the positive y direction. The net momentum of the system is P Pi P j net = + 1 2 h h h i j    = +2 2 1 2 1 1 1    = + 1 2 2 2 1 2      = + 132. (d) 1.1 eV KEmax = E - 