Tiêu đề 8.ELECTROMAGNETIC WAVES - Explanations.pdf ✅

1 (a) Here, amplitude of electric filed, E0 = 100 Vm−1 ; amplitude of magnetic field, B0 = 0.265 Am−1 . We know that the maximum rate of energy flow S = E0 × B0 = 100 × 0.265 = 26.5 Wm−2 2 (d) Energy of a photon E = hc λ ∴ Wavelength λ = hc E = 6.6 × 10−34 × 3 × 108 13.2 × 103 × 1.6 × 10−19 = 0.9375 × 10−10 m = 1 Å Wavelength range of X-rays is from 10-11 m to 10- 8 m (0.1 Å to 100 Å). Therefore, the given electromagnetic radiation belongs to the X-ray region of electromagnetic spectrum. 3 (b) Energy flowing per sec per unit area from a face is = 1 μ0 [E⃗ × B⃗ ]. It will be in the negative z- direction. It shows that the energy will be flowing infaces parallel to x − y plane and is zero in all other faces. Total energy flowing per second from a face in x − y plane= 1 μ0 (EB sin 90°)a 2 = EBa 2 μ0 4 (c) The relation between electric field vector E, the displacement vector D and the polarization vector P for a dielectric placed in electric field E is given byD = ε0E + P. 6 (d) Ultraviolet is absorbed by the ozone layer. 8 (b) E0 = cB0 = 3 × 108 × 10−4 = 3 × 104 Vm−1 9 (c) t = 2s c = 2 × 38400 × 1000 3 × 108 = 2.5 s 10 (c) Given, By = 3 × 10−7 sin(103x + 6.28 × 1012t). Comparing with the general equation By = B0 sin(kx + ωt) we get k = 103 or 2π λ = 103 ⇒ λ = 2π 103 = 6.28 × 10−3m = 0.63 cm 11 (c) Power = I × area = (1.4 × 103 ) × 5 Force F = Power c = 1.4×103 ×5 3×108 = 2.33 × 10−5 N 15 (b) Radioactive source, X-ray tube, sodium vapour lamp, crystal oscillator 17 (d) Generally, temperature of human body is 37°C, corresponding to which IR and microwave radiations are emitted from the human body 18 (c) Velocity of light, c = 1 √μ0ε0 or μ0ε0 = 1 c 2 = 1 (ms−1)2 = s 2m−2 20 (a) As v of charged particle is remaining constant, it means force acting on charged particle is zero. So, q(v × B) = qE ⇒ v × B = E ⇒ v = E × B B2 22 (c) The frequency of Electromagnetic Waves produced by the oscillator is equal to the frequency of the oscillating particle ie, 109 Hz. 23 (b) All the component of electromagnetic spectrum have same velocity, ie, 3 × 108 ms−1 . 24 (a) Here, λ = c v = 3×108 8.2×106 = 36.6 m 25 (b) ψ(x,t) = 103 sinπ(3 × 106 x − 9 × 1014 t) = 103 sin3 × 106 π(x − 3 × 108 t) Comparing it with the relation ψ(x,t) = a sin 2π λ (x − ct); We note that c = 3 × 108 ms −1 26 (c) Equation second shows that the electromagnetic wave travels along the positive x-axis 27 (d) In electromagnetic wave, the average value of electric filed or magnetic field is zero 28 (c) i = dq dt = d dt (q0 sin 2πft) = q02πf cos 2πft 29 (b) Diffraction takes places when the wavelength of wave is comparable with the size of the obstacle in path. The wavelength of radio waves is greater than the wavelength of light waves. Therefore, radio waves are diffracted around building 30 (b) Refractive index= c0 c = 1/√μ0ε0 1/√με = √ με μ0ε0 31 (c) Using Ampere circuit law,
∮ B⃗ ∙ dl ⃗⃗⃗ = μ0iD or B2πr = μiD or B = μ0iD/2πr 32 (d) d = √2hR Population covered = πd 2 ×population density = 3.114 × (2 × 0.1 × 6.37 × 103 ) × 1000 ≈ 40 lakh 33 (b) Initial momentum of surface Pi = E C Where, c = velocity of light (constant). Since, the surface is perfectly, reflecting, so the same momentum will be reflected completely. Final momentum Pf = E c (negative value) ∴Change in momentum Δp = pf − pi = − E c − E c = − 2E c Thus , momentum transferred to the surface is Δp ′ = |∆p| = 2E c 34 (d) The wavelength order of the given types of waves are given below Waves Wavelength Range (in meter) Gamma rays 10−14 − 10−10 IR-rays 7 × 10−7 = 10−3 UV-rays 10−9 − 4 × 10−7 Microwave 10−4 − 100 Hence, statements (A) and (D) are correct. 36 (b) The ozone layer absorbs ultraviolet radiations 37 (b) Wavelength, λ = c/v = 3 × 108 /2 × 1010 = 1.5 × 10−2 m 39 (d) Given refractive index n = 1.5 Permeability μ0 = 5 × 10−7 n = √μrεr εr = n 2 μr or ρ = n 2μ0 μ (∵ μr = μ μ0 ) or εr = (1.5) 2 × 4π × 10−7 5 × 10−7 or εr = 6 40 (b) According to Daun-Hunt law, the wavelength of X-rays lies between minimum to certain limit 41 (a) The amplitude of the electric and magnetic fields in free space are related by E0 B0 = c Here, E0 = 30 Vm−1 , c = 3 × 108 ms−1 ∴ B0 = E0 c = 30 3 × 108 = 10−7 T 42 (c) Here, kx = θ or k = θ/x And ωt = θ0 or ω = θ0/t ∴ k/ω = t/x = 1/(x/t) = 1 v , where v is the velocity of electromagnetic wave, which is independent of wavelength of wave but depends upon the nature of medium of propagation of wave 43 (b) vferrite = c √μrεr = 3 × 108 √10 × 1033 = 3 × 106 ms −1 λferrite = vferrite v = 3×106 90×106 = 3.33 × 10−2 m 44 (a) The frequency of Electromagnetic Wave along y −direction v = 30 MHz The electric field component of the wave along y- direction. E = 6 Vm−1 In Electromagnetic, the ratio of the amplitudes of electric and magnetic field is always constant and it is equal to velocity of the Electromagnetic Waves. ie, E B = c or B = E c = 6 3 × 108 or B = 2 × 10−8 T 45 (b) Energy contained in a cylinder U = avearge energy density × volume = 1 2 ε0E0 2 × Al = 1 2 × (8.85 × 10−12) × (50) 2 × (10 × 10−4) × 1 = 1.1 × 10−11 J 46 (a) d = √2hR or h = d 2 /2R 47 (d) B0 = E0 c = 9 × 103 3 × 108 = 3 × 10−5 T 48 (b)
eV = hc/λ Or λ = hc/eV ie, λ ∝ 1/V 49 (c) C = ε0KA d = (8.85 × 10−12) × 10 × 1 10−3 = 8.85 × 10−8 F i = d dt (CV) = C dV dt = 8.85 × 10−8 × 25 = 2.2 × 10−6 A 52 (d) Given , frequency of EM waves v = 8.196 × 106 Hz Velocity of EM waves (v) = 3 × 108 m/s ∴Wavelength of EM waves λ = v V = 3 × 108 8.196 × 106 = 36.60 m = 3660 cm 53 (a) E = E0 sin(kx − ωt) Comparing with standard equation we will get Wavelength = k ω 54 (b) Velocity of light in vacuum c = 1 √μ0ε0 velocity of light in medium v = 1 √μ ε ∴ μ = c v = ( με μ0ε0 ) 1/2 55 (c) A changing electric field produces a changing magnetic field and vice − versa which gives rise to a transverse wave known as Electromagnetic Wave. The time varying electric and magnetic fields are mutually perpendicular to each other and also perpendicular to the direction of propagation of this wave. 57 (a) Consider a loop of radius r(< R) between the two circular plates, placed coaxially with them. The area of the loop = πr 2 By symmetry magnetic field is equal in magnetic at all points on the loop. If iD ′ is the displacement current crossing the loop and iD is the total displacement current between plates iD ′ = iDr πR2 × πr 2 . Using Ampere Maxwell’ law we have, ∮ B⃗ ∙ dI ⃗⃗ = μ0iD ′ or 2πr = μ0iD πr 2 πR2 or B = μ0iDr 2πR2 59 (b) For an EM wave, E0 B0 = c or E0 = cB0 = 3 × 108 × 510 × 10−9 NC−1 = 153 NC−1 60 (d) Electromagnetic Waves are not deflected in electric and magnetic fields. 61 (b) c = 1 √μ0ε0 and c = E0/B0 62 (d) The equation of electric field occurring in Y- direction Ey = 66 cos 2π × 1011 (t − x c ) Therefore, for the magnetic field in Z-direction Bz = Ey c = ( 66 3 × 108 ) cos 2π × 1011 (t − x c ) = 22 × 10−8 cos 2π × 1011 (t − x c ) = 22 × 10−7 cos 2π × 1011 (t − x c ) 63 (d) The earth’s atmosphere above the height of 80 km up to 400 km is called Ionosphere 64 (a) d = √2hR or d ∝ h 1/2 65 (b) B = μ0 4π 2iD r = μ0 4π × ε0 dφE dt = μ0 2π 2iD r = μ0 4π 2 r × ε0 dφE dt = μ0ε0πr 2dE 2πrdt = μ0ε0r 2 dE dt 66 (d) Intensity of electromagnetic wave is I = Pav 4π×r 2 = E0 2 2μ0c or E0 = √ μ0cPav 2πr 2 = √ (4π × 10−7) × (3 × 108) × 800 2π × (4) 2 = 54.77 Vm−1 67 (b) When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly transmitted. Q = Qa + Qr + Qt And Qa Q + Qr Q + Qt Q = a + r + t = 1 ⇒ 15 150 + 0.6 + x = 1 or 0.1 + 0.6 + x = 1
or x = 0.3 Transmitting power, t = Qt Q Or 0.3 = Qt 150 ⇒ Qt = 45 J 70 (b) The wavelength of X-rays is of the order of 1 Å to 100 Å. The wavelength of radiowaves is of the order of 109Å to 101.4Å. The wavelength of microwaves is of the order of 107Å to 109Å. Thus, λX < λM < λR The waves with less wave length will have more energy. Hence, EX > EM > ER 71 (a) Intensity or power per unit area of the radiations, P = pv ⇒ p = P v = 0.5 3 × 108 = 0.166 × 10−8 Nm−2 72 (a) From Maxwell’s Electromagnetic theory, the Electromagnetic Wave propagation contains electric and magnetic fields vibrating perpendicularly to each other . Hence, changing of electric field gives rise to magnetic field. 73 (c) According to Maxwell, a changing electric field is a source of magnetic field 74 (c) c = 1 √μ0ε0 or 1 μ0ε0 = c 2 = [M0LT−1 ] 2 = [M0L 2T −2 ] 75 (b) Let E = Energy falling on the surface per second = 10 J Momentum Of photons p = h λ = h (c1v) = hv c = E C On reflection, Change in momentum per second = 2p = 2E c We know that, Change in momentum per second = force F = 2E c = 2 × 10 3 × 108 = 6.7 × 10−8 N 76 (b) Range, R = √2hr where r is the radius of earth so R ∝ h 1/2 77 (b) Infrared radiations are detected by pyrometer 78 (c) The direction of propagation of Electromagnetic Wave is in the plane perpendicular to both E and B ie, along E × B. 79 (a) i = dQ dt = d dt (CV) = C dV dt = 2 × 10−12 × 1012 = 2 A 80 (c) Speed of Electromagnetic Waves in vacuum = 1 √μ0ε0 = costant 81 (c) Number of oscillator in coherence length = l λ = 0.024 5900 × 10−10 = 4.068 × 106 82 (a) For an Electromagnetic Wave (in vacuum), Velocity c = 1 √μ0ε0 = 3 × 108 ms−1 Air acts almost as vacuum, hence a = 3(approx) 84 (c) In vacuum, ε0=1 In medium, ε = 4 So, refractive index μ = √ε/ε0 = √4/1 = 2 wavelength λ ′ = λ μ = λ 2 and wave velocity v = c μ = c 2 Hence, it is clear that wavelength and velocity will become half but frequency remains uncharged when the wave is passing through any medium. 85 (b) In an electromagnetic wave, the average energy density of magnetic field μB = average energy density of electric filed vE = 1 4 ε0E0 2 = 1 4 × (8.85 × 10−12) × 1 2 = 2.21 × 10−12 Jm−3 86 (b) The density of air in mesosphere with height decreases from 1/ 103 to 1/105 times that due to the surface of earth 87 (d) The electron placed in the path of electromagnetic wave will experience force due to electric field vector and not due to magnetic field vector 88 (c) Radiation force=momentum transferred per sec by electromagnetic wave to the mirror = 2SavA c = 2 × (10) × (20 × 10−4 ) (3 × 108)