Title 10-electromagnetic-waves-.pdf ✅

Electromagnetic waves 1. (C) As λ = hc E Where the symbols have their usual meanings Here, E = 15keV = 15 × 103 V And hc = 1240eVnm ∴ λ = 1240eVnm 15 × 103eV = 0.083 nm As the wavelength range of X-rays in from 1 nm to 10−3 nm, so this wavelength belongs to X-rays. 2. (D) Energy of radiation, E = hv = hC λ Also, its momentum p = h λ = E C = p1pr = −pi = − E C So, momentum transferred to the surface = pi − pr = E C − (− E C ) = 2E C 3. (B) Here, Energy flux, I = 25 × 104Wm−2 Area, A = 15 cm2 = 15 × 10−4 m2 Speed of light, c = 3 × 109 ms−1 For a perfectly reflecting surface, the average force exerted on the surface is F = 2IA c = 2 × 25 × 104Wm−2 × 15 × 10−4 m2 3 × 108 ms−1 = 250 × 10−8 N = 2.50 × 10−6 N 4. (C) In microwave oven the frequency of the microwaves must match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of the molecules. 5. (B) Frequency of electromagnetic wave does not change with change in medium but wavelength and velocity of wave changes with change in medium. Velocity of electromagnetic wave in vacuum c = 1 √μ0ε0 = vλvacuum Velocity of electromagnetic wave in the medium vmedium = 1 √μ0μr∈0∈r = c √μrεt Where μ, and εr , be relative permeability and relative permittivity of the medium. For dielectric medium, μr = 1 ∴ vmedium = c √εr Here, εr = 4.0 ∴ vmedium = c √4 = c 2 Wavelength of the wave in medium λmedium = vmedium v = c 2v = λvacuum 2 (Using (i) and (ii)) 6. (A) Compare the given equation with E = E0cos (kz − ωt) We get, ω = 6 × 108 s −1 We get, ω = 6 × 108 −1 Wave vector, k = ω c = 6×108 s −1 3×108 ms−1 = 2 m−1 7. (B) The amplitude of magnetic field and electric field for an electromagnetic wave propagating in vacuum are related as E0 = B0C Where c is the speed of light in vacuum. ∴ B0 E0 = 1 c 8. (A) The electromagnetic wave is propagating along the +z axis. Since the electric and magnetic fields are perpendicular to each other and also perpendicular to the direction of propagation of wave. Also, E⃗ × B⃗ given the direction of wave propagation. ∴ E⃗ = E0iˆ, B = B0jˆ (∵ iˆ × jˆ = kˆ ) 9. (A) The decreasing order of wavelength of the given electromagnetic waves is as follows: λMicrowave > λInfrared > λUltraviolet > λGamma rays 10. (C) In an electromagnetic wave both electric and magnetic vectors are perpendicular to each other as well as perpendicular to the direction of propagation of wave. 11. (D) As given, E = 10cos (107 t + kx) Comparing it with standard equation of e.m. wave, E = E0cos (ωt + kx) Amplitude E0 = 10 V/m and ω = 107 rad/s ∵ c = νλ = ωλ 2π Or λ = 2πc ω = 2π×3×108 107 = 188.4 m; Also, c = ω k or k = ω c = 107 3×108 = 0.033 The wave is propagating along y direction. 12. (B) Ey = 2.5 [(2π × 106 rad m ) t − (π × 10−2 rad x ) x] E2 = 0 The wave is moving in the positive direction of x. This is the from Ey = E0(ωt − kx)ω = 2π × 106 2πv = 2π × 106 ⇒ v = 106 Hz
2π λ = k ⇒ 2π λ = π × 10−2 ⇒ λ = 2π π × 10−2 = 2 × 102 = 200 m 13. (A) The velocity of electromagnetic radiation in vacuum is 1 √μ3ε0 , where μ0 and ε0 are the permeability and permittivity of vacuum. 14. (C) In electromagnetic wave, electric and magnetic field are in phase and perpendicular to each other and also perpendicular to the direction of the propagation of the wave. 15. (B) λm > λv > λx. In spectrum X-rays has minimum wavelength and microwave has maximum wavelength 16. (A) Every body at all time. All al temperatures emit radiation (except at T = 0 ), which fall in the infrared region. 17. (C) 18. (B) According to Maxwell, the electromagnetic waves are those waves in which there are sinusoidal variation of electric and magnetic field vectors at to the direction of wave propagation. If the electric field (E⃗ ) and magnetic field (B⃗ ) and vibrating along Y and Z direction, propagation of electromagnetic wave will be along the X-axis. Therefore, the velocity of electromagnetic wave is parallel to E⃗ × B⃗ . 19. (A) As the electro- magnetic radiations from sun pass through the atmosphere, some of them are absorbed by it while other reach the surface of earth, the range of wavelength which reaches earth lies in infrared region. This part of the radiation from the sum has shorter wavelength and can penetrate through the layer of gases line CO2 and reach earth surface. But the radiation from the earth being of longer wavelength can't escapes through this layer. As a result the earth surface gets warm. This is known as green house effect. 20. (A) The ozone layer absorbs the harmful ultraviolet rays coming from sun. 21. (A) 22. (D) The range is from 380 mm to even 200 mm to 120 mm 23. (B) λ = 3×108 100 Hz = 3 × 106 m 24. (B) c = 1 √μ0ε0 (free space) v = 1 √με (medium) ∴ μ = c v = √ με μ0ε0 25. (B) 26. (D) 27. (B) X-rays are used for the investigation of structure of solids. 28. (B) The wave length of radiation used should be less than the size of the particle Size of particle = λ = c v 3 × 10−4 = 3×1010 v or v = 1014 hertz However, when frequency is higher than this, wavelength is still smaller, Resolution becomes better. 29. (D) Rays wave length [Range in m ] X-rays 1 × 10−11 to 3 × 10−8 γ - rays 6 × 10−14 to 1 × 10−11 Microwaves 10−3 to 0.3 Radiowaves 10 to 104 30. (A) Eradiowave < Eyellow < Eblue < Ex− ray (D) (B) (A) (C) 31. (A) 32. (D) An electromagnetic wave propagating in +x direction means electric field and magnetic field should be function of x and t. Also, E⃗ ⊥ B⃗ or Eˆ ⊥ Bˆ i.e. ; (yˆ − zˆ) ⋅ (yˆ + zˆ) = yˆ ⋅ yˆ − zˆ ⋅ zˆ = 0 33. (D) v = 2 × 1014 Hz E0 = 27Vm−1 We know, E0 B0 = c ; So B0 = 27 3×108 = 9 × 10−8 T λ = c v = 3 × 108 2 × 1014 = 1.5 × 10−6 m B = B0sin 2π ( x λ − vt) ; B = (9 × 10−8T)sin 2π ( x 1.5 × 10−6 2 × 1014t)
Oscillation of B can be along either jˆ or kˆ direction. 34. (B) For electromagnetic wave, direction of propagation, E⃗ and B⃗ are transverse in nature. According to questions, E⃗ × B⃗ = direction of propagation = +z direction. Only option (b) satisfies both conditions (i) E ̅ ⋅ B ̅ = 0 (ii) (E ̅ × B ̅ ) directed along the z-axis. 35. (2) For partially absorbed and reflected F = 1 c ( dE dt ) (2r + a) But r + a = 1 ⇒ r = 1 − a F = 1 c dE dt [2(1 − a) + a] = 1 c dE dt [2 − a] Radiation pressure = E A = SA c (2−f) A = S c (2 − f) 36. (13) P = 100 W r = 2 m D = Diameter of Pupil = 4 × 10−3 m λ = 600 × 10−9 m D = Diameter of Pupil = 4 × 10−3 m λ = 600 × 10−9 m P = n hc λ ; n = no. of photons /sec No. of photons per unit by source = P A hc λ = λP Ahc ∴ No. of photons entering the eye = λP Ahc × ( πD 2 4 ) Here, A = π(2) 2 N = no. of photons entering the eye = 600×10−9×100 π(2) 2×6.62×10−34×3×108 × π(4×10−3 ) 2 4 = 7.55 × 1013 37. (19) Energy absorbed = mCdT n hc λ = mCdT Here, n = no. of photons n hc λ = Vol.× ρ × CdT n = Vol.× ρ × CdT × λ hc = 10−6 × 1000 × 1 × 1 × 575 × 10−9 6.62 × 10−34 × 3 × 108 n = 1.2 × 1019 38. (2) Rad. Of capacitor = 0.12 Capacitance, C = 5 × 10−6 F Resistance, R = 4 × 105Ω ID = ε0dφE dt = ε0A d dV dt = CdV dt = 5 × 10−6 × 2000 = 0.01 A 39. (5) I = 220 W/m2 R = 0.3 m F = dP dt = 1 c dE dt (2r + a) = IA c (2r + a) dp = IA c (2r + a) × dt = 220 × π(0.3) 2 3 × 108 [2 × 0.4 + 0.6] × 5 × 60 = 8.71 × 10−5 kg m/s 40. (3) r = distance = 6 m E0 = electric field amplitude = 10 V/m E0 B0 = c ⇒ B0 = E0 c = 10 3 × 108 = 3.33 × 10−8T 41. ( 500) P 4πR2 = 1 2 ε0E 2 c R = √ 2P 4πε0Ec 2 On substitution, we get R ∼ 5.00 m 42. (714) Erms = √μ0CI 43. (18) F = Projected Area × radiation pressure = πR 2 I C 44. (13) Number of photons Area s = Pλ 4πhcr 2 = 6 × 1013 cm−2 s −1 45. (2) E = 12400 λÅ eV = 12400 6000 = 2eV 46. (2) E = 2.5cos (2π × 106 t − π × 10−2x) Comparing with E = E0cos (ωt − kx) k = π × 10−2 2π λ = π × 10−2 λ = 200 m = (100x) meter Where, x = 2 47. (2) Iav = ( 1 2 ε0E0 2 ) c = 8.854×10−12×(38.8) 2×3×108 2 ≃ 2watt/m2 48. (4) λ = c J = 3×108 7.5×106 = 40 m 49. (4) B0 = E0 c = 120 3×108 = 4 × 10−7T ∴ n = 4