Title 08. ELECTROMAGNETIC WAVES(H).pdf ✅

NEET REVISION 08. ELECTROMAGNETIC WAVES(H) NEET REVISION Date: March 18, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on 2. () : Explana on 3. () : Explana on gives the clue that energy flowing per second in faces parallel to plane and zero in all others. 4. () : Explana on From a diode antenna, the electromagne c waves are radiated outwards. Intensity is directly propor onal to square of amplitude . The amplitude of electric field vector which transports significant energy from the source falls off inversely as the distance from the antenna, i.e., . 5. () : Explana on 6. () : Explana on As we know, and should be along -direc on As Hence op on (2) is the correct op on. 7. () : Explana on pressure 8. () : Explana on Microwaves to Ultraviolet to -rays to Infrared to . So, A - S, B - P, C - Q, D - R 9. () : Explana on Given that, Amplitude magne c field is given by Since wave is travelling in direc on and electric field is along axis. So, is along - axis. General equa on of is given by and So, P = = ⇒ n = nE t nhc λt pλt hc = 0.5 × 10 3 × = 100 × 10 18 = 10 20 660 × 10−9 × 60 × 10−3 6.6 × 10 −34 × 3 × 10 8 E0 = B0C = (3 × 10−8) (3 × 10 8) sin(ωt + kx) E0 = 9 sin(1.6 × 10 3x + 48 × 10 10 t) → P = [ → E × → B ] 1 μ0 = a 2 EB 2μ0 xy IαE 2 oα 1 r 2 (E0) (r) E0 ∝ 1 r B = ( ) μ0ε0 2πR πd 2 4 dE dt = 2×10−7×8.85×10−12×3.14×0.01×20 4×0.15 = 1.85 × 10 −18 T → E ⋅ → B = 0 ∵ [ → E ⊥ → B ] → E × → B Z (−2 ^i − 3 ^j) × (−3 ^i − 2 ^j) = 5k^ P = = ΔPmomentum Δt × A P = × = [E = mv 2] E V 1 A × Δt l V P = [V = ] 6 × 10 6 c/μ c μ P = × 3 6 × 10 6 3 × 10 8 = 6 N/m2 −0.1 m 1 mm −400 nm 1 nm; X −1 nm 10 −3 nm −1 mm 700 nm λ = 4 mm, E0 = 60 V /m B0 = = = 2 × 10 E0 −7 c 60 3 × 10 8 +x +y B z B→ B→ = B0 sin[k(x − vt]k^T f = = = × 10 11 c λ 3 × 10 8 4 × 10−3 3 4 ω = 2πf = 2π × × 10 11 = × 10 3c 3 4 π 2 k = = = × 10 3 2π λ 2π 4 × 10−3 π 2 −→ Bz = 2 × 10 −7 sin( × 10 3 (x − 3 × 10 8 t))k^T π 2
NEET REVISION 10. () : Explana on should give direc on of wave propaga on op on-1 op on-2 and 4 do not sa sfy this. wave propaga on vector should be along So correct op on is 3 . 11. () : Explana on Velocity of wave is given by From the given wave eq's velocity in air velocity in medium 12. () : Explana on Area Change in momentum in Momentum delivered to the surface in 13. () : Explana on From the given equa on, Volume Average energy density of an em wave Energy stored volume 14. () : Explana on We are given that, Capaci ve reactance, If is the value of the conduc on current, The conduc on current is equal to the displacement even in case of . Let be the peak value (amplitude) of the conduc on current Clearly, Or Obviously, peak value (amplitude) of is also the same, . Consider a circular path of radius parallel to the plates of the capacitor and with its cen‐ tre on the axis of the plates. According to the Ampere's law, Now circumference of the circle of ra‐ dius As there is no conduc on current in the plates, Further, From Eqs. (1), (2), (3) and (4), Or 15. () : Explana on Gauss's law in electrosta cs Faraday's law Gauss's law in magne sm, , as the net magne c pole strength enclosed by the Gaussion sur‐ face is zero. Ampere's Maxwell law, Eˆ × Bˆ ⇒ Kˆ × BˆII ^i + ^j √2 Kˆ × ( ) = = ∥ ^i + ^j √2 ^j − (−^i) √2 ^i + ^j √2 ^i + ^j √2 Kˆ ^i + ^j √2 EM 1 √με = c ω k c 2 = = 2 ⇒ = 1 √μ0εr1 1 εr2 c ( ) c 2 εr1 εr2 1 4 F = Prad × = × Area I c = × 30 = 2 × 10 −6 N 20 3×10 8 = 1sec ∴ 30 min = 2 × 10−6 × 30 × 60 = 36 × 10−4 kg m s−1 E→ = 20 sin ω (t − ) ^jN/C x c E0 = 20 N/C = 5 × 10−4 m3 = ε0E 2 0 1 2 = ( ε0E2 0)( 1 2 ) = × 8.85 × 10−12 × (20) 2 × (5 × 10−4) J 1 2 = 8.85 × 10−13 J C = 100 pF = 100 × 10−12 F = 10−10 F, Vrms = 230 V , ω = 300 rad/s. XC = = = Ω 1 ωC 1 300 × 10−10 10 8 3 Irms rms Irms = = = 6.9 × 10 −6 A = 6.9μA Vrms XC 230 10 8/3 ac I0 (I) Irms = I0 √2 I0 = Irms × √2 = (6.9μA) × 1.414 = 9.76μA Id i. e. , 9.76μA 3.0 cm(= 0.03 m) (O) ∮ B→ ⋅ → dl = μ0 (I + Id) (1) ∮ B→ ⋅ → dl = B× 0.03 m = B(2π × 0.03) (2) I = 0 (3) Id = × π(0.03) 2 9.76(μA) π(0.06) 2 = 2.44 μA = 2.44 × 10−6 A (4) B(2π × 0.03) = μ0 (2.44 × 10−6) B = ( ) μ0 2π 2.44×10 −6 0.03 = (2 × 10−7) (81.3 × 10−6) = 1.63 × 10−11 T ⇒ φ = ∮ E→ ⋅ ds→ = qnet ε0 ⇒ ∮ E→ ⋅ d→l = − dφB dt ∮ B→ ⋅ d →A = 0 ∮ B→ ⋅ d→l = μ0 ic + μ0ε0 dφE dt A − S, B − P, C − Q, D − R
NEET REVISION 16. () : Explana on Given, Total average density of wave Here, we will use, 17. () : Explana on Charge on -par cle . Maximum force on -par cle due to electric field Maximum force on - par cle due to magne c field 18. () : Explana on For equilibrium, the force exerted by the light beam should balance the weight of plate. 19. () : Explana on and energy density , momentum density br>Where - change in mo‐ mentum of waves present in the volume . 20. () : Explana on Area of plates 21. () : Explana on The bulb, as a point source, radiates light in all direc‐ ons uniformly. At a distance of , the surface area of the surrounding sphere is The intensity at this distance is 22. () : Explana on Power, lies in range of Gamma rays 23. () : Explana on Total energy transferred to a surface, Total momentum transferred to a surface for absorp‐ on is given by 24. () : Explana on Mean intensity 25. () : Explana on 26. () : Explana on Given, and Intensity of wave is given by, Intensity, So, correct op on is (3). 27. () : Explana on phase difference between two fields is zero. f = 5 × 10 10 Hz, E0 = 50 V /m EM uT = ε0 1 2 E 2 0 = 1 2 B2 0 μ0 uT = ε0E 2 0 1 2 = × 8.85 × 10−12 × 2500 = 1.106 × 10−8 1 Jm−3 2 q = α (q) = 2 × e α FE = q × E0. = 3.2 × 10−19 × 1.5 FE = 4.80 × 10 −19 N α FB = qvB0 = 3.2 × 10−19 × 3 × 10 8 × 10−9 × 5 FB = 4.80 × 10 −19 N. Fphoton = mg (Fphoton = = , where power P = IA) IA c P c ⇒ = 10 × 10 −3 × 10 P c ⇒ P = 30 × 10 6 W = 30 MW I = ε0E 2 0c 1 2 = I c = = I c 2 ΔP V ΔP EM V = = 2.78 × 10−14 kgm−2 s 2500 −1 9×10 16 A = πr 2 = 3.14 × (0.1) 2 m2 = 5 × 10 dE 3 dt V m×s 1d = ε0 A [ ] {∵ = A } dE dt dφE dt dE dt = 8.85 × 10−12 × 3.14 × (0.1) 2 m2 × 5 × 10 C 13 2 Nm2 V m×s 1d = 13.9 A 3 cm A = 4πr 2 = 4π(3) 2 = 113 m2 I = = Power Area 100 W×2.5% 113 m2 = 100 W × = 0.022 W/m2 113 m2 2.5 100 I = εoE 2 1 2 o c ⇒ Eo = √ 2I εoc or Eo = √ 2×0.022 = 4.07 V m−1 8.85×10−12×3×10 8 P = 15 kW P = = E t nhc λ λ = = nhc P 10 16 × 6.63 × 10 −34 × 3 × 10 8 15 × 10 3 λ = 1.32 × 10−13 m λ < 10 −12 E = 6.4 × 10 5 J p = E c p = 6.48×10 5 J 3×10 8 ms−1 ⇒ p = 2.16 × 10−3 kg ms−1 = c B2 rms μ0 ⇒ B 2 rms = ⇒ Brms ≈ 10−4 T 10 8×4π×10−7 3×10 8 UB = B2 2μ0 Ey = 600 sin(ωt − kx) ε0 = 9 × 10 −12C 2/Nm2 EM I = ε0E 2 0 c = c 1 2 1 2 B 2 0 μ0 I = ε0E2 0 c 1 2 = × 9 × 10−12 × 600 2 × 3 × 10 8 1 = 486 W/m2 2 c = ⇒ E = cB0 = cB0 E0 Bo E = cB0 sin(kx + ωt)k^
NEET REVISION 28. () : Explana on We have, Comparing with the standard equa on , we get The wavelength range of microwaves is to The wavelength of this wave lies between to , so the equa on represents a microwave. 29. () : Explana on Average energy density of electric field and magne c field are same. 30. () : Explana on where coefficient of reflec on 31. () : Explana on Plane of oscilla ons of electric field is the plane of po‐ lariza on. Direc on of propaga on is always present along the direc on of . 32. () : Explana on ( energy Power me Momentum gained by the object is, 33. () : Explana on 34. () : Explana on Length of antenna 35. () : Explana on The two antennas have same height , distance between them: , where radius of earth. 36. () : Explana on -rays are produced by bombarding a metal target with high energy electrons. 37. () : Explana on 38. () : Explana on The descending order of energy for following waves 39. () : Explana on Average intensity of wave in terms of Electric field is given by where Speed of wave in vaccum Amplitude of electric field 40. () : Explana on Force on a non-reflec ng surface at normal incidence is given by 41. () : Explana on Energy density By = 2 × 10−7 sin(0.5 × 10 3x + 1.5 × 10 11 t) By = B0 sin(kx + ωt) k = 0.5 × 10 3 ⇒ λ = 2π k = 2π 0.5×10 3 = 0.01256 m 10−3 0.1 m. 10−3 0.1 m Prad = (1 + ρ) I c ρ = F = (1 + ρ) × A I C ∴ F = (1 + 0.6) × A = 1.07 × 10 −6 N 200 3×10 8×A K ̄ E ̄ × B ̄ U )= × = (20 × 10−3) × (300 × 10−9 s) = 6 × 10−9 joules p = = U c 6 × 10 −9 3 × 10 8 = 2 × 10 −17 kg ms−1 = − dE dz dB dt = −2E0K sin kz cos ωt = − dB = +2E0K sin kz cos ωtdt dE dz dB dt B = +2E0K sin kz ∫ cosωtdt = +2E0 sin kz sin ωt k ω = = c E0 B0 ω k B = ˆj sin kz sin ωt 2E0 c ≥ λ 4 ∴ λ ≤ 4 × 25 ⇒ λ ≤ 100 m. ∵ (H) D = 2√2RH R = ⇒ H = = km = 39.55 m D2 8R 45 2 8×6400 X B = |→ p | = = E c EB μ0 E2 cμ0 = = 26.5 W m−1 10 4 3×10 8×4π×10 −7 Eγ > Ex > Euv > Evissible > EIR > EMW > ER EM I = cε0E 2 0 1 2 c = EM E0 = = × 3 × 10 8 × 8.85 × 10−12 × (200) 1 2 2 Z = 53.1 W/m2 F = IA c ∴ A = = Fc I 2.4×10 −4×3×10 8×10 −4 360 = 0.02 m2 u = ε0E 2m = 1.73 × 10−8 1 J/m3 2