Tiêu đề 08. ELECTROMAGNETIC WAVES.pdf ✅

08. ELECTROMAGNETIC WAVES NEET PREPARATION (MEDIUM PHYSICS PAPER) Date: March 12, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on Power Area 2. () : Explana on Electromagne c waves have linear momentum as well as energy. This concludes that they can exert ra‐ dia on pressure by falling beam of electromagne c radia on on an object. So op on (3) is correct. 3. () : Explana on Micro waves are in range to . Radio waves are in range to . So correct op‐ on is (2). 4. () : Explana on 5. () : Explana on Proper es of em waves are Microwaves are used in air cra naviga on. (A)-(S) UV rays are used in lasik eye surgery. (B)-(R) Infrared lamps used in the treatment of muscular complaints. (C)-(P) -rays used in the treatment of cancer as -rays can pass through flesh but not through bones in medical diagnosis. (D)-(Q) 6. () : Explana on 7. () : Explana on (or) total power / velocity of light 8. () : Explana on Energy density of an electromagne c wave, So, the magne c energy density is 9. () : Explana on All the men oned laws and respec ve equa ons are given by Ampere - Maxwell law is given by, Gauss's law for electricity is given by, Gauss's law for magne sm is, Faraday's law is So, correct op on is (3). 10. () : Explana on The speed of an electromagne c wave in vacuum is It is independent of amplitude, frequency and wave‐ length of the electromagne c wave. 11. () : Explana on Satellites in polar orbits at 879 kilometers al tude pass over the same Earth loca on at a specific local me, ensuring consistent ligh ng and atmospheric condi ons. So correct op on is (1). = I = 1.4 × 10 3 × 5 W F = = = 2.33 × 10 −5 N Power c 1.4×10 3×5 3×10 8 1 GHz 300 GHz 3 KHz 1 GHz I = P 2πr 2 I = = ε0E2 0 c 2 P 2πr 2 E0 = √ ε P = √ 0cπr 2 50×10 3 8.85×10 −12×3×10 8×10 10×3.14 = 2.45 × 10 −2 V m−1 X X = 3.5 mm λ 2 λ = 7.0 mm c = fλ = 2.2 × 10 10 × 0.7 × 10 −2 = 1.54 × 10 8 ms−1 F = ProdA = = = 5.33 × 10 −4 N IA c 10 3×160 3×10 8 F = U = ε0E2 + 1 2 1 2 B2 μ0 . B2 2μ0 ∮ B→d →l = μ0ic + μ0ε0 dφE dt ∮ E→ ⋅ dA→ = Q ε0 ∮ B→ ⋅ dA→ = 0 ∮ E→ ⋅ d →l = dφB dt c = = 3 × 10 8 m s 1 −1 √μ0ε0
12. () : Explana on Change in momentum in Momentum delivered to the surface in 13. () : Explana on 14. () : Explana on The magnitude of is 15. () : Explana on Radio waves are produced by the accelerated mo on of charges in conduc ng wires. They are used in radio and television communica on systems. Cellular phones use radio waves to transmit voice communi‐ ca on in the Ultra High Frequency ( ) band. 16. () : Explana on 17. () : Explana on 18. () : Explana on Surface wave propaga on is favored in medium-wave radio ( to ) broadcas ng for reli‐ able, quality coverage over long distances, near the Earth's curvature, making it cost-effec ve and prac ‐ cal for urban and regional broadcas ng. So correct op on is (1). 19. () : Explana on 20. () : Explana on The magne c field of an wave propaga ng along -axis is given by, On comparing the above equa on with the given electromagne c wave equa on we get, The electric field equa on is given by, Also, Rela on between and is given by 21. () : Explana on 22. () : Explana on Suppose the charge on the capacitor at me is . The electric field between the plates of the capacitor is . The flux through the area considered is The displacement current is 23. () : Explana on (1)Comparing the given equa on with we get, , and 24. () : Explana on C and D only.The speed of light is always independent of the mo on of source as . The speed of light in a medium is independent of in‐ tensity as , where is the refrac ve index of the medium. 25. () : Explana on Standard equa on: F = Prad × Area = × Area I c = × 30 = 2 × 10 −6 N 20 3×10 8 = 1 sec ∴ 30 min = 2 × 10 −6 × 30 × 60 = 36 × 10 −4 kg m s−1 I = = = 2.7 × 10 P 9 W/m2 A P λ2 B B = E C = = 2.1 × 10 −8 T 6.3 V /M 3×10 8 m/s UHF Prad = = = 4.7 × 10 −6 Pa I c 1.4×10 3 3×10 8 P = IA = IπR2 = 19.6 mW 500 KHz 1500 KHz I = E At E = IAt = × 20 × 10 −4 × 60 20 10 −4 = 24 × 10 3 J EM x By = B0 sin (kx + ωt) T. B0 = 3.5 × 10 −7 J. E = E0 sin(kx + ωt)V /m E B = c E B ∴ E0 = B0c = 3 × 3.5 × 10 8 × 10 −7 = 105 V /m−1 E = E0 sin(1.5 × 10 3x + 0.5 × 10 11 t)V /m−1 Ez = 105 sin(1.5 × 10 3x + 0.5 × 10 11 t)V /m−1 irms = , Brms = Vrms XC μ0 irms 2πr Brms = ( ) = = 44π × 10 −13T μo 2πr Vrms XC μo2πvc 2πr t Q E = Q ε0A ΦE = ⋅ = Q ε0A A 2 Q 2ε0 id = ε0 = ε0 ( ) = dΦE dt 1 2ε0 dQ dt i 2 By = B0 sin( x − 2πft) 2π λ λ = 2π m = 1.26 cm 0.5×10 3 = υ = (1.5 × 10 11) /2π = 23.9 GHz 1 T c = 1/√μ0 ∈0 v = c/n n E = 3.1 cos[(1.8y + 5.4 × 10 8] iN/C E = E0 cos[kx − ωt]N/C ∴ k = = 1.8 λ = = 3.49 ≃ 3.5. 2π λ 2π 1.8

NEET PREPARATION (MEDIUM PHYSICS PAPER) NEET PREPARATION 37. () : Explana on . If decreases, also decreases. Correct op on is (2). 38. () : Explana on If is the poten al difference across the wire, then and magne c field at the surface of wire . Hence poin ng vector, directed radi‐ ally inwards, is given by Hence the correct answer will be (4). 39. () : Explana on Hertz observed maximum spark produc on when the detector gap was parallel to the source gap, indicat‐ ing the electric vector's alignment perpendicular to both gaps and the radia on propaga on. So correct op on is (3). 40. () : Explana on Magne c field amplitude Angular frequency Wave length Angular wave number Hence op on (3) is correct. 41. () : Explana on -rays are absorbed by Earth's atmosphere, if used in ground-based observatories. Placing - ray tele‐ scopes in space satellites eliminates atmospheric in‐ terference, enabling discoveries in -ray astronomy. So correct op on is (3). 42. () : Explana on In free space, the energy density of a sta c field is and the energy density of magne c field is 43. () : Explana on 44. () : Explana on Radio waves are used for op cal communica on have shorter wavelength than microwaves. Hence, asser‐ on is incorrect. Infrared rays have higher frequency and energy than microwaves. Hence, reason is true. 45. () : Explana on The equa on signifies that the electric field's magnitude is much greater than the magne c field's magnitude in electro‐ magne c waves. Despite this, both fields share en‐ ergy equally during wave propaga on, emphasizing their collabora ve role in transpor ng electromag‐ ne c energy. So correct op on is (4). iC = iD = sin ωt V0 XC iC = iD = (V0ωC)sin ωt ⇒ iC = iD = V0(2πf)C sin ωt ⇒ iC = iD = V (2πf)C sin ωt ⇒ f iC V E = = V l IR l B = μ0I 2πa S = = × = EB μ0 IR μ0l μ0I 2πa I 2R 2πal B0 = = E0 c 120 3 × 10 8 = 400nT ω = 2πf = 2π × 50 × 10 6 = π × 10 8rad/s λ = c = = 6 m f 3×10 8 50×10 6 k = = 2π λ 2π 6 ≈ 2.1 rad/m X X X E uE = ε0E2 , 1 2 uB = 1 B2 2μ0 B = μ0i 2πr ⇒ i = = 0.2A B(2πr) μ0 = c = 3 × 10 8 m/s E B (E) (B)