Title 08. ELECTROMAGNETIC WAVES.pdf ✅

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
26. () : Explana on 27. () : Explana on The speed of electromagne c wave in a medium is- 28. () : Explana on Let the electric field is The energy density due to electric field is Again, . Thus, So, the frequency of and is double the fre‐ quency of electromagne c wave. 29. () : Explana on Electromagne c wave equa on is given by Speed of electromagne c wave, Given, equa on is Comparing Eqs. (1) and (2), we get Here, wave factor, 30. () : Explana on Comparing the given equa on With the standard equa on we get Moreover, we know that As the direc on of field of electromagne c wave is in -direc on so, the wave is moving along posi ve - direc on with frequency and wavelength . 31. () : Explana on From modified Ampere's law For Loop and For Loop and Due to 32. () : Explana on 33. () : Explana on For a electromagne c wave, amplitude of electric and magne c field are related by, 34. () : Explana on 35. () : Explana on The electromagne c wave is propaga ng along -axis and component is along -axis. So B component should be along -axis. The correct op on is (3). 36. () : Explana on The total energy falling on the surface is Therefore, the total momentum delivered (for com‐ plete absorp on) is The average force exerted on the surface is B = = 6 × 10 −8 T μ0ir 2 2πR2 v = = = 1 √με 1 √μ0ε0μrεr c √μrεr v = = 1 × 10 8 m/s. 3×10 8 √4×2.25 E = E0 sin(ωt − kx) UE = ε0E2 = ε0E 2 0 sin 2(ωt − kx) 1 2 1 2 UE = ε0E2 0 [1 − cos 2(ωt − kx)] 1 4 [cos 2θ = 1 − 2 sin 2 θ] B = B0 sin(ωt − kx) UB = [1 − cos 2(ωt − kx)] 1 4 B2 0 μ0 UE UB E = E0 cos(kz − ωt) (1) v = ω k E = ^i40 cos(kz − 6 × 10 8 t) (2) ω = 6 × 10 8 and E0 = 40^i k = ω = = 2 m−1 v 6×10 8 3×10 8 Ey = 2.5 cos[(2π × 10 6 ) t − (π × 10 −2 ) x] N C rad m rad sec Ey = E0 cos(ωt − kx) ω = 2πf = 2π × 10 6 ∴ f = 10 6 Hz = k = π × 10 −2m−1 ⇒ λ = 200m 2π λ y x 10 6HZ 200 m ∮ B. dl = μ0 (IC + ID) L1 IC ≠ 0 ID = 0 L2 IC = 0 ID ≠ 0 KCL IC = ID = I vferrite = = = 3 × 10 6 ms c 1 √εrμr 3×10 8 √10 3×10 λferrite = = = 3.33 × 10 −2 m vferrite f 3×10 6 90×10 6 = c Em Bm ∴ Em = Bm × C = 6 × 10 −7 × 3 × 10 8 = 18 × 10; E = 180V m−1 B = i = [qo sin ωt] μo 2πR μo 2πR d dt B = qoω cos ωt μo 2πR B = qoω (qo = CVo) Bmax = 2μT μo 2πR x E y z U = (18 W/cm2) × 20 cm2 × (30 × 60)∞ = 6.48 × 10 5 J p = = = 2.16 × 10 −3 kg/ms U c 6.48×10 5 J 3×10 8 m/s F = = = 1.2 × 10 −6 N p t 2.16×10 −3 0.18×10 4
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)