Tiêu đề 5-electromagnetic-induction-.pdf ✅

slope. ∴ di dt = constant ⇒ e = +ve and constant For T 2 ≤ t ≤ 3T 4 For 3T 4 ≤ t ≤ T, T is zero ∴ di dt = 0 ⇒ e = 0 From this analysis, the variation of induced emf with time as shown in figure below. 15. (C) i = i0e −t/τL 18 = 20e −2/τL Solving this, we get : τL = 2 ln (10/9) 16. (B) [ Magnetic flux Electric flux ] = [ Magnetic field Electric field ] = [ B E ] [ E B ]has the dimensions of velocity. i.e., [LT −1 ] ∴ [ magnetic flux electric flux ] = [ 1 velocity ] = [TL−1 ] 17. (B) During charging τL1 = L 2R and during discharging τL2 = L 3R ∴ τL1 τL2 = L/2R L/3R = 3 2 18. (B) L = μ0N2A l 19. (A) For r ≤ R: ∮ E ̅ ⋅ dl̅ = | dφ dt | or E(2πr) = (πr 2 ) dB dt or E ∝ r. i.e., E − r graph is a straight line passing through origin. For x ≥ R φE ̅ ⋅ dl̅ = | dφ dt| or E(2πr) = (πR 2 ) ( dB dt ) i.e., E ∝ 1 r or E-r graph is a rectangular hyperbola. 20. (ABD) 21. (A) e = − dφ dt ∴ Volt = Weber S = T − m2 /s 22. (A) 23. (ABC) 24. (D) 25. (C) During the entry and exit of the bar magnet in the coil, the polarity charges. The only option is (C). 26. (C) 27. (A) 28. (D) 29. (B) The flux associated with coil of area A and magnetic induction B is : φ = BAcos θ = 1 2 Bπr 2 cos ωt [∵ A = 1 2 πr 2 ] ∴ einduced = − dφ dt = − d dt ( 1 2 Bπr 2 cos ωt) = 1 2 Bπr 2ωsin ωt Power P = einduced 2 R = B 2π 2 r 4ω 2 sin2 ωt 4R Hence, Pmean = ⟨P⟩
= B 2π 2 r 2ω 2 4R ⋅ 1 2 [∵< sin ωt >= 1 2 ] = (Bπr 2ω) 2 8R 30. (A) Maximum field lines will pass in situation (i). 31. (A) I1 is in the direction ba and I2 in the direction dc 32. (C) 33. (B) 34. (D) M = μ0N1×N2×A l where N1 = 300 turns and N2 = 400 turns, A = 10 cm2 and l = 20 cm Substituting the values in the given formula, we get M = 2.4π × 10−4H. 35. (B) Here, B = B0e −t/τ Area of the circular loop, A = πr 2 . Flux linked with the loop at any time t. φ = BA = πr 2B0e −t/τ Emf induced in the loop, ε = − dφ dt = πr 2B0 1 τ e −t/τ Net heat generated in the loop = ∫0 ∞ ε 2 R dt = π 2 r 4B0 2 τ 2R ∫0 ∞ e − 2t τ dt = π 2 r 4B0 2 τ 2R × 1 (− 2 τ ) × [e − 2t τ ] 0 ∞ = −π 2 r 4B0 2 2τ 2R × τ(0 − 1) = π 2 r 4B0 2 2τR 36. (D) length of the line plane, I = 20 m Wing span, l ′ = 15 m Height of plane, h = 5 m Velocity of plane = 240 m s −1 towards east sin θ = 2 3 , B = 5 × 10−5T, VB =? , Vw =? VB = Voltage developed between lower and upper side of the plane = vhBcos θ = 240 × 5 × 10−5 × √5 3 = 44.72 × 10−5 × √5 3 = 44.72 × 10−3V = 45mV Bv = Bsin θ = 5 × 10−5 × 2 3 = 1 3 × 10−4 T Vw = Voltage developed between tips of the wings = Wv l ′v = 1 3 × 10−4 × 15 × 240 = 1200 × 10−4 = 120mV BV 37. (B) Magnetic force on electron in the metal sheet, F⃗m = −e(v⃗ × B⃗⃗) At equilibrium, Fm = Fe (induced) and σ2 = −σ1 evB = e σ ε0 = e σ ε0 ⇒ σ = ε0vB = σ1 σ2 = −ε0vB 38. (A) For RC circuit : i = E R e −t/RC
For RL circuit : i = E R (1 − e −t(L/R) 39. (2) Use φ = B ̅ ⋅ A ̅ 40. (8) i1 = 10 6+4 = 1 A (iL = 0) i2 = 10 6 + 2 = 10 8 A ∴ i1 i2 = 0.8 41. (104) In step up, VS VP = NS NP = 8 ∴ VS = 8VP ⇒ PS = IS(960) or IS = 104 960 = 10.4 A 42. (10) 43. (16) Potential difference across coil is V = L di dt or V = (2) (4) = 8V Now energy stored per unit time = Power = Vi = (8)(2) = 16 J/s 44. (512) v = L di dt ⇒ 4t = L di dt 1 L ∫0 4 4tdt = ∫0 i di ⇒ i = 32 A Energy stored, U = 1 2 Li 2 = 1 2 × 1 × 322 = 512 J 45. (4) The emf induced across the rod AB is e = Bv ⊥l Here, v⊥ = vsin 30∘ = component of velocity perpendicular to length ∴ e = Bvlsin 30∘ = (2)(4)(1) ( 1 2 ) = 4V The free electrons of the rod shift towards right due to the force q(v⃗ × B⃗⃗). Thus, the left side of the rod is at higher potential. or VA − VB = 4V 46. (51) eAB = ∫0.07 0.1 (2)(10r)dr = 0.051V 47. (10) εi = − dφ dt = −20t + 50 At t = 3s, εi = −10 V 48. (1) Use : I = I0(1 − e −t/τ ) 49. (167) I1 = 5 A,I2 = 2 A ΔI = 2 − 5 = −3 A Δt = 0.1 s ε = 50 V As, ε = −L ΔI Δt ; 50 = −L ( −3 0.1 ) ; 50 = 30L L = 5 3 = 1.67H 50. (8) VA − IR + E − L dI dt = VB ⇒ VA − 5 × 1 + 8 − 5 × 10−3 (−103 ) = VB ⇒ VB − VA = 8V 51. (6) From Energy conservation : 1 2 CV0 2 = 1 2 CV 2 + + 1 2 LI 2 ⇒ 1 2 × 2 × 10−6 × 122 = 1 2 × 10−6 × 6 2 + 1 2 × 6 × 10−6 I 2 ⇒ I = 6 A 52. (4) e = M dI dt = M(4t) − 20t i = e R = 20t 10 = 2t, q = ∫ Idt = ∫0 2 2tdt = 4C 53. (4) I = 2sin t 2A L = 2H dI = 2cos t 2 ⋅ 2tdt = 4tcos t 2dt When I = 0,t = 0 When I = 2,t = √ π 2 E = ∫0 2 LIdI = L∫0 2 2sin t 2dt = L∫0 √π/2 8tsin t 2 cos t 2dt Let t 2 = x, so 2tdt = dx E = L∫0 π/2 4sin xcos xdx = 2L∫0 π/2 sin 2xdx = 2L ( −cos 2x 2 ) 0 π/2 = 2L 2 = 4J