Tiêu đề 06. ELECTROMAGNETIC INDUCTION.pdf ✅

8. () : Explana on Mo onal , length Magne c field, As rod is held perpendicular to the magne c field, so In a metal rod, induced is given by, When and are mutually perpendicular, where is velocity of moving rod or 9. () : Explana on Induced currents are clockwise. Therefore, induced magne c field is into the plane of the paper. As it op‐ poses the increasing external field, the field must be out of the plane of the paper. 10. () : Explana on The back emf but and 11. () : Explana on Mutual inductance between two coils Peak current, Angular frquency, Current, Hence maximum induced emf is given by 12. () : Explana on When switch is OFF and , if there are induc ve loads (such as transformers and motors) are in circuit, there can be 'sudden varia ons' (called surges) in the circuit. This can cause counter EMF 's and when po‐ ten all difference across bulb is more than the usual of supply) then bulb will 'Fuse' Both Asser on and Reason are correct and Reason ex‐ plains Asser on.So, correct op on is (1). 13. () : Explana on 14. () : Explana on induced, 15. () : Explana on Total resistance of the circuit Current flowing . Voltage across load 16. () : Explana on As the plane of coil is perpendicular to the magne c field, so the angle between area vector and magne c field is zero. The magne c flux changes when there is change in magne c field, change in area and change in angle be‐ tween the magne c field and area vector. So, appro‐ priate answers are and . 17. () : Explana on 18. () : Explana on Back emf is actually the induced emf produced in mo‐ tor. When motor is just switched on, the magnitude of back emf is quite small due to the low speed of the motor. 19. () : Explana on Here, Inductance, , current Magne c energy stored in the solenoid is 20. () : Explana on A change in magne c flux produces current in a coil. So op on (4) is correct. 21. () : Explana on 22. () : Explana on Here, , distance between the ends of wings , , angle of dip, Mo onal emf, 23. () : Explana on - Total number of turns in the coil. - Length of the coil So op on (2) is correct Emf, e = 0.08 V l = 10 cm = 0.1 m B = 0.4T θ = 90 ∘ emf e = ( →V × B→) ⋅ →l →V , B→ →l v v = = e Bl 0.08 0.4 × 0.1 ∴ v = = 2 m/s 0.08 0.04 ε = −L di dt ε = 4V = di dt (0−1) 10 −3 ∴ (−L) = 4 ⇒ L = 4 × 10−3henry −1 10 −3 M = 0.001H I0 = 20 A ω = 100π rad s−1 I = I0 cos ωt ⇒ ∣ ∣ ∣max = (I0 cos ωt) = I0 sin ωt ⋅ ω ⇒ ∣ ∣ ∣max = 20 × 1 × 100π = 2000π dI dt d dt dI dt e = M × = 0.001 × 2000 × π = 2πV di dt ON 220 V(= rms AC ⇒ ∣εarg ∣ = L ⇒ 1 = ⇒ L = 25 mH Δi Δt L×[10−(−10)] 0.5 Emf E = Bl 2ω 1 2 = × 0.5 × 4 × 50 = 50 V 1 2 = 4000 + 400 = 4400 W i = = = 0.1 amp V R 440 4400 = Ri = 4000 × 0.1 = 400 volt. φ = BA cos θ A, B C |e| = L ⇒ 1 = ⇒ L = 25mH di dt L × {10 − (−10)} 0.5 L = 2H I = IA U = LI 2 = × 2 × (1) 2 = 1 J 1 2 1 2 φ = 5t 2 − 30t + 150 E = = −(10t − 30) = −10t + 30 −dφ dt Et=2 = −20 + 30 = 10 V . v = 400 ms−1 l = = 20 m B = 4 × 10−4 T δ = 30 ∘ ε = Bvvl = 4 × 10−4 × sin 30 ∘ × 400 × 20 = 3.2 × = 1.6 V 1 2 L = μ0N 2A l N l
24. () : Explana on Induced emf is given by 25. () : Explana on According to Faraday's law there will be change in flux across the rectangular coil which induces a current in the coil as well as an emf will also be induced. Now, according to Lenz's law, magnet will experience a force in opposite direc on of its mo on. Hence ac‐ celera on of the magnet will be less than . So correct op on is (2) 26. () : Explana on If there is flux change , then 27. () : Explana on EMF Inductance Rate of change of current) 28. () : Explana on Magne c flux linked with magne c field and area is given by So, Induced emf, 29. () : Explana on 30. () : Explana on Induced e.m. 31. () : Explana on When a magnet falls through a long solenoid, the changing magne c flux induces a current that gener‐ ates a magne c field opposing the magnet's mo on. This opposing force acts in a direc on to reduce the magnet's accelera on, in accordance with Lenz's law, which states that induced currents always oppose the changes that produced them. So, correct op on is (1). 32. () : Explana on Presence of magne c flux cannot produce current. Only a charge in flux induces current. So op on (3) is correct. 33. () : Explana on If the radius is at a me , then the intantaneous magne c flux is given by: Now, induced e.m.f. is given by 34. () : Explana on Conceptual Ques on 35. () : Explana on The asser on correctly describes the occurrence of sparks between the poles of a switch when it is opened as flow of current is suddenly interupted due to a high-voltage arc across the switch contacts, The reason is true. Current flowing through a conductor generates a magne c field, according to Ampère's cir‐ cuital law. But 'Reason' is not explana on for 'Asser on'. So correct op on is (2) 36. () : Explana on Maximum efficiency in an electric motor is achieved when the back electromo ve force (back emf) ap‐ proaches the applied electromo ve force (applied emf). Specifying an exact ra o of "half" is an approxi‐ ma on and might not hold true for all motor designs and applica ons. The reason is inaccurate in sta ng that motor efficiency depends solely on the magni‐ tude of back emf. In reality, motor efficiency is influ‐ enced by various other factors, such as resis ve losses, mechanical fric on losses. Asser on is true but reason is false. So, correct op on is (3). 37. () : Explana on where is angle between and 38. () : Explana on Given: Induced emf in a inductor is Change in current in a coil, e = Bvl = 0.1 × 10 × 4 ⇒ e = 4V g Δφ q = = × (1.35 − 0.79) = 0.08 Δφ R 1 7 = −( × ε = −L = −6 × 10−3 × dI dt 3−4 40−20 = 3 × 10−4 V φ B A φ = B ⋅ A = |B||A| cos θ (θ = 0 ∘) φ = BA = Bπr 2 |ε| = ∣ ∣ ∣ ∣ ∣ ∣ = Bπ(2r) −dφ dt dr dt = 0.025 × π × 2 × 2 × 10−2 × 1 × 10−3 = πμV Nφ = Li ⇒ = ⇒ NB = Ndφ dt Ldi dt dA dt Ldi dt ⇒ = L × ( ) ⇒ L = 10H 1 × 1 × 5 10−3 2 − 1 2 × 10−3 f = Blv = 0.3 × 10−4 × 10 × 5 = 1.5 × 10−3 V = 1.5 mV r t f φ = πr 2B ε ε = − = − (πr 2B) = πB (2r ) ε = 2πrB ( ) dφ dt d dt dr dt dr dt Δm = 0.005311 amu θ A→ B→ ∴ θ = 30 ∘ Eavg = = √3 V 4×(25×2) cos 30 ∘ 10 dt = 0.2 s, e = 0.1 V ε = −L di dt di = 2 − (−2) = 4 A e = L di dt ⇒ 0.1 = L × ;L = 5 mH 4 0.2
NEET PREPARATION (MEDIUM PHYSICS PAPER) NEET PREPARATION 39. () : Explana on At 40. () : Explana on But and henry 41. () : Explana on Self-inductance, 42. () : Explana on Energy So op‐ on (2) is correct. 43. () : Explana on The magnitude of induced e.m. is given by tesla 44. () : Explana on Given, Here magne c field is decreasing at steady rate So op on (4) is correct. 45. () : Explana on In L-R circuit, the growing current at me is given by Charge passed through the ba ery in one me con‐ stant is i = i0 (1 − e ) −Rt L = −i0 (− ) e = ⋅ e di dt R L −Rt L i0R L −Rt L t = 0 = = = = 2.5 amp/ sec di dt i0R L E L 5 2 e = −L di dt e = 4 V = = −1/10 di −3 dt 0−1 10 −3 ∴ (−L) = 4 ⇒ L = 4 × 10 −1 −3 10 −3 L = ⇒ = ( ) 2 ( ) = ( ) 2 × ( ) = = 1 : 2 μ0N 2πr 2 l L1 L2 r1 r2 l2 l1 1 2 2 1 L1 L2 1 2 = Li 2 = × 2 × (10) 2 = 100J 1 2 1 2 f |Vind| = Blv v = 300 m/ min = 5 m/s ∴ B = = = 0.8 |Vind| lv 2 0.5×5 R = cm = × 10−2 m; 10 √π 10 √π B = 0.5T ∴ = − [ ] = −1T/s ΔB Δt 0.5 − 0 0.5 Vind = Induced emf, = − = −A = −A(−1) = A = πR 2 Δφ Δt ΔB Δt = π × ( ) 2 = 10−2 V 10×10 −2 √π = 10 mV t i = i0 [1 − e−t/τ ] where, i0 = and τ = E R L R ∴ q = ∫ τ 0 idt = ∫ τ 0 i0 (1 − e−t/τ) dt q = i0τ + i0τ [e−1 − 1] = q = = = i0τ e i0τ e (E/R)(L/R) e EL eR2