2. THE ONE WHICH IS MOVED RAPIDLY
3. (i) CLOCKWISE IN LOOP 1 AND ANTICLOCKWISE IN LOOP 2
(ii) EMF INDUCED IS MORE IN LOOP 2 THAN IN LOOP 1
5. (i) As magnet approaches the coil the rate of change of magnetic flux increases and
the induced emf also increases. As soon as one pole of magnet enters into the coil the emf
decreases due to the other pole effect and also the induced emf polarity reverses due to the
(ii) The longer peak is due to increase in the rate of change of magnetic flux as the magnet comes out of the coil.
7.IN CASE OF (b) IT IS MORE
8.(i) DECREASES (ii) INCREASES. THERE IS AN OPPOSITION FOR THE CURRENT IN COIL DUE TO SELF INDUCTION SO THE BULB GETS MORE ELECTRIC CURRENT INITIALLY.
9. DECREASES, INCREASES, INCREASES
10. THE RATE OF CHANGE OF AREA IS MORE INITIALLY AND DECREASES WITH TIME AND SO THE INDUCED EMF.
11. HIGH FREQUENCY AC PRODUCES CHANGING MAGNETIC FLUX AND THE LARGE EDDY CURRENTS PRODUCE HEAT.
12. (i) DECREASES FROM X TO Z (ii) DECREASES FROM X TO Z IN CASE OF X QUALITY FACTOR IS MORE, IMPEDANCE DECREASES FROM X TO Z
13. L SHOULD DECREASE AND C SHOULD INCREASE
15. (i) X IS RESISTOR (ii) Y IS CAPACITOR
16. (i) BRIGHTNESS OF THE BULB INCREASES SLOWLY (ii) BRIGHTNESS REMAINS SAME
17. (a) 8.24A, 11.7A (b) VL=207V, VC=437V (c) zero (d) zero (e) zero.
18. (a) 4167 rad s-1 , 1.41A (b) 2300 W (c) 648Hz, 678Hz, I0=10A(d)21.7
19. (a)1 J,(b)159Hz (c)electrical at t =
20. L= 2W / t2 ; Self-inductance is defined as double the work done against the induced emf in producing unit current in the coil itself.
21. The woodden block sinks when current flows through the circuit, as parallel wires carrying currents in the opposite directions repel. The given wave form shows the input current of a transformer.
22. 1800 phase difference due to Lenz’s law
23. No Ans
24. No Ans
27. THE RING MOVESAWAY FROM THE SOLENOID
28. Just to miss the opposite plate, the particle must move in a circular path with radius d so that Bqv = mv2/d, B = (2mK)1/2/(qd)
29. No Ans
30. According to the Flemings Right hand Rule, the magnetic force ILB is directed upward. Equilibrium in the vertical direction yields 2T + ILB = Mg, so that T = (Mg – ILB)/2
31. -Due to Flemings Right hand Rule the magnetic force ILB is directed downward. This constant force shifts the equilibrium position downward by a displacement = (ILB)/2K
32. Wires A and B are in series. IA = IB = I/3, IC = 2I/3.Wire C makes a contribution to the field at O whose magnitude is twice that of A or B. By Flemings Right hand Rule, directions of field due to Wire A and B are directed down into the page. That due to wire C is upward. Net field at O is zero
33.Due to Lenz’s law, the magnetic field produced by the induced current must counteract the decrease in flux and hence it must be directed into the plane of the figure (within the loop).So the induced current must be clock –wise.
34. In both the cases the induced emf doubles
35. E = L I2 = 16 H and e = 320 V
36. -With resistor, current is same both for 100 Hz and 100 kHz. With inductor, the current is 11.9 A and 11.9 mA respectively
37. Resultant force F = 2 attractive forces + 2 repulsive forces = 520 mN (attractive)
38. The induced e.m.f in coil A decreases due to large copper plate introduced between the two coils as Cu is diamagnetic material
39. Due to Lenz’s law, end A will behave as South Pole and end B will behave as North Pole. The end face A will have clock wise direction of current and end face B will have anti clock wise direction of current when seen from the magnet side.
40. Calculate L = 16 H. e = L di/dt = 320 V
41. Zero Induced emf.
42. Induced emf will be same in the both but Induced Current will be more in Copper loop.
43. A large Induced emf is setup across the gap in the switch.
44. To cancel the effect of self Induced emf in the coil.
45. A galvanometer measures mean value of a.c., which is zero over a cycle.
46. Xc = 1 / 2 π θ c = ∞
47. Magnetic flux linked with Primary coil does not vary with time so no Induced emf in secondary.
49. Voltages across different elements of the LCR circuit are not in same phase.
50. Less than that due to gravity.
51. -For Coil AB: Anticlockwise. For Coil CD: Anticlockwise.
52. -(i) The bulb B2 will light up earlier. (ii) The bulb B1 will grow more brightly.
53. -i. L α n2 => L is decreased. ii. L will Increase.
55. -(a) Bulb will grow more brightly. (b) Brightness of the bulb will become maximum.
56. L=0.066 H, R=12Ω
57. -(a). Series combination of a register and a capacitor.
(b). Power factor = cosΦ = 0.81
(c). Pav = EvIvCosΦ = 72.9W
58. -(i) No effect (ii) current will decrease (iii)Current will Increase.
59. -(i) Current and Voltage are in the same phase.
(ii) Current leads voltage by Phase angleΦ .
(iii) Current lags behind voltage by Phase angleΦ .
CBSE Physics (Chapter Wise With Hint / Solution) Class XII (By Mr. Sreekumaran Nair)