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# prgmNEW1

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1. :STOP:NIFTY
2. :INDUCTION//FLUX
3.
4. (Ε)=BCOS(Θ)A=BACOS(Θ)
5. SI UNIT: 1T*M^2= 1 WEBER
7. Ε=-N * (ΔΦ)//ΔT)
8.
9. A CONDUCTING LOOP IS HELD NEAR A PERSON'S HEAD.
10. WHEN THE CURRENT IN THE LOOP IS CHANGED RAΠDLY,
11. THE MAGNETIC FIELD IT CREATES CAN CHANGE AT THE RATE
12. OF RAΠDLY CHANGING MAGNETIC FIELD
13. 3 X ₁₀^(4T/S. {IF THE MAGNETIC FIELD CHANGES OVER AN AREA OF 1.13*₁₀^(2 M^2, WHAT IS THE INDUCED EMF?}
14. Ε=ΔΦ//ΔT= (A*ΔB)//ΔT =A(ΔB//ΔT)
15. =(1.13*₁₀^(-2 M^2)(3*₁₀^(4 T//S)
16. (Ε)=1.13*1-^-2 M^2  *  3 *₁₀^(4 T//S
17. (Ε)=339V
18.
19. A BAR MAGNET IS MOVED RAΠDLY TOWARD A 40 TURN
20. CIRCULAR COIL OF WIRE. AS THE MAGNET MOVES, THE
21. AVERAGE VALUE OF BCOS(Θ) OVER THE AREA OF THE COIL
22. INCREASES FROM O. 0125 T TO 0.45 T IN O. 25S IF
23. THE RADIUS OF THE COIL IS 3.05 CM, & RESISTANCE OF
24. ITS WIRE IS 3.55 Ω, {FIND THE MAGNITUDE OF (A) THE
25. INDUCED EMF.}
26. A) <PIPE>(Ε)<PIPE>=<PIPE>((Φ)INITIAL-ΦFINAL)//ΔT))<PIPE>
27. ΦINTITIAL= (BCOSΘ)INITIAL*A= 0.0125T*Π*(0.305^2)= 3.65*₁₀^(-5 T*M^2
28. ΦFINAL=(BCOS(Θ)_*A=(0.45T)*Π*(0.0305M)^2= 1.32 * ₁₀^(-3 T*M^2
29. Ε<PIPE>=(40)<PIPE>((132E-3 - 3.65E-5)T*M^2))//0.25S))<PIPE> = 0.205V
30. {B) FIND THE MAG OF THE INDUCED CURRENT.}
31. I=V//R
32. LENZ'S LAW
33. Ε=-N(ΔFLUX//ΔTIME)
34. AN INDUCED CURRENT ALWAYS FLOWS IN A DIRECTION THAT OPPOSES THE CHANGE THAT CAUSED IT.
35. Ε=-N(ΔFLUX//ΔTIME)
36. Φ=BACOSΘ=1.7T*(1.2M)^2
37.
38. 1. THE MAGNETIC FIELD PRODUCED BY AN MRI SOLENOID 2.6 M LONG AND 1.8 M IN DIAMETER IS 2.1 T.
39. {FIND THE MAGNITUDE OF THE MAGNETIC FLUX THROUGH THE CORE OF THIS SOLENOID.}
40. Φ=BACOSΘ=(1.7)((Π/4)(1.2)^2))COS0=1.9WB
41.
42. 2. A SINGLE-TURN SQUARE LOOP OF SIDE L IS CENTERED ON THE AXIS OF A LONG SOLENOID. IN ADDITION, THE PLANE OF THE SQUARE LOOP IS PERPENDICULAR TO THE AXIS OF THE SOLENOID. THE SOLENOID HAS 1340 TURNS PER METER AND A DIAMETER OF 6.00 CM , AND CARRIES A CURRENT OF 2.60 A.
43. {A. FIND THE MAGNETIC FLUX THROUGH THE LOOP WHEN L= 2.90 CM.}
44. E IS CONFINED WITHIN THE SOLENOID. AREA WILL BE THE SMALLER OF THE CROSS-SECTIONAL AREA OF THE SOLENOID OR AREA OF THE RECTANGLE.
45. Φ=BA=(3.926MT)(0.0290M)^2
46. {B. FIND THE MAGNETIC FLUX THROUGH THE LOOP WHEN L= 6.00 CM}
47. C. Φ=BA=(3.926MT)Π(0.0290M)^2
48. {FIND THE MAGNETIC FLUX THROUGH THE LOOP WHEN L= 11.0 CM.}
49. Φ IS THE SAME AS THAT FOUND IN PART B
50.
51. 3.A COIL WITH MANY LOOPS IS ORIENTED PERPENDICULAR TO A CHANGING MAGNETIC FIELD.
52. THE AREA OF A 120 TURN COIL ORIENTED WITH ITS PLANE PERPENDICULAR TO A 0.33 T MAGNETIC FIELD IS 4.1×₁₀^(-2 M2.
53. {FIND THE AVERAGE INDUCED EMF IN THIS COIL IF THE MAGNETIC FIELD REVERSES ITS DIRECTION IN 0.25 S}
54. A. COIL WITH MANY LOOPS PERPENDICULAR TO A CHANGING MAGNETIC FIELD
55. <PIPE>Ε<PIPE>=N*<PIPE>ΔΦ//ΔT<PIPE>=N*<PIPE>(BA-(-BA)//ΔT)<PIPE>=N<PIPE>(2BA/ΔT)<PIPE>=120(2(0.33T)((0.050M^2))//0.11=36V
56.
57. 4.A SQUARE LOOP OF WIRE IS IMMERSED IN A MAGNETIC FIELD. THE SQUARE IS RAΠDLY CHANGED INTO A CIRCLE WITH THE SAME PERIMETER.
58. AN EMF IS INDUCED IN A CONDUCTING LOOP OF WIRE 1.48 M LONG AS ITS SHAPE IS CHANGED FROM SQUARE TO CIRCULAR.
59. {FIND THE AVERAGE MAGNITUDE OF THE INDUCED EMF IF THE CHANGE IN SHAPE OCCURS IN 0.165 S AND THE LOCAL 0.531-T MAGNETIC FIELD IS PERPENDICULAR TO THE PLANE OF THE LOOP.}
60. A. C=2ΠR R=1.40M//2Π=0.223M
61. B. ΦCIRCLE=BA=BΠR^2=(0.335T)Π(0.223M)^2=0.0710WB
62. C. P=4S S=P//4=1.40//3=0.350M
63. D.ΦSQUARE=BA=BS^2=(0.455T)(0.350M)^2=0.0557WB
64. E. Ε=N<PIPE>(ΦCIRCLE-ΦSQUARE)//ΔT)<PIPE>=(1)(0.0710-0.0557WB)//(0.125S)=0.122V
65.
66. 5. A CIRCUIT CONTAINS A RESISTOR AND CAPACITOR. A MAGNETIC FIELD POINTS INTO THE PAGE AND BECOMES STRONGER OVER A SHORT TIME INTERVAL.
67. SHOWS A CIRCUIT WITH AN AREA OF 0.070 M2 CONTAINING A R=1.0Ω RESISTOR AND A C=280ΜF UNCHARGED CAPACITOR. POINTING INTO THE PLANE OF THE CIRCUIT IS A UNIFORM MAGNETIC FIELD OF MAGNITUDE 0.17T. IN 1.0×102S THE MAGNETIC FIELD STRENGTHENS AT A CONSTANT RATE TO BECOME 0.90T POINTING INTO THE PLANE.
68. WHAT MAXIΜM CHARGE (SIGN AND MAGNITUDE) ACCΜLATES ON THE UPPER PLATE OF THE CAPACITOR IN THE DIAGRAM?
69. 1. {CALCULATE THE EMF}
70. E=N<PIPE>ΔΦ//ΔT=(1)ΔBA//ΔT=(0.80-0.15T)(0.060M^2)/0.010S=3.9V
71. 2. {CALCULATE THE CHARGE ON THE UPPER PLATE OF THE CAPACITOR}
72. Q=-CV=C<PIPE>Ε<PIPE>=-(250*₁₀^(-6F)(3.9)=-975ΜC=-.98MC
73.
74. 6. A FRICTIONLESS ROD SLIDES ACROSS TWO CONDUCTING RAILS THAT ARE PERPENDICULAR TO A UNIFORM MAGNETIC FIELD.
75. A ZERO-RESISTANCE ROD SLIDING TO THE RIGHT ON TWO ZERO-RESISTANCE RAILS SEPARATED BY THE DISTANCE L = 0.362 M . THE RAILS ARE CONNECTED BY A 10.7 Ω RESISTOR, AND THE ENTIRE SYSTEM IS IN A UNIFORM MAGNETIC FIELD WITH A MAGNITUDE OF 0.680 T .
76. A) {FIND THE FORCE THAT ΜST BE EXERTED ON THE ROD TO MAINTAIN A CONSTANT CURRENT OF 0.154 A IN THE RESISTOR.}
77. F=IBL -.0656N
78. B)WHAT IS THE RATE OF ENERGY DISSIPATION IN THE RESISTOR?
79. P=I^2R=.306W
80. C)WHAT IS THE MECHANICAL POWER DELIVERED TO THE ROD?
81. P=FV=0.306W
82.
83. 7. A FRICTIONLESS ROD SLIDES ACROSS TWO CONDUCTING RAILS THAT ARE PERPENDICULAR TO A UNIFORM MAGNETIC FIELD. THE INDUCED EMF DRIVES
84. A CURRENT THROUGH A LIGHTBULB. THE LIGHTBULB IN THE CIRCUIT SHOWN IN THE FIGURE HAS A RESISTANCE OF 10 Ω AND CONSUMES 4.9 W OF POWER; THE ROD IS 1.34 M LONG AND MOVES TO THE LEFT WITH A CONSTANT SPEED OF 3.4 M/S . THE STRENGTH OF THE MAGNETIC FIELD IS 1.5 T .
85. A) {FIND THE CURRENT THAT FLOWS IN THE CIRCUIT.}
86. I=√(P/R)=.64A
87. B) {WHAT SPEED ΜST THE ROD HAVE IF THE CURRENT IN THE CIRCUIT IS TO BE 1.1 A ?}
88. V=Ε/BL=IR/BL=4.1M/S
89. :STOP
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