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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
  6. FARADAY:
  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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