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WKU PHYS 260 - Chapters 27-Examples

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Chapters 27--ExamplesProblemF=qv x BSlide 4Prep WorkSlide 6Slide 7CyclotronsPart B)Slide 10Free Body Diagram1Chapters 27--Examples2ProblemA electron that has a velocity v= (2x106 m/s) i + (3 x 106 m/s) j move through a magnetic field, B=(0.030)i-(0.15)j.a) Find the force on the electronb) Recalculate for a proton.3F=qv x Bq=-1.602 x 10-19F=(-1.602x10-19)*(-3.9 x 105) kF=(6.2x10-14 N) kIf a proton, then q=1.602 x 10-19 soF=(-6.2x10-14 N) k (opposite direction of e)  zEzEEEEzyxBvˆ.ˆ)*.(.*..6390630300150620150030006362 4ProblemA 150-g ball containing 4.0 x 106 excess electrons is dropped into a 125-m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 T and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball as it enters the field.5Prep Workq=4 x 106 e = 4 x 106 * 1.602 x 10-19q=6.408 x 10-11No air resistance so mechanic energy, E=KE+U is conserved. (KE2-0)=-(0-U1)½ m v2=mgyV2=2gy (y=125 m)V=sqrt(2*9.8*125)=49.5 m/s6F=qv x B||F||=qvB=6.408x10-11*49.5*0.250||F||=7.93 x 10-10 NDirectionForce is northNWESBUpDownvF7ProblemIn a certain cyclotron, a proton moves in a circle of radius 0.5 m. The magnitude of the magnetic field is 1.2 Ta) What is the cyclotron frequency?b) What is kinetic energy of the proton in electron volts?8CyclotronsHzmqBffmqBRvqBmvRqvBrvm719192108.1)1067.1(22.1*10602.1229Part B)        eVJKEJmqBRKEmqBRmvKEmqBRmvqBRmvqBRmvqvBrmvrvmeV7106.11121227219222222222107.11076.21076.21067.1*25.0*2.1*106.12212111910ProblemA wire of 62.0 cm length and 13.0 g mass is suspended by a pair of flexible leads in a magnetic field of 0.44 T. What are the magnitude and direction of the current required to remove the tension in the supporting leads?X X X X X X XX X X X X X XX X X X X X XX X X X X X XX X X X X X X0.62 mB field into page11Free Body DiagramX X X X X X XX X X X X X XX X X X X X XX X X X X X XX X X X X X X0.62 mmgBased on the RH rule the current must go CCW, so i flowsFBF=iLB (if L perpendicular to B and it is)mg=iLB or (mg)/(LB)=ii=(.013*9.8)/(0.62*0.44)=0.467 ATension vanisheswhen FB


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