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Chapters 27--Examples

Chapters 27--Examples. Problem. A electron that has a velocity v = (2x10 6 m/s) i + (3 x 10 6 m/s) j move through a magnetic field, B =(0.030) i -(0.15) j . Find the force on the electron Recalculate for a proton. F =q v x B. q=-1.602 x 10 -19 F =(-1.602x10 -19 )*(-3.9 x 10 5 ) k

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Chapters 27--Examples

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  1. Chapters 27--Examples

  2. Problem A 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. • Find the force on the electron • Recalculate for a proton.

  3. F=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)

  4. Problem A 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.

  5. Prep Work • q=4 x 106e= 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/s

  6. F=qv x B • ||F||=qvB=6.408x10-11*49.5*0.250 • ||F||=7.93 x 10-10 N • Direction • Force is north N Down v F E W B Up S

  7. Problem In a certain cyclotron, a proton moves in a circle of radius 0.5 m. The magnitude of the magnetic field is 1.2 T • What is the cyclotron frequency? • What is kinetic energy of the proton in electron volts?

  8. Cyclotrons

  9. Part B)

  10. X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X 0.62 m Problem A 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? B field into page

  11. X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X 0.62 m Free Body Diagram FB Tension vanishes when FB =mg mg Based on the RH rule the current must go CCW, so i flows F=iLB (if L perpendicular to B and it is) mg=iLB or (mg)/(LB)=i i=(.013*9.8)/(0.62*0.44)=0.467 A

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