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Magnetic Forces on Charged Particles: Calculations for Electrons and Protons

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This document presents detailed examples of calculating magnetic forces on charged particles, specifically an electron and a proton moving through a magnetic field. It includes steps to compute the force acting on each particle, utilizing the principles of electromagnetism and the Lorentz force equation. The examples also cover a scenario involving a ball with excess electrons dropping in a magnetic field, providing insights into conservation of mechanical energy and force calculation. Various applications of cyclotrons and current in magnetic fields are also discussed.

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Magnetic Forces on Charged Particles: Calculations for Electrons and Protons

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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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