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Forces on Charged particles Derivation Example Whiteboards Motion in a circle Direction

Forces on Charged particles Derivation Example Whiteboards Motion in a circle Direction Northern Lights Whiteboards. Recall F = IlBsin  Derive F = qvB sin  F = force on moving particle q = charge on particle (in C) (+ or -???) v = particle’s velocity  = angle twixt v and B

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Forces on Charged particles Derivation Example Whiteboards Motion in a circle Direction

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  1. Forces on Charged particles • Derivation • Example • Whiteboards • Motion in a circle • Direction • Northern Lights • Whiteboards

  2. Recall • F = IlBsin • Derive • F = qvBsin • F = force on moving particle • q = charge on particle (in C) (+ or -???) • v = particle’s velocity •  = angle twixt v and B • Can a magnetic force increase a charged particle’s speed?

  3. Example - proton going 4787.81 m/s in earth’s magnetic field (5.0 x 10-5 T) What force? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . p+ v = 4787.81 m/s q = 1.602 x 10-19 C F = qvBsin F = qvBsin F = (1.602 x 10-19 C)(4787.81 m/s)(5.0 x 10-5 T)sin(90o) F = 3.8 x 10-20 N

  4. Consider the path of particle Force is centripetal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F F F F If at 90o qvB = ma = mv2/r qvB = mv2/r qB = mv/r, r = mv/(qB) r = (1.673 x 10-27 kg)(4787.81 m/s)/(1.602 x 10-19 C)/(5.0 x 10-5 T) r = 1.0000 m = 1.0 m

  5. What is the path of the proton in the B field? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . p+ CW

  6. What is the path of the electron in the B field? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e- ACW

  7. What is the path of the electron in the B field? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e- ACW

  8. What is the path of the proton in the B field? p+ 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 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 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 x x x x x x x x x x x x x x x ACW

  9. What is the path of the electron in the B field? 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 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 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 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 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 e- CW

  10. Superconducting magnets Energy of particles Why we have particle accelerators

  11. Whiteboards - Force: F = qvBsin 1 | 2 | 3 | 4

  12. What is the force acting on a proton moving at 2.5 x 108 m/s to the North in a .35 T magnetic field to the East? q = 1.602 x 10-19 C F = qvBsin F = qvBsin F = (1.602 x 10-19 C)(2.5 x 108 m/s)(.35 T)sin(90o) = 1.4 x 10-11 N 1.4 x 10-11 N, Downward

  13. What magnetic field would exert 1.2 x 10-12 N on an alpha particle going 17% the speed of light? alpha = 2p2n q = 2(1.602 x 10-19 C) = 3.204 x 10-19 C v = .17x3.00 x 108 m/s = 5.1 x 107 m/s F = qvBsin F = qvBsin 1.2 x 10-12 N = (3.204 x 10-19 C)(5.1 x 107 m/s)Bsin(90o) B = 0.073437615 T = .073 T .073 T

  14. If the electron is going 1.75 x 106m/s, and the magnetic field is .00013 T, what is the radius of the path of the electron? 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 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 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 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 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 m = 9.11 x 10-31 kg q = 1.602 x 10-19 C F = qvBsin F = ma, a = v2/r F = qvBsin F = ma, a = v2/r qvB = mv2/r r = mv/qB r = 0.07655 m = 7.7 cm e- 7.7 cm

  15. What B-Field do you need to make a proton going 2.13 x 107 m/s go in a 3.2 cm radius circle? m = 1.673 x 10-27 kg q = 1.602 x 10-19 C F = qvBsin F = ma, a = v2/r F = qvBsin F = ma, a = v2/r qvB = mv2/r B = mv/qr B = 6.951 T = 7.0 T 7.0 T

  16. Explain angle to B-Field (vertical B field, velocity angled upward) Earth’s Magnetic field

  17. e- e+ Bubble Chamber - particles make tracks in B field. Directions, spiral

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