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Unformatted text preview: RECITATION 9 1. (a) We use Eq. 293: ( 29 ( 29 ( 29 ( 29 mag 19 18 sin 3.2 10 C 550m s 0.045 T sin52 6.2 10 N. F q vB φ = = × = × o (b) The acceleration has a magnitude of mag 18 8 2 27 6.2 10 N 9.5 10 m s . 6.6 10 kg F a m × = = = × × (c) Since it is perpendicular to mag , v F r r does not do any work on the particle. There is therefore no change in the particle's kinetic energy, and the speed does not change. 9. (a) From 2 1 2 e K m v = we get ( 29 ( 29 3 19 7 31 2 1.20 10 eV 1.60 10 J eV 2 2.05 10 m s. 9.11 10 kg e K v m × × = = = × × (b) Solving Eq. 297 for the magnetic field strength yields ( 29 ( 29 ( 29 ( 29 31 7 4 19 2 9.11 10 kg 2.05 10 m s 4.67 10 T. 1.60 10 C 25.0 10 m e m v B qr × × = = = × × × (c) The frequency is ( 29 7 7 2 2.07 10 m s 1.31 10 Hz. 2 2 25.0 10 m v f r π π × = = = × × (d) The period is ( 29 1 7 8 1 1.31 10 Hz 7.63 10 s. T f = = × = × 11. (a) The frequency of revolution is ( 29 ( 29 ( 29 19 6 5 31 1.60 10 C 35.0 10 T 9.78 10 Hz.9....
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This note was uploaded on 06/01/2008 for the course PHYS 232 taught by Professor Goksu during the Spring '08 term at Millersville.
 Spring '08
 Goksu
 Physics, Acceleration

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