30_Chapter 19 HomeworkCH19 Magnetism

30_Chapter 19 HomeworkCH19 Magnetism - 170 CHAPTER 19 19.65...

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170 CHAPTER 19 19.65 Note: We solve part (b) before part (a) for this problem. (b) Since the magnetic force supplies the centripetal acceleration for this particle, 2 qvB mv r = or the radius of the path is rm v q B = . The speed of the particle may be written as () 2 vK E = m , so the radius becomes ( )( ) ( ) 27 6 19 -19 2 1.67 10 kg 5.00 10 eV 1.60 10 J eV 2 1.60 10 C 0.050 0 T 6.46 m mKE r qB −− ×× × == × = Consider the circular path shown at the right and observe that the desired angle is 11 1.00 m 1.00 m sin sin 8.90 6.46 m r α  === °   ± ± ± ± ±²³³´µ ±² ± ² (a) The constant speed of the particle is 2 E = m , so the vertical component of the momentum as the particle leaves the field is
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