170CHAPTER 19 19.65Note: We solve part (b) before part (a) for this problem. (b) Since the magnetic force supplies the centripetal acceleration for this particle, 2qvBmvr=or the radius of the path is rmvqB=. The speed of the particle may be written as ()2vKE=m, so the radius becomes ( )( )( )27619-192 1.6710kg5.0010 eV1.6010J eV21.6010C0.050 0 T6.46 mmKErqB−−×××==×=Consider the circular path shown at the right and observe that the desired angle is 111.00 m1.00 msinsin8.906.46 mrα===°±±±±±²³³´µ±²±²(a) The constant speed of the particle is 2E=m, so the vertical component of the momentum as the particle leaves the field is
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