Question

012 10.0 points What is the radius of the smallest possible circular orbit that a proton with the kinetic energy equal to 2 MeV can have in a 1 T magnetic field? The mass of a proton is 1.67 x 10-27 kg and its charge is 1.609 x 10-19 C. Answer in units of m. 013 10.0 points Two species of singly charged positive ions of masses 2.1x10-26 kg and 3.3 x 10-26 kg enter a magnetic field at the same location with a speed of 48000 m/s. The charge on the ion is 1.60218 x 10-19 C. If the strength of the field is 0.31 T, and they move perpendicularly to the field, find their distance of separation after they com- plete one half of their circular path. Answer in units of cm.

the answer for #12 is not 2.694576674

the answer for #13 is not 11.14481

Answer 1

12). E=2 Mev = 2x100 X1.6x10197 E-32 2103 J B=15 m= 1.67 X10 27 kg 2=1.602 21019 conservation of energy E mines 29- 2E - 72x3:2X16 13 J V 1.67x1027 kg 0-1.9576x107 mig radius r-me-1.67X16379 21:5576x107 m ITX1.609 210 12 0.203185 m Reme 13). n = 2.1. 210-26 kg Me= 3.3 210-26 kg 2,= , = 48000 W 2= 1.60218 X109 B = 0.31 7 The distance separation is one half d=R-R, mut d=21hz-R.) = 2 mg z LEB d=20 (mm): 24 48000 m +(3-3810 y. 2.140 g 2B 21 = 1.60218 2109 20:32 T d=0.0231942 m/or 12=2.31942 cm