Question

Tutorial Exercise A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.575 T over a region of radius 1.30 m. (a) What is the cyclotron frequency? (b) What is the maximum speed acquired by the protons? Part 1 of 4 - Conceptualize Cyclotrons can be used to produce radioactive substances used in diagnosis and treatment. We expect the angular frequency to be on the order of millions of radians per second. We expect the maximum speed of the proton to be a substantial fraction of the speed of light. Part 2 of 4 - Categorize We apply the equations of angular frequency to the proton that accelerates in the magnetic field of the cyclotron. Part 3 of 4 - Analyze (a) Cyclotron frequency refers to the angular frequency or angular speed of the accelerated particle. For the angular speed of a proton, we have the following. W = 9B (1.60 x 10-19) (No Response) (1.67 x 10-27 kg) 0.575) = (No Response) 5.51 x 107 rad/s Part 4 of 4 - Analyze The path radius is given by RB9 Just before the protons escape, their speed is V = BOR m 1)(1.60 x 10-19 )( m)

Answer 1

Given soin magnetic field =0.575T Radius of cyclotron = R=1.30 m Charge on Proton q = 1.6610lac mass of proton m= 1.678102 kg freauery of cyclotron w = QB = 4.6x1099) (0.575) 551100 ha - cai 167 16 bi maximum speed acquired by podtons V= BIR = 10:575) (1.68 109 ) 11:30) - 1. byx1027 V = 71.6Ixiom ) sec gf you are saticfy with us then please appreciate us with like. gf not then tell us in comment box. we are happy to help you again. Thank you.