A proton moves in a circular orbit of radius 60 cm perpendicular to a uniform ma

Question

A proton moves in a circular orbit of radius 60 cm perpendicular to a uniform magnetic fieldof magnitude 0.80 T. (a) What is the period for this motion?
ns

(b) Find the speed of the proton.
m/s

(c) Find the kinetic energy of the proton.
MeV


PLEASE GIVE ME AN ANSWEAR AT THE END. YOU DON"T HAVE TO SHOW ME THEWORK.. (a) What is the period for this motion?
ns

(b) Find the speed of the proton.
m/s

(c) Find the kinetic energy of the proton.
MeV


PLEASE GIVE ME AN ANSWEAR AT THE END. YOU DON"T HAVE TO SHOW ME THEWORK..

Explanation / Answer

given mass of the proton m = 1.67*10-27kg charge of proton q = 1.6*10-19C magnetic field B = 0.80T radius of circular orbit of the prton r =60*10-2m a) the period for this motion is     T = 2m/Bq         = -------- b) the radius of the circular path is r = mv/Bq                     or                                   v = Bqr/m                                      = ----------------- c) The kinetic energy of the proton is KE =1/2*mv2                                                            =---------------                                                            =---------------

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