(8%) Problem 9: A proton moves in a circular path of the same radius as a cosmic
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Question
(8%) Problem 9: A proton moves in a circular path of the same radius as a cosmic ray electron moving at 8.5 x 102 m/s to the Earth's magnetic field at an altitude where the field strength is 1.0 x 10-ST. 25% Part (a) what will the speed of the proton be in mis? as 25% Part (b) What would the radius of the path be in meters if the proton had the same speed as the electron? ?? ? 25% Part (c) what would the radius be in meters if the proton had the same kinetic energy as the electron? 25% Part (d) What would the radius be in meters if the proton had the same momentum as the electron?Explanation / Answer
a) radius of path of electron, r = m*v/(B*q)
= 9.1*10^-31*8.5*10^6/(1*10^-5*1.6*10^-19)
= 4.83 m
proton speed, v = ?
use, r = m*v/(B*q)
==> v = B*q*r/m
= 1*10^-5*1.6*10^-19*4.83/(1.67*10^-27)
= 4.63*10^3 m/s <<<<<<<<<<--------------------------Answer
b) r = m*v/(B*q)
= 1.67*10^-27*8.5*10^6/(1*10^-5*1.6*10^-19)
= 8.87*10^3 m or 8.87 km <<<<<<<<<<--------------------------Answer
c) KE_proton = KE_electron
(1/2)*mp*vp^2 = (1/2)*me*ve^2
vp = ve*sqrt(me/mp)
= 8.5*10^6*sqrt(9.1*10^-31/(1.67*10^-27))
= 1.98*10^5 m/s
r = m*vp/(B*q)
= 1.67*10^-27*1.98*10^5/(1*10^-5*1.6*10^-19)
= 207 m <<<<<<<<<<--------------------------Answer
d)
P_proton = P_electron
mp*vp = me*ve
vp = ve*(me/mp)
= 8.5*10^6(9.1*10^-31/(1.67*10^-27))
= 4631 m/s
r = m*vp/(B*q)
= 1.67*10^-27*4631/(1*10^-5*1.6*10^-19)
= 4.83 m <<<<<<<<<<--------------------------Answer
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