A proton synchrotron accelerates protons to a kinetic energy of 533 GeV. At this energy, calculate (a) the Lorentz factor, (b) the speed parameter, and (c) the magnetic field for which the proton orbit has a radius of curvature of 764 m.
we know that kinetic energy of relativistic particle is
K.E = mc2 ( - 1)
or
= K.E / mc2 + 1
where mc2 is rest energy of proton = 938.257 MeV
= 533e9 / 938.257e6 + 1
= 569.1
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speed parameter is found as
= sqrt ( 1 - 1 / 2)
= sqrt ( 1 - 1 / 569.12)
= 0.99999845
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we know for charged particle in field
qvB = mv2 / r
r = mv / qB
in relativistic terms
r = mv / qB
where v = *
so,
B = mc2 * sqrt (2 - 1) / qcr
B = 938.257e6 * 1.6e-19 * sqrt (569.12 - 1) / 3e8 * 1.6e-19 * 764
B = 2.329 T
A proton synchrotron accelerates protons to a kinetic energy of 533 GeV. At this energy, calculate...
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