Question

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.

Answer 1

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