 Solution

ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/section-a/a4/solution

r = p / (e B)

. At this energy the proton is ultrarelativistic, giving

p c ≃ E

and

r ≃ \frac{7 \times 1 0 ^{12} eV}{ec ( 9 T )} = 2.59 \times 1 0^{3} m ≃ 2.6 km .

The electronvolt-to-joule conversion cancels the elementary-charge factor numerically. This uses external CODATA light speed and charge definitions.^[ref] If the stated beam energy means total energy,

p c = E^{2} - m_{p}^{2} c^{4}

p c = T (T + 2 m_{p} c^{2})

. Either interpretation gives the quoted radius at the precision of the supplied numbers.

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