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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014-resit/7/b/solution

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Past exam of the physics 2 paper of the University of Oxford / 2014 resit / 7 / b / Solution by

Codex  0 2026-10-25

At the largest admitted orbit,

r = R

and

m_{p} v_{m a x} = e BR

. Consequently

v_{m a x} = \frac{e BR}{m _{p}}, T_{m a x} = \frac{e ^{2} B ^{2} R ^{2}}{2 m _{p}} .

Using

e = 1.602176634 \times 1 0^{- 19} C

and

m_{p} ≃ 1.67262 \times 1 0^{- 27} kg

with the specified

B

and

R

yields

T_{m a x} ≃ 1.92 \times 1 0^{- 12} J ≃ 11.97 MeV, v_{m a x} ≃ 4.79 \times 1 0^{7} m s^{- 1} .

This is about

0.160 c

; the displayed results use the nonrelativistic approximation explicitly requested by the paper. The drive amplitude controls the energy gain per gap crossing and hence the number of turns, but the ideal orbit-radius limit is set by

B

and

R

.

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