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Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 4 / 9B / a / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2011 ia Paper 4 9B a
Created 2026-09-24 Updated 2026-10-14  0 By others on same topic  0 Discussions Create my own version
Take the scalar product of Newton's second law with the velocity r˙. The cross product is perpendicular to its first factor, so
dtd​(21​m∣r˙∣2)=r˙⋅F=−r˙⋅∇ϕ+r˙⋅(r˙×B)=−dtd​ϕ(r).
(1)
Consequently
E=21​m∣r˙∣2+ϕ(r)is constant.​
(2)
This is conservation of energy: the sum of kinetic energy and potential energy stays fixed. The magnetic-type Lorentz force r˙×B does no work, because it is perpendicular to the instantaneous velocity. It may turn the trajectory but cannot directly change its speed, regardless of the spatial variation of B.

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