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Asked in GSEB Board July 2017 · Lorentz force and work done
Given: an electron going round a circle of radius r in a field B perpendicular to its velocity.
Idea: the magnetic force ⃗F=-e(⃗v×⃗B) is perpendicular to ⃗v at every instant.
The work done over a small step of the path is therefore ⃗F· d⃗s=F ds cos 90°=0.
Adding zeros over half a revolution — or over any part of one — still gives zero.
By the work-energy theorem the kinetic energy gained is zero: the field turns the electron without speeding it up or slowing it down.
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