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Asked in NEET 2013 · Crossed electric and magnetic fields
Given: at rest the proton accelerates west with a₀; moving north at v₀ it accelerates west with 3a₀. Take east as ̂x, north as ̂y, up as ̂z.
Idea: the field that acts at rest is the electric one; whatever extra appears when the proton moves is magnetic.
At rest: eE=ma₀, so E=(ma₀)/e, and the proton is positive, so ⃗E points west, the way it accelerates.
Moving north, the extra acceleration is 3a₀-a₀=2a₀, all of it magnetic.
ev₀B=2ma₀, so B=(2ma₀)/(ev₀).
Direction: the force e⃗v×⃗B must point west, -̂x. With ⃗v=v₀̂y, taking ⃗B=-B̂z gives ̂y×(-̂z)=-̂x, which is west.
So ⃗B points down: (ma₀)/e west and (2ma₀)/(ev₀) down.
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