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Asked in GUJCET 2014 · Circular motion: radius, period and energy
Given: an α-particle (m_α=4mₚ, q_α=2e) and a proton (mₚ, e) circling in the same field B.
Idea: the period of the circular motion is T=(2π m)/(qB). The speed cancels out, so only m/q matters.
T_α=(2π(4mₚ))/((2e)B)=2×(2π mₚ)/(eB)=2Tₚ.
So T_α : Tₚ=2 : 1.
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