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Asked in GSEB Board August 2020 · Circular motion: radius, period and energy
Given: m=9×10⁻³¹ kg, e=1.6×10⁻¹⁹ C, B=4×10⁻⁴ T perpendicular to the velocity.
Idea: the time for one revolution is T=(2π m)/(eB), which carries no speed in it, so the 3×10⁷ m s⁻¹ is not needed. The frequency is f=1/T=(eB)/(2π m).
Top: 1.6×10⁻¹⁹×4×10⁻⁴=6.4×10⁻²³.
Bottom: 2π×9×10⁻³¹=5.65×10⁻³⁰.
f=(6.4×10⁻²³)/(5.65×10⁻³⁰)=1.132×10⁷ Hz=11.32 MHz.
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