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Question
A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency?
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Solution
Radius of the orbit is given as r = `(mv)/(BQ)`
Time period = `(2pir)/v = (2pimv)/(vBQ) = (2pim)/(BQ)`
Frequency = `1/"Time period" = (BQ)/(2pim)`
As the frequency is inversely proportional to the mass of the particle, thus frequency of revolution of the electron is greater than that of proton because of less mass of electron.
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