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.
Concept: Motion in Combined Electric and Magnetic Fields - Cyclotron
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