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?
Advertisement Remove all ads
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
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads
