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InterviewSolution
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Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. |
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Answer» Solution :When a particle of mass m and charge q performs uniform circular motion with CONSTANT speed v in a plane perpendicular to uniform magnetic field B, radius of its circular path is, `R=(mv)/(Bq)` `thereforer=(m(romega))/(Bq)""(becausev=romega)` `thereforeomega=(Bq)/m` (Where `OMEGA` = angular frequency) `therefore2pif=(Bq)/m` (Where f = frequency of revolution = no. of revolutions in unit time) `thereforef=(Bq)/(2pim)` `thereforef=((6.5xx10^(-4))(1.6xx10^(-19)))/((2)(3.14)(9.1xx10^(-31)))` `thereforef=18.2xx10^(6)Hz` (= 18.2 MHz) `thereforef=1.82xx10^(7)Hz` |
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