InterviewSolution
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InterviewSolution
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A monoenergetic (18 keV) electron beam Initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (m_e= 9.11 xx 10^(-31) kg). (Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth's magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.] |
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Answer» Solution :Given : `(K.E)_(e) = 18 xx10^(3)xx1.6xx10^(-19) J = 28.8 xx10^(-16) `J. `B_(H) = 0.04 G = 4XX 10^(-6) T.` SINCE electronsdescribe circular path Bev sin `theta = (mv^(2))/( R) and theta = 90^(@)` ` R = (mv)/(EB)` and `"" KE = 1//2 mv^(2)` ` v = sqrt((2)/(m)(KE))` and `mv = sqrt(2m (KE))` i.e., `"" R= (1)/(eB) sqrt(2m(KE))` i.e., `"" R = sqrt(2xx9.11xx10^(31)xx28.8xx 10^(-16))/(1.6xx10^(-19)xx4xx10^(-6))` R= 11.3 m sin `theta = 0.3//11.3 = 0.2655 ` `cos theta = sqrt(1-sin^(2)theta) = sqrt(1-0.007)= 0.9996` Up or down deflection - R (1-cos0) = 11.3 (1-0.9996) = 0.00398 m `= 3.98 xx 10^(-3)`m `~~ 4.0 mm` |
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