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 0.04G 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 exercies are so chosen that the answer will give you an idea of the effect of earth's magnetic field onthe motion of the electron beam from electron gun to the screen in a TV set]. |
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Answer» Solution :Here energy E `= 18 KEV = 18 XX 1.6 xx 10^(-19)J` `B = 0.40 G = 0.40 xx 10^(-4)T` `x = 30 cm = 0.3 m` As `E = (1)/(2) mv^(2) :. v = sqrt((2E)/(m))` In magnetic field electron beam is deflected along a circular arc or radius r, such that `BeV = (mv^(2))/(r)` or `r = (mv)/(Be)` `r = (m)/(Be)sqrt((2E)/(m)) = (1)/(Be) sqrt(2Em) = 11.3 m` If y is the deflection at the END of the path it is clear from FIG. `theta = (x)/(r) = (y)/(x//2) = 2(y)/(x)` or `y = (x^(2))/(2r) = (0.30 xx (0.30))/(2 xx 11.3)m = 0.004 m = 4mm` 
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