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A monoenergetic (18 KeV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.40 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^(-19) C) . |
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Answer» Solution :Here `m= 9.11xx10^(-31) kg, e= 1.6xx10^(-19) C, B = 0.40 G = 4 xx 10^(-5) T ` and kinetic energy of ELECTRON beam K = 18 keV `= 18 xx 1.6xx10^(-16) J = 2.88 xx 10^(-15) J`. Radius of circular PATH of electron beam `r =(mv)/(eB) = (sqrt(2Km))/(eB) = sqrt(2 xx 2.88 xx 10^(-15) xx 9.11 xx 10^(-31))/(1.6 xx 10^(-19) xx 4 xx 10^(-5)) = 11. 32 m ` Fig. shows and electron covering a path PQ = 30 cm = 0.3 m and radius of the circle OP = OQ = 11.32 m . It is clear that electron beam will undergo a deflection PA in the downward direction. If `theta` be the angle subtended by PQ at centre point O, then `theta = (PQ)/(OP) =(0.3)/(11.32) = 0.0265 "rad" =1^@ 31.` `therefore PA = OP - OA =r-rcos theta = r(1-cos theta ) = 11.32 (1-cos 1^@ 31.)` `=11.32 (1-0.996) =4 xx 10^(-3) m = 4` mm 
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