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(a) A monoenergetic electron beam with electron speed of 5.20 xx 10^(6) m s^(-1) is subject to a magnetic field of 1.30 xx10^(-4) T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 xx 10^(11)C kg^(-1) . (b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified? |
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Answer» Solution :Here velocity `v=5.20xx10^(6)ms^(-1)` Magnetic field B`=1.30xx10^(-4)`T `(e )/(m)=1.76xx10^(11)C kg^(-1)` Kinetic energy of electron BEAM K=20 MEV Mass of electron m`=9.1xx10^(-31)Kg` (a) Centripetal fornce provided by magnetic field to electron, `(mv^(2))/(r)=bev` `therefore r=(mv)/(be)=(v)/(B((e)/(m)))` `therefore r=(5.20xx10^(6))/(1.30xx10^(-4)xx1.76xx10^(11))` `therefore r=2.2727xx10^(-1)` `therefore r~~0.227 m=22.7cm` (b)K=20 MeV `therefore (1)/(2)mv^(2)=20xx10^(6)xx1.6xx10^(-19)J` `therefore v=sqrt((2xx20xx10^(6)xx1.6xx10^(-19))/(m))` `therefore v=sqrt((64xx10^(-13))/(9.1xx10^(-31)))=sqrt(7.033xx10^(18))` `therefore v=2.65xx10^(9)m//s` No. `implies`Here speed of electron v`GT` speed of light c `therefore` This suggest that there is something incorrect in calculation .HEnce equation `r=(m_(0)v)/(eB)`,not valid.Hence we should use equation of RELATIVITY `r=(mv)/(eB)` but `m=(m_(0))/(sqrt(1-(v^(2))/(c^(2))))` `therefore` Equation `r=(m_(0)v)/(eBsqrt(1-(v^(2))/(c^(2))))` should be used. |
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