(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?

Explanation

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.