(a) A monoenergetic electron beam with electron speed of `5.20xx10^(6)ms^(-1)` is subjected to a magnetic field of `1.30xx10^(-4)T`, normal to the beam velocity. What is the radius of the circle traced by the beam, given `e//m` for electron equal `1.76xx10^(11)C.kg^(-1)`.
(b) Is the formula you employ in (a) valid for calculating radius of the path of 20 MeV electrons beam? If not, in what way is it modified!

Explanation

(a) Here, `v=5.20xx10^(6)ms^(-1), B=1.30xx10^(-4)T, e//m=1.76xx10^(11) Ckg^(-1), theta = 90^(@)`
Force exerted by the magnetic field on the electron
`F=e|vecvxxvecB|=evBsin theta=evB ( :'sin 90^(@)=1)`
Since, the normal magnetic field provides the centripetal force,
`:. evB=(mv^(2))/r or r=(mv)/(eB) =v/((e//m)B)=(5.20xx10^(6))/(1.76xx10^(11)xx1.30xx10^(-4))=0.277m=22.7cm`
(b) Energy, `E=20 MeV =20xx1.6xx10^(-13)J=1/2mv^(2) :. v=((2xx20xx1.6xx10^(-13))/(9xx10^(-31)))^(1//2)=2.67xx10^(9)m//s`
Which is greater than the velocity of light.
Therefore, the formula `r=mv//eB` is not valid for calculating the radius of the path of 20 MeV electron beam because electron with such a high energy has velocity in the relativistic domain (i.e., comprable with the velocity of light). For this we use relativistic formula as follows.
`r=(mv)/(eB)=((m_(0))/(sqrt(1-v^(2)//c^(2))))v/(eB)`