(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)Speed of an electron, `v=5.20xx10^6` m/s
Magnetic field experienced by the electron , `B=1.30xx10^(-4)`T
Specific charge of an electron , e/m =`1.76xx10^11 C kg ^(-1)`
Where, e=Charge on the electron =`1.6xx10^(-19) C`
m=Mass of the electron =`9.1xx10^(-31) kg ^(-1)`
The force exerted on the electron is given as :
`F=e|vecv xx vecB|`
`=ev B sin theta`
`theta` = Angle between the magnetic field and the beam velocity .
The magnetic field is normal to the direction of beam. ltbr `therefore theta=90^@`
F=evB ... (1)
The beam traces a circular path of radius , r. It is magnetic field, due to its bending nature, that provides the centripetal force `(F=(mv^2)/r)` for the beam.
Hence, equation (1) reduces to :
`evB=(mv^2)/r`
`therefore r =(mv)/(eB)=v/((e/m)B)`
`=(5.20xx10^6)/((1.76xx10^11)xx1.30xx10^(-4))=0.227 m = 22.7 cm`
Therefore , the radius of the circular path is 22.7 cm
(b)Energy of the electron beam, `E=20 Mev=20xx10^6 xx 1.6xx 10^(-19) J`
The energy of the electron is given as : `E=1/2 mv^2` `therefore v=((2E)/m)^(1/2)`
`=sqrt((2xx20xx10^6 xx 1.6xx10^(-19))/(9.1xx10^(-31)))=2.652xx10^9` m/s
This result is incorrect because nothing can move faster than light. In the above formula, the expression `(mv^2//2)` for energy can only be used in the non-relativistic limit, i.e., for v lt lt c
When very high speeds are concerned , the relativistic domain comes into consideration .
In the relativistic domain , mass is given as :
`m=m_0[1-v^2/c^2]^(1/2)` Where, `m_0` = mass of the particle of rest
Hence, the radius of the circular path is given as : r=mv/eB
`=(m_0v)/(eBsqrt((c^2-v^2)/c^2))`