(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))`
