(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) Given, speed of electron , `v=5.20xx10^6ms^(-1)`
Magnetic field , `B=1.30xx10^(-4)T`
`e//m=1.76xx10^(11)C kg ^(-1)`
Magnetic force exerted electron will provide the necessary centripetal force for the electron to move in the circular path of radius r. Then
`evB=(mv^2)/r`
`impliesr=(mv)/(eB)`
or `r=v/(e//mxxB)`
`:.r=(5.20xx10^6)/(1.76xx10^11xx1.30xx10^(-14))`
`=0.227m = 22.7 cm`
(b) Given , `E=20 MeV = 20xx1.6xx10^(-19)J`
Since ,
`E=1/2mv^2`
`impliesv=((2E)/m)^(1//2)`
`=((2xx20xx1.6xx10^(-19))/(9.1xx10^(-31)))^(1/2)`
`=2.65xx10^6m//s`
This is comparable to velocity of light . So, the formula is not valid for calculating radius of path of 20 Me V electron beam.
The relativistic formula is given by
`r=p/(eB)=(mv)/(eB)`
or `r=(m_0v)/(eBsqrt(1-v^2/c^2))`