(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) 22.7 cm
(b) No. As explained above, a 20 MeV electron moves at relativistic speed. Consequently, the non-relativistic formula `R = (m_(0) v//e B)` is not valid. The relativistic formula is
`R=p//eB=mv//eB or R=m_(0)v//(eBsqrt(1-v^(2)//c^(2)))`