(a) A monoenergetic electron beam with electron speed of` 5.20 xx 10^(6) m s^(-1)` is subject to a magnetic field of `1.30 xx10^(-4) T` normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals `1.76 xx 10^(11)C kg^(-1)` .
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron 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)))`