$(a)$ $A$ monoenergetic electron beam with electron speed of $5.20 \times 10^{6} \;m s^{-1}$ is subject to a magnetic field of $1.30 \times 10^{-4} \;T$ normal to the beam velocity. What is the radius of the circle traced by the beam,given $e/m$ for electron equals $1.76 \times 10^{11} \;C \;kg^{-1}$?
$(b)$ Is the formula you employ in $(a)$ valid for calculating the radius of the path of a $20 \;MeV$ electron beam? If not,in what way is it modified?

(N/A) Given: Speed of electron $v = 5.20 \times 10^{6} \;m s^{-1}$,Magnetic field $B = 1.30 \times 10^{-4} \;T$,Specific charge $e/m = 1.76 \times 10^{11} \;C \;kg^{-1}$.
The magnetic force provides the necessary centripetal force for circular motion: $evB = \frac{mv^2}{r}$.
Rearranging for radius $r$: $r = \frac{mv}{eB} = \frac{v}{(e/m)B}$.
Substituting the values: $r = \frac{5.20 \times 10^{6}}{(1.76 \times 10^{11}) \times (1.30 \times 10^{-4})} = \frac{5.20 \times 10^{6}}{2.288 \times 10^{7}} \approx 0.227 \;m = 22.7 \;cm$.
$(b)$ No,the formula is not valid for a $20 \;MeV$ electron beam. At $20 \;MeV$,the kinetic energy is much larger than the rest mass energy of the electron $(0.511 \;MeV)$,meaning the electron speed $v$ approaches the speed of light $c$.
In this relativistic regime,the mass $m$ is not constant but is given by $m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$,where $m_0$ is the rest mass.
The modified formula for the radius is $r = \frac{mv}{eB} = \frac{m_0 v}{eB \sqrt{1 - v^2/c^2}}$.