Q.(a) A monoenergetic electron beam with electron speed of $5.20 \times 10^{6}\ \text{m s}^{-1}$ is subject to a magnetic field of $1.30 \times 10^{-4}\ \text{T}$ normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals $1.76 \times 10^{11}\ \text{C kg}^{-1}$.

Equating the magnetic force to the centripetal force needed for circular motion gives the radius directly from the beam's speed, the field, and $e/m$; but at relativistic energies the electron's momentum is no longer simply $m_0v$, so the same formula needs modification.

Step 1 — Radius from the balance of magnetic and centripetal force.

For a charge moving in a circle under a perpendicular magnetic field, $evB = \dfrac{mv^2}{r}$, so

(a) With $v = 5.20\times10^{6}\ \text{m/s}$, $B=1.30\times10^{-4}\ \text{T}$, $e/m=1.76\times10^{11}\ \text{C/kg}$:

(b) Is this valid at 20 MeV?

The electron's rest mass energy is only $m_0c^2=0.511\ \text{MeV}$. A 20 MeV beam has kinetic energy about 39 times this rest energy, so the electron is moving at a speed extremely close to $c$ and is strongly relativistic. The formula $r=mv/(eB)$ used above assumed $m$ is the fixed rest mass $m_0$ — but the correct relation for a relativistic charge is

i.e. the rest mass $m_0$ in the formula must be replaced by the relativistic momentum $p$ (equivalently, by $\gamma m_0$, the relativistically increased effective inertia). So the formula in (a) is not valid as-is for the 20 MeV beam; it must be modified by using the relativistic momentum in place of $m_0v$.