Hint: \(r=\frac{mv}{eB}\)

Step 1: Find the radius of the circle traced by the beam.
The radius of the circle traced by the beam is given by:
\(r=\frac{mv}{eB}\)

$\begin{array}{l}\begin{array}{rl}& =\frac{5.20\times {10}^6}{(1.76\times {10}^{11})\times 1.30\times {10}^{-4}}=0.227\mathrm{m}=22.7\mathrm{cm}\end{array}\end{array}$

Therefore, the radius of the circular path is 22.7 cm.

(b)
Hint: \(\mathrm{v}=\left(\frac{2 \mathrm{eV}}{\mathrm{m}}\right)^{\frac{1}{2}}\)

Step 1: Find the velocity of the electron beam carrying the energy of 20 MeV.
$\begin{array}{l}\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2\\ \therefore \mathrm{v}={\left(\frac{2\mathrm{E}}{\mathrm{m}}\right)}^{\frac{1}{2}}\\ =\sqrt{\frac{2\times 20\times {10}^6\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}=2.652\times {10}^9\mathrm{m}/\mathrm{s}\end{array}$

Step 2: Identify the mistake.
This result is incorrect because nothing can move faster than light. In the above formula, the expression (mv²/2) for energy can only be used in the non-relativistic limit, i.e for v  << c.

When very high speeds are concerned, the relativistic domain comes into consideration. In the relativistic domain, mass is given as:

$\mathrm{m}={\mathrm{m}}_0{[1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}]}^{\frac{-1}{2}}$

Where,

m₀ = Mass of the particle at rest

Hence, the radius of the circular path is given as:

$\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{eB}}=\frac{{\mathrm{m}}_0\mathrm{v}}{\mathrm{eB}\sqrt{\frac{{\mathrm{c}}^2-{\mathrm{v}}^2}{{\mathrm{c}}^2}}}$