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+### Statement
+
+$14.4.18.$ [Insert the problem statement]
+
+### Solution
+
+Data:
+
+Arc radius: $R = 0.3\ \text{m}$.
+Magnetic flux density: $B = 1\ \text{T}$.
+Elementary charge: $e = 1.6 \times 10^{-19}\ \text{C}$
+
+The kinetic energy of protons and electrons that pass through the magnet following this arc is requested.
+
+ Momentum–magnetic field relation
+
+A particle of charge e moving perpendicular to a uniform magnetic field B describes a circular arc. The magnetic force $ F = e v B $ acts as the centripetal force. In relativity, the relation is:
+
+$p = e B R$
+
+where p is the linear momentum
+
+We calculate the momentum :
+
+$p = (1.6 \times 10^{-19}\ \text{C}) \times (1\ \text{T}) \times (0.3\ \text{m}) = 4.8 \times 10^{-20}\ \text{kg·m/s}$
+
+To work with energies in MeV, we use pc
+
+$pc = p \times c = 4.8 \times 10^{-20} \times 3 \times 10^8\ \text{m/s}
+ = 1.44 \times 10^{-11}\ \text{J}$
+
+Convert to $ MeV (1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J})$:
+
+$pc = \frac{1.44 \times 10^{-11}}{1.6 \times 10^{-13}}\ \text{MeV} \approx 90\ \text{MeV}$.
+
+Kinetic energy
+
+The total energy E and momentum p are related to the rest mass $ m_0 c^2$:
+
+$E^2 = (pc)^2 + (m_0 c^2)^2, \qquad K = E - m_0 c^2$.
+
+We apply this to each particle.
+
+Proton $(m_p c^2 \approx 938\ \text{MeV})$:
+
+$K_p = \sqrt{90^2 + 938^2} - 938
+ = \sqrt{8100 + 879844} - 938
+ \approx \sqrt{887944} - 938
+ \approx 942.3 - 938
+ \approx 4.3\ \text{MeV}$
+
+The non‑relativistic approximation could also be used $(pc \ll m_p c^2)$:
+
+$K_p \approx \frac{p^2}{2m_p} = \frac{(pc)^2}{2m_p c^2} = \frac{90^2}{2 \times 938} \approx 4.3\ \text{MeV}$
+
+Electron $(m_e c^2 \approx 0.511\ \text{MeV})$
+
+$K_e = \sqrt{90^2 + 0.511^2} - 0.511
+ \approx \sqrt{8100 + 0.261} - 0.511
+ \approx 90.0015 - 0.511
+ \approx 89.5\ \text{MeV}$
+
+The electron is ultra‑relativistic:
+$K_e \approx pc - m_e c^2 \approx 90 - 0.5 \approx 89.5\ \text{MeV}$
+
+3. Final result
+
+$\boxed{K_p \approx 4.3\ \text{MeV}}, \qquad $
+
+$\boxed{K_e \approx 89.5\ \text{MeV}}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]