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

$\boxed{K_p \approx 4.3\ \text{MeV}}, \qquad$

$\boxed{K_e \approx 89.5\ \text{MeV}}$