| Data: | | Data: |
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| Arc radius: $R = 0.3\ \text{m}$. | | Arc radius: $R = 0.3\ \text{m}$. |
| Magnetic flux density: $B = 1\ \text{T}$. | | Magnetic flux density: $B = 1\ \text{T}$. |
| Elementary charge: $e = 1.6 \times 10^{-19}\ \text{C}$ | | Elementary charge: $e = 1.6 \times 10^{-19}\ \text{C}$ |
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| The kinetic energy of protons and electrons that pass through the magnet following this arc is requested. | | The kinetic energy of protons and electrons that pass through the magnet following this arc is requested. |
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| Momentum–magnetic field relation | | Momentum–magnetic field relation |
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| 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: | | 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: |
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| $p = e B R$ | | $p = e B R$ |
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| where p is the linear momentum | | where p is the linear momentum |
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| We calculate the momentum : | | We calculate the momentum : |
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| $p = (1.6 \times 10^{-19}\ \text{C}) \times (1\ \text{T}) \times (0.3\ \text{m}) = 4.8 \times 10^{-20}\ \text{kg·m/s}$ | | $p = (1.6 \times 10^{-19}\ \text{C}) \times (1\ \text{T}) \times (0.3\ \text{m}) = 4.8 \times 10^{-20}\ \text{kg·m/s}$ |
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| To work with energies in MeV, we use pc | | To work with energies in MeV, we use pc |
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| $pc = p \times c = 4.8 \times 10^{-20} \times 3 \times 10^8\ \text{m/s} | | $pc = p \times c = 4.8 \times 10^{-20} \times 3 \times 10^8\ \text{m/s} |
| = 1.44 \times 10^{-11}\ \text{J}$ | | = 1.44 \times 10^{-11}\ \text{J}$ |
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| Convert to $ MeV (1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J})$: | | Convert to $ MeV (1\ \text{MeV} = 1.6 \times 10^{-13}\ \text{J})$: |
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| $pc = \frac{1.44 \times 10^{-11}}{1.6 \times 10^{-13}}\ \text{MeV} \approx 90\ \text{MeV}$. | | $pc = \frac{1.44 \times 10^{-11}}{1.6 \times 10^{-13}}\ \text{MeV} \approx 90\ \text{MeV}$. |
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| Kinetic energy | | Kinetic energy |
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| The total energy E and momentum p are related to the rest mass $ m_0 c^2$: | | The total energy E and momentum p are related to the rest mass $ m_0 c^2$: |
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| $E^2 = (pc)^2 + (m_0 c^2)^2, \qquad K = E - m_0 c^2$. | | $E^2 = (pc)^2 + (m_0 c^2)^2, \qquad K = E - m_0 c^2$. |
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| We apply this to each particle. | | We apply this to each particle. |
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| Proton $(m_p c^2 \approx 938\ \text{MeV})$: | | Proton $(m_p c^2 \approx 938\ \text{MeV})$: |
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| $K_p = \sqrt{90^2 + 938^2} - 938 | | $K_p = \sqrt{90^2 + 938^2} - 938 |
| = \sqrt{8100 + 879844} - 938 | | = \sqrt{8100 + 879844} - 938 |
| \approx \sqrt{887944} - 938 | | \approx \sqrt{887944} - 938 |
| \approx 942.3 - 938 | | \approx 942.3 - 938 |
| \approx 4.3\ \text{MeV}$ | | \approx 4.3\ \text{MeV}$ |
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| The non‑relativistic approximation could also be used $(pc \ll m_p c^2)$: | | The non‑relativistic approximation could also be used $(pc \ll m_p c^2)$: |
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| $K_p \approx \frac{p^2}{2m_p} = \frac{(pc)^2}{2m_p c^2} = \frac{90^2}{2 \times 938} \approx 4.3\ \text{MeV}$ | | $K_p \approx \frac{p^2}{2m_p} = \frac{(pc)^2}{2m_p c^2} = \frac{90^2}{2 \times 938} \approx 4.3\ \text{MeV}$ |
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| Electron $(m_e c^2 \approx 0.511\ \text{MeV})$ | | Electron $(m_e c^2 \approx 0.511\ \text{MeV})$ |
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| $K_e = \sqrt{90^2 + 0.511^2} - 0.511 | | $K_e = \sqrt{90^2 + 0.511^2} - 0.511 |
| \approx \sqrt{8100 + 0.261} - 0.511 | | \approx \sqrt{8100 + 0.261} - 0.511 |
| \approx 90.0015 - 0.511 | | \approx 90.0015 - 0.511 |
| \approx 89.5\ \text{MeV}$ | | \approx 89.5\ \text{MeV}$ |
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