Magnetic Rigidity Calculator

Find magnetic rigidity Bρ = p/q, the key quantity accelerator engineers use to design bending magnets for a given particle beam.

🧲 Magnetic Rigidity Calculator
GeV/c
Magnetic rigidity (Bρ)
Step-by-step working

🧲 What is the Magnetic Rigidity Calculator?

This magnetic rigidity calculator finds Bρ=p/q, the key quantity accelerator engineers use to design bending magnets for a charged particle beam. Enter the beam's momentum in GeV/c and its charge state, and it returns the magnetic rigidity in tesla-metres.

Bρ is the product of a bending magnet's field strength B and the radius of curvature ρ of the particle's circular path through it, directly linking beam momentum, charge, and the magnet system needed to steer it.

This calculator's default LHC-like example (7000 GeV/c, Z=1) gives a magnetic rigidity of roughly 23,349 T·m, closely matching the real LHC's actual 8.33 T dipole field combined with its roughly 2,804 m bending radius.

This calculator is useful for accelerator physics students and engineers designing beamlines, ring circumferences, and magnet systems for a target particle momentum.

📐 Formula

Bρ  =  p / q
p = particle momentum (entered in GeV/c)
q = particle charge = Z × e
Bρ[T·m] = p[GeV/c] / (0.299792458 × Z)
Example: LHC-like p=7000 GeV/c, Z=1: Bρ ≈ 23,349.49 T·m.

📖 How to Use This Calculator

Steps

1
Enter the particle momentum.
2
Enter the charge state.
3
Read the magnetic rigidity.

💡 Example Calculations

Example 1 - LHC-like proton beam

1
p=7000 GeV/c, Z=1
2
Bρ = 7000 / (0.299792458 × 1)
3
Bρ = 23,349.49 T·m, matching the real LHC's ~8.33T × ~2804m dipole system
Bρ = 23,349.49 T·m
Try this example →

Example 2 - Lower-energy synchrotron

1
p=10 GeV/c, Z=1
2
Bρ = 10 / (0.299792458 × 1)
3
Bρ = 33.36 T·m, needing a much smaller ring than the LHC
Bρ = 33.36 T·m
Try this example →

Example 3 - Heavy ion beam (fully stripped lead)

1
p=1000 GeV/c, Z=82 (lead ion)
2
Bρ = 1000 / (0.299792458 × 82)
3
Bρ = 40.68 T·m, far lower than a proton at the same momentum due to the high charge state
Bρ = 40.68 T·m
Try this example →

❓ Frequently Asked Questions

What is magnetic rigidity?+
Magnetic rigidity, Bρ, is the product of a bending magnet's field strength B and the radius of curvature ρ of a charged particle's circular path through it. It equals the particle's momentum divided by its charge (Bρ = p/q), making it the single quantity accelerator engineers use to design bending magnet systems.
What is the formula for magnetic rigidity?+
Bρ = p/q, where p is the particle's momentum and q is its electric charge. In practical accelerator-physics units, this is often written Bρ[T·m] = p[GeV/c] / (0.299792458 × Z), where Z is the charge state in units of the elementary charge.
Why do accelerator physicists use momentum in GeV/c rather than kg·m/s?+
Particle momenta at accelerator energies are enormous in SI units but conveniently sized when expressed in GeV/c (giga-electronvolts per speed of light), since particle energies are already routinely quoted in GeV or TeV. This calculator performs the conversion to SI internally.
How is magnetic rigidity used in beamline design?+
Given a beam's magnetic rigidity and a chosen dipole magnet field strength B, the required bending radius follows directly as ρ = Bρ/B. This lets engineers size ring circumferences, choose magnet field strengths, and design beam transport lines for a target particle momentum.
Why does the LHC need such a large ring?+
The LHC's 7 TeV/c proton beams have an enormous magnetic rigidity, requiring either extremely strong bending magnets or a very large bending radius (or, in practice, both) to keep the beam circulating. Its 27 km circumference and 8.3 T superconducting dipole magnets are a direct consequence of the magnetic rigidity needed at its design energy.
How does charge state affect magnetic rigidity?+
Magnetic rigidity is inversely proportional to charge state Z, so a heavy ion carrying multiple units of charge (like a fully stripped lead nucleus with Z=82) has much lower magnetic rigidity than a proton at the same momentum, allowing the same magnet ring to handle far higher-momentum heavy ion beams than protons.
Is magnetic rigidity the same as momentum?+
No, though closely related, momentum is a property of the particle's motion alone, while magnetic rigidity specifically combines momentum with charge to describe how that particle behaves in a magnetic field, the quantity actually relevant to accelerator magnet design.
What units is magnetic rigidity usually expressed in?+
Tesla-metres (T·m), reflecting its definition as the product of magnetic field strength (tesla) and bending radius (metres). This calculator reports the result in T·m.
Does magnetic rigidity apply to both circular and linear accelerators?+
It is most directly relevant to circular accelerators and any beamline section using bending (dipole) magnets, since it describes circular motion in a magnetic field. Linear accelerators (linacs) do not bend the beam, so magnetic rigidity is not directly applicable to their main accelerating sections, though it still matters for any beam transport or steering magnets they include.
How is magnetic rigidity measured or verified experimentally?+
It can be inferred by measuring a particle's actual bending radius in a known magnetic field (ρ measured directly from the beam trajectory, B measured from the magnet), then computing Bρ, providing an experimental cross-check against the value calculated from the beam's known momentum and charge.

What is magnetic rigidity?

Magnetic rigidity, Bρ, is the product of a bending magnet's field strength B and the radius of curvature ρ of a charged particle's circular path through it. It equals the particle's momentum divided by its charge (Bρ = p/q), making it the single quantity accelerator engineers use to design bending magnet systems.

What is the formula for magnetic rigidity?

Bρ = p/q, where p is the particle's momentum and q is its electric charge. In practical accelerator-physics units, this is often written Bρ[T·m] = p[GeV/c] / (0.299792458 × Z), where Z is the charge state in units of the elementary charge.

Why do accelerator physicists use momentum in GeV/c rather than kg·m/s?

Particle momenta at accelerator energies are enormous in SI units but conveniently sized when expressed in GeV/c (giga-electronvolts per speed of light), since particle energies are already routinely quoted in GeV or TeV. This calculator performs the conversion to SI internally.

How is magnetic rigidity used in beamline design?

Given a beam's magnetic rigidity and a chosen dipole magnet field strength B, the required bending radius follows directly as ρ = Bρ/B. This lets engineers size ring circumferences, choose magnet field strengths, and design beam transport lines for a target particle momentum.

Why does the LHC need such a large ring?

The LHC's 7 TeV/c proton beams have an enormous magnetic rigidity, requiring either extremely strong bending magnets or a very large bending radius (or, in practice, both) to keep the beam circulating. Its 27 km circumference and 8.3 T superconducting dipole magnets are a direct consequence of the magnetic rigidity needed at its design energy.

How does charge state affect magnetic rigidity?

Magnetic rigidity is inversely proportional to charge state Z, so a heavy ion carrying multiple units of charge (like a fully stripped lead nucleus with Z=82) has much lower magnetic rigidity than a proton at the same momentum, allowing the same magnet ring to handle far higher-momentum heavy ion beams than protons.

Is magnetic rigidity the same as momentum?

No, though closely related, momentum is a property of the particle's motion alone, while magnetic rigidity specifically combines momentum with charge to describe how that particle behaves in a magnetic field, the quantity actually relevant to accelerator magnet design.

What units is magnetic rigidity usually expressed in?

Tesla-metres (T·m), reflecting its definition as the product of magnetic field strength (tesla) and bending radius (metres). This calculator reports the result in T·m.

Does magnetic rigidity apply to both circular and linear accelerators?

It is most directly relevant to circular accelerators and any beamline section using bending (dipole) magnets, since it describes circular motion in a magnetic field. Linear accelerators (linacs) do not bend the beam, so magnetic rigidity is not directly applicable to their main accelerating sections, though it still matters for any beam transport or steering magnets they include.

How is magnetic rigidity measured or verified experimentally?

It can be inferred by measuring a particle's actual bending radius in a known magnetic field (ρ measured directly from the beam trajectory, B measured from the magnet), then computing Bρ, providing an experimental cross-check against the value calculated from the beam's known momentum and charge.