Charged-particle motion in magnetic fields
Use the magnetic Lorentz force to predict straight, circular, and helical particle motion with correct direction and sign logic.
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Guided steps
- Find the velocity cross magnetic field direction first, then apply the charge sign. Force magnitude is nonnegative. In the stated right-handed axes, x cross z is negative y. Changing charge sign at the same velocity reverses the force; a fixed-conventional-current comparison is a different constraint.
- Split the velocity into parallel and perpendicular components relative to the field. The parallel component has zero cross product with the field and stays constant. The transverse component turns at constant magnitude. Those two motions together distinguish a helix from a circle or a straight path.
- Balance magnetic force with radial acceleration of the transverse motion. Only perpendicular speed enters radius. Dividing perpendicular speed by that radius gives a rotation rate depending on charge, mass and field, not speed. During one turn the parallel velocity produces axial advance. The figure’s helix is a projection of a three-dimensional path, not a field line the particle must follow.
- The speed-independent period assumes non-relativistic motion in a uniform field. Purely parallel entry has no actual orbit to time. Magnetic force is perpendicular to the instantaneous velocity and does no work. After the field ends with no other force present, the entire instantaneous velocity continues unchanged.
- For perpendicular entry, use the given speed in magnetic force and radial acceleration, then obtain the period from one circumference. Keep charge magnitude separate from turning direction and retain guard digits between calculations. Check the non-relativistic and uniform-field assumptions before interpreting the result.
- An oblique entry needs two component decisions before substitution. Use the transverse component for force and radius, one common rotation period, and the parallel component for axial pitch. Total speed times that period is distance along the helix, not distance along its axis.
- Find the initial signed force component, transverse radius, rotation period and axial advance per turn. Keep the stated axes and distinguish seconds from microseconds. Compare the component and energy explanations separately.
- Explain why different velocity components set the force, radius and pitch. Show why magnetic force changes direction without doing work. Compare your cross product and power argument with the public model yourself; the four numerical values do not mark that reasoning. Paper work is fine.
- Distinguish two separately prepared entry states from a particle being accelerated by a magnetic field. Track which component is held fixed before predicting radius or pitch. At a force-free exit, use the full instantaneous velocity rather than the old field axis.
- Compare radius, period and axial pitch when only the parallel entry component doubles. Then choose the force-free exit path. Explain the initial-state and energy distinction in your separate response.
- Explain the parameter factors, why the two initial energies may differ without magnetic work, and why exit motion is tangent to the old helix. Compare those arguments yourself; the checked factors and path choice do not mark the explanation.
- Predict the magnetic-force result for a particle moving parallel to the field. Use the cross product rather than assuming that every charged particle in a field must turn.
- Find the electron orbit radius using this particle’s mass and the stated perpendicular speed. Give a positive radius in metres; the charge sign controls turning direction separately.
- Predict the radius change when only field magnitude doubles. Keep mass, charge magnitude and perpendicular speed fixed.
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A uniform magnetic field can turn a velocity vector without changing its magnitude. Separate the velocity along the field from the velocity across it: that decision determines whether the path is straight, circular or helical, and which speed belongs in the radius formula.
Use radial acceleration in circular motion. The cross-product direction is developed below; a wire-force formula is a later application, not a prerequisite. Assume a nonzero uniform magnetic field, no electric field or other force, constant mass and charge, and non-relativistic speeds.
Fix the force direction before choosing a path
The magnetic part of the Lorentz force is F = qv × B, F = |q|vB sin θ, where θ is the angle between velocity and field. The magnitude is nonnegative; the charge sign belongs in the vector direction. For nonparallel vectors, point your right-hand fingers along v and curl them towards B; your thumb gives v × B. Reverse that direction for a negative charge.
For example, in right-handed axes x hat × z hat = -y hat. A positive charge initially moving along +x in a field along +z has force along −y; a negative charge with the same velocity has force along +y. This comparison holds velocity fixed, unlike a comparison at fixed conventional current in the Hall effect.
Separate parallel and perpendicular motion
Resolve the initial velocity into v_∥ along the field and v_⊥ in the plane normal to it. If their magnitudes are written v_∥ and v_⊥, the cross product depends only on the perpendicular component. An axial direction can be retained separately when stating the advance.
The magnetic force has no field-parallel component, so the parallel velocity stays constant. The transverse velocity turns at constant magnitude. Pure parallel or antiparallel entry therefore gives straight motion, pure perpendicular entry gives a circle, and two nonzero components give a helix.
Use one rotation and one axial drift
For perpendicular entry the magnetic force supplies radial acceleration: |q|v_⊥ B = mv_⊥²/r, r = mv_⊥/(|q|B). The same equation determines the transverse radius of a helix. Its angular speed and rotation period are ω = v_⊥/r = |q|B/m, T = (2π m)/(|q|B). During one rotation, uniform parallel motion advances by p = v_∥ T. Here p denotes the axial distance per turn, not distance along the helical curve. If the parallel velocity points opposite to the chosen axis, state that direction as well as the pitch magnitude.
At every instant F⊥v, so P = F · v = 0. The velocity direction changes while speed and kinetic energy stay constant. This is the same component model used in OpenStax’s treatment of magnetic particle motion.
The helix panel is a projection of a three-dimensional path. Its axis follows the parallel motion; its visible transverse oscillation represents rotation around that axis. A field line is not a mechanical rail constraining the particle.
Check the model’s limits
The circular and helical formulas require a uniform field over the relevant path. The speed-independent period uses non-relativistic constant mass; do not extend it to speeds where relativistic dynamics matter. When v_⊥ = 0, the actual path is straight and there is no completed orbit to time, even though the parameter 2π m/(|q|B) remains definable. When the magnetic field ends and no other force acts, the full instantaneous velocity continues unchanged.
Circular motion of a proton
Worked example 1
Orbit radius and period
Problem
A proton (mₚ = 1.67 × 10⁻²⁷ kg, q = +1.60 × 10⁻¹⁹ C) enters a uniform B = 0.20 T field with perpendicular speed v = 3.0 × 10⁶ m s⁻¹. With no other forces, find its orbit radius and period using the non-relativistic model.
Show full solution
Find the radius
Method
Set qvB = mv²/r and solve for r.Reason
The magnetic force is perpendicular to velocity and provides the centripetal force without changing speed.Working
Answer
r = mv/(qB); = ((1.67 × 10⁻²⁷)(3.0 × 10⁶))/((1.60 × 10⁻¹⁹)(0.20)) ≈ 1.57 × 10⁻¹ m.Find the orbital period
Method
Use T = 2π r/v = 2π m/(qB).Reason
One orbit covers circumference 2π r; substituting the radius cancels the speed.Working
Answer
T = (2π m)/(qB); = (2π(1.67 × 10⁻²⁷))/((1.60 × 10⁻¹⁹)(0.20)) ≈ 3.28 × 10⁻⁷ s.State the orbit quantities
Working
Answer
r ≈ 0.16 m and T ≈ 3.3 × 10⁻⁷ s. The period is independent of speed within this model.
Quick check
An electron at the same speed would circle the other way, with a radius about 1840 times smaller.
Combine the components in an oblique entry
Guided practice 1
One rotation, two velocity components
Problem
A proton enters a uniform field B = (0.200 T)z hat with initial velocity v = (3.00 × 10⁶ m/s)x hat + (4.00 × 10⁶ m/s)z hat. Use m = 1.67 × 10⁻²⁷ kg and q = +1.60 × 10⁻¹⁹ C. The magnetic force is the only force and the motion is non-relativistic.
Find the initial force component Fy, helix radius, rotation period in µs and axial advance per turn. Explain which component determines each quantity and why a changing velocity direction does not imply changing kinetic energy.
Resolve the motion before calculating
Hints
Hint 1
Choose the active component
The z component is parallel to the field. Only the x component contributes to the initial cross product and transverse rotation.Hint 2
Use the same period twice
Find T = 2π m/(qB), then use p = v_∥ T. Pitch is distance along the axis, not distance along the full helical path.Show solution step by step
Resolve direction and force
Method
Separate v_⊥ = 3.00 × 10⁶ m/s from v_∥ = 4.00 × 10⁶ m/s.Reason
The parallel component has zero cross product with the field; x hat × z hat = -y hat.Working
Answer
Fy = -qv_⊥ B = -9.60 × 10⁻¹⁴ N. There is initially no x or z force component.
Find the transverse rotation
Method
Use the perpendicular speed for the radius and the same charge-to-mass ratio for the period.Reason
The radial acceleration belongs to transverse circular motion, while axial motion is uniform.Working
Answer
r = mv_⊥/(qB) = 0.157 m and T = 2π m/(qB) = 0.328 μs, retaining guard digits for the next calculation.
Advance along the axis and check energy
Method
Multiply v_∥ by the unrounded period.Reason
During each transverse turn the unchanged parallel component carries the particle along z.Working
Answer
p = v_∥ T = 1.31 m. The total speed is 5.00 × 10⁶ m/s and remains constant. The force is always perpendicular to the instantaneous velocity, so F · v = 0 and kinetic energy is unchanged. Compare this component and work argument yourself; the checker verifies the four values only.
Quick check
A radius based on total speed would be too large, while total speed times period would give distance along the helix rather than its axial pitch.
Apply the complete model, then check short limits
Check your understanding 1
Change the entry condition, then remove the field A second beam has the same particle mass, charge, uniform field and perpendicular entry component as the oblique-entry example, but twice its parallel component. Compare the radius, rotation period and axial pitch using second/original factors. These are two different initial preparations, not a particle gaining energy from the magnetic field.
After the second particle leaves the field region, no forces act on it. What path follows, and why? Explain both the parameter comparison and the exit motion without treating a field line as a track.
Show hint
Show answer
Radius and period are unchanged: their factors are 1 and 1. The pitch factor is 2 because p = v_∥ T. The two beams have different initial kinetic energies, but each keeps its own energy while the magnetic force acts.
After exit the particle follows a straight line tangent to its helical path at the exit point. Its transverse velocity does not vanish; continuing along the field axis or maintaining the helix would require a change of velocity. Compare the explanations yourself; the checker verifies the three factors and the selected exit path.
Check your understanding 2
A charged particle moves parallel to a magnetic field. What magnetic force acts on it?
Show hint
Evaluate the cross product for parallel vectors.
Show answer
Zero: F = |q|vB sin 0° = 0. A magnetic field acts only on the velocity component perpendicular to it.
Check your understanding 3
An electron enters a uniform 0.10 T field with perpendicular speed 2.0 × 10⁶ m s⁻¹. Find its orbit radius using mₑ = 9.11 × 10⁻³¹ kg and e = 1.60 × 10⁻¹⁹ C.
Show hint
Use the charge magnitude in the radius formula; the negative sign affects the turning direction.
Show answer
To two significant figures, r = 1.1 × 10⁻⁴ m. A radius is positive.
Check your understanding 4
If the magnetic field magnitude doubles while m, |q| and v_⊥ stay fixed, what happens to the orbit radius?
Show hint
Compare the denominators in r = mv_⊥/(|q|B).
Show answer
The radius halves because r ∝ 1/B for these fixed quantities. A stronger field supplies greater centripetal force at the same speed.
Keep direction, radius and axial advance distinct
Use the signed cross product for force direction, the perpendicular speed for radius, and the parallel velocity for axial advance. Magnetic force alone does no work. Check the uniform-field and non-relativistic assumptions before using the circular or helical model; after a force-free exit, use the instantaneous velocity rather than the old field direction.
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