The Lorentz Force: A Moving Charge in a Magnetic Field

1 · Predict

A charged particle moves through a uniform magnetic field region. Does the magnetic force on it depend on the particle's speed, or only on the field strength?

2 · Set Up

  1. Open the qv-cross-b preset and press Reset. The field region has a fixed Bz = 0.12 T; the test charge is q = 3×10⁻¹⁰ C.
  2. Enable the force-magnitude readout on the test charge.
  3. Set the test charge's velocity components for each trial and record the force magnitude.

3 · Collect Data

Velocity x-component v_x (m/s)Velocity y-component v_y (m/s)Force magnitude F (nN)
200000150000
100000200000
3000000

Plot force magnitude F (y-axis) against speed |v| = √(v_x² + v_y²) (x-axis) for your three trials.

4 · Analyze

  1. For one trial, compute |F| = |q|·|B|·|v| using q = 3×10⁻¹⁰ C, B = 0.12 T, |v| = √(v_x² + v_y²). Compare to the table.
  2. Explain why the magnetic force magnitude depends on the particle's total speed |v|, not on the direction of its velocity — even though the force's direction does depend on which way it's moving.

5 · Extend

  1. The magnetic force is always perpendicular to the velocity (F = qv×B), so it can change a particle's direction but never its speed. Explain why a magnetic field alone can never speed up or slow down a charged particle, only steer it.
  2. A charged particle moving perpendicular to a uniform magnetic field travels in a circle, because the magnetic force always points toward the circle's center. Explain, using F = qvB, why a faster particle in the same field traces a bigger circle.

The Physics Behind This Experiment

Lorentz Force (Magnetic)

A charge q moving with velocity v through a magnetic field B experiences a force F = qv×B, perpendicular to both its velocity and the field. Its magnitude is |F| = |q||v||B| when v is perpendicular to B.

← Back to experiment