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Magnetic Force on Moving Charges

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A force built from a cross product

A magnetic field B (units: tesla, 1 T = 1 N·s/(C·m)) exerts a force on a moving charge given by the cross product F = qv × B. The magnitude is F = qvB·sinθ, where θ is the angle between v and B; the direction comes from the right-hand rule: fingers point along v, curl toward B, thumb gives F for a positive charge — and the opposite direction for a negative charge. A stationary charge, or one moving parallel to B, feels no force at all.

Magnetic force on a charge
→F = q→v × →B, F = qvB·sinθ
F is always perpendicular to both v and B. Maximum force at θ = 90°; zero force for motion along the field.

Perpendicular force means circular motion

Because F is always perpendicular to v, a magnetic force does what a string does for a whirling ball: it turns the velocity without changing its magnitude. A charge fired perpendicular to a uniform B moves in a circle. Setting the magnetic force equal to the centripetal requirement, qvB = mv²/r, gives the radius r = mv/(qB). The period T = 2πr/v = 2πm/(qB) is independent of speed — faster particles trace bigger circles in exactly the same time. This “cyclotron” property powers particle accelerators and mass spectrometers.

Circular motion in a magnetic field
r = mv/(qB), T = 2πm/(qB)
Radius grows with momentum mv; the period depends only on m, q, and B — not on how fast the particle moves.
Watch out

Magnetic forces never do work on a charge. F is perpendicular to v at every instant, so W = ∫F·ds = 0: the field changes the direction of motion, never the speed or kinetic energy. When a magnetic setup appears to “speed something up,” an electric field (often induced) is doing the actual work.

Worked example

A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) moves at 2.0 × 10⁶ m/s perpendicular to a 0.50 T field. Find the force on it and the radius of its circular path.

  1. 1.With θ = 90°, F = qvB = (1.6 × 10⁻¹⁹)(2.0 × 10⁶)(0.50).
  2. 2.Multiply: F = 1.6 × 10⁻¹³ N — tiny in newtons, but enormous relative to the proton’s weight (∼10⁻²⁶ N).
  3. 3.Radius: r = mv/(qB) = (1.67 × 10⁻²⁷)(2.0 × 10⁶) / [(1.6 × 10⁻¹⁹)(0.50)].
  4. 4.Numerator: 3.34 × 10⁻²¹ kg·m/s. Denominator: 8.0 × 10⁻²⁰ C·T. So r = 4.2 × 10⁻² m ≈ 4.2 cm.
  5. 5.Check the physics: the force stays perpendicular to v, so the proton circles at constant speed with this radius.
Answer: F = 1.6 × 10⁻¹³ N, r ≈ 4.2 cm
Worked example

In a velocity selector, an electric field E = 2.0 × 10⁴ N/C and a magnetic field B = 0.10 T are arranged perpendicular to each other and to the beam, with their forces on a charge opposing. What speed passes through undeflected, and what happens to faster particles?

  1. 1.Undeflected means the two forces balance: qE = qvB.
  2. 2.The charge cancels — the selector works for any q, positive or negative: v = E/B.
  3. 3.v = (2.0 × 10⁴)/(0.10) = 2.0 × 10⁵ m/s.
  4. 4.A faster particle has a larger magnetic force (qvB grows with v) while the electric force is unchanged, so the magnetic force wins and the particle deflects toward the magnetic-force side.
Answer: v = E/B = 2.0 × 10⁵ m/s; faster particles are deflected by the now-dominant magnetic force
Checkpoint

How much work does the magnetic force do on a proton during one full circular orbit in a uniform field?

Checkpoint

A charged particle circling in a uniform magnetic field has its speed doubled. Its orbital period:

Answer the 2 checkpoints as you read.

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