Magnetic fields & forces
The idea
Magnetic fields exert a force only on moving charges, and the force law has an unusual geometry: F = qvB sinθ in magnitude, directed perpendicular to both the velocity and the field (right-hand rule, reversed for negative charges). Because the force is always perpendicular to v, a magnetic field can never change a particle's speed or kinetic energy — it only steers. A charge moving perpendicular to a uniform field is steered into a circle, with the magnetic force supplying exactly the centripetal force.
Setting qvB = mv²/r gives the radius r = mv/(qB): momentum per unit charge per unit field. Stiffer (higher-momentum) particles curve less; stronger fields curl them tighter. The orbital period T = 2πm/(qB) is independent of speed — faster particles travel proportionally larger circles in the same time — which is the principle behind the cyclotron and the reason mass spectrometers sort ions so cleanly.
Be careful with the energy intuition: since magnetic forces do no work, any speeding-up you observe near magnets is done by electric fields (often induced ones) or by other forces. The same v × B geometry, applied to the drifting charges in a wire, yields the force on a current-carrying conductor, F = BIL for a straight segment perpendicular to the field.
Worked example
A proton (mass 1.67 × 10⁻²⁷ kg, charge 1.6 × 10⁻¹⁹ C) moves at 3.0 × 10⁶ m/s perpendicular to a uniform 0.50 T magnetic field. Find the radius of its circular path and the time for one revolution.
- Recognize the geometry: with v perpendicular to B, the magnetic force qvB is constant in magnitude and always perpendicular to the motion, so it acts as a centripetal force and the path is a circle.
- Set the magnetic force equal to the required centripetal force: qvB = mv²/r, and solve for the radius: r = mv/(qB).
- Substitute: r = (1.67 × 10⁻²⁷ × 3.0 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.50) = (5.01 × 10⁻²¹)/(8.0 × 10⁻²⁰) ≈ 0.063 m, about 6.3 cm.
- Compute the period from circumference over speed: T = 2πr/v = (2π × 0.063)/(3.0 × 10⁶) ≈ 1.3 × 10⁻⁷ s — equivalently 2πm/(qB), with the speed canceling out.
- Note the cancellation: a proton at twice the speed would orbit a circle of twice the radius in exactly the same 0.13 μs, which is why a cyclotron can use a fixed driving frequency while particles spiral outward.
Answer. The proton circles with radius about 6.3 cm, completing each revolution in about 1.3 × 10⁻⁷ s (0.13 μs).
Check your understanding
- Why can a magnetic field never change a charged particle's kinetic energy, no matter how strong it is?
- How would the path change if the proton's velocity had a component along the field as well as across it?
- Why is the orbital period independent of speed, and which technology exploits that fact?
- How would an electron's trajectory differ from the proton's in the same field at the same speed, and why?
Build the foundations first
Magnetic fields & forces builds on these concepts. If any feel shaky, start there.