Path of a Charged Particle in a Magnetic Field

Path of a Charged Particle in a Magnetic Field: MDCAT Physics notes

Path of a Charged Particle in a Magnetic Field for MDCAT: circular motion, r = mv/qB, time period and cyclotron frequency, straight and helical paths.

Unit: Electromagnetism · Updated

Three possible paths

The magnetic force $F = qvB\sin\theta$ depends on the angle $\theta$ between the velocity and a uniform field. This decides the path.

Direction of entryForcePath
Parallel or antiparallel to $B$ZeroStraight line, speed unchanged
Perpendicular to $B$$qvB$, always perpendicular to $v$Circle
At some other angleFrom the perpendicular componentHelix (spiral) about the field lines

Inside a long current-carrying solenoid the field is along the axis. A proton fired along the axis moves parallel to $B$, so it continues in a straight line with the same velocity: it is neither accelerated nor deflected.

Circular motion

When $\vec{v}\perp\vec{B}$ the magnetic force is always at right angles to the velocity, so it provides the centripetal force:

$$qvB = \frac{mv^2}{r} \quad\Rightarrow\quad r = \frac{mv}{qB} = \frac{p}{qB}$$

  • A heavier particle or a faster one moves in a larger circle.
  • A stronger field or a larger charge gives a smaller circle.
  • The speed and kinetic energy stay constant, because the force does no work.

Time period and frequency

$$T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \qquad f = \frac{qB}{2\pi m}$$

$f$ is the cyclotron frequency. It is independent of speed and radius: faster particles move in bigger circles but take the same time per revolution. Doubling the mass halves the frequency; doubling $B$ doubles it.

Helical path

If $\vec{v}$ makes an angle $\theta$ with $\vec{B}$, split it into $v\cos\theta$ along the field (unaffected, uniform motion) and $v\sin\theta$ across it (circular motion of radius $mv\sin\theta/qB$). The combination is a helix.

Worked examples

1. A proton ($m = 1.67\times10^{-27}$ kg, $q = 1.6\times10^{-19}$ C) moves at $2\times10^6$ m s$^{-1}$ perpendicular to a 0.5 T field. $r = \dfrac{1.67\times10^{-27}\times2\times10^6}{1.6\times10^{-19}\times0.5} \approx 0.042$ m.

2. If the proton's speed doubles, $r$ doubles but the period is unchanged.

Applications

The radius of the circular path lets us measure $e/m$ of the electron ($e/m = 2V/B^2r^2$) and separate ions of different mass in a mass spectrometer. The fixed cyclotron frequency is the principle of the cyclotron accelerator.

Key formulas

  • $r = mv/qB$
  • $T = 2\pi m/qB$
  • $f = qB/2\pi m$

Common MDCAT traps

  • Entering parallel to the field gives a straight path, not a circle.
  • The path in a uniform perpendicular field is circular, not parabolic; a parabola is the path in a uniform electric field.
  • Cyclotron frequency does not depend on speed.
  • $f \propto 1/m$: doubling mass halves the frequency.

Quick revision

  • Perpendicular entry gives a circle.
  • Oblique entry gives a helix.
  • Radius is proportional to momentum.
  • Kinetic energy is constant in a magnetic field.

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