D.3.2 Charged particle motion in magnetic fields Notes

Circular Motion of Charged Particles in a Magnetic Field

When a charged particle moves through a uniform magnetic field at an angle, it experiences a magnetic force that is always perpendicular to its velocity.

This force acts as a centripetal force, causing the particle to move in a circular path.

Deriving the Radius of the Circular Path

1. The magnetic force acting on a charged particle is given by: $$F = qvB$$ where:
   1. $q$ is the charge of the particle.
   2. $v$ is the velocity of the particle.
   3. $B$ is the magnetic field strength.
2. This force provides the centripetal force required to keep the particle in a circular path: $$F = \frac{mv^2}{r}$$
3. Equating the two expressions for force: $$qvB = \frac{mv^2}{r}$$
4. Solving for the radius $r$: $$r = \frac{mv}{qB}$$

Time for One Revolution

1. The time $T$ for the particle to complete one full revolution is: $$T = \frac{2\pi r}{v}$$
2. Substituting the expression for $r$: $$T = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB}$$

Helical Motion: Combination of Circular and Linear Motion

If a charged particle enters a magnetic field with a velocity that has both perpendicular and parallel components to the field, its motion becomes helical.

Components of Motion

1. Perpendicular Component ( $v \perp$):
   1. Causes the particle to move in a circular path.
   2. The radius of the circle is given by: $$r = \frac{mv_\perp}{qB}$$
2. Parallel Component ($v \|$):
   1. Causes the particle to move in a straight line along the direction of the magnetic field.
   2. This results in a helical path.

Pitch of the Helix

The pitch of the helix is the distance the particle travels along the field in one complete revolution:

$$p = v_\parallel T = v_\parallel \cdot \frac{2\pi m}{qB}$$