KAHIBARO Bookshelves
Discord Login Register
☰

5.6.3. Motion of Charged Particles

Table of Contents

Magnetic Force and Particle Motion

A charged particle moving through a magnetic field experiences a force that can change its motion. This force is called the magnetic force. It acts only when the particle is moving, and it depends on the particle's charge, its velocity, and the magnetic field.

For a particle of charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$, the magnetic force is

$$
\vec{F} = q \, \vec{v} \times \vec{B}
$$

This means the force is a cross product, so its direction is perpendicular to both the velocity and the magnetic field. Because of this, the magnetic force does not usually make the particle speed up or slow down. Instead, it bends the path.

Important formula:
$$
\vec{F} = q \, \vec{v} \times \vec{B}
$$
Magnitude:
$$
F = |q|vB\sin\theta
$$
where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

When the Force Is Zero or Maximum

The size of the magnetic force depends on the angle between the velocity and the magnetic field.

If the particle moves parallel or antiparallel to the field, then $\theta = 0^\circ$ or $180^\circ$, so

$$
F = |q|vB\sin\theta = 0
$$

In that case, the magnetic field does not deflect the particle.

If the particle moves perpendicular to the field, then $\theta = 90^\circ$, so

$$
F = |q|vB
$$

This gives the maximum force.

The table below summarizes this.

Angle between $\vec{v}$ and $\vec{B}$$\sin\theta$Force magnitude
$0^\circ$$0$$0$
$90^\circ$$1$$qvB$
$180^\circ$$0$$0$

Direction of Motion for Positive and Negative Charges

The direction of the magnetic force depends on the sign of the charge. For a positive charge, use the right hand rule for $\vec{v} \times \vec{B}$. For a negative charge, the force points in the opposite direction.

This means two particles moving in the same magnetic field with the same velocity, but with opposite charges, curve in opposite directions.

Opposite deflection of positive and negative charges

Motion Perpendicular to a Uniform Magnetic Field

If a charged particle enters a uniform magnetic field with velocity perpendicular to the field, the magnetic force is always perpendicular to the velocity. A force of this kind changes only the direction of motion, not the speed. The result is uniform circular motion.

The magnetic force acts as the centripetal force:

$$
F_B = F_c
$$

so

$$
|q|vB = \frac{mv^2}{r}
$$

Solving for the radius gives

$$
r = \frac{mv}{|q|B}
$$

This radius is sometimes called the cyclotron radius or gyroradius.

For motion perpendicular to a uniform magnetic field:
$$
|q|vB = \frac{mv^2}{r}
$$
Therefore,
$$
r = \frac{mv}{|q|B}
$$
A larger mass or speed gives a larger circular path. A larger charge magnitude or magnetic field gives a smaller circular path.

The time for one full revolution is the period:

$$
T = \frac{2\pi r}{v}
$$

Substituting $r = \frac{mv}{|q|B}$ gives

$$
T = \frac{2\pi m}{|q|B}
$$

So the period does not depend on the speed.

The frequency of revolution is

$$
f = \frac{1}{T} = \frac{|q|B}{2\pi m}
$$

Motion at an Angle to the Magnetic Field

If the particle's velocity is not exactly perpendicular to the magnetic field, it is useful to think of the velocity as having two parts. One part is parallel to the field, and one part is perpendicular to the field.

The parallel component, $v_\parallel$, is unaffected by the magnetic field. The perpendicular component, $v_\perp$, produces circular motion. Together, these create a helical path, which is like a spiral wrapped around the field direction.

Helical motion in a uniform magnetic field

The radius of the helix depends only on the perpendicular part of the velocity:

$$
r = \frac{m v_\perp}{|q|B}
$$

The period is still

$$
T = \frac{2\pi m}{|q|B}
$$

During one period, the particle moves forward along the field direction. The distance advanced in one turn is called the pitch:

$$
p = v_\parallel T
$$

So,

$$
p = v_\parallel \frac{2\pi m}{|q|B}
$$

Why the Speed Stays Constant

Since the magnetic force is perpendicular to the velocity, it does no work on the particle. Work depends on the component of force along the displacement. Here, the force has no component in the direction of motion at any instant.

As a result, the kinetic energy stays constant:

$$
K = \frac{1}{2}mv^2
$$

The magnetic field can change the direction of $\vec{v}$, but not its magnitude.

A magnetic field alone cannot change the speed of a charged particle.
It changes direction, not kinetic energy, because the magnetic force is always perpendicular to the velocity.

Comparing Different Particles

The path of a particle in a magnetic field depends on $m$, $q$, $v$, and $B$. This is very useful in physics because particles with different masses or charges curve differently.

Quantity increasedEffect on radius $r = \frac{mv}{qB}$
Mass $m$Radius increases
Speed $v$Radius increases
Charge magnitude $q$Radius decreases
Magnetic field $B$Radius decreases

For example, a heavier particle curves less than a lighter one if both have the same speed and charge magnitude. A stronger magnetic field bends particles more sharply.

Special Case of Straight-Line Motion

If the particle moves exactly along the magnetic field direction, then the force is zero. In that case, the particle continues in a straight line.

This is consistent with the force formula, because when $\vec{v}$ is parallel to $\vec{B}$, the angle is zero and $\sin 0 = 0$.

Common Physical Picture

A uniform magnetic field acts like a guide that bends moving charges sideways. If the particle enters sideways to the field, it goes in a circle. If it enters with some forward component along the field, it follows a helix. If it moves exactly along the field, it keeps going straight.

Three possible motions in a uniform magnetic field

Summary Equations

For a particle in a uniform magnetic field, the most important results are collected below.

SituationResult
General magnetic force$\vec{F} = q \, \vec{v} \times \vec{B}$
Force magnitude$F =qvB\sin\theta$
Perpendicular entryCircular motion
Circular radius$r = \frac{mv}{qB}$
Period of circular motion$T = \frac{2\pi m}{qB}$
Frequency$f = \frac{qB}{2\pi m}$
Angled entryHelical motion
Helix radius$r = \frac{m v_\perp}{qB}$
Helix pitch$p = v_\parallel T$

Core ideas to remember:
$$
\vec{F} = q \, \vec{v} \times \vec{B}
$$
$$
F = |q|vB\sin\theta
$$
$$
r = \frac{mv}{|q|B}
$$
A magnetic field bends the path of a moving charge, but by itself it does not change the particle's speed.

Views: 0

Comments

Please login to add a comment.

Don't have an account? Register now!

© 2025 Kahibaro
Visitors: 0 / 24h
Impressum / Imprint
Datenschutz / Data Policy