Lorentz Force: How to Determine its Direction and Understand Formula

by Alexander FufaeV

Table of contents

Lorentz force $ F $ is a force experienced by a particle in an electromagnetic field. Mathematically, Lorentz force is the sum of electric force $ \class{gray}{F_{\text e}} $ and magnetic force $ \class{green}{F_{\text m}} $.

$$ \begin{align} F ~=~ \class{gray}{F_{\text e}} ~+~ \class{green}{F_{\text m}} \end{align} $$

In this lesson we consider only the case where the electric force on the particle is zero: $ \class{gray}{F_{\text e}} = 0 $. The particle is only in a magnetic field, but NOT in an electric field.

If a particle (e.g. an electron) with charge $q$ and velocity $\class{blue}{v} $ moves through a magnetic field $ \class{violet}{B} $, then it experiences a magnetic force $\class{green}{F_{\text m}}$ (Lorentz force).

In principle, there can be 3 cases how the charge can move. The charge moves...

perpendicular to the magnetic field: $ \class{blue}{v} $ ⊥ $ \class{violet}{B} $
parallel to the magnetic field: $ \class{blue}{v} $ || $ \class{violet}{B} $
at an angle to the magnetic field: $ \angle \left( \class{blue}{v},~\class{violet}{B} \right) $.

Charge moves orthogonal to the magnetic field

In the experiment, we find that both the charge, velocity and magnetic field are proportional to the Lorentz force. So we can write down an equation with which we can calculate the Lorentz force on a particle:

Formula: Lorentz force on a charge (perpendicular motion)

$$ \begin{align} \class{green}{F} ~=~ q \, \class{blue}{v} \, \class{violet}{B} \end{align} $$

You can deduce three important informations from the formula. For Lorentz force $ \class{green}{F} $ on a particle to occur at all, the following properties must be fulfilled:

The particle must move - otherwise velocity would be $ \class{blue}{v} ~=~ 0$ and thus also the Lorentz force:

Lorentz force - when velocity equals zero
Formula anchor $$ \begin{align} \class{green}{F} ~=~ q ~\cdot~ \class{blue}{0} ~\cdot~ \class{violet}{B} ~=~ 0 \end{align} $$
The particle must be electrically charged - neutral particles (like neutrons) have the charge $ q = 0$ and consequently experience no Lorentz force:

Lorentz force on neutral particles
Formula anchor $$ \begin{align} \class{green}{F} ~=~ 0 ~\cdot~ \class{blue}{v} ~\cdot~ \class{violet}{B} ~=~ 0 \end{align} $$
The particle must be in a magnetic field - if the particle does not move through a magnetic field: $ \class{violet}{B} = 0 $, then it does not experience a Lorentz force:

Lorentz force - when magnetic field is zero
Formula anchor $$ \begin{align} \class{green}{F} ~=~ q ~\cdot~ \class{blue}{v} ~\cdot~ \class{violet}{0} ~=~ 0 \end{align} $$

Lorentz force is also zero, if the particle moves parallel to magnetic field lines. But more about this later.

Cause of the Lorentz force

The cause of the Lorentz force is a moving charge in the magnetic field.

Determine the direction of the Lorentz force

When a positive charge moves through a homogeneous magnetic field that is perpendicular to the direction of motion (see Illustration 1), the positive charge in the magnetic field is deflected upward. Lorentz force acts upward on the particle.

Hover the image!

A positive charge in the magnetic field (into the plane) is deflected upward.

If a negative charge moves through a homogeneous magnetic field that is perpendicular to the direction of motion (see illustration 2), then the negative charge is deflected downward. Lorentz force acts downward on the particle.

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A negative charge in the magnetic field (into the plane) is deflected downward.

How do we determine the direction of the Lorentz force?

Or, in other words, how do I know that the charge will be deflected downward or upward? For this, you use the so-called left hand rule.

For positive charges you use the right hand. For negative charges you use the left hand.

Thumb - points in the direction of the motion of the charge, that is in the direction of the velocity $ \class{blue}{v} $.
Index finger - points in the direction of the magnetic field $ \class{violet}{B} $ (to the south pole of the magnet).
Middle finger - tells you the direction of deflection, i.e. Lorentz force direction $ \class{green}{F_{\text m}} $, if you have aligned your fingers correctly (as shown in illustration 3 or 4).

However, the electric charge is not simply deflected downward or upward, there is something else going on - provided we give the charge enough space inside magnetic field.

Since the Lorentz force $ \class{green}{F_{\text m}} $ is always perpendicular to the velocity $ \class{blue}{v} $ (nature dictates this), the charge is forced into a circular path!

Do you know which force is responsible for keeping a particle on a circular path? The centripetal force $ F_{\text z} $!

$$ \begin{align} F_{\text z} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{r} \end{align} $$

Here $ \class{brown}{m} $ is the mass of the particle and $r$ is the radius of the circular path. In this case, the Lorentz force takes over the job of the centripetal force. In other words: Lorentz force IS here at the same time the centripetal force.

The qvB formula 2 for Lorentz force alone does not help us much, because it is not easy to determine the velocity $ \class{blue}{v} $ of the charge experimentally. With the centripetal force, however, this task becomes much easier because we can use it to formulate an equation for the velocity.

Equate the Lorentz force formula with the centripetal force formula 6:

$$ \begin{align} q \, \class{blue}{v} \, \class{violet}{B} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{r} \end{align} $$

You may cancel the velocity $ \class{blue}{v} $ once on both sides, so that it is eliminated on the left side:

$$ \begin{align} q \, \class{violet}{B} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}}{r} \end{align} $$

Bring the radius $r$ to the other side (multiply both sides by $r$). And bring mass $ m $ to the other side (divide both sides by $ \class{brown}{m} $), then you get a formula for the velocity $ \class{blue}{v} $:

$$ \begin{align} \class{blue}{v} ~=~ \frac{ q \, \class{violet}{B} \, r }{ \class{brown}{m} } \end{align} $$

Of course, you can also calculate the radius of the circular path by equating the centripetal force and Lorentz force:

Formula: Radius of the circular path

$$ \begin{align} r ~=~ \frac{\class{brown}{m} \, \class{blue}{v}}{q \, \class{violet}{B}} \end{align} $$

Formula 10 reveals two interesting facts:

The larger the velocity $ \class{blue}{v} $ and the mass $ \class{brown}{m} $ of the particle, the larger is the traversed circle.
The larger the external magnetic field $ \class{violet}{B} $ and the electric charge $ q $ of the particle, the smaller is the traversed circle.

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An electron shot out of a cannon traverses a circular path. Lorentz force points to the center of the circle.

Another interesting question we can ask ourselves is:

How much time does the particle need to make exactly ONE round trip?

The time it takes the particle to make exactly one revolution is the period $ T $.

The distance covered by the particle within this time is the circumference $ U = 2\,\pi\,r $ of the circle. Distance $U$ PER time $ T $ is exactly equal to velocity $ \class{blue}{v}$:

$$ \begin{align} \frac{2 \, \pi \, r}{T} ~=~ \class{blue}{v} \end{align} $$

Substitute only the velocity from 9 into 11 and rearrange the equation for the period $ T $:

Formula: Period of the circular motion

$$ \begin{align} T ~=~ 2 \, \pi \frac{ \class{brown}{m} }{ |q| \, \class{violet}{B} } \end{align} $$

Here we have used the magnitude $ | q | $ of the charge (i.e. without sign), so that we are not tempted to use a negative charge. Because then we would get a negative time, which makes no sense.

From here, we can easily calculate the frequency $ f $ with which the particle orbits. The frequency indicates the number of revolutions per second and is the reciprocal of the period $ T $: $ f = \frac{1}{T} $. To do this, swap the denominator with the numerator on both sides of Eq. 12 and you get the frequency:

$$ \begin{align} f ~=~ \frac{1}{2\pi} \, \frac{ |q| \, \class{violet}{B} }{ \class{brown}{m} } \end{align} $$

The frequency of a circling charge in the magnetic field is also called cyclotron frequency. In most cases, the cyclotron frequency is not given by the frequency $ f $, but by the so-called circular frequency $ \omega $ ("omega"). It is defined as $ \omega = 2\pi \, f $. So bring the factor $ 2 \pi $ to the other side in Eq. 13 and you get:

Formula: Cyclotron frequency of the charged particle

$$ \begin{align} \omega ~=~ \frac{ |q| \, \class{violet}{B} }{ \class{brown}{m} } \end{align} $$

Charge moves AT AN ANGLE relative to the magnetic field

The homogeneous magnetic field viewed from the side. For the qvB formula (without sine) the angle must be 90 degrees.

Of course, the charge may NOT be moving perfectly perpendicular to the magnetic field lines. Thus the angle, let us call this $ \alpha$, between the velocity $ \class{blue}{v} $ and the magnetic field $ \class{violet}{B} $ is not 90 degrees.

To account for this, we need to multiply the qvB formula by the sine of the angle (you'll learn why if you know about vectors and cross products):

Formula: Lorentz force with angle

$$ \begin{align} \class{green}{F_{\text m}} ~=~ q \, \class{blue}{v} \, \class{violet}{B} \, \sin(\alpha) \end{align} $$

You can always decompose the velocity $ \class{blue}{v} $, which is oblique to the magnetic field $ \class{violet}{B} $, into a parallel $ \class{blue}{v_{||}} $ and an perpendicular $ \class{blue}{v_{\perp}} $ component to the magnetic field.

The parallel component of the velocity (see illustration 8), in contrast to the perpendicular component, has no influence on the Lorentz force and thus this component is not responsible for the deflection of the electron in the magnetic field. Vertical component $ \class{blue}{v_{\perp}} $ includes an angle of 0 degrees with the magnetic field $ \class{violet}{B} $, therefore the force for this component vanishes (because of $ \sin(0^{\circ}) ~=~ 0 $).

By a partial movement parallel and a partial movement perpendicular to the magnetic field, a cylindrical spiral path, a so-called helix, is generated. Its axis is parallel to the magnetic field. It has a radius $r$ and pitch $h$. Where pitch is simply a distance parallel to the magnetic field covered within a period $T$.

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Due to the movement of the electron diagonally to the magnetic field, a helical path is created.

Charge moves PARALLEL to the magnetic field

If the charge moves exactly parallel to the magnetic field, the angle is $ \alpha = 0 $. Then $ \sin(0) = 0 $, therefore no Lorentz force acts on the charge.

So we can state that charges moving parallel to the magnetic field are NOT deflected!

In the next lesson, we will look at an important experiment related to the Lorentz force, namely the teltron tube experiment.