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# Lorentz Force: How to Determine its Direction and Understand Formula

## Important Formula

What do the formula symbols mean?

## Magnetic force

Unit
Magnetic force acts on a charge $$q$$ when it moves with velocity $$\class{blue}{v}$$ through magnetic field $$\class{violet}{B}$$. The requirement is that the magnetic field $$\class{violet}{B}$$ is perpendicular to the velocity $$\class{blue}{v}$$, that is: the two directions are orthogonal to each other.

This formula represents the magnetic component of the Lorentz force. (Lorentz force is the sum of electric and magnetic force).

## Magnetic flux density (B-field)

Unit
Magnetic flux density indicates how strong the magnetic field is in which the charge moves. The greater the magnetic flux density, the greater the magnetic force.

## Velocity

Unit
Velocity of the charged particle. The greater the velocity of the charged particle, the greater the magnetic force.

## Electric charge

Unit
Electric charge can be repulsive or attractive (proton, electron). The greater the electric charge, the greater the magnetic force.

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 $$\class{green}{F_{\text m}}$$.

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:

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:

• The particle must be electrically charged - neutral particles (like neutrons) have the charge $$q = 0$$ and consequently experience no Lorentz force:

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

### 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.

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.

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.

1. Thumb - points in the direction of the motion of the charge, that is in the direction of the velocity $$\class{blue}{v}$$.

2. Index finger - points in the direction of the magnetic field $$\class{violet}{B}$$ (to the south pole of the magnet).

3. 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 $$F_{\text z}$$!

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:

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

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 $$\class{blue}{v}$$:

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

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.

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 $$\class{blue}{v}$$:

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

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:

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:

## Charge moves AT AN ANGLE relative to the magnetic field

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):

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 $$T$$.

## 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.

## Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

### Exercise #1: An Ion Orbiting Around The Equator in The Magnetic Field

The earth has a radius of 6370 kilometers. The magnetic field at the equator has a value of $$30 \, \mu\text{T}$$ and is directed to the north. You want to get an ion with mass $$352 \, u$$ and charge $$e$$ to orbit the Earth at the equator at a height of 30 kilometers.

1. In what direction along the equator must the ion fly to orbit the Earth?
2. At what velocity must the ion move so that it remains on the circular path? Consider also the gravitational force. You can calculate classically and assume a homogeneous gravitational field.
3. Reason why the gravitational force on the ion plays only a minor role.
4. The earth is also negatively charged electrically and has a average homogeneous magnetic field of $$130 \, \frac{ \mathrm V }{ \mathrm m }$$. Does this E-field have to be taken into account?

Tip: Consider the forces acting on the ion and write an equation for the equilibrium of forces. To determine the velocity, you have to solve a quadratic equation in any case.

#### Solution to exercise #1.1

To find out the direction of motion of the ion, use the right-hand rule. We are dealing with a positive charge here. Therefore you use your right hand for this.

• The earth's magnetic field $$B$$ points north at the equator. Point your index finger to the north.
• The Lorentz force $$F_{\text m}$$ (magnetic force) on the ion must point towards the earth so that it orbits around the earth. Point your middle finger towards the center of the earth.
• The extended thumb then shows you the direction in which the ion must move.

Thus the ion must fly to the west, so that it is deflected around the equator. If the ion would fly to the east, it would be deflected away from the earth.

#### Solution to Exercise #1.2

In order for the ion to remain on a stable circular orbit with the radius (6370 kilometers Earth radius + 30 kilometers height), the Lorentz force... $$\class{green}{F} ~=~ q \, \class{blue}{v} \, \class{violet}{B}$$ and the gravitational force... $$F_{\text g} ~=~ m \, g$$ must correspond exactly to the centripetal force $$F_{ \text z} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{ r }$$

Centripetal force is a force acting to the center of the earth and tells us how to get a stable circular orbit with radius $$r$$ and velocity $$v$$ of a particle of mass $$m$$. The role of the centripetal force is taken over here by the Lorentz force and the gravitational force, because they are also radially directed.

So we have the following equation of forces: \begin{align}F_{\text z} &~=~ F_{\text m} ~+~ F_{\text g} \\\\ \frac{m}{r} \, v^2 &~=~ q \, v \, B ~+~ m \, g\end{align}

We have to rearrange this force equation with respect to the velocity $$v$$. Here you have a quadratic equation that you cannot rearrange uniquely with respect to $$v$$. This equation has two solutions as possible velocities. We determine the two solutions $$v_1$$ and $$v_2$$ with the pq formula.

For this purpose, the last equation is converted into a suitable form: \begin{align}\frac{m}{r} \, v^2 ~-~ q \, v \, B ~-~ m \, g & ~=~ 0 \\\\ v^2 ~-~ \frac{r}{m} \, q \, v \, B ~-~ \frac{r}{\cancel m} \, \cancel{m} \, g & ~=~ 0 \\\\ v^2 ~-~ \frac{r \,q\, B}{m} \, v ~-~ r \, g & ~=~ 0 \end{align}

We put everything on the left side in the second step and in the third step we multiplied the whole equation by $$\frac{r}{m}$$ to eliminate the factor before the $$v^2$$. Now we have the p-value $$- \frac{r \,q\, B}{m}$$ and q-value $$- r \, g$$ for the pq-formula: \begin{align}v_{1,2} & ~=~ -\frac{p}{2} ~\pm~ \sqrt{ \left(\frac{p}{2}\right)^2 ~-~ q } \\\\ v_{1,2} & ~=~ \frac{r \,q\, B}{2m} ~\pm~ \sqrt{ \left(\frac{r \,q\, B}{ 2m}\right)^2 ~+~ r \, g }\end{align}

Now we just have to insert the concrete values to determine $$v_1$$ and $$v_2$$.

• The radius $$r$$ is equal to the earth radius plus the height of the ion above the earth: $$r ~=~ 6370 \, \text{km} ~+~ 30 \, \text{km} ~=~ 6400 \cdot 10^3 \, \text{m}$$
• The charge $$q$$ corresponds to the charge of the ion: $$q ~=~ e ~=~ 1.60 \cdot 10^{-19} \, \text{C}$$
• The mass $$m$$ corresponds to the mass of the ion: $$m ~=~ 352 \, u ~=~ 352 ~\cdot~ 1.66 \cdot 10^{-27} \, \text{kg}$$
• The magnetic field $$\class{violet}{B}$$ corresponds to the constant earth magnetic field at the equator: $$\class{violet}{B} ~=~ 30 \, \mu\text{T} ~=~ 30 \cdot 10^{-6} \, \text{T}$$
• The gravitational acceleration $$g$$ on earth is: $$g ~=~ 9.8 \, \frac{\text m}{ \text{s}^2 }$$

If all these values are used, the result is: $$v_{1,2} ~=~ 26 \,286\,966.05 \, \frac{\text m}{ \text s } ~\pm~ 26\,286 \,967 \frac{\text m}{ \text s }$$

We obtain the first solution of the quadratic equation by adding the two values in the pq formula: \begin{align}v_1 &~=~ 52\,573\,933 \, \frac{\text m}{ \text s } \\\\ &~\approx~ 5.2 \cdot 10^7 \, \frac{\text m}{ \text s }\end{align}

We get the second solution if we subtract the two values in the pq formula: $$v_2 ~=~ -1.19 \, \frac{\text m}{ \text s }$$

The solution $$v_2$$ is not a valid solution. You can check this by substituting $$v_2$$ into the pq formula. However, $$v_1$$ is a valid solution. Therefore, the required speed is: $$v ~=~ 5.2 \cdot 10^7 \, \frac{\text m}{ \text s }$$

#### Solution to exercise #1.3

To justify why the gravitational force hardly plays a role, you can look at the ratio of the gravitational force $$F_{\text g}$$ to the Lorentz force $$F_{\text m}$$: $$\frac{ F_{\text g} }{ F_{\text m} } ~=~ \frac{ m \, g }{ q \, v \, B }$$

Inserting the values (including the calculated velocity in exercise b), results in: \begin{align}\frac{ F_{\text g} }{ F_{\text m} } &~=~ \frac{ 5.73\cdot 10^{-24}\, \text{N} }{ 2.5\cdot 10^{-16} \, \text{N} } \\\\ &~\approx~ 2 \cdot 10^{-6} \, \%\end{align}

So the gravitational force is just 0.000002 % of the Lorentz force and can be neglected.

#### Solution to exercise #1.4

To check whether the electric field of the earth has to be taken into account, we can calculate the ratio of the electric force $$F_{\text e} = e \, E$$ on the ion to the Lorentz force in analogy to the previous exercise: $$\frac{ F_{\text e} }{ F_{\text m} } ~=~ \frac{ e \, E }{ q \, v \, B }$$

The electric field $$E ~=~ 130 \, \frac{ \text V }{ \text m }$$ is given in the exercise. Inserting the E-field results in: \begin{align}\frac{ F_{\text e} }{ F_{\text m} } &~=~ \frac{ 2.08 \cdot 10^{-17} \, \text{N} }{ 2.5\cdot 10^{-16} \, \text{N} } \\\\ &~\approx~ 8.3 \, \%\end{align}

The electric force accounts for 8.3% of the Lorentz force and can be taken into account for a more accurate result.

### Exercise #2: Flying Aircraft in a Magnetic Field

An aircraft with a wingspan $$b = 20 \, \text{m}$$ is flying with a velocity of $$v = 800 \, \text{km}/\text{h}$$ in the north direction. The vertical component of the Earth's magnetic field in the area of the aircraft is $$B = 10 \, \mu \text{T}$$.

1. What is the magnitude of the Lorentz force $$F_{\text m}$$ on an electron in the aircraft?
2. What is the magnitude of the voltage $$U$$ between the wingtips?
3. Is the voltage calculated in Exercise 2.2 even measurable?

Tips: For Exercise 2.1: How is the Lorentz force defined? For Exercise 2.2: Consider here the Lorentz force and the electric force. How are electric field and voltage related? For Exercise 2.3: Consider the reference system of the voltage measuring device.

#### Solution for Exercise #2.1

The magnetic component of the Lorentz force is: $F_{\text m} = q \, v \, B$

Substituting the given values yields the following Lorentz force: \begin{align} F &~=~ - 1.6 \cdot 10^{-19} \, \text{C} ~\cdot~ 800 \cdot \frac{10^3}{3600} \, \frac{\text m}{\text s} ~\cdot~ 10 \cdot 10^{-6} \, \text{T} \\\\ &~=~ - 3.6 \cdot 10^{-22} \, \text{N} \end{align}

#### Solution for Exercise #2.2

The electric force and the electric field $$E$$ are related as follows: $F_{\text e} = q \, E$

In force equilibrium, the deflecting magnetic force (right side) and the oppositely acting electric force (left side) are equal: $q \, E = q \, v \, B$

Together with the electric voltage $$U = E \, b$$ follows: $U = b \, v \, B$

Substituting the given values yields the following voltage between the two aircraft wings: $U = 0.04 \, \text{V} = 40 \, \text{mV}$

#### Solution for Exercise #2.3

The voltage calculated in Exercise 2.2 of $$40 \, \text{mV}$$ is not measurable in the aircraft, because from the perspective of a voltage measuring device carried with the aircraft, the charge carriers of the aircraft are at rest. The voltage measuring device moves with the aircraft, consequently there is no current flow due to the Lorentz force and therefore no voltage can be measured.