# Derivation Lorentz force on current-carrying wires

**Level 2**requires school mathematics. Suitable for pupils.

## Table of contents

- Lorentz force on a current-carrying conductor Here you will learn how to derive the Lorentz force formula for a current-carrying wire that is perpendicular or oblique to an external magnetic field.
- Lorentz force between two current-carrying wires Here you will learn the force with which two current-carrying conductors attract or repel each other.

In the following, we will derive a formula for Lorentz force experienced by one or two current-carrying wires in an external magnetic field.

## Lorentz force on a current-carrying conductor

### Current-carrying wire perpendicular to the magnetic field

We have the following situation:

An

**electric current**\( \class{blue}{I} \) flows through a straight electric conductor (wire). It does not matter whether \(\class{blue}{I}\) is a current of positive or negative charges.The wire has the

**length**\(L\).The conductor is in a

**homogeneous magnetic field**\( \class{violet}{B} \) which is*perpendicular*to the current \( \class{blue}{I} \).

The electric charges flowing through the wire, experience a **magnetic force** (**Lorentz force**) \(\class{green}{F}\).

**Formula: Lorentz force on the charge inside wire**

Since the velocity \(\class{blue}{v}\) is unknown, we want to use the current \(\class{blue}{I}\) instead. We can find out the current value quite easily with an ammeter. The current \(\class{blue}{I}\) is here the total amount of charge \(\class{blue}{Q}\), which flows along the distance \(L\) per time \(t\):

The individual charges in the wire move through the wire with a certain **average velocity** \(\class{blue}{v}\). Each individual charge experiences a Lorentz force. We can write the total Lorentz force on the wire as the Lorentz force acting on the **total charge** \(\class{blue}{Q}\). The total charge here is sum of the individual charges moving through the considered piece of the wire:

**Definition: Current due to movement of charges**

A charged particle travels the length \(L\) of the wire within the **time** \(t\). "Distance per time" is exactly the definition of velocity. In this case, it is the speed with which a charge travels through the wire:

**Formula: Speed is length per time**

Let us now rearrange the velocity 3

for the time \(t\): \(t ~=~ \frac{L}{\class{blue}{v}}\) and insert the time into the definition 2

of the current:

**Current expressed with speed**

&~=~ \frac{\class{blue}{Q} \, \class{blue}{v}}{L} \end{align} $$

Then we rearrange the current 4

for the unknown velocity \(\class{blue}{v}\):

**Velocity is current times length divided by charge**

Very nice, because now we can insert this relation into the Lorentz force formula 1

and thus eliminate the unknown velocity:

**Lorentz force with eliminated velocity**

&~=~ \class{blue}{Q}\, \frac{\class{blue}{I} \, L}{\class{blue}{Q}} \, \class{violet}{B} \end{align} $$

The unknown charge \(\class{blue}{Q}\) is thereby eliminated. The formula to be derived is thus:

**Formula: Lorentz force on current-carrying wire in perpendicular B-field**

### Current-carrying wire oblique to the magnetic field

If the magnetic field \(\class{violet}{B}\) is *oblique* (at an angle not 90 degrees) to the current \(\class{blue}{I}\), then we need to make a small correction to the derived formula.

The Lorentz force \( \class{green}{F} \) on the charge \(\class{blue}{Q}\) moving *not perpendicular* to the homogeneous magnetic field is:

**Formula: Lorentz force in the non-perpendicular B-field using velocity**

Here \(\alpha\) is the **angle** between the velocity direction (velocity vector) and the magnetic field direction (magnetic field vector).

Inserting the rewritten velocity 5

results in the following formula:

**Formula: Lorentz force on current-carrying wire oblique to the B-field**

## Lorentz force between two current-carrying wires

For a single wire, the magnetic field \(\class{violet}{B}\) was generated by some external source. If we now add a second current-carrying wire, we can use the magnetic field generated by this wire and see how another current-carrying wire behaves in this magnetic field. So we have the following setup:

The first wire has the

**length**\(L\) and a**current**\(\class{blue}{I_1}\) flows through it. The conductor generates a circular**magnetic field**\(\class{violet}{B_1}\) concentrically around the wire.The second wire also has the

**length**\(L\) and a possibly different**current**\(\class{blue}{I_2}\) flows through it. For example, this current may flow in the opposite direction or have a different magnitude. The second conductor also generates a circular**magnetic field**\(\class{violet}{B_2}\) concentrically around itself.

The magnetic field of a straight wire can be derived with the *Ampere's law*. We take the corresponding formula as given. The first wire produces the following magnetic field \(\class{violet}{B_1}\):

**Formula: Generated magnetic field of the first wire**

Here \(\mu_0\) is the **magnetic field constant** and \(\pi = 3.14... \) a mathematical constant. More important here is the **distance** \(r\) from the wire. So the magnetic field \(\class{violet}{B_1}\) generated by the first wire depends on the magnitude of the current \(\class{blue}{I_1}\) and the distance \(r\) from the wire.

If we now place the second wire in the magnetic field \(\class{violet}{B_1}\) perpendicular to it (that is, \(\class{violet}{B_1}\) and \(\class{blue}{I_2}\) are perpendicular to each other), then we can use the previously derived formula 7

for the Lorentz force acting on a current-carrying wire:

We just need to adjust the formula a bit.

The Lorentz force \(\class{green}{F}\) in this case corresponds to the Lorentz force \(\class{green}{F_2}\) on the second wire.

The current \(\class{blue}{I}\) here is the current \( \class{blue}{I_2}\) flowing through the second wire.

We also place the second wire at a distance \(r\) from the first wire. At this distance \(r\) the magnetic field has the value \(\class{violet}{B_1}\) generated by the first wire.

**Formula: Lorentz force on the second wire due to the magnetic field of the first wire**

Now we can substitute the formula 9

for the magnetic field \(\class{violet}{B_1}\) into the Lorentz force formula 11

:

**Magnetic field of the first wire inserted into the Lorentz force formula**

&~=~ \frac{\mu_0 \, L}{2\pi} \frac{\class{blue}{I_1} \, \class{blue}{I_2} }{r} \end{align} $$

We can analogously determine the Lorentz force \(\class{green}{F_1}\) on the first wire, which is in the magnetic field \(\class{violet}{B_2}\) of the second wire:

**Formula: Lorentz force on the first wire**

As you can see, both conductors experience the same magnitude of Lorentz force: \( \class{green}{F_2} ~=~ \class{green}{F_1}\). Therefore, we can also omit the numbering of the forces and simply write \( \class{green}{F} \):

**Formula: Lorentz force on a wire in the magnetic field of another wire**