# Derivation: Lorentz Force on Current-Carrying Wires 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 A current-carrying wire in an external magnetic field with field lines perpendicular to the wire.

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
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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
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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
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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
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Then we rearrange the current 4 for the unknown velocity $$\class{blue}{v}$$:

Velocity is current times length divided by charge
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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
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The unknown charge $$\class{blue}{Q}$$ is thereby eliminated. The formula to be derived is thus:

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### 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
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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
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## 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: A current-carrying wire that is in the magnetic field of the other wire and vice versa.
• 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
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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:

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We just need to adjust the formula a bit.

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

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

3. 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
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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
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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
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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 $$\class{green}{F}$$:

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