Derivation: Lorentz Force on Current-Carrying Wires
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:
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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.
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The wire has the length \(L\).
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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) \(\class{green}{F}\).
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:
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:
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:
&~=~ \frac{\class{blue}{Q} \, \class{blue}{v}}{L} \end{align} $$
Then we rearrange the current 4
for the unknown velocity \(\class{blue}{v}\):
Very nice, because now we can insert this relation into the Lorentz force formula 1
and thus eliminate the unknown 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:
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:
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:
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:
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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.
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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 \(\class{violet}{B_1}\):
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.
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The Lorentz force \(\class{green}{F}\) in this case corresponds to the Lorentz force \(\class{green}{F_2}\) on the second wire.
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The current \(\class{blue}{I}\) here is the current \( \class{blue}{I_2}\) flowing through the second wire.
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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.
Now we can substitute the formula 9
for the magnetic field \(\class{violet}{B_1}\) into the Lorentz force formula 11
:
&~=~ \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:
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 \( \class{green}{F} \):
