Applying Lorentz Force correctly

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SUMMARY

The discussion centers on the application of the Lorentz Force law, specifically the equation $$\vec{F} = q \cdot ( \vec{E} + \vec{v} \times \vec{B} )$$ for point charges and its extension to continuous charge distributions. Participants clarify that the electric field (E) and magnetic field (B) considered in these equations are external fields, not those generated by the charge itself. When integrating force density over a volume, it is established that contributions from the fields produced by other volume elements do not need to be included, as they result in a net force of zero. This principle also applies to current-carrying wire loops and the derivation of Maxwell's stress tensor.

PREREQUISITES
  • Understanding of the Lorentz Force law
  • Familiarity with electric and magnetic fields
  • Knowledge of continuous charge distributions
  • Basic concepts of Maxwell's equations
NEXT STEPS
  • Study the derivation and implications of the Lorentz Force law in various contexts
  • Explore the concept of force density in continuous charge distributions
  • Investigate the role of Maxwell's stress tensor in electromagnetic theory
  • Learn about the interactions of current-carrying conductors and their self-generated fields
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Physicists, electrical engineers, and students studying electromagnetism who seek to deepen their understanding of the Lorentz Force and its applications in continuous charge systems and current-carrying conductors.

Pau Hernandez
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Hello,

I am confused about the Lorentz Force. For a point charge it is: $$\vec{F} = q \cdot ( \vec{E} + \vec{v} \times \vec{B} )$$
Now from what I understand the E and the B field are external fields. So we are not talking about the fields created by the charge itself.
Now where it gets confusing. If we extend this concept to continuous charge distributions, then: $$d\vec{F} = dq \cdot ( \vec{E} + \vec{v} \times \vec{B} ) = \rho \cdot dV \cdot ( \vec{E} + \vec{v} \times \vec{B} ) = \rho \cdot dV \cdot \vec{E} + \vec{J} \times \vec{B} \cdot dV )$$ For the force density f, it follows $$\vec{f} = \rho \cdot \vec{E} + \vec{J} \times \vec{B}$$ Now here, again the E and the B field are external fields. Suppose I have a charged body. To get the net force I integrate the force density over the whole volume of the body. So lets pick one volume element as an example. In this case I would have to consider the contribution to the E and B field that stem from other volume elements (because there are charges) but actually ignore the E and B field that are produced in the volume element that I am currently 'looking' at?

Where it gets even more confusing is by considering a current carrying wire loop. The force law states: $$\vec{F} = I \cdot \int { d\vec{l} \times \vec{B} }$$ Do I have to consider the B-field that is 'produced' by other wire elements dl?

Last but not least. If we take Lorentz Force law and do some algebra, we end up with Maxwell's stress tensor. Same question here. What are the E and the B fields that have to be considered in Maxwell's stress tensor?
 
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Pau Hernandez said:
Do I have to consider the B-field that is 'produced' by other wire elements dl?
You don’t have to do it. Integrating the contribution of all the parts we will have zero net force on the wire.
Not static, not stationary cases require consideration of interaction with its own field. You may find topics in advanced textbooks.
 

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