I Applying Lorentz Force correctly

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The discussion centers on the application of the Lorentz Force equation, particularly in the context of continuous charge distributions and current-carrying wires. It clarifies that while external electric (E) and magnetic (B) fields are considered, the fields generated by the charge or current itself can often be ignored for net force calculations. However, in non-static scenarios, the interaction with its own fields may need to be accounted for. The conversation also touches on the implications for Maxwell's stress tensor, emphasizing that the self-generated fields do not contribute to the net force when integrated over the entire system. Understanding these nuances is essential for correctly applying the Lorentz Force in various physical situations.
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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