Integrating over a cross product?

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Discussion Overview

The discussion centers around the mathematical manipulation of integrals involving the cross product in the context of calculating the force on a wire segment in a uniform magnetic field. Participants explore whether the integral of the cross product can be simplified under certain conditions, specifically when the magnetic field is constant in magnitude and direction.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes that the force can be expressed as F = I [ (∫dl) × B] if the magnetic field B is constant.
  • Another participant agrees with this proposal and suggests that it can be proven.
  • A different participant reiterates the need for a proof and relates the situation to the distributive property of the cross product, indicating that the same logic applies to the sum of segments of the wire.
  • Some participants emphasize that while the analogy to the distributive property clarifies the idea, it does not constitute a formal proof.
  • There is a suggestion that the assumption of B being constant allows it to be factored out of the integral.

Areas of Agreement / Disagreement

Participants express varying levels of agreement regarding the manipulation of the integral, with some supporting the idea that the simplification is valid under the stated conditions, while others seek a formal proof and clarify that the analogy used does not serve as a proof.

Contextual Notes

The discussion does not resolve whether the proposed simplification is universally applicable, as participants have not reached a consensus on the proof or validity of the approach.

Jonathan K
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Lets look at the force on a wire segment in a uniform magnetic field

F = I∫(dl×B)

I am curious if, from this, we can say:

F = I [ (∫dl) × B] since B is constant in magnitude and direction
 
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Jonathan K said:
I am curious if, from this, we can say:
Yes, we can.
 
blue_leaf77 said:
Yes, we can.
can you offer a proof?
 
As you said, if you suppose ##\mathbf{B}## to be constant in direction and uniform, then it can be taken outside the integral.
 
Jonathan K said:
can you offer a proof?
In essence, it is the same as doing this:
L1×B + L2×B + ... = (L1 + L2 + ... )×B
Which is the distrubative property of cross product.
(That's not a proof of course; I just want to make sure the idea is clear.)
 
Nathanael said:
In essence, it is the same as doing this:
L1×B + L2×B + ... = (L1 + L2 + ... )×B
Which is the distrubative property of cross product.
(That's not a proof of course; I just want to make sure the idea is clear.)
This is how I arrived at my original idea, just wasn't sure if was applicable to integration.
 

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