Integrating over a cross product?

[Jonathan K]

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

 

[blue_leaf77]

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

Yes, we can.

 

[Jonathan K]

Yes, we can.

can you offer a proof?

 

[blue_leaf77]

As you said, if you suppose $\mathbf{B}$ to be constant in direction and uniform, then it can be taken outside the integral.

 

[Nathanael]

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.)

 

[Jonathan K]

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.