Why do I have to use integration and components?

  • #1
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Homework Statement


in the picture.

Homework Equations


F=ILcrossB
Along with the original question, the other picture contains the solution with different numbers. I was just wondering why when finding the force on the wire instead of making dl= R*dtheta and going through the integration and separating the force into to components. Why isn't it possible to simply solve the equation for F=ILBsin90 with L= 2*pi*r/4?
 

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  • #2
BvU
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  • #3
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Homework Statement


in the picture.

Homework Equations


F=ILcrossB
Along with the original question, the other picture contains the solution with different numbers. I was just wondering why when finding the force on the wire instead of making dl= R*dtheta and going through the integration and separating the force into to components. Why isn't it possible to simply solve the equation for F=ILBsin90 with L= 2*pi*r/4?
From what you've learned (using the right-hand rule), what is the direction of the differential force on each small section of the wire?
 
  • #4
haruspex
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F=ILcrossB
That equation is only for a straight L.
There are various ways to generalise it. E.g. if we take ##\vec I## and ##\vec B## as functions of position s along a wire then ##\vec F=\int\vec I\times\vec B.ds##. Or we could take current as a scalar I and make the path element the vector: ##\vec F=\int I \vec {ds}\times\vec B##. Same result.
Since the path changes direction, this cannot be reduced to ##IL\times B##.
 

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