Why do I have to use integration and components?

[Mariah]

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?

 

[BvU]

in the picture.

Not really

 

[Chestermiller]

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?

 

[haruspex]

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 $IL\times B$.