Calculating magnetic force on semicircle conductor

[shirobon]

Homework Statement

Given the figure below[/B]

I need to calculate the total magnetic force on the semicircle section of the conductor.
Current is I, Radius is R, and the Magnetic Field is B.

Homework Equations

d\vec{F} = Id\vec{l} \times \vec{B}[/B]

The Attempt at a Solution

[/B]
dl is equal to Rdθ

Since d\vec{l} and \vec{B} are perpendicular, the magnitude of the force on the segment d\vec{l} is equal to I dl B = I(Rdθ)B, and the components of these forces is I(Rdθ)Bcosθ and I(Rdθ)Bsinθ. Integrating these separately from 0 to π and adding the result gives IB(2R+L)j.

However, I am trying to solve it using the cross product to obtain d\vec{F} and then solve it from there.

The magnetic field vector is 0i + 0j + Bk
And the vector for d\vec{l} I got was Rdθcosθ i + Rdθsinθ j + 0k

Calculating the cross product yields
-IBRdθsinθi + IBRdθcosθj + 0k

Integrating to from 0 to π to obtain \vec{F} yields
-2IBRi + IBRj + 0k

This vector is obviously different than what was obtained before, and I am wondering where I went wrong.
I think that a place I could have gone wrong is when I found d\vec{l} as Rdθcosθ i + Rdθsinθ j + 0k, but I don't know of any other way I could get the components for the vector.

I'd appreciate if somebody could enlighten me as to where I went wrong. Thank you.

 

[mfb]

If j is the vertical axis and your angle goes from 0 to pi, the vector dl needs the cosine for the j component (and sine for i): the cable starts alone the j direction.

The integration over the cosine should lead to 0 due to symmetry.