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.