In this post, we will look at an example.
The conductor has a straight segment with length L perpendicular to the plane of the diagram on the right, with the current opposite to $\vec{B}$, which is followed by a semicircle with radius R and another straight segment with length L. The conductor carries a current of $I$. The magnetic field $\vec{B}$ is uniform and perpendicular to the plane of the diagram, directed out of the plane of the diagram. Find the total magnetic force on these three segments of wire.
Recall that:
$$d \vec{F} = I \, d\vec{l} \times \vec{B}$$
OR $$dF = BI \, dl$$
Notice that due to the symmetry of the problem, the $F_{x}$ on the right half of the semicircle is cancelled out by the $F_{x}$ on the left half of the semicircle. Hence, we will only need to deal with $F_{y}$.
$$\begin{aligned} dF_{y} &= BI \, dl \, \sin{\theta} \\ &= BIR \sin{\theta} \, d\theta \\ F_{y} &= BIR \int\limits_{0}^{\pi} \! \sin{\theta} \, d\theta \\ F_{y} &= 2 BIR \end{aligned}$$
Note: $\sin{\theta}$ comes from $F_{y} = |\vec{F}| \sin{\theta}$ and $dl = R \, d\theta$.
Hence, the total force is given by: (straight + curved)
$$\vec{F} = BI (L + 2R) \hat{j}$$
Note: Only one of the straight conductor experience a force.
Next: Force & Torque On Current Loop In Magnetic Field
