Force on a current carrying wire

$$d\vec{F} = Id\vec{L} \times \vec{B}.$$

Note that if $\vec{B}$ is constant then this expression can be integrated immediately to get $\vec{F} = I\vec{L} \times \vec{B}$, which is the form in which we will usually use it. $\vec{L}$ has length equal to the physics length of the wire and points in the current direction.

Note that for a closed circuit in a constant a field we have

$$\begin{eqnarray*} \vec{F} &=& \int_i^f Id\vec{L} \times \vec{B}\\ &=& I\left(\... ... &=& I\left(\vec{L_f}-\vec{L_i} \right)\times \vec{B}\\ &=& 0, \end{eqnarray*}$$

where the last line follows since a closed loop ends up where it starts.