Give the expression for the force on a current-carrying conductor in a magnetic field.

(N/A) As shown in the figure,consider a conductor $PQ$ of length $l$ and cross-sectional area $A$,carrying current $I$ along the positive $y$-direction. The magnetic field $\overrightarrow{B}$ acts along the positive $z$-direction.
The electrons drift towards the left with drift velocity $\overrightarrow{v_{d}}$.
Each electron experiences a force along the positive $x$-axis,which is given by:
$\vec{f} = -e(\overrightarrow{v_{d}} \times \overrightarrow{B})$
If $n$ is the number of free electrons per unit volume,then the total number of electrons in the conductor is:
$N = n \times \text{Volume} = nAl$
The total force on the conductor is:
$\overrightarrow{F} = N \vec{f} = nAl[-e(\overrightarrow{v_{d}} \times \overrightarrow{B})]$
$= nAe[-(l \overrightarrow{v_{d}} \times \overrightarrow{B})]$
Since $I\vec{l}$ represents a current element vector in the direction of current,we can write:
$\vec{v}_{d} = v_{d} \vec{l}$
$\therefore \overrightarrow{F} = nAe(v_{d} \vec{l} \times \overrightarrow{B}) = nAev_{d}(\vec{l} \times \overrightarrow{B})$
Since $nAev_{d} = I$ (current),
$\therefore \vec{F} = I(\vec{l} \times \overrightarrow{B})$
The magnitude is $F = IlB \sin \theta$,where $\theta$ is the angle between $\overrightarrow{B}$ and $\vec{l}$.
For a wire of arbitrary shape,we calculate the force by considering it as a collection of linear strips $d\vec{l}$ and summing:
$\overrightarrow{F} = \sum I(d\vec{l} \times \overrightarrow{B})$