Derivation of force on current carrying conductor in uniform magnetic field:
Let us consider:
The length of the conductor  $l$
The crosssection area of the current carrying conductor  $A$
The current flow in a conductor $i$
The drift or average velocity of the free electrons  $v_{d}$
The current carrying conductor is placed in a magnetic field  $B$
The total number of free electrons in the crrent carrying conductor  $N$

Force on current carrying conductor in the uniform magnetic field 
Now the magnetic force on one free electron in a conductor 
$F'= ev_{d}B sin\theta \qquad(1)$
The net force on the conductor is due to all the free electrons present in the conductor
$F=N\: F' \qquad(2)$
Let $n$ is the number of free electrons per unit volume of conductor. So the total number of free electrons in the $Al$ volume of the conductor will be
$N=nAl \qquad(3)$
Now substitute the value of $N$ and $F'$ from above equation $(1)$ and equation $(3)$ in equation $(2)$
$F=neAlv_{d}B\: sin\theta$
Where $i=neAv_{d}$
So from the above equation
The Vector Form of the above equation:
$F=i \left(l \times B \right) $
