Force on current carrying conductor in uniform magnetic field

Derivation of force on current carrying conductor in uniform magnetic field:

Let us consider:
• The length of the conductor - $l$

• The cross-section 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
 $F=ilB \: sin\theta$
The Vector Form of the above equation:
 $F=i \left(l \times B \right)$