Deduce an expression for the force on a current carrying conductor placed in a magnetic field. Derive on expression for the force per unit length between two parallel current carrying conductors?

Flemings left hand rule: Stretch the left hand such that the thumb, the forefinger and the middle finger are mutually parpendicular to each other. If fore finger indicates the direction of magnetic field, the middle finger current direction then the thumb indicates direction of force on the conductor.
The force on a current carrying conductor in a magnetic field:
1) Consider a straight wire of length '$l$' of cross sectional area A and carrying a current 'i' placed in a uniform magnetic field induction 'B' as shown in fig.

2) Due to the potential difference, electrons move with drift velocity. '$v_d$'
3) We know that the force 'f' acting on charge 'q' in the magnetic field of induction B is
$f = q{V}_{d}B \sin \theta ......\left(1\right)$
4) If 'n' is no of free electrons per unit volume of the conductor, the current in it is
$i = nq{v}_{d}A...... \left(2\right)$
5) The total force F acting on '$nlA$' no of electrons in length of conductor is

$F = \left(nlA\right) f = nlA. q{V}_{d}B\sin \theta$

$F = nq V_{d}A \left(lB\sin \theta \right) \left[\because nq{V}_{d}A\right]$

$F = ilB\sin \theta$

Force between two parallel conductors carrying current :

1) Consider two long parallel conductors I & II of length $l$ carrying current $i_1{i}_2$ separated by a distance r.

2) The magnetic indiction field strength due to conductor I at distance 'r' is

$B_1^{\otimes }=\frac{{\mu }_0}{2\pi }\frac{i_1}{r}\stackrel{ }{\to }\left(1\right)$

3) conductor II is in '$B_1$'. So the force on it is $F_2=i_{2}l{B}_1$ (towards first conductor)

$F_2=\frac{{\mu }_0}{2\pi }\frac{i_1{i}_{2}l}{r}\stackrel{ }{\to }\left(2\right)$

The magnetic induction due to conductor II at a distance 'r' is

$B_1^{\odot }=\frac{{\mu }_0}{2\pi }\frac{i_2}{r}\stackrel{ }{\to }\left(3\right)$

Conductor 'I' is in '$B_2$'. So the force on it is $F_1=i_{1}l{B}_1$ (towards second conductor)

$F_1=\frac{{\mu }_0}{2\pi }\frac{i_1{i}_2}{r}l\stackrel{ }{\to }\left(4\right)$

$\therefore F_1=F_2=\frac{{\mu }_0}{2\pi }\frac{i_1{i}_2}{r}l$