Derive an expression for the force acting between two very long parallel current-carrying conductors and hence define the Ampere.

Explanation

Force between two parallel conductors carrying current :

Let two long parallel conductors A and B seperated by a distance r carry currents `i_(1) and i_(2)` in the same directions.
The current `i_(1)` produces magnetic induction `B_(1)` around the conductor A and the current `i_(2)` produces magnetic induction `B_(2)` around the conductor B.
If l is the length of each conductor, The magnetic induction `B_(1)` at distance r from conductor A is
`B_(1)=(mu_(0)i_(1))/(2pi r)rarr(1)`
The force on conductor B is given by
`F_(2)=i_(2)l B_(1) rarr (2)`
The direction of force is given by Fleming left hand rule.
`F_(2)=i_(2) l xx (mu_(0)i_(1))/(2pir)=(mu_(0)i_(1)i_(2)l)/(2pi r)rarr(3)`
the direction of force `F_(2)` is towards conductor A. Similary the magnetic induction `B_(2)` at distance r from conductor B is
`B_(2)=(mu_(0) i_(2))/(2pi r) rarr(4)`
The force on conductor A is given by
`F_(1)=i_(1) l B_(2) rarr (5)`
Substituting `B_(2)` value in equation (5)
`F_(1)=(i_(1)l xx mu_(0) i_(2))/(2pi r)`
`F_(1)=(mu_(0) i_(1) i_(2)l)/(2 pi r) rarr (6)`
If can be seen that `|F_(1)| = |F_(2)|=(mu_(0) i_(1) i_(2) l)/(2 pi r)`
the force per unit length of the conductor is given by
`(F)/(l)=(mu_(0) i_(1) i_(2))/(2 pi r) rarr (7)`
Definition of Ampere :
If `i_(1)=i_(2)=1A, r=1m`
`F//l =(4 pi xx 10^(-7)xx1xx1)/(2pi xx1)=2xx10^(-7) Nm^(-1)`
''When two infinitely long parallel conductors, carrying the same current are seperated by `2xx10^(-7) Nm^(-1)`, then the current flowing through each conductor is said to be one Ampere''.