Derive the expression for force per unit length between two long straight parallel current carrying conductors. Hence define one ampere.

Explanation

On (2),`B` due to (i) is `=(mu_(0)I_(1))/(2pid)ox`
`F` on (2) on `1m` length
`I_(2).(mu_(0)I_(1))/(2pid)` towards left it is attractive
`(mu_(0)I_(1)I_(2))/(2pid)` (hence proved)
Similarly on the other wire also.
•Definition of ampere (fundamental unit of current) using the above formula.
If `I_(1)=I_(2)=1A,d=1m` then `F=2xx10^(-7) N`
`therefore` "When two very long wires carrying equal currents and separated by `1m` distance exert on each other a magnetic force of `2xx10^(-7)N` on `1m` length then the current is `1` ampere."
•If the currents are in the oppostive direction then the magnetic force on the wires will be repulsive.
OR
The electron of charge `(-e)(e=+1.6xx10^(19)C)` performs uniform circular motion around a stationary heavy nucleus of charge `+Ze`.This constitutes a current `I`, where.
`I=e/T`
and `T` is the time period of revolution.Let `r` be the orbital radius of the electron and `v` the orbital speed. Then, `T=(2pir)/v`
Substituting in Equation, we have `I=ev//2pir`.
There will be a magnetic moment usually denoted by `mu`, associated with this circulating current.From equation its magnitude is `mu=lpir^(2)=evr//2`.
The direction of this magnetic moment is into the plane of the paper in figure.[This follows from the right hand rule discussed earlier and the fact that the negatively charged electron is moving anti-clockwise, leading to a clockwise current.]Multiplying and dividing the right-hand side of the above expression by the electron mass `m_(e)`, have,
`mu_(1)=e/(2m_(e))(m_(e)vr)=e/(2m_(e))L`
Here, `L` is the magnitude of the angular momentum of the electron about the central nucleus ("orbital angular momentum).
`mu_(1)=e/(2m_(e))L`
The negatives sign indicates that the angular momentum of the electron is opposite in direction the magnetic moment.Instead of electron with charge `(-e)`.If we had taken a particle with charge `(+q)`, the angular momentum and magnetic moment would be in the same direction.
The ratio `(mu_(1))/L=e/(2m_(e))` is called the gyromagnetic ratio and is a constant. Its values is `8.8xx10^(10)C//kg` for an electron, which has been verified by experiments.
Bohr hypothesised that the angular momentum assumes a discrete set of values, namely ,`L=(nh)/(2pi)` where `n` is a natural number, `n=1,2,3,..` and `h` is constant named after max planck (Planck's constant) with a value `h=6.626xx10^(-34) Js`. Our aim here is merely to use it to calculate the elementary dipole memont.Take the value `n=1`, we
`(mu_(L))_min=e/(4pim_(e))h=(1.60xx10^(-19)xx6.63xx10^(-34))/(4xx3.14xx9.11xx10^(-31))=9.27xx10^(-24) Am^(2)`
This value is called the Bohr magneton.