Derive and expression for the force per unit length acting on the two straight parallel current carrying conductors.In which condition will this force be attractive and repulsive?Define the standard unit of current.
OR
An electron is revolving with speed `v` in a circular orbit of radius `r`.Obtain the expession of gyromagnetic ratio.What is a Bohr magneton?Write its value.

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.