What is the Landé $g$ factor? What is the Landé $g$ factor?
I know that it gives the relation between magnetic moment and angular moment, but i wanted to know why are those magnitudes related to each other and why is the magnetic moment operator defined proportional to he angular moment and how is the gyromagnetic ratio calculated theoretically.
 A: 
I know that it gives the relation between magnetic moment and angular moment

The relation is between the magnetic moment $\vec{\mu}$ and the angular momentum $\vec{L}$. They are two different physical quantities.

why are those magnitudes related to each other and why is the magnetic moment operator defined proportional to he angular moment

The magnetic moment arise due to the motion of a charged particle. Also the particle has mass. The motion of particle is measured as angular momentum. So a charge particle in motion has a magnetic moment as well as angular momentum. It is to be noted that the moment is defined only when the motion is bounded as in the case of an electron around a nucleus as shown below.


why is the magnetic moment operator defined proportional to he angular moment
how is the gyromagnetic ratio calculated theoretically.

$\vec{\mu}$ is connected with Amperian loops; Amperian loop is a bounded flow of current $J$ flowing around a loop with area $A$ constitutes Amperian loop. The magnetic moment $\vec{\mu}$ is defined as
$$\vec{\mu} = J \vec{A}.$$
The area has a vectorial character with the direction given by the right hand thumb rule. It is to be noted that moment is a quantum mechanical outcome. The classical picture of Amperian loop only helps to visualize it.
For the electron orbiting around the nucleus with a frequency $\omega$ the current is
$$J =  - \frac{|e| \omega}{2\pi}.$$
$$\therefore \vec{\mu} = - \frac{|e| \omega \vec{A}}{2\pi}.$$
As the angular velocity around the nucleus increases, the current increases thereby increasing the moment.
Also, the electron has mass. A mass under motion has angular momentum. The angular momentum
$$\vec{L} = I \omega,$$
is also a vector quantity with magnitude proportional to angular velocity where $I=m r^{2}$ is the moment of inertia of the electron with mass $m$ orbiting around the nucleus with a radius $r$. The direction is again perpendicular to the orbit.
If the orbit has radius, then
$$I = m r^{2} = \frac{m A}{\pi}.$$
$$\therefore \vec{L} = \frac{m \omega \vec{A} }{\pi}.$$
From the above it is evident that $\vec{\mu}$ is proportional to $\vec{L}$ and antiparallel to each other due to the negative sign of the charge of the electron.
$$\vec{\mu} = -\gamma \vec{L}. $$
$$\gamma = \frac{e}{2m}.$$
The proportionality constant $\gamma$ is called the gyromagnetic ratio and can be experimentally determined from the Einstein-de Haas experiment
Land$\acute{e}$ factor
Experimentally, it was observed the gyromagnetic ratio is twice of that expected from theory. This extra factor is the so called g-factor. This was addressed by Dirac's relativistic equation. The underlying basis for the factor is that in addition to the orbital angular momentum, the electron has an intrinsic spin angular momentum. For a group of electrons with both total orbital angular momentum as well as total spin angular momentum, the correction factor to the gyromagnetic ratio becomes the Land$\acute{e}$ g-factor.
For further information refer to Stephen Blundell's Magnetism in Condensed Matter by Oxford University Press
A: Think about an electron in a circular orbit about a nucleus, at some radius $r$ and with velocity $v$.
It has a mass $m$ so it has an angular momentum $L=mvr$.
It has a charge $e$ so one turn, in which it travels $2\pi r$, is equivalent to a current $I=ev/2\pi r$. 
The magnetic moment of a circular loop is $\mu= I A={ev \over 2 \pi r} \pi r^2={evr\over 2}$
So both the angular momentum and the magnetic moment depend on the velocity and the radius - which is obvious - but it turns out that they do so in the same way, through the product $vr$. We can combine the two equations to get the relation
$\mu = {e \over 2 m} L$.
This can be extended to noncircular (i.e. elliptical) orbits and it still holds. 
So far this has all been classical.  In quantum mechanics the component of $L$ along an axis (usually taken as $\hat z$) is quantised: $L_z=m \hbar$, so the component $\mu_z$, which you need for the Zeeman effect and such,  is just ${e \hbar \over 2 m } m$.    ( ${e \hbar \over 2 m }$ is so handy it's known as the Bohr Magneton and given the symbol $\mu_B$)
Now it gets complicated by the intrinsic spin of the electron. This has half a unit of angular momentum, $S_z=\pm {1 \over 2} \hbar$, but punches above its weight for the magnetic moment, by a factor 2. This was regarded as anomalous and weird until Dirac discovered the Dirac equation, from which it emerges naturally.
So for the electrons in an atom the link between angular momentum and magnetic moment is $\mu_z=\mu_B m$ for orbital angular momentum (where $m=-\ell ... \ell$) and $\mu_z=2 \mu_B m_s$ for spin angular momentum (where $m_s=\pm {1 \over 2}$). Spin gives you twice as much magnetic moment as orbital motion, for the same angular momentum.
The final link in the argument is that in many atoms the relevant quantity is not $\vec L$ or $\vec S$ but the total $\vec J$. The atom has a defined value of $J_z$ with $\vec L$ and $\vec S$ precessing around $\vec J$. The  magnetic moment depends on the relative amounts of $L$ and $S$ that make up this $J$. So we have to write
$\mu_z=g \mu_B m_j$
where $g$ is the $g$ factor: 1 for pure orbital, 2 for pure spin, and Lande showed 
(for a derivation see Wikipedia) to be given by ${3 \over 2 }+{s(s+1)-\ell(\ell+1)\over 2 j(j+1)}$ in the general case (where $\ell, j, s$ are the quantum numbers)
