(N/A) Consider a long solenoid with $n$ turns per unit length carrying a current $I$.
The magnetic field in the interior of the solenoid due to the current is:
$\vec{B}_{0} = \mu_{0} n I \quad \dots (1)$
If the interior of the solenoid is filled with a magnetic material,the total magnetic field $\vec{B}$ inside the solenoid is the sum of the field due to the current $(\vec{B}_{0})$ and the field due to the magnetization of the material $(\vec{B}_{m})$:
$\vec{B} = \vec{B}_{0} + \vec{B}_{m} \quad \dots (2)$
The additional field $\vec{B}_{m}$ is proportional to the magnetization $\vec{M}$ of the material:
$\vec{B}_{m} = \mu_{0} \vec{M} \quad \dots (3)$
We define the magnetic intensity $\vec{H}$ as:
$\vec{H} = \frac{\vec{B}}{\mu_{0}} - \vec{M} \quad \dots (4)$
Rearranging this equation,we get the relation for the total magnetic field $\vec{B}$:
$\vec{B} = \mu_{0}(\vec{H} + \vec{M}) \quad \dots (5)$
Here,$\vec{H}$ represents the contribution from external currents,and $\vec{M}$ represents the contribution from the magnetic material.