# Magnetizing field intensity and intensity of magnetization | magnetism and matter class 12

In this short piece of article, we will discuss magnetizing field intensity and intensity of magnetization. This is the topic of chapter magnetism and matter class 12. So let’s get started… [latexpage]

## Magnetising field intensity

Definition: It is the ability of the magnetising field to magnetize a magnetic material when it is placed in any external magnetising field ${\vec{B_0}}$. It is expressed by a vector $\vec{H}$ called magnetising field intensity or magnetic intensity.

Its magnitude may defined as the ampere-turns $(nI)$ flowing through the unit length of the solenoid required to produce the following magnetising field. Thus $$H=nI$$

$$\therefore \qquad B_0=\mu_0 nI=\mu_0 H$$

The dimensions of magnetising field intensity is $[AL^{-1}]$. Its SI unit is ampere per meter $(Am^{-1})$. It is equivalent to $Nm^{-2}T^{-1}$ or $Jm^{-1}{Wb}^{-1}$.

## Intensity of magnetisation

As we know that when a magnetic material is placed in an external magnetizing field, it becomes magnetized.

Definition: The magnetic moment developed per unit volume of a material when it is placed in external magnetising field called intensity of magnetisation or simply magnetisation.

Thus, mathematically, the intensity of magnetization is given as $$\overrightarrow{M}=\frac{\vec{m}}{V}$$

If $I_{M}$ is the surface magnetization current which has been set up in a solenoid of cross-sectional area $A$ and has $n$ turns per unit length, then the magnetic moment is developed at per unit length of the solenoid is $nI_{M}A$. Therefore, the magnetic moment developed per unit volume of a solenoid of magnetization $\vec{M}$ is given as $$M=\frac{m}{V}=\frac{nI_{M}A}{A}=nI_{M}$$

Hence,

$$B_{M}=\mu_0 nI_{M}=\mu_0 M$$ Now, consider a bar of magnetic material having cross-sectional area $a$ and length $2l$. Its volume is given by $$V=a\times 2l$$ Let’s suppose that the bar developed a pole strength $q_{m}$ at its end, when placed in external magnetising field, then the magnetic moment so developed is given as: $$m=q_m \times 2l$$ $$\therefore\quad M=\frac{m}{V}=\frac{q_m \times 2l}{a\times 2l}=\frac{q_m}{a}$$

The total magnetic field or magnetic induction $\vec{B}$ inside a magnetic material is the sum of magnetizing field $\vec{B_0}$ and the field $\vec{B_M}$ arise due to the magnetisation of the material. Therefore, $$B=B_0+B_M = \mu_0 H+\mu_0 M$$

SI unit of both $H$ and $M$ is the same $Am^{-1}$.