Magnetic Boundary conditions

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Magnetic Boundary conditions:-

In order to find out the $\overrightarrow{B}$ , $\overrightarrow{H}$ and $\overrightarrow{M}$ at the interface between two different magnetic materials boundary conditions are required.

Consider two magnetic materials having permeabilities $\mu _{1}$ and $\mu _{2}$ as shown in the figure

We will find out these boundary conditions by using

$\oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0$ – Gauss’s law in Magneto statics.
$\oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I$ – Ampere’s law.

In order to apply $\oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0$ , a surface is required which is nothing but Gaussian surface (or) a Pillbox.

$\oint_{s} \overrightarrow{B} .\overrightarrow{ds} = \oint_{side} \overrightarrow{B} .\overrightarrow{ds} + \oint_{top} \overrightarrow{B} .\overrightarrow{ds} + \oint_{bottom} \overrightarrow{B} .\overrightarrow{ds}$ .

( $\oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0$ -Because on the boundary $\Delta h=0$ , so there exists no side of surface).

$B_{1n}\Delta S -B_{2n}\Delta S =0$ .

$B_{1n}=B_{2n}-----------(1)$

since $\overrightarrow{B} = \mu \overrightarrow{H}$ , $\mu _{1} H_{1n}= \mu _{2} H_{2n}$ .

$H_{1n}= \frac{\mu _{2}}{\mu _{1}} H_{2n} ---------(2)$ .

From equations (1) and (2) the normal component of $\overrightarrow{B}$ is continuous at the boundary and the normal components of $\overrightarrow{H}$ is discontinuous at the boundary.

we know that $\overrightarrow{B} = \mu _{o}(1+ \chi _{m})\overrightarrow{H}$ .

$\overrightarrow{B} = \mu _{o}\overrightarrow{H}+ \mu _{o}\overrightarrow{M}$ .

$\overrightarrow{M} = \chi _{m}\overrightarrow{H}$ .

$M_{2n} = \chi _{m2} H_{2n}$ .

$M_{1n} = \chi _{m1} H_{1n}$ .

$M_{2n} = \chi _{m2} H_{2n}$ since $H_{2n} = \frac{\mu _{1}}{\mu _{2}} H_{1n}$ .

$M_{2n}=\chi _{m2} \frac{\mu _{1}}{\mu _{2}} H_{1n}$ .

$M_{2n}=\frac{\chi _{m2}}{\chi _{m1}} \frac{\mu _{1}}{\mu _{2}} M_{1n}$ .

The magnetisation normal components are also discontinuous.

To apply $\oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I$ a path is required, which is a closed one in a plane normal to the boundary surface.

Here a closed path is abcda , which encloses a surface current density k on the surface of the boundary

Then $\oint_{l} \overrightarrow{H} .\overrightarrow{dl} = I$ .

$H_{1t}\Delta w -H_{1n}\frac{\Delta h}{2}-H_{2n}\frac{\Delta h}{2}-H_{2t}\Delta w+H_{2n}\frac{\Delta h}{2}+H_{1n}\frac{\Delta h}{2}=I$ .

(since $\Delta h=0$ On the boundary)

$H_{1t}\Delta w-H_{2t}\Delta w=k \Delta w$ .

$H_{1t}-H_{2t}=k$ .

$\frac{B_{1t}}{\mu _{1}}-\frac{B_{2t}}{\mu _{2}}=k$ , So the tangential components of $\overrightarrow{B}$ and $\overrightarrow{H}$ are discontinuous.

$(\overrightarrow{H_{1t}} -\overrightarrow{H_{2t}})X \overrightarrow{a_{n12}} = \overrightarrow{k}$ .

(or) $(\overrightarrow{H_{1t}} -\overrightarrow{H_{2t}}) = \overrightarrow{k}X \overrightarrow{a_{n12}}$ .

When $\overrightarrow{k}=0$ , $\overrightarrow{H_{1t}}=\overrightarrow{H_{2t}}$ and $\frac{\overrightarrow{B_{1t}}}{\mu _{1}} =\frac{\overrightarrow{B_{2t}}}{\mu _{2}}$ .