Magnetic Boundary conditions

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  $\mu&space;_{1}$   and  $\mu&space;_{2}$ as shown in the figure

We will find out these boundary conditions by using

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

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

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

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

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

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

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

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

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 $\overrightarrow{B}&space;=&space;\mu&space;_{o}(1+&space;\chi&space;_{m})\overrightarrow{H}$ .

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

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

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

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

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

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

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

The magnetisation normal components are also discontinuous.

To apply    $\oint_{l}&space;\overrightarrow{H}&space;.\overrightarrow{dl}&space;=&space;I$  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 $\oint_{l}&space;\overrightarrow{H}&space;.\overrightarrow{dl}&space;=&space;I$ .

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

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

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

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

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

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

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

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

Similar to normal components of magnetisation the tangential components are

$M_{2t}&space;=&space;\chi&space;_{m2}&space;H_{2t}$ .

$M_{1t}&space;=&space;\chi&space;_{m1}&space;H_{1t}$.

$M_{2t}&space;=&space;\chi&space;_{m2}&space;H_{2t}$ since  $H_{2t}&space;=&space;H_{1t}-k$ .

$M_{2t}=\chi&space;_{m2}&space;(H_{1t}-k)$ .

$M_{2n}=\frac{\chi&space;_{m2}}{\chi&space;_{m1}}&space;M_{1t}-\chi&space;_{m2}&space;k$.

Obtain the results for magnetisation by using the same procedure as that of  normal components.

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