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 _{1}   and  \mu _{2} as shown in the figure

We will find out these boundary conditions by using

  1. \oint_{s} \overrightarrow{B} .\overrightarrow{ds} =0  –  Gauss’s law in Magneto statics.
  2. \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}} .

Similar to normal components of magnetisation the tangential components are

M_{2t} = \chi _{m2} H_{2t} .

M_{1t} = \chi _{m1} H_{1t}.

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

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

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

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

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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.