# Application of Ampere’s circuit law-infinite sheet of current

consider an infinite sheet in the z=0 plane, which has uniform current density  $\overrightarrow{k}&space;=&space;k_{y}\overrightarrow{a_{y}}$    A/m .

Let us suppose the current is flowing in the positive y direction.

the sheet of current is assumed to be in rectangular co-ordinate system

$\overrightarrow{H}&space;=&space;H_{x}\overrightarrow{a_{x}}+H_{y}\overrightarrow{a_{y}}+H_{z}\overrightarrow{a_{z}}$

Let us suppose the conductor is carrying a current I , by right hand thumb rule magnetic field is produced around the conductor is right angles to the direction of I.

In this case of infinite sheet , the current is in the y-direction there is no component of H along the direction of y  and also the z components cancel each other because of opposite direction of the fields produced so only x components of H exists.

$\overrightarrow{H}&space;=&space;\left\{\begin{matrix}&space;H_{o}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z>0\\&space;-H_{o}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z<0&space;\end{matrix}\right.$

from Ampere’s Circuit law  $\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;\int&space;\left&space;(\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}&space;+\overrightarrow{H}_{2-3}.&space;\overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}.&space;\overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}.&space;\overrightarrow{dl}_{4-1}&space;\right&space;)$ .

the component $\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;\int&space;\left&space;(\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}&space;+\overrightarrow{H}_{2-3}.&space;\overrightarrow{dl}_{2-3}+\overrightarrow{H}_{3-4}.&space;\overrightarrow{dl}_{3-4}+\overrightarrow{H}_{4-1}.&space;\overrightarrow{dl}_{4-1}&space;\right&space;)$

$\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}=&space;\overrightarrow{H}_{1-0}.&space;\overrightarrow{dl}_{1-0}+\overrightarrow{H}_{0-2}.&space;\overrightarrow{dl}_{0-2}$

$\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}=&space;H_{o}&space;\overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}})-&space;H_{o}&space;\overrightarrow{a_{x}}.(\frac{a}{2})(-\overrightarrow{a_{z}})$ .

$\overrightarrow{H}_{1-2}.&space;\overrightarrow{dl}_{1-2}=0$ .

similarly $\overrightarrow{H}_{3-4}.&space;\overrightarrow{dl}_{3-4}&space;=0$ .

$\therefore&space;\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;\int&space;\left&space;(\overrightarrow{H}_{2-3}.&space;\overrightarrow{dl}_{2-3}+\overrightarrow{H}_{4-1}.&space;\overrightarrow{dl}_{4-1}&space;\right)$ .

$\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;-(H_{o}\&space;\overrightarrow{a_{x}}).(b&space;\&space;-\overrightarrow{a_{x}})+(H_{o}\&space;\overrightarrow{a_{x}}).(b&space;\&space;\overrightarrow{a_{x}})$ .

$\oint&space;\overrightarrow{H}.&space;\overrightarrow{dl}&space;=&space;2H_{o}\&space;b$ .

$I=&space;2H_{o}&space;\&space;b$ .

$k_{y}\&space;b&space;=&space;2H_{o}&space;\&space;b$ .

$H_{o}&space;=\frac{1}{2}&space;k_{y}$ .

as   $\overrightarrow{H}&space;=&space;\left\{\begin{matrix}&space;H_{o}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z>0&space;\\&space;-H_{o}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z<0&space;\end{matrix}\right.$

this will be changed to $\overrightarrow{H}&space;=&space;\left\{\begin{matrix}&space;\frac{1}{2}&space;k_{y}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z>0&space;\\&space;-\frac{1}{2}&space;k_{y}\&space;\overrightarrow&space;{a_{x}}&space;\&space;for&space;\&space;z<0&space;\end{matrix}\right.$

In general for a finite sheet of current density  $\overrightarrow{k}$  A/m  Magnetic field is generalised as     $\overrightarrow{H}&space;=&space;\frac{1}{2}&space;(\overrightarrow{k}&space;X&space;\overrightarrow{a_{n}})$ .

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