Inconsistensy in Ampere’s law (or) Displacement Current density

Faraday’s experimental law has been used to obtain one of Maxwell’s equations in differential form \overrightarrow{\bigtriangledown }X \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} , which shows that a time-varying Magnetic field produces an Electric field.

From Ampere’s  Circuital law which is applicable to Steady Magnetic fields

\overrightarrow{\bigtriangledown } X \overrightarrow{H}=\overrightarrow{J}

By taking divergence of Ampere’s law the Ampere’s law is not consistent with time-varying fields

\overrightarrow{\bigtriangledown }.(\overrightarrow{\bigtriangledown } X \overrightarrow{H})=\overrightarrow{\bigtriangledown }.\overrightarrow{J}

\overrightarrow{\bigtriangledown } . \overrightarrow{J}=0 ,since \overrightarrow{\bigtriangledown }.(\overrightarrow{\bigtriangledown } X \overrightarrow{H})=0---------Equation(1)

the divergence of the curl is identically zero which implies \overrightarrow{\bigtriangledown }.\overrightarrow{J}=0------Equation(2), but from the continuity equation \overrightarrow{\bigtriangledown }.\overrightarrow{J} = -\frac{\partial \rho _{v}}{\partial t}-------Equation(3) which is not equal to zero, as \frac{\partial \rho _{v}}{\partial t}\neq 0 is an unrealistic limitation(i.e we can not assume \frac{\partial \rho _{v}}{\partial t} as zero) .

\therefore to make a compromise between the above two situations we must add an unknown term \overrightarrow{G} to Ampere’s Circuital law

i.e, \overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\overrightarrow{G}

then by taking the Divergence of the above equation

\overrightarrow{\bigtriangledown }.(\overrightarrow{\bigtriangledown }X\overrightarrow{H}) = \overrightarrow{\bigtriangledown }.\overrightarrow{J}+\overrightarrow{\bigtriangledown }.\overrightarrow{G}------Equation(4)

from Equation(1),Equation(4) becomes     \overrightarrow{\bigtriangledown }.\overrightarrow{J}+\overrightarrow{\bigtriangledown }.\overrightarrow{G}=0

\overrightarrow{\bigtriangledown }.\overrightarrow{G}=-\overrightarrow{\bigtriangledown }.\overrightarrow{J}

thus \overrightarrow{\bigtriangledown }.\overrightarrow{G} = \frac{\partial \rho _{v}}{\partial t}---------Equation(5)

from Maxwell’s first Equation \overrightarrow{\bigtriangledown }.\overrightarrow{D}=\rho _{v} 

then Equation (5) becomes \overrightarrow{\bigtriangledown }.\overrightarrow{J} = \frac{\partial }{\partial t} (\overrightarrow{\bigtriangledown }.\overrightarrow{D})

\overrightarrow{\bigtriangledown }.\overrightarrow{G} =\overrightarrow{\bigtriangledown }. \frac{\partial \overrightarrow{D}}{\partial t}

then   \overrightarrow{G} = \frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

This is the equation obtained which does not disagree with the continuity equation. It is also consistent with all other results. This is a second Maxwell’s Equation is time-varying fields so the term \frac{\partial \overrightarrow{D}}{\partial t} has the dimensions of current density Amperes/Square-meter. Since it results from a time-varying electric flux density (\overrightarrow{D} ) , Maxwell termed it as displacement current density \overrightarrow{J_{D}}.

\overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\frac{\partial \overrightarrow{D}}{\partial t}

\overrightarrow{\bigtriangledown }X\overrightarrow{H} = \overrightarrow{J}+\overrightarrow{J_{D}}

\overrightarrow{J_{D}}=\frac{\partial \overrightarrow{D}}{\partial t}

up to this point three current densities are there \overrightarrow{J}=\sigma \overrightarrow{E} , \overrightarrow{J}=\rho _{v} \overrightarrow{v} and \overrightarrow{J_{D}}= \frac{\partial\overrightarrow{D} }{\partial t}.

when the medium is Non-conducting medium \overrightarrow{\bigtriangledown }X \overrightarrow{H}=\frac{\partial\overrightarrow{D} }{\partial t}

the total displacement current crossing any given surface is expressed by the surface integral I_{d} = \oint_{s} \overrightarrow{J_{D}}.\overrightarrow{ds}

I_{d} = \oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

from Ampere’s law \oint_{s}(\overrightarrow{\bigtriangledown }X \overrightarrow{H}).\overrightarrow{ds}=\int_{s} \overrightarrow{J}.\overrightarrow{ds} +\oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=\int_{s} \overrightarrow{J}.\overrightarrow{ds} +\oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=I +\oint_{s} \frac{\partial \overrightarrow{D}}{\partial t}.\overrightarrow{ds}

\oint_{l}\overrightarrow{H}.\overrightarrow{dl}=I +I_{d}

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Author: Lakshmi Prasanna Ponnala

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