Question

state and explain with diagram maxwell's modification of amperecircuital law

Answer 1

$$\begin{gathered} \mathrm{To}\mathrm{remove}\mathrm{inconsistency},\mathrm{Maxwell}\mathrm{suggested}\mathrm{the}\mathrm{existence}\mathrm{of}\mathrm{an}\mathrm{additional}\mathrm{current}.\mathrm{This}\mathrm{additional}\mathrm{current}\mathrm{is}\mathrm{called}\mathrm{displacement}\mathrm{current}. \\ \mathrm{It}\mathrm{is}\mathrm{due}\mathrm{to}\mathrm{time}\mathrm{varying}\mathrm{electric}\mathrm{field}. \\ \mathrm{The}\mathrm{displacement}\mathrm{current}\mathrm{will}\mathrm{be}\mathrm{given}\mathrm{by}, \\ {\mathrm{i}}_{\mathrm{d}}={\in }_0\frac{{\mathrm{d\varphi }}_{\mathrm{E}}}{\mathrm{dt}} \\ \mathrm{Thus},\mathrm{Ampere}\text{'}\mathrm{s}\mathrm{law}\mathrm{will}\mathrm{be}\mathrm{restated}\mathrm{as}, \\ \oint \overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{dl}}={\mathrm{\mu }}_0[{\mathrm{i}}_{\mathrm{c}}+{\mathrm{i}}_{\mathrm{d}}] \\ \mathrm{where},{\mathrm{i}}_{\mathrm{c}}=\mathrm{Conduction}\mathrm{current} \\ \mathrm{and}{\mathrm{i}}_{\mathrm{d}}=\mathrm{Displacement}\mathrm{current} \\ \mathrm{Finally}, \\ \mathbf{\oint }\overrightarrow{\mathbf{}\mathbf{B}}\mathbf{.}\mathbf{}\overrightarrow{\mathbf{dl}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{}\mathbf{[}{\mathbf{i}}_{\mathbf{c}}\mathbf{}\mathbf{+}{\mathbf{\in }}_{\mathbf{0}}\frac{{\mathbf{d\varphi }}_{\mathbf{E}}}{\mathbf{dt}}\mathbf{]} \\ \mathrm{This}\mathrm{is}\mathrm{called}\mathrm{Ampere}-\mathrm{Maxwell}\mathrm{law}. \end{gathered}$$