Question

Explain the Maxwell’s modification of Ampere’s circuital law.

Answer 1

Ampere’s circuital law is ${\displaystyle \oint \overrightarrow{\text{B}}\cdot \text{d}\overrightarrow{\text{s}}={\mu }_{0}l}$

Modification by J.C. Maxwell on Ampere’s circuital law

1. Due to the external source applied between the plates, the increasing current flowing through the capacitor produce an increasing electric field between the plates.
2. This change in the electric field between the capacitor plate produce a current between the plates.
3. This change in the electric field between the capacitor plate produce a current between the plates.
   By Gauss Law ${\displaystyle {\varphi }_{\text{E}}=\oint \overrightarrow{\text{E}}\cdot \overrightarrow{\text{dA}}=\text{EA}=\frac{\text{q}}{{\epsilon }_0}}$
   Change in electric flux
   ${\displaystyle \frac{\text{d}{\varphi }_{\text{E}}}{\text{dt}}=\frac{1}{{\epsilon }_0}\frac{\text{dq}}{\text{dt}}=\frac{1}{{\epsilon }_0} {\text{I}}_{\text{d}}}$
   ${\displaystyle \therefore {\text{I}}_{\text{d}}={\epsilon }_0\frac{\text{d}{\varphi }_{\text{E}}}{\text{dt}}}$
   

   I_d is the displacement current.

4. The displacement current is defined as the current which comes into play in the region in which the electric field and the electric flux are changing with time.
5. So, Maxwell modified Ampere’s Law
   ${\displaystyle \oint \overrightarrow{\text{B}}\cdot cec\text{ds}={\mu }_0\text{I}={\mu }_0({\text{I}}_{\text{c}}+{\text{I}}_{\text{d}})}$
   

   Total current I = conduction current I_c + displacement current I_d