Question

Maxwell's modified form of Ampere's circuital law is:
A. $\oint\limits_s {\mathop B\limits^ \to } .\mathop {ds}\limits^ \to = 0$
B. $\oint\limits_s {\mathop B\limits^ \to } .\mathop {dl}\limits^ \to = {\mu _0}I$
C. $\oint\limits_s {\mathop B\limits^ \to } .\mathop {dl}\limits^ \to = {\mu _0}I + \dfrac{I}{{{\varepsilon _0}}}\dfrac{{dq}}{{dt}}$
D. $\oint\limits_s {\mathop B\limits^ \to } .\mathop {dl}\limits^ \to = {\mu _0}I + {\mu _0}{\varepsilon _0}\dfrac{{d{\phi _E}}}{{dt}}$

Answer 1

When charge is at rest it produces only electric fields but when charge is under motion it produces both electric and magnetic fields. So a current carrying conductor produces a magnetic field. The magnetic field produced will be generally found by using modified ampere’s circuital law

Formula used:
$\oint\limits_s {\mathop B\limits^ \to } .\mathop {dl}\limits^ \to = {\mu _0}{I_{total}}$

Complete answer:
Electricity and magnetism are interdependent on each other. Motion of the magnet in the conducting loop generates the eddy currents while motion of charge in the conductor produces the magnetic field around it.4
In case of current carrying a long conductor we will consider an circular ampere loop which will be in a plane perpendicular to the current carrying direction. On any point of the loop the magnitude of the magnetic field will be the same.
In a long conductor, conduction current will be passed, but if that conductor is connected to a capacitor, then between the plates there will be displacement current too. Then we will use a modified form of ampere circuital law. Total current is the sum of conduction current and displacement current. First version of ampere law contains only conduction current. Later modified versions have displacement current too.
\eqalign{ & \oint\limits_s {\mathop B\limits^ \to } .\mathop {dl}\limits^ \to = {\mu _0}{I_{total}} \cr & \therefore \oint\limits_s {\mathop B\limits^ \to } .\mathop {dl}\limits^ \to = {\mu _0}I + {\mu _0}{\varepsilon _0}\dfrac{{d{\phi _E}}}{{dt}} \cr}
Where ${\mu _0}{\varepsilon _0}\dfrac{{d{\phi _E}}}{{dt}}$ is the displacement current. Where ${\phi _E}$ is the electric flux that we had studied in gauss law.

Hence option D is the correct answer.
Note:
The displacement current which we had got, is obtained from gauss law only. That displacement current is produced from the time varying electric field between the plates of the capacitor. Due to that electric field, there will be electric flux. So if electric flux is constant then displacement current will be zero.