Maxwell’s modified form of Ampere’s circuital law is:

Question

Question

\(\oint \vec B.d\vec S = 0\)
\(\oint \vec B.d\vec l = {\mu _0}i\)
\(\oint \vec B.d\vec l = {\mu _0}i + \frac{1}{{{\epsilon_0}}}\frac{{dq}}{{dt}}\)
\(\oint \vec B.d\vec l = {\mu _0}i + {\mu _0}{\epsilon_0}\frac{{d{\phi _E}}}{{dt}}\)

Answer 1

CONCEPT:

Amperes law: The line integral of the magnetic field around any closed curve is equal to μ_o times the net current I threading through the area enclosed by the curve.

\(\oint \vec B \cdot \overrightarrow {dl} = {\mu _0}I\)

Ampere’s law in this form is not valid if the electric field at the surface varies with time.

Displacement current (I_D): It is that current that comes into existence, in addition to the conduction current, whenever the electric field and hence the electric flux changes with time.

EXPLANATION:

To modify Ampere’s law, Maxwell followed a symmetry consideration.
By Faraday’s law, a changing magnetic field induces an electric field, hence a changing electric field must induce a magnetic field. As currents are the usual sources of the magnetic field, a changing electric field must be associated with the current. Maxwell called that current as displacement current.
To maintain the dimensional consistency, the displacement current is added in ampere’s law:

\(\Rightarrow \oint \vec B \cdot \overrightarrow {dl} = {\mu _0}I + {\mu _0}{\epsilon_0}\left( {\frac{{d{{\rm{\Phi }}_E}}}{{dt}}} \right)\)

Where, \({\epsilon_0}\left( {\frac{{d{{\bf{\Phi }}_E}}}{{dt}}} \right)\) is the displacement current. So option 4 is correct.