Matd1 List - I with List - II.
atd1 List - I with List - II.
List - I List - II
Relation Law
A. $$\oint\overrightarrow{E}.\overrightarrow{dl}=-\frac{d}{dt}\oint\overrightarrow{B}.\overrightarrow{da}$$ I. Ampere's circuital law
B. $$\oint\overrightarrow{B}.\overrightarrow{dl}=\mu_{\circ}\left(I+\epsilon_{\circ}\frac{d\phi_{E}}{dt}\right)$$ II. Faraday's laws of electromagnetic induction
C. $$\oint\overrightarrow{E}.\overrightarrow{da}=\frac{1}{\epsilon_{\circ}}\int_{v}^{} \rho dv$$ III. Ampere - Maxwell law
D. $$\oint\overrightarrow{B}.\overrightarrow{dl}=\mu_{\circ}I$$ IV. Gauss's law of electrostatics
Choose the correct answer from the options given below :

We need to match Maxwell's equations (List I) with their corresponding laws (List II).

$$\oint \vec{E} \cdot \vec{dl} = -\frac{d}{dt}\oint \vec{B} \cdot \vec{da}$$ This is the integral form of Faraday's law of electromagnetic induction — a changing magnetic flux induces an electric field. → II (Faraday's law)

$$\oint \vec{B} \cdot \vec{dl} = \mu_0\left(I + \epsilon_0\frac{d\phi_E}{dt}\right)$$ This is Ampere's law with Maxwell's displacement current correction — magnetic fields are produced by both conduction current and changing electric flux. → III (Ampere-Maxwell law)

$$\oint \vec{E} \cdot \vec{da} = \frac{1}{\epsilon_0}\int_V \rho \, dv$$ This is Gauss's law for electricity — the electric flux through a closed surface equals the enclosed charge divided by $$\epsilon_0$$. → IV (Gauss's law of electrostatics)

$$\oint \vec{B} \cdot \vec{dl} = \mu_0 I$$ This is the original Ampere's circuital law (without the displacement current term). → I (Ampere's circuital law)

The correct answer is Option (3): A-II, B-III, C-IV, D-I.