Match List I with List II.
$$\begin{array}{|l|l|l|l|}\hline & \text{List I} & & \text{List II} \\ \hline \text{A.} & \oint \vec B\cdot d\vec l = \mu_0 i_c + \mu_0\varepsilon_0\dfrac{d\phi_E}{dt} & \text{I.} & \text{Gauss' law for electricity} \\ \hline \text{B.} & \oint \vec E\cdot d\vec l = -\dfrac{d\phi_B}{dt} & \text{II.} & \text{Gauss' law for magnetism} \\ \hline \text{C.} & \oint \vec E\cdot d\vec A = \dfrac{Q}{\varepsilon_0} & \text{III.} & \text{Faraday law} \\ \hline \text{D.} & \oint \vec B\cdot d\vec A = 0 & \text{IV.} & \text{Ampere–Maxwell law} \\ \hline \end{array}$$
Choose the correct answer from the options given below:
Answer: (C) A – IV, B – III, C – I, D – II
A: the line integral of $\vec B$ with conduction and displacement current is the Ampere–Maxwell law (IV).
B: the emf equals the rate of change of magnetic flux: Faraday's law (III).
C: the flux of $\vec E$ through a closed surface equals $Q/\varepsilon_0$: Gauss's law for electricity (I).
D: the flux of $\vec B$ through a closed surface is zero: Gauss's law for magnetism (II).
Solution by Sreeraj P, M.Sc Physics