Match List - I with List - II
Q. Match List - I with List - II.

List - I (Relation)

A. $\displaystyle \oint \vec{E}\cdot d\vec{l} = -\frac{d}{dt}\oint \vec{B}\cdot d\vec{a}$

B. $\displaystyle \oint \vec{B}\cdot d\vec{l} = \mu_0\left(I+\varepsilon_0\frac{d\phi_E}{dt}\right)$

C. $\displaystyle \oint \vec{E}\cdot d\vec{a} = \frac{1}{\varepsilon_0}\int_v \rho\,dv$

D. $\displaystyle \oint \vec{B}\cdot d\vec{l} = \mu_0 I$

List - II (Law)

I. Ampere's circuital law
II. Faraday's laws of electromagnetic induction
III. Ampere - Maxwell law
IV. Gauss's law of electrostatics

Choose the correct answer from the options given below :
A. A-I, B-IV, C-III, D-II
B. A-II, B-III, C-IV, D-I
C. A-IV, B-I, C-II, D-III
D. A-II, B-III, C-I, D-IV
Correct Answer: A-II, B-III, C-IV, D-I

Explanation

Relation A represents the line integral of electric field equal to the negative rate of change of magnetic flux.

This is exactly the mathematical form of Faraday's laws of electromagnetic induction.

Hence,

$$ A \rightarrow II $$

Relation B includes both conduction current and displacement current terms.

This is the modified Ampere’s law introduced by Maxwell, known as the Ampere–Maxwell law.

Thus,

$$ B \rightarrow III $$

Relation C gives the surface integral of electric field equal to charge enclosed divided by $\varepsilon_0$.

This is Gauss’s law of electrostatics.

Hence,

$$ C \rightarrow IV $$

Relation D is the original Ampere’s circuital law without displacement current.

Therefore,

$$ D \rightarrow I $$

So the correct matching is:

$$ \boxed{A-II,\; B-III,\; C-IV,\; D-I} $$

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