Match List I and List II A Gauss’s Law in Electrostatics I $\oint \vec{E}\cdot \vec{dl}=-\frac{d{\phi }_{B}}{dt}$ B Faraday's Law II $\oint \vec{B}\cdot d\vec{A}=0$ C Gauss’s Law in Magnetism III $\oint \vec{B}\cdot d\vec{l}={\mu }_{0}{i}_{c}+{\mu }_{0}{\in }_{0}\frac{d{\phi }_{E}}{dt}$ D Ampere-Maxwell Law IV $\oint \vec{E}\cdot d\vec{s}=\frac{q}{{\in }_{0}}$ Choose the correct answer from the options given below :

(A) Gauss law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. So, $\phi =\oint \vec{E}\cdot d\vec{s}=\frac{q}{{\in }_{0}}$ (B) Faraday’s law explains that the direction of the current will be opposite to the direction of flux that produced it, so $\oint \vec{E}\cdot d\vec{l}=-\frac{d{\phi }_{B}}{dt}$ (C) Gauss's law for magnetism states that the magnetic flux B across any closed surface is zero, so $\oint \vec{B}\cdot d\vec{A}=0$ (D) Ampere’s Maxwell law is stated as $\oint \vec{B}\cdot d\vec{l}={\mu }_{0}{i}_{c}+{\mu }_{0}{\in }_{0}\frac{d{\phi }_{E}}{dt}$ Where ${i}_{c}=$Conduction current and ${\in }_{0}\frac{d{\phi }_{E}}{dt}=$Displacement current Thus, A-IV, B-I, C-II, D-III

Match List I and List II A Gauss’s Law in… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Electromagnetism

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Match List I and List II

A	Gauss’s Law in Electrostatics	I	$\displaystyle \oint \vec{E}\cdot \vec{dl}=-\frac{d{\phi }_{B}}{dt}$
B	Faraday's Law	II	$\displaystyle \oint \vec{B}\cdot d\vec{A}=0$
C	Gauss’s Law in Magnetism	III	$\displaystyle \oint \vec{B}\cdot d\vec{l}={\mu }_{0}{i}_{c}+{\mu }_{0}{\in }_{0}\frac{d{\phi }_{E}}{dt}$
D	Ampere-Maxwell Law	IV	$\displaystyle \oint \vec{E}\cdot d\vec{s}=\frac{q}{{\in }_{0}}$

Choose the correct answer from the options given below :

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

Options

A
A-IV, B-I, C-II, D-III
B
A-I, B-II, C-III, D-IV
C
A-III, B-IV, C-I, D-II
D
A-II, B-III, C-IV, D-I

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Solution

(A) Gauss law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. So, $\displaystyle \phi =\oint \vec{E}\cdot d\vec{s}=\frac{q}{{\in }_{0}}$

(B) Faraday’s law explains that the direction of the current will be opposite to the direction of flux that produced it, so $\displaystyle \oint \vec{E}\cdot d\vec{l}=-\frac{d{\phi }_{B}}{dt}$

(C) Gauss's law for magnetism states that the magnetic flux B across any closed surface is zero, so $\displaystyle \oint \vec{B}\cdot d\vec{A}=0$

(D) Ampere’s Maxwell law is stated as

$\oint \vec{B}\cdot d\vec{l}={\mu }_{0}{i}_{c}+{\mu }_{0}{\in }_{0}\frac{d{\phi }_{E}}{dt}$

Where ${i}_{c}=$ Conduction current and

${\in }_{0}\frac{d{\phi }_{E}}{dt}=$ Displacement current

Thus, A-IV, B-I, C-II, D-III

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