Question

Match List I with List II

LIST I		LIST II
A	Gauss's Law in Electrostatics	I	\(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\)
B	Faraday's Law	II	\(\oint \vec{B} \cdot d \vec{A}=0\)
C	Gauss's Law in Magnetism	III	\(\oint \vec{B} \cdot d \vec{l}=\mu_0 i_c+\mu_0 \in_0 \frac{d \phi_E}{d t}\)
D	Ampere-Maxwell Law	IV	\(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\)

Choose the correct answer from the options given below:

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

Answer 1

The Correct Option is B

The correct answer is (B) : A-IV, B-I, C-II, D-III
Gauss's Law of electrostatic
$\varphi =\oint \vec{E}\cdot d\vec{s}=\frac{q}{{ϵ}_0}$
Faraday's law $\oint \vec{E}\cdot \overrightarrow{dl}=\frac{-d{\varphi }_B}{dt}$
Gauss's law of magnetism $\oint \vec{B}\cdot d\vec{A}=0$
Ampere's Maxwell law
$\oint \vec{B}.d\vec{l}={\mu }_0{i}_C+{\mu }_0{\in }_0\frac{d{\varphi }_E}{dt}$
Where $i_C$ : Conduction current
${ϵ}_0\frac{d{\varphi }_E}{dt}$ : Displacement current

Answer 2

Applying the definitions of the laws: 

Gauss's Law in Electrostatics: \( \oint \mathbf{E} \cdot d\mathbf{s} = \frac{q}{\epsilon_0} \). 

Faraday's Law: \( \oint \mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi_B}{dt} \). 

Gauss's Law in Magnetism: \( \oint \mathbf{B} \cdot d\mathbf{A} = 0 \). 

Ampere-Maxwell Law: \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt} \). 

Final Answer: A-IV, B-I, C-II, D-III