Match List I and List II A Gauss’s Law in Electrostatics I ∮ E → · d l → = - d ϕ B d t B Faraday's Law II ∮ B → · d A → = 0 C Gauss’s Law in Magnetism III ∮ B → · d l → = μ 0 i c + μ 0 ∈ 0 d ϕ E

Question

Match List I and List II

A	Gauss’s Law in Electrostatics	I	$\oint \vec{E} \cdot \vec{d l} = - \frac{d ϕ_{B}}{d t}$
B	Faraday's Law	II	$\oint \vec{B} \cdot d \vec{A} = 0$
C	Gauss’s Law in Magnetism	III	$\oint \vec{B} \cdot d \vec{l} = μ_{0} i_{c} + μ_{0} \in_{0} \frac{d ϕ_{E}}{d t}$
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 :

TestHub · Accepted Answer

(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, $ϕ = \oint \vec{E} \cdot d \vec{s} = \frac{q}{\in_{0}}$

(B) Faraday’s law explains that the direction of the current will be opposite to the direction of flux that produced it, so $\oint \vec{E} \cdot d \vec{l} = - \frac{d ϕ_{B}}{d t}$

(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

$\oint \vec{B} \cdot d \vec{l} = μ_{0} i_{c} + μ_{0} \in_{0} \frac{d ϕ_{E}}{d t}$

Where $i_{c} =$ Conduction current and

$\in_{0} \frac{d ϕ_{E}}{d t} =$ Displacement current

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

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