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 d t D Ampere-Maxwell Law IV ∮ E → · d s → = q ∈ 0 Choose the correct answer from the options given below :

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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 d t D Ampere-Maxwell Law IV ∮ E → · d s → = q ∈ 0 Choose the correct answer from the options given below :, A-IV, B-I, C-II, D-III, A-I, B-II, C-III, D-IV, A-III, B-IV, C-I, D-II, A-II, B-III, C-IV, D-I

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(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 = ∮ E → · d s → = q ∈ 0 (B) Faraday’s law explains that the direction of the current will be opposite to the direction of flux that produced it, so ∮ E → · d l → = - d ϕ B d t (C) Gauss's law for magnetism states that the magnetic flux B across any closed surface is zero, so ∮ B → · d A → = 0 (D) Ampere’s Maxwell law is stated as ∮ B → · d l → = μ 0 i c + μ 0 ∈ 0 d ϕ E d t Where i c = Conduction current and ∈ 0 d ϕ E d t = Displacement current Thus, A-IV, B-I, C-II, D-III

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