Match List I and List II A Gauss’s Law in Electrostatics I ∮ → E · → dl = - d ϕ _ B dt 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 dt D

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Quantrex Academy · Accepted Answer

Correct answer: A. A-IV, B-I, C-II, D-III

(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, $ϕ = ∮ \overset{\to}{E} \cdot d\overset{\to}{s} = \dfrac{q}{{\in}_{0}}$

(B) Faraday’s law explains that the direction of the current will be opposite to the direction of flux that produced it, so $∮ \overset{\to}{E} \cdot d\overset{\to}{l} = - \dfrac{d{ϕ}_{B}}{dt}$

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

(D) Ampere’s Maxwell law is stated as

$∮ \overset{\to}{B} \cdot d\overset{\to}{l} = {\mu}_{0}{i}_{c} + {\mu}_{0}{\in}_{0}\dfrac{d{ϕ}_{E}}{dt}$

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

${\in}_{0}\dfrac{d{ϕ}_{E}}{dt} =$ Displacement current

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

Match List I and List II A Gauss’s Law in Electrostatics I ∮ → E ·...

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More from Electromagnetic Waves

A	Gauss’s Law in Electrostatics	I	$∮ \overset{\to}{E} \cdot \overset{\to}{dl} = - \dfrac{d{ϕ}_{B}}{dt}$
B	Faraday's Law	II	$∮ \overset{\to}{B} \cdot d\overset{\to}{A} = 0$
C	Gauss’s Law in Magnetism	III	$∮ \overset{\to}{B} \cdot d\overset{\to}{l} = {\mu}_{0}{i}_{c} + {\mu}_{0}{\in}_{0}\dfrac{d{ϕ}_{E}}{dt}$
D	Ampere-Maxwell Law	IV	$∮ \overset{\to}{E} \cdot d\overset{\to}{s} = \dfrac{q}{{\in}_{0}}$