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JEE Main 202325 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
25 January
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2023 · 25 January · Evening Shift · Physics · Electromagnetic Waves

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

Match List I and List II

A Gauss’s Law in Electrostatics I ∮E→⋅dl→=−dϕBdt∮ \overset{\to}{E} \cdot \overset{\to}{dl} = - \dfrac{d{ϕ}_{B}}{dt}
B Faraday's Law II ∮B→⋅dA→=0∮ \overset{\to}{B} \cdot d\overset{\to}{A} = 0
C Gauss’s Law in Magnetism III ∮B→⋅dl→=μ0ic+μ0∈0dϕEdt∮ \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 ∮E→⋅ds→=q∈0∮ \overset{\to}{E} \cdot d\overset{\to}{s} = \dfrac{q}{{\in}_{0}}

Choose the correct answer from the options given below :

Options

  1. AA-IV, B-I, C-II, D-III
  2. BA-I, B-II, C-III, D-IV
  3. CA-III, B-IV, C-I, D-II
  4. DA-II, B-III, C-IV, D-I

Correct answer

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

Step-by-step 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, ϕ=∮E→⋅ds→=q∈0ϕ = ∮ \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 ∮E→⋅dl→=−dϕBdt∮ \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 ∮B→⋅dA→=0∮ \overset{\to}{B} \cdot d\overset{\to}{A} = 0

(D) Ampere’s Maxwell law is stated as

∮B→⋅dl→=μ0ic+μ0∈0dϕEdt∮ \overset{\to}{B} \cdot d\overset{\to}{l} = {\mu}_{0}{i}_{c} + {\mu}_{0}{\in}_{0}\dfrac{d{ϕ}_{E}}{dt}

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

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

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

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