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QJEEM24Ph0c3b: JEE Main 2024 Physics - Class 12 JEEM24Ph0c3b Solution: Match List I with List II .tg…

Match List I with List II

List - I List - II
(A) ∮B⃗⋅dl→=μoic+μoεodϕEdt\oint \vec{B} \cdot \overrightarrow{d l}=\mu_o i_c+\mu_o \varepsilon_o \frac{d \phi_E}{d t} (I) Gauss' law for electricity
(B) ∮E⃗⋅dl→=dϕBdt\oint \vec{E} \cdot \overrightarrow{d l}=\frac{d \phi_B}{d t} (II) Gauss' law for magnetism
(C) ∮E⃗⋅dA→=Qεo\oint \vec{E} \cdot \overrightarrow{d A}=\frac{Q}{\varepsilon_o} (III) Faraday law
(D) ∮B⃗⋅dA→=0\oint \vec{B} \cdot \overrightarrow{d A}=0 (IV) Ampere - Maxwell law

Choose the correct answer from the options given below:

Multiple Choice Options

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

Step-by-Step Verified Solution

Ampere-Maxwell law

→∮B→⋅dl→=μ0ic+μ0ε0dϕEdt\rightarrow \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dl}}=\mu_0 \mathrm{i}_{\mathrm{c}}+\mu_0 \varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}

Faraday law →∮E⃗⋅dl→=dϕBdt\rightarrow \oint \vec{E} \cdot \overrightarrow{d l}=\frac{d \phi_{\mathrm{B}}}{d t}

Gauss' law for electricity →∮E→⋅dA→=Qε0\rightarrow \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dA}}=\frac{\mathrm{Q}}{\varepsilon_0}

Gauss ' law for magnetism →∮B⃗⋅dA→=0\rightarrow \oint \vec{B} \cdot \overrightarrow{d A}=0

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