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NVAEasyJEE 2024Displacement Current

JEE Physics 2024 Question with Solution

Match List I with List II: List I A. ∮B⃗⋅dl⃗=μ0ic+μ0ε0dΦEdt\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} B. ∮E⃗⋅dl⃗=−dΦBdt\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt} C. ∮E⃗⋅dA⃗=Qε0\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\varepsilon_0} D. ∮B⃗⋅dA⃗=0\oint \vec{B} \cdot d\vec{A} = 0 List II I. Gauss' law for electricity II. Gauss' law for magnetism III. Faraday's law IV. Ampere–Maxwell law

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: Four electromagnetic equations in List I and four named laws in List II.

Find: The correct matching between the two lists.

From the given equations:

  • A corresponds to Ampere–Maxwell law.
  • B corresponds to Faraday's law.
  • C corresponds to Gauss' law for electricity.
  • D corresponds to Gauss' law for magnetism.

Therefore, the matching is A-IV, B-III, C-I, D-II. Hence, the correct answer is 33.

Common mistakes

  • Confusing Ampere's law with Ampere–Maxwell law is a common mistake. The displacement current term μ0ε0dΦEdt\mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} identifies the Maxwell-corrected form, so equation A must match IV.

  • Students often interchange Gauss' law for electricity and Gauss' law for magnetism. The equation ∮E⃗⋅dA⃗=Qε0\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\varepsilon_0} involves enclosed charge, so it is electricity, whereas ∮B⃗⋅dA⃗=0\oint \vec{B} \cdot d\vec{A} = 0 expresses absence of magnetic monopoles.

  • Another mistake is not recognizing Faraday's law because of the negative sign. In ∮E⃗⋅dl⃗=−dΦBdt\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}, the minus sign reflects Lenz's law, and this equation must match III.

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