Question:

Match List I with List II
List IList II
A. \( \oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt} \)I. Gauss' law for electricity
B. \( \oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt} \)II. Gauss' law for magnetism
C. \( \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0} \)III. Faraday law
D. \( \oint \vec{B} \cdot d\vec{A} = 0 \)IV. Ampere – Maxwell law

Choose the correct answer from the options given below

Updated On: Jan 13, 2026
  • A-IV, B-I, C-III, D-II
  • A-II, B-III, C-I, D-IV
  • A-IV, B-III, C-I, D-II
  • A-I, B-II, C-III, D-IV
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The Correct Option is C

Approach Solution - 1

The question requires matching the given Maxwell's equations with their respective laws. Let's break down each equation and identify which concept it belongs to: 

  1. \(\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}\) represents the Ampere-Maxwell Law. This law combines Ampere's circuital law with Maxwell's addition of the displacement current, which accounts for a time-varying electric field. It is used to describe how a magnetic field is generated by electric currents and changing electric fields. \(\rightarrow\) IV. Ampere-Maxwell law.
  2. \(\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}\) describes Faraday's Law of Electromagnetic Induction. This law states that the line integral of the electric field around a closed loop is equal to the negative rate of change of the magnetic flux through the loop. It explains how a changing magnetic field can induce an electric field. \(\rightarrow\) III. Faraday law.
  3. \(\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}\) corresponds to Gauss's Law for electricity. This law states that the electric flux through a closed surface is proportional to the enclosed electric charge. \(\rightarrow\) I. Gauss' law for electricity.
  4. \(\oint \vec{B} \cdot d\vec{A} = 0\) represents Gauss's Law for magnetism. This law indicates that the net magnetic flux through any closed surface is zero, implying that magnetic monopoles do not exist. \(\rightarrow\) II. Gauss' law for magnetism.

Matching List I with List II gives us:

  • A-IV: Ampere-Maxwell Law is described by equation A.
  • B-III: Faraday's Law corresponds to equation B.
  • C-I: Gauss's Law for electricity is represented by equation C.
  • D-II: Gauss's Law for magnetism fits with equation D.

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

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Approach Solution -2

Ampere–Maxwell Law: - \( \oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d \Phi_E}{dt} \). - This matches with A-IV.

Faraday’s Law of Electromagnetic Induction: - \( \oint \vec{E} \cdot d\vec{l} = -\frac{d \Phi_B}{dt} \). - This matches with B-III.

Gauss’ Law for Electricity: - \( \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0} \). - This matches with C-I.

Gauss’ Law for Magnetism: - \( \oint \vec{B} \cdot d\vec{A} = 0 \). - This matches with D-II.

So, the correct option is : (3) A-IV, B-III, C-I, D-II

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