Q61 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
Match List - I with List - II.
$\begin{array}{l} \text{List - I} & \text{List - II} \\ \text{Relation} & \text{Law} \\ \text{A. } \oint \vec{E} \cdot \overrightarrow{d l}=-\frac{d}{d t} \oint \vec{B} \cdot \overrightarrow{d a} & \text{I. Ampere's circuital law} \\ \text{B. } \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0}\left(I+\epsilon_{0} \frac{d \phi_{E}}{d t}\right) & \text{II. Faraday's laws of} \\ & \text{electromagnetic induction} \\ \text{C. } \oint \vec{E} \cdot \overrightarrow{d a}=\frac{1}{\epsilon_{0}} \int_{\mathrm{V}} \rho \mathrm{dv} & \text{III. Ampere - Maxwell law} \\ \text{D. } \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I & \text{IV. Gauss's law of electrostatics} \end{array}$
Choose the correct answer from the options given below :
$\begin{array}{l} \text{List - I} & \text{List - II} \\ \text{Relation} & \text{Law} \\ \text{A. } \oint \vec{E} \cdot \overrightarrow{d l}=-\frac{d}{d t} \oint \vec{B} \cdot \overrightarrow{d a} & \text{I. Ampere's circuital law} \\ \text{B. } \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0}\left(I+\epsilon_{0} \frac{d \phi_{E}}{d t}\right) & \text{II. Faraday's laws of} \\ & \text{electromagnetic induction} \\ \text{C. } \oint \vec{E} \cdot \overrightarrow{d a}=\frac{1}{\epsilon_{0}} \int_{\mathrm{V}} \rho \mathrm{dv} & \text{III. Ampere - Maxwell law} \\ \text{D. } \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I & \text{IV. Gauss's law of electrostatics} \end{array}$
Choose the correct answer from the options given below :
- AA-II, B-III, C-I, D-IV
- Correct answerBA-II, B-III, C-IV, D-I
- CA-I, B-IV, C-III, D-II
- DA-IV, B-I, C-II, D-III
Solution
Theoretical
A-II, B-III, C-IV, D-I
A. $\oint \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \oint \vec{B} \cdot d\vec{a}$ → Faraday's law of electromagnetic induction (II)
B. $\oint \vec{B} \cdot d\vec{l} = \mu_0 \left(I + \epsilon_0 \frac{d\Phi_E}{dt}\right)$ → Ampere-Maxwell law (III)
C. $\oint \vec{E} \cdot d\vec{a} = \frac{1}{\epsilon_0} \int \rho \, dv$ → Gauss's law of electrostatics (IV)
D. $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$ → Ampere's circuital law (I)
A-II, B-III, C-IV, D-I
A. $\oint \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \oint \vec{B} \cdot d\vec{a}$ → Faraday's law of electromagnetic induction (II)
B. $\oint \vec{B} \cdot d\vec{l} = \mu_0 \left(I + \epsilon_0 \frac{d\Phi_E}{dt}\right)$ → Ampere-Maxwell law (III)
C. $\oint \vec{E} \cdot d\vec{a} = \frac{1}{\epsilon_0} \int \rho \, dv$ → Gauss's law of electrostatics (IV)
D. $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$ → Ampere's circuital law (I)
Hence, the correct answer is option: 'B'.