Maxwell's modified form of Ampere's Circuital law is:
1. \(\oint{\overrightarrow{{B}}}.\overrightarrow{ds}={0}\)
2. \(\oint{\overrightarrow{{B}}}.\overrightarrow{dl}={\mathit{\mu}}_{0}{I}\)
3. \(\oint{\overrightarrow{{B}}}.\overrightarrow{dl}={\mathit{\mu}}_{0}{I}+\dfrac{1}{{\varepsilon}_{0}}\dfrac{dq}{dt}\)
4. \(\oint{\overrightarrow{{B}}}.\overrightarrow{dl}={{\mu}}_{0}{I}+{{\mu}}_{0}{\varepsilon}_{0}\dfrac{d{\mathit{\phi}}_{E}}{dt}\)
Subtopic:  Maxwell's Equations |
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Maxwell's equation \(\oint\vec E\cdot d\vec l=\dfrac{-d\phi_B}{dt}\) is a statement of:
1. Ampere's law
2. Faraday's law of induction
3. Gauss' law of electricity
4. Gauss' law of magnetism

Subtopic:  Maxwell's Equations |
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Match List-I with List-II (the symbols carry their usual meaning).
List-I List-II
A. \( \oint \vec{E} \cdot d \vec{A}=\dfrac{Q}{\varepsilon_0}\) I. Ampere-Maxwell's Law
B. \( \oint \vec{B} \cdot d \vec{A}=0 \) II. Faraday's Law
C. \( \oint \vec{E} \cdot d\vec{ l}=\dfrac{-d(\phi)}{d t} \) III. Gauss Law of electrostatics
D. \( \oint \vec{B} \cdot d\vec{l}=\mu_0 i_c+ \mu_0 \varepsilon_0 \dfrac{d\left(\phi_E\right)}{d t}\) IV. Gauss's law of magnetism

Choose the correct answer from the options given below:
1. A-III, B-IV, C-II, D-I
2. A-IV, B-III, C-II, D-I
3. A-III, B-II, C-IV, D-I
4. A-IV, B-I, C-III, D-II
Subtopic:  Maxwell's Equations |
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NEET - 2024
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