Match the following Maxwell’s equations: (The symbols used here have their usual meanings)
List-$I$List-$II$
$(a)$ Gauss’ law for electrostatics$(i)$ $\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$
$(b)$ Gauss’ law for magnetism(ii) $\oint \vec{B} \cdot d\vec{l} = \mu_0 [i_c + \epsilon_0 \frac{d\phi_E}{dt}]$
$(c)$ Faraday’s law(iii) $\oint \vec{B} \cdot d\vec{A} = 0$
$(d)$ Ampere-Maxwell’s law(iv) $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$

  • A
    $a-i, b-iii, c-iv, d-ii$
  • B
    $a-ii, b-iii, c-i, d-iv$
  • C
    $a-i, b-ii, c-iii, d-iv$
  • D
    $a-ii, b-iii, c-iv, d-i$

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Similar Questions

Match List-$I$ with List-$II$:
List-$I$ List-$II$
$A$. Gauss's Law in Electrostatics $I$. $\oint \vec{E} \cdot d \vec{l} = -\frac{d \phi_B}{d t}$
$B$. Faraday's Law $II$. $\oint \vec{B} \cdot d \vec{A} = 0$
$C$. Gauss's Law in Magnetism $III$. $\oint \vec{B} \cdot d \vec{l} = \mu_0 i_C + \mu_0 \epsilon_0 \frac{d \phi_E}{d t}$
$D$. Ampere-Maxwell Law $IV$. $\oint \vec{E} \cdot d \vec{s} = \frac{q}{\epsilon_0}$

Choose the correct answer from the options given below:

What is called displacement current?

The capacitance of a capacitor is $2 \ pF$. If the electric field inside the capacitor changes at a rate of $10^{12} \ V \ s^{-1}$,the displacement current is ..... $A$.

If the rate of change of electric field across the plates of a parallel plate capacitor is $E$ and the displacement current is $I$,then the area of one plate of the capacitor is ($\varepsilon_{0}$ is permittivity of free space).

"Changing electric field produces magnetic field". Explain the importance of the given statement.

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