(C) The displacement current density $J_d$ is uniform across the cross-section of the capacitor plates.$J_d = \frac{I}{A} = \frac{6 \ A}{16 \ cm^2}$.T

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(C) The displacement current density $J_d$ is uniform across the cross-section of the capacitor plates.$J_d = \frac{I}{A} = \frac{6 \ A}{16 \ cm^2}$.The displacement current $I_d$ through a smaller area $A_0$ is given by $I_d = J_d \times A_0$.$I_d = \left( \frac{6}{16} \right) \times 3.2 \ A$.$I_d = 0.375 \times 3.2 \ A = 1.2 \ A$.Since $1 \ A = 1000 \ mA$,the displacement current is $1.2 \times 1000 \ mA = 1200 \ mA$.

Class 12 Physics Electromagnetic waves Maxwell's equations , Concept of displacement current and Hertz experiment

Medium

$A$ parallel plate capacitor of area $A = 16 \ cm^2$ and separation between the plates $10 \ cm$ is charged by a $DC$ current. Consider a hypothetical plane surface of area $A_0 = 3.2 \ cm^2$ inside the capacitor and parallel to the plates. At an instant,the current through the circuit is $6 \ A$. At the same instant,the displacement current through $A_0$ is . . . . . . $mA$.

JEE MAIN 2025 All JEE MAIN

A
$1600$
B
$1400$
C
$1200$
D
$1900$

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Class 12

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Electromagnetic waves

Subtopic

Maxwell's equations , Concept of displacement current and Hertz experiment

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In the charging of a capacitor,what is the contradiction in Ampere's circuital law?

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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:

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	$IV$. $\oint \vec{E} \cdot d \vec{s} = \frac{q}{\epsilon_0}$