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A parallel plate capacitor is charged by connecting it to a battery through a resistor. If $I$ is the current in the circuit, then in the gap between the plates:
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The correct answer is:
Displacement current of magnitude equal to I flows in the same direction as I
According to modified Ampere's law
$\oint B . d l=\mu_0\left(I_C+I_D\right)$
For Loop $L_1 \quad I_C \neq 0$ and $I_D=0$
For Loop $L_2 \quad I_C=0$ and $I_D \neq 0$
Due to $\mathrm{KCL} \quad I_C=I_D$
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