A parallel plate capacitor is charged by conn | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to the modified Ampere's circuital law,the total current is the sum of conduction current $(I_C)$ and displacement current $(I_D)$: $\oi

Vedclass · Accepted Answer

Displacement current of magnitude equal to $I$ flows in the same direction as $I$.

Vedclass · Answer

(A) According to the modified Ampere's circuital law,the total current is the sum of conduction current $(I_C)$ and displacement current $(I_D)$: $\oint B \cdot dl = \mu_0(I_C + I_D)$.In the wire,there is only conduction current $I_C = I$,and the displacement current $I_D = 0$.In the gap between the plates,there is no conduction current $(I_C = 0)$,but the changing electric field produces a displacement current $I_D$.By the principle of continuity of current,the displacement current in the gap must be equal to the conduction current in the wire,i.e.,$I_D = I_C = I$.This displacement current flows in the same direction as the conduction current to maintain the continuity of the total current in the circuit.

$A$ 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:

Similar Questions

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

Which is the missing term in Ampere's circuital law?

Consider a parallel plate capacitor which is maintained at a potential of $200 \, V$. The separation distance between the plates of the capacitor and the area of the plates are $1 \, mm$ and $20 \, cm^2$, respectively. Calculate the displacement current in $1 \, \mu s$. (in $ \, mA$)

$A$ magnetic field can be produced by

An insulator plate is passed between the plates of a capacitor. The displacement current

Vedclass Test Series

Exam Paper Generator

Online Exam Module