Displacement Current

Displacement Current

(i)Let us consider a circuit containing the capacitor as shown in fig.

(ii) According to Ampere's circuital law, the line integral of magnetic field along any closed path is $\mu$0 times the total current enclosed by the closed path. Mathematically

$\oint \vec{B}.\overrightarrow{dl}={\mu }_0\mathrm{I}$

In this law it is assumed that conduction current flows through the connecting wires charging the condenser plates but no current flows in the space in between the plates. Actually it is not true.

Illustration 1: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency 2 x 1010 Hz and amplitude 48 V/m. The amplitude of oscillating magnetic field will be:

(A) $\frac{1}{16}\times {10}^8$ Wb/m2 (B) 16 x 10–8 Wb/m2

(C) 12 x 10–7 Wb/m2 (D) $\frac{1}{12}\times {10}^7$ Wb/m2

Solution: (B) Oscillating magnetic field

B = $\frac{E}{c}=\frac{48}{3\times {10}^8}$ = 16 x 10–8 Wb/m2

(iii) When the circuit is closed, conduction current flows from the plate P of the capacitor to the other plate Q through the conducting wires. Maxwell suggested that due to time varying electric field between the plates, an electric current, called displacement current (ID), also flows across the space between the plates of the capacitor.

(iv) Thus, there is a continuous flow of current in a capacitive circuit alos, through the conducting wire there is flow of conduction current IC and through the space across the plates of capacitor, there is flow of displacement current ID.

(v) Maxwell pointed out that in Ampere's circuital law, the current I should be treated as total current i.e., the sum of the conduction current IC and displacement current ID and modified the law as

$\oint \vec{B}.\overrightarrow{dl}={\mu }_0\left({\mathrm{I}}_C+{\mathrm{I}}_D\right)$ It is called the Ampere-Maxwell's circuital law.

(vi) The displacement current is defined as

ID = ${\epsilon }_0\frac{d{\varphi }_E}{dt}$

where ${\varphi }_E$ is the electric flux linked between the plates of the capacitor at any instant. Therefore Ampere-Maxwell circuital law may be expressed as

$\oint \vec{B}.\overrightarrow{dl}={\mu }_0\left({\mathrm{I}}_C+{\epsilon }_0\frac{d{\varphi }_E}{dt}\right)$

Illustration 2: A parallel plate capacitor of plate separation 2 mm is conncected in an electric circuit having source voltage 400V. If the plate area is 60 cm2, then the value of displacement current for 10–6 sec. will be:

(A) 1.062A (B) 1.062 x 10–2A

(C) 1.062 x 10–3A (D) 1.062 x 10–4A

Solution: (D) ID = ${\epsilon }_0\frac{\partial {\varphi }_E}{\partial t}={\epsilon }_0\frac{EA}{t}=\frac{{\epsilon }_{0}VA}{dt}$

or ID = $\frac{8.85\times {10}^{12}\times 400\times 60\times {10}^4}{2\times {10}^3\times {10}^6}$

= 1.062 x 10–4A