CBSE · Class 12 · Physics · Electromagnetic WavesWhat is displacement current? Why was it introduced by Maxwell? Derive the expression for displacement current and write the modified Ampere's circuital law.

Introduction and Need for Displacement Current

• Displacement Current Definition: It is that current which comes into play in the region where the electric field and hence the electric flux is changing with time. It is represented by $I_d$.
• Maxwell's Modification: Ampere's circuital law states that the line integral of magnetic field $\vec{B}$ around a closed loop is equal to $\mu_0$ times the total current $I$ passing through the surface bounded by the loop ($\oint \vec{B} \cdot d\vec{l} = \mu_0 I$).
• Maxwell found a logical inconsistency in Ampere's circuital law while applying it to a charging capacitor. During the charging of a capacitor, conduction current flows through the connecting wires, but no actual charge carriers move through the space between the capacitor plates. However, a magnetic field exists in that region. To resolve this contradiction and make equations symmetric, Maxwell introduced the concept of displacement current.

Derivation of Displacement Current Expression\nConsider a parallel plate capacitor being charged by a battery. Let $q(t)$ be the charge on the plates at any instant $t$. \nThe electric field $E$ between the plates of area $A$ is:

$$E = \frac{q}{\varepsilon_0 A}$|\nThe electric flux $\Phi_E$ linked with the capacitor is: $$\Phi_E = E \cdot A = \left(\frac{q}{\varepsilon_0 A}\right) A = \frac{q}{\varepsilon_0}$|\nRearranging for charge $q$: $$q = \varepsilon_0 \Phi_E$$ \nDifferentiating both sides with respect to time $t$ gives the rate of change of charge, which is the current: $$I_d = \frac{dq}{dt} = \varepsilon_0 \frac{d\Phi_E}{dt}$|\nThis is the required expression for displacement current.

Modified Ampere's Circuital Law (Maxwell-Ampere Law)\nIncluding both conduction current ($I_c$) and displacement current ($I_d$), the total current is $(I_c + I_d)$.\nThe modified Ampere's circuital law is given by:

$$\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I_c + I_d \right) = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$$