Question

Explain displacement current and write the formula related to Ampere-Maxwell's law.

Hint:

A changing electric field between capacitor plates acts like a current, \( I_d = \varepsilon_0\, d\Phi_E/dt \); add it to the conduction current in Ampere's law.

Answer 1

Step 1: Why displacement current was needed.
The original Ampere's circuital law says the line integral of the magnetic field around a closed loop equals \( \mu_0 \) times the conduction current linked with the loop:
\[ \oint \vec{B}\cdot d\vec{l} = \mu_0 I_c \]
Maxwell noticed a problem while charging a capacitor. In the region between the two capacitor plates there is no flow of charge, so the conduction current is zero there, yet a magnetic field still exists around that region. Thus the simple Ampere's law fails between the plates.

Step 2: Concept of displacement current.
While the capacitor is charging, the electric field between its plates keeps changing with time. Maxwell proposed that a changing electric field also produces a magnetic field, and behaves like a current. This equivalent current, produced by the time-varying electric field, is called the displacement current \( I_d \). It is given by
\[ I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \]
where \( \Phi_E \) is the electric flux through the loop and \( \varepsilon_0 \) is the permittivity of free space. So the displacement current keeps the total current continuous across the gap between the plates.

Step 3: Ampere-Maxwell law.
Maxwell modified Ampere's law by adding the displacement current to the conduction current:
\[ \oint \vec{B}\cdot d\vec{l} = \mu_0 (I_c + I_d) \]
\[ \oint \vec{B}\cdot d\vec{l} = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} \]
This is the Ampere-Maxwell law. It shows that a magnetic field is produced both by a conduction current and by a changing electric field.

\[\boxed{\ \oint \vec{B}\cdot d\vec{l} = \mu_0 I_c + \mu_0\varepsilon_0 \dfrac{d\Phi_E}{dt}\ }\]