Question

What is the displacement current? Why was this concept introduced? Explain. Also, obtain the Ampere-Maxwell equation.

Answer 1

Hint: - Displacement current is defined as the rate of change of the electric displacement field and it is the quantity that comes in Maxwell’s equations.
Formula Used: -
$E = \dfrac{Q}{{{\varepsilon _ \circ }A}}$
${i_d} = \dfrac{{dQ}}{{dt}} = {\varepsilon _ \circ }\dfrac{{d{\phi _E}}}{{dt}}$
$\smallint \mathop B\limits^ \to \cdot \mathop {dl}\limits^ \to = {\mu _ \circ }({i_c} + {i_d})$

Complete step-by-step solution -
Displacement current is defined as the rate of change of electric displacement field and its unit is the same as that of electric current density.
This concept was introduced to make the Ampere circuit law logically consistent. As we know that Ampere circuit law is a law which states the relationship between current and magnetic field. This magnetic field is created by the current in the circuit.
When we charge or discharge a capacitor current flows in the circuit. There is no actual charge transfer in the insulted region between capacitors and this is contradictory to the flow of current. Therefore, we can say that due to changing electric flux there is current in the insulated region and this is known as displacement current.
Ampere-Maxwell circuit law states that
$\smallint \mathop B\limits^ \to \cdot \mathop {dl}\limits^ \to = {\mu _ \circ }({i_c} + {i_d})$
Here $B$is the magnetic field, ${\mu _ \circ }$is magnetic permeability in free space, ${i_c}$is Conduction current and ${i_d}$ is Displacement current and $dl$ is the length for which magnetic field is being calculated.
In a circuit displacement current is generated due to varying electric fields. Therefore, it can be written as
${i_d} = \dfrac{{dQ}}{{dt}} = {\varepsilon _ \circ }\dfrac{{d{\phi _E}}}{{dt}}$, as we know that $Q = {\varepsilon _ \circ }{\phi _E}$
Here $dQ$is the small current and $dt$is the small amount of time. ${\varepsilon _ \circ }$ is electrical permittivity of free space
Hence, our equation becomes
$\smallint \mathop B\limits^ \to \cdot \mathop {dl}\limits^ \to = {\mu _ \circ }({i_c} + {\varepsilon _ \circ }\dfrac{{d\phi }}{{dt}})$
As per Maxwell’s Ampere circuital law, the line of magnetic field along a closed path is equal to ${\mu _ \circ }$ times the total current.

Note: - The total current involved in the Ampere-Maxwell equation consists of free current and bound current, although all currents are essentially the same from a microscopic perspective. Treating free current and bound current differently offers physical insights to the Ampere-Maxwell equation in different contexts.